\section{Introduction}

Let $A$ be a finite dimensional algebra of the form $kQ_A/I_A$, where $k$ is a field, $Q_A$ is a finite quiver and $I_A$ is an admissible ideal in the path algebra $kQ_A$. Let $B$ be a finite dimensional algebra such that there is a radical embedding $\phi: B \to A$, that is, $\phi$ is an algebra monomorphism with $\phi(\mathrm{rad}B)=\mathrm{rad}A$, where rad$A$ (resp. rad$B$) denotes the (Jacobson) radical of $A$ (resp. $B$). The radical embedding is a common construction in the studies of finite dimensional algebras and their representation theory (see, for example, \cite{EHIS,KL,X}). By Xi's observation in \cite{X} we can without loss of generality assume that $B$ is a subalgebra of $A$ and obtained from $A$ by repeatedly gluing two idempotents. Recall that if $B$ is obtained from $A$ by gluing two idempotents $e_1$ and $e_2$, then $B$ has the form $kQ_B$ quotient by the admissible ideal $I_B$, where the quiver $Q_B$ is obtained from $Q_A$ by identifying $e_1$ and $e_2$ with a new vertex $f_1$, $I_B$ is generated by the elements in $I_A$ plus all newly formed paths, if any, (of length $2$) through the vertex $f_1$. By a result of the first author jointly with Koenig (see \cite[Theorem 4.10]{KL}), when we glue a source and a sink we obtain a stable equivalence $\underline{\mathrm{mod}}A\stackrel{\sim}{\longrightarrow} \underline{\mathrm{mod}}B$ between the stable module categories modulo projective modules.

The main aim of this note is to study the behaviour of Hochschild cohomology and of the fundamental groups when we glue two idempotents from a monomial algebra $A$.

It is well known that Hochschild cohomology is not functorial, that is, if we have an algebra homomorphism
$\phi: B \to A$ then
we do not know how to construct a map from
$\HH^*(A)$ to
$\HH^*(B)$ or from
$\HH^*(B)$ to
$\HH^*(A)$. This makes Hochschild cohomology difficult to compute since it is not possible in general to reduce the study of Hochschild cohomology to some  smaller, and potentially easier, algebras. However, when we have a radical embedding $\phi: B \to A$ which is obtained by gluing two idempotents of a monomial algebra $A$ there is some sort of functoriality, at least for $\HH^1$. In fact very often we can construct a map (even a Lie algebra homomorphism) from $\HH^1(B)$ to $\HH^1(A)$ or from  $\HH^1(A)$ to $\HH^1(B)$. We will distinguish two types of gluings: gluing two idempotents from the same block of $A$ and from different blocks of $A$. It turns out that the behaviour of the latter is much simpler than  of the former. In both types we will study the case of gluing a source and a sink, that is, when we obtain a stable equivalence.


Our main result in this direction is Theorem \ref{Lie-strucutre-of-hh1}, which states that under some mild condition on the characteristic of the field $k$ (Assumption \ref{assump}), there is a close connection between the Lie algebras $\HH^1(A)$ and $\HH^1(B)$, once there is a radical embedding $\phi: B \to A$ which is obtained by gluing two idempotents
of a monomial algebra $A$. In particular, we can explicitly compare the dimensions of $\HH^1(A)$ and $\HH^1(B)$ in terms of some combinatorial datum, see Theorem \ref{hh1-quotient-isomorphism-for-monomial-algebras} for the details.


Of particular interest is the case of gluing a source and a sink, in which case we get a stable equivalence $F: \underline{\mathrm{mod}}A\stackrel{\sim}{\longrightarrow} \underline{\mathrm{mod}}B$, and Theorem \ref{Lie-strucutre-of-hh1} and Theorem \ref{hh1-quotient-isomorphism-for-monomial-algebras} take their nicest forms, see Corollary \ref{quotient-lie-algebra-source-and-sink}. According to \cite{KL}, we know that in contrast to the situation for a stable equivalences of Morita type, the above stable equivalence $F$ is induced by bimodules that are only projective on one side, but not on the other. For this reason, we do not know how to set up a (restricted) Lie algebra homomorphism between $\HH^*(A)$ and $\HH^*(B)$ under such a stable equivalence using the above mentioned bimodules, as it was set in \cite{KLZ} \cite{BR3}.  Nevertheless, we can compare $\HH^*(A)$ and $\HH^*(B)$ directly using the algebra monomorphism (that is, the radical embedding) $\phi: B \to A$ and Strametz's description of Hochschild cochain complex of monomial algebras. This allows us to `control' the behaviours of Hochschild cohomologies far beyond the scope of stable equivalences, although in stable equivalence case we get the best results.

The radical embedding $\phi: B \to A$ also gives a close connection between the centers $Z(A)$ and $Z(B)$. More precisely, if $A$ is an indecomposable monomial algebra and $\phi: B \to A$ is a radical embedding which is obtained by gluing two idempotents of $A$, then there is an algebra monomorphism $Z(A)\hookrightarrow Z(B)$ and the difference between $\mathrm{dim}_kZ(A)$ and $\mathrm{dim}_kZ(B)$ can be described by some combinatorial data (Proposition \ref{center-from-the-same-block}); if $A$ is a monomial algebra with two blocks and $\phi: B \to A$ is a radical embedding which is obtained by gluing two idempotents from different blocks, then $\phi: B \to A$ restricts to a radical embedding $Z(B)\hookrightarrow Z(A)$, in particular $\mathrm{dim}_kZ(A)=\mathrm{dim}_kZ(B)+1$ (Proposition \ref{center-from-different-block}).

We also study the relation between the fundamental groups and the idempotent gluings. More
precisely, we consider the $\pi_1$-$\mathrm{rank}(A)$ which has been introduced in \cite{BR}.  The $\pi_1$-$\mathrm{rank}(A)$ is the maximal dimension of a dual fundamental group for some (minimal) presentation of $A$. For a monomial algebra $A$ the $\pi_1$-$\mathrm{rank}(A)$ coincides with the first Betti number of the Gabriel quiver $Q_A$ of $A$ which intuitively count the number of holes of $Q_A$. We  compare $\pi_1$-$\mathrm{rank}(A)$ and $\pi_1$-$\mathrm{rank}(B)$ for the two types of gluings, see Lemma \ref{pi-one-rank}.


We also compare higher Hochschild cohomology groups for radical square zero algebras. More precisely, we show that when we glue a source and a sink, there is always an injective map from cocycles (respectively coboundaries) of $A$ to cocycles (respectively coboundaries) of $B$, see Proposition \ref{highhochgluing}.


Rather interestingly, our results can be easily generalized to the situation of gluing arrows. We will discuss this and other generalizations in a subsequent paper.

{\bf Outline.} In Section \ref{pre} we give some notations and terminologies
which we keep throughout the paper. It also provides some background on various topics. In Section \ref{HH1sec} we focus on the
study of the behaviour of the first Hochschild cohomology under gluing idempotents. The main results in this section are Theorem
\ref{hh1-quotient-isomorphism-for-monomial-algebras} and its strengthened form Theorem \ref{Lie-strucutre-of-hh1}. Some applications  of our main results are presented in Corollaries \ref{quotient-lie-algebra-source-and-sink}, \ref{hh1-radical-square-zero-case} and \ref{hh1-radical-square-zero-case-structure}. Moreover, in Subsection \ref{rszssec} we give an interpretation on S\'{a}nchez-Flores' description of the first Hochschild cohomology for radical square zero algebras by inverse gluing operations. In Section \ref{eg}
we give various examples to illustrate our definitions and results in Section \ref{HH1sec}. In Section
\ref{fundsec} we study the relation between gluing idempotents and the $\pi_1$-$\mathrm{ranks}$. Finally in Section \ref{HHsec} we study how gluing changes the center and higher Hochschild cohomology groups.


\section{Preliminaries}
\label{pre}

\medskip

{\bf Bound quivers.} All algebras considered are
finite dimensional algebras which are isomorphic to $kQ/I$, where $k$ is a field of arbitrary characteristic, $Q$ is a finite quiver and $I$ is an admissible ideal
in the path algebra $kQ$. Any homomorphism between two algebras is requested to send the identity to the identity. For all $n \in \mathbb{N}$, let $Q_{n}$ be the set of paths of length $n$ of $Q$ and let $Q_{\ge n}$ be the set of paths of length greater than or equal to  $n$. Note that $Q_{0}$ is the set of vertices and  $Q_{1}$ is the set of arrows of $Q$. We denote by $s(\gamma)$ the source vertex of an (oriented) path $\gamma$ of $Q$ and by $t(\gamma)$ its terminal vertex. The path algebra $kQ$ is the $k$-linear span of the set of paths of $Q$ where the multiplication of $\beta \in Q_{i}$ and $\alpha \in Q_{j}$ is provided by the concatenation $\beta\alpha \in Q_{i+j}$ if $t(\alpha)=s(\beta)$ and 0 otherwise. A path $p$ of length $l\geq 1$ is said to be an oriented cycle (or an oriented $l$-cycle) if $s(p)=t(p)$. An oriented $1$-cycle is called a loop. Two paths $\epsilon,\gamma$ of $Q$ are called parallel if $s(\epsilon)=s(\gamma)$ and $t(\epsilon)=t(\gamma)$, denoted by $\epsilon \| \gamma$. If $\epsilon$ and $\gamma$ are not parallel, we denote by $\epsilon \nparallel \gamma$. If $X,Y$ are sets of paths of $Q$, we denote by $X \| Y$ the set of parallel paths consisting of the couples $\epsilon \| \gamma$ with $\epsilon \in X$ and $\gamma \in Y$, and denote by $k(X \| Y)$ the $k$-vector space with basis $X \| Y$.

We now fix a finite dimensional $k$-algebra $A=kQ_{A}/I_{A}$ (where $I_A$ is an admissible ideal in $kQ_{A}$) and denote the vertices of $Q_A$ by $e_1,\cdots,e_n$.  A vertex $e_i$ is isolated if it does not exist any arrow $\alpha$ such that $s(\alpha)=e_i$ or $t(\alpha)=e_i$. By a source vertex $e_i$ of $Q_A$, we mean that there is no arrow $\alpha$ with $t(\alpha)=e_i$; by a sink vertex $e_j$ of $Q_A$, we mean that there is no arrow $\alpha$ with $s(\alpha)=e_j$. By abuse of notation, we denote by $e_1,\cdots,e_n$ the corresponding primitive idempotents in the algebra $A$, and for a path $p$ in $Q_A$, we use the same notation to denote its image $\overline{p}=p+I_A$ in $A$. If $A=A_1\times \cdots \times A_s$ is a decomposition of $A$ into a product of indecomposable algebras, then $A_i$'s are called blocks of $A$. Note that such a decomposition of $A$ is unique and if $s=1$ then $A$ is an indecomposable algebra.

{\bf Radical embeddings.} Let $B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_{1}$ and $e_{n}$ of $A$. That is, $B$ is identified as a subalgebra of $A$ generated by $f_1:=e_1+e_n,f_2:=e_2,\cdots,f_{n-1}:=e_{n-1}$ and all arrows in $Q_A$. By abuse of notation,  we denote the vertices of $B$ by $f_1,\cdots,f_{n-1}$.
Then the algebra $B$ has the form $kQ_{B}/I_{B}$, where the quiver $Q_B$ is obtained from $Q_A$ by identifying the vertices $e_1$ and $e_n$, and the admissible ideal $I_B$ is generated by the elements in $I_A$ plus all newly formed paths of the form $\cdot\rightarrow f_1\rightarrow\cdot$ (each of such a path has length $2$). In particular, we have $\mathrm{dim}_kB=\mathrm{dim}_kA-1$.

Note that there is an obvious bijection between the arrows
of $A$ and the arrows of $B$. For each arrow $\alpha$ in
$Q_{A}$, we denote the corresponding arrow in $Q_{B}$ by
$\alpha^{*}$. We define a quiver morphism $$\varphi:Q_{A}
\to Q_{B}$$
as follows: let $\varphi(e_{i})=f_{i}$ for
$2\le i \le n-1$ and
$\varphi(e_{1})=\varphi(e_{n})=f_{1}$, and let
$\varphi(\alpha)=\alpha^{*}$. It is possible also to define $\varphi_n: (Q_A)_n \to (Q_B)_n$  by extending the map $\varphi:Q_{A}\to Q_{B}$. More precisely, let $p=a_n\dots a_1$ be a path in $(Q_A)_n$. Then $\varphi_n(p)=p^*=a^*_n\dots a^*_1$. It is known that the above radical embedding induces a stable equivalence (modulo projective modules) $\underline{\mathrm{mod}}
A\rightarrow \underline{\mathrm{mod}} B$ if and only if $B$ is obtained from
$A$ by gluing a sink
vertex and a source vertex, at least when we modulo the
Auslander-Reiten conjecture for stable equivalences (see \cite{KL}).

{\bf Monomial algebras.} We denote by $\Lambda$ a finite dimensional monomial $k$-algebra, that is, a finite dimensional $k$-algebra which is isomorphic to a quotient $kQ/I$ of a path algebra where the two-sided ideal $I$ of $kQ$ is generated by a set $Z$ of paths of length $\ge$ 2. We shall assume that $Z$ is minimal, that is, no proper subpath of a path in $Z$ is again in $Z$. Let $\mathcal{B}=\mathcal{B}_{\Lambda}$ be the set of paths of $Q$ which do not contain any element of $Z$ as a subpath. It is clear that the (classes modulo $I$ of) elements of $\mathcal{B}$ form a basis of $\Lambda$. We shall denote by $\mathcal{B}_{n}$ the subset $Q_{n} \cap \mathcal{B}$ of $\mathcal{B}$ formed by the paths of length $n$. Moreover, we shall use $E \simeq kQ_{0}$ to denote the separable subalgebra of $\Lambda$ generated by the (classes modulo $I$ of the) vertices of $Q$.

{\bf Hochschild cohomology of monomial algebras.} The Hochschild cohomology $\mathrm{HH}^{*}(\Lambda,\Lambda)=\mathrm{Ext}^{*}_{\Lambda^{e}}(\Lambda,\Lambda)$ of the $k$-algebra $\Lambda$ can be computed using different projective resolutions of $\Lambda$ over its enveloping algebra $\Lambda^{e}=\Lambda \otimes_{k} \Lambda^{op}$.  We usually denote H$^{*}$($\Lambda,\Lambda$) by $\mathrm{HH}^{*}$($\Lambda$). The zero-th Hochschild cohomology group $\mathrm{HH}^{0}(\Lambda)$ is identified as the center $Z(\Lambda)$ of the algebra $\Lambda$. In particular, $Z(\Lambda)$ is a commutative subalgebra of $\Lambda$. In order to compute the first Hochschild cohomology groups $\mathrm{HH}^{1}(\Lambda)$ of a finite dimensional monomial algebra $\Lambda$, one can use the minimal projective resolution of the $\Lambda$-bimodule $\Lambda$ given by Bardzell in \cite{Ba}. The part of this resolution $\mathcal{P}_{min}$ in which we are interested is the following:
\begin{center}
	\begin{tikzcd}
	\cdots \arrow[r] & \Lambda \otimes_{E} kZ \otimes_{E} \Lambda \arrow[r, "\delta_{1}"] & \Lambda \otimes_{E} kQ_{1} \otimes_{E} \Lambda \arrow[r, "\delta_{0}"] & \Lambda \otimes_{E} \Lambda \arrow[r, "\pi"] & \Lambda \arrow[r] & 0,
	\end{tikzcd}
\end{center}
where the $\Lambda$-bimodule morphisms are given by
\begin{equation*}
\begin{split}
\pi(\lambda \otimes_{E} \mu) =& \lambda\mu \\
\delta_{0}(\lambda \otimes_{E} a \otimes_{E} \mu) =& \lambda a \otimes_{E} \mu - \lambda \otimes_{E} a\mu \text{ and} \\
\delta_{1}(\lambda \otimes_{E} p \otimes_{E} \mu) =& \sum_{d=1}^{n} \lambda a_{n}\cdots a_{d+1} \otimes_{E} a_{d} \otimes_{E} a_{d-1} \cdots a_{1} \mu
\end{split}
\end{equation*}
for all $\lambda,\mu \in \Lambda, a,a_{n},\cdots,a_{1} \in Q_{1}$ and $p=a_{n}\cdots a_{1} \in Z$ (with the conventions $a_{n+1}=t(a_{n})$ and $a_{0}=s(a_{1})$).

Applying the contravariant functor Hom$_{\Lambda^{e}}(-,\Lambda)$ to $\mathcal{P}_{min}$ we obtain the following cochain complex $\mathcal{C}_{min}$ (see \cite[Section 2]{STR}):
\begin{center}
		\begin{tikzcd}
		0 \arrow[r] & Hom_{E^e}(kQ_{0},\Lambda) \arrow[r, "{\delta_{0}}^*"] & Hom_{E^e}(kQ_{1},\Lambda) \arrow[r, "{\delta_{1}}^*"] & Hom_{E^e}(kZ,\Lambda) \arrow[r, "{\delta_{2}}^*"] & \cdots,
		\end{tikzcd}	
	\end{center}
where the differentials are given by
	\begin{equation*}
	\begin{split}
	({\delta_{0}}^*f)(a)=af(s(a))-f(t(a))a \quad and\\
	({\delta_{1}}^*g)(p)=\sum_{d=1}^{n}a_n\cdots a_{d+1}g(a_d)a_{d-1}\cdots a_1.
	\end{split}
	\end{equation*}
where $f\in$ Hom$_{E^e}(kQ_{0},\Lambda)$, $a,a_n,\cdots,a_1\in Q_1$, $g\in$ Hom$_{E^e}(kQ_{1},\Lambda)$ and $p=a_n\cdots a_1\in Z$.
In particular, we have $\mathrm{HH}^{1}(\Lambda)\simeq\mathrm{Ker}({\delta_{1}}^*)/\mathrm{Im}({\delta_{0}}^*)$ as $k$-spaces, where the elements in $\mathrm{Ker}$(${\delta_{1}}^*$) can be interpreted as the $E^e$-derivations of $kQ_1$ into $\Lambda$ and the elements in $\mathrm{Im}$(${\delta_{0}}^*$) can be interpreted as the inner $E^e$-derivations of $kQ_1$ into $\Lambda$. It is well-known that $\mathrm{Ker}$(${\delta_{1}}^*$) is a Lie algebra under the Lie bracket $[f,g]=f\circ g-g\circ f$ and $\mathrm{Im}$(${\delta_{0}}^*$) is a Lie ideal of $\mathrm{Ker}$(${\delta_{1}}^*$), so that $\mathrm{HH}^{1}$($\Lambda$) has a Lie algebra structure.

A practical way of computing $\mathrm{HH}^{1}(\Lambda)$ of a monomial algebra $\Lambda$ is given by Strametz in \cite{STR}.

\begin{Prop}\label{Strametz-proposition2.6} {\rm(\cite[Proposition 2.6]{STR})}
\label{straCmon}
	Let $\Lambda$ be a finite dimensional monomial algebra. Then the above cochain complex $\mathcal{C}_{min}$ is isomorphic to the following cochain complex $\mathcal{C}_{mon}$:
	\begin{center}
		\begin{tikzcd}
		0 \arrow[r] & k(Q_{0}\| \mathcal{B}) \arrow[r, "\delta^{0}"] & k(Q_{1}\| \mathcal{B}) \arrow[r, "\delta^{1}"] & k(Z\| \mathcal{B}) \arrow[r, "\delta^{2}"] & \cdots,
		\end{tikzcd}	
	\end{center}
	where the differentials are given by
	\begin{equation*}
	\begin{split}
	\delta^{0}:k(Q_{0}\| \mathcal{B}) &\to k(Q_{1}\| \mathcal{B})\\
	e\| \gamma &\mapsto \sum_{s(a)=e,a\gamma \in \mathcal{B}}a\| a\gamma - \sum_{t(a)=e,\gamma a \in \mathcal{B}}a\| \gamma a,\\
	\delta^{1}:k(Q_{1}\| \mathcal{B}) &\to k(Z \| \mathcal{B})\\
	a\| \gamma &\mapsto \sum_{r \in Z}r\| r^{a \| \gamma},
	\end{split}
	\end{equation*}
where $r^{a \| \gamma}$ denotes the sum of all paths in $\mathcal{B}$ obtained by replacing each appearance of the arrow $a$ in $r$ by the path $\gamma$. In particular, we have $\mathrm{HH}^{1}$($\Lambda$)$\simeq$ $\mathrm{Ker}$($\delta^{1}$)/$\mathrm{Im}$($\delta^{0}$) as $k$-spaces, where this isomorphism is induced by the following map: send each $f$ in Hom$_{E^{e}}(kQ_{1},k\mathcal{B})$ to be an element $\underset{a\| \gamma\in Q_1\| \mathcal{B}}{\sum}\lambda_{a,\gamma}(a\| \gamma)$ in $k(Q_{1}\| \mathcal{B})$, where $f(a)=\underset{\gamma\in \mathcal{B}}\sum \lambda_{a,\gamma}\gamma$. Moreover, the inverse of the above isomorphism is induced by the following map: send an element $a \| \gamma$ in $k(Q_{1}\| \mathcal{B})$ to be a map $f$ in Hom$_{E^{e}}(kQ_{1},k\mathcal{B})$ with $f(a)=\gamma$ and $f(b)=0$ for $a\neq b\in Q_1$.
\end{Prop}

\begin{Thm}\label{Strametz-theorem2.7} {\rm(\cite{STR})}
Let $\Lambda$ be a finite dimensional monomial algebra. Then the bracket $$\left[a \| \gamma,b \| \epsilon \right] = b\| \epsilon^{a \| \gamma}-a\| \gamma^{b \| \epsilon}\quad \quad (a\| \gamma,  b\| \epsilon \in Q_{1} \| \mathcal{B})$$
induces a Lie algebra structure on $\mathrm{Ker}(\delta^{1})/\mathrm{Im}(\delta^{0})$ such that the natural isomorphism
$$\mathrm{HH}^{1}(\Lambda)\simeq \mathrm{Ker}(\delta^{1})/\mathrm{Im}(\delta^{0})$$
is a Lie algebra isomorphism.
\end{Thm}

If the field $k$ has positive characteristic $p$, then $\mathrm{HH}^1(A)$ is a restricted Lie algebra, that is, it is a Lie algebra endowed with a map called $p$-power map that satisfies some compatibility properties with respect to the Lie algebra structure. For further background on
restricted Lie algebras see for example \cite[Chapter 2]{FS}. The $p$-power map of a derivation $f$ is  defined by composing $f$ with itself $p$-times. The inner derivations form a restricted Lie ideal of space of derivations, therefore $\mathrm{HH}^1(A)$ is a restricted Lie algebra.For monomial algebras, it is easy to describe the $p$-power map using the chain map from  $\mathcal{C}_{min}$ to  $\mathcal{C}_{mon}$ and its inverse chain map. For example, for $p=3$, the $p$-power map of $a\| \gamma$  is $(a \| \gamma^{ a \| \gamma})^{ a \| \gamma}$. We remark that some of results in the present paper, such as Proposition \ref{Kernel-delta-one-injective} and Corollary \ref{quotient-lie-algebra-source-and-sink}, can be easily generalized from `Lie' level to `restricted Lie' level.

\begin{Rem}\label{center-ker-zero} The center $Z(\Lambda)$ of $\Lambda$ is naturally isomorphic to $\mathrm{Ker}(\delta^{0})$. For a concrete map between $\mathrm{Ker}(\delta^{0})$ and $Z(\Lambda)$, see the proof of Proposition \ref{center-from-the-same-block}.
\end{Rem}

According to \cite[Section 4]{STR}, the Lie algebra $\mathrm{HH^1}(\Lambda)$ of a monomial algebra $\Lambda=kQ/\langle Z \rangle$ has a natural graduation. Actually, if $a\|\gamma \in Q_1\| \mathcal{B}_n$ and $b\|\epsilon \in Q_1\| \mathcal{B}_m$, then the Lie bracket in Theorem \ref{Strametz-theorem2.7} shows that $\left[ a \| \gamma,b \| \epsilon \right] \in k(Q_1\| \mathcal{B}_{n+m-1})$. Thus, we have a graduation on the Lie algebra $k(Q_1\| \mathcal{B})=\oplus_{i\in \mathbb{N}} k(Q_1\| \mathcal{B}_i)$ by considering that the elements of $k(Q_1\| \mathcal{B}_i)$ have degree $i-1$ for all $i\in \mathbb{N}$. It is clear that the Lie subalgebra $\mathrm{Ker}(\delta^{1})$ of $k(Q_1\| \mathcal{B})$ preserves this graduation and that $\mathrm{Im}(\delta^{0})$ is a graded ideal, which induces a graduation on the Lie algebra $\mathrm{HH^1}(\Lambda)\simeq \mathrm{Ker}(\delta^{1})/\mathrm{Im}(\delta^{0})$. More precisely, if we set $$L_{-1}:=k(Q_1\|Q_0)\cap \mathrm{Ker}(\delta^{1}),$$ $$L_0:=k(Q_1\|Q_1)\cap \mathrm{Ker}(\delta^{1})/\langle \delta^0(e\|e)\ |\ e\in Q_0 \rangle \ and$$  $$L_i:=k(Q_1\|\mathcal{B}_{i+1})\cap \mathrm{Ker}(\delta^{1})/\langle \delta^0(e\|p)\ |\ e\|p \in Q_1\| \mathcal{B}_i \rangle$$
for all $i\geq 1,i\in \mathbb{N}$, then $\mathrm{HH^1}(\Lambda)=\bigoplus_{i\geq -1}L_i$ and $\left[ L_i,L_j \right]\subset L_{i+j}$ for all $i,j\geq -1$, where $L_{-2}=0$.

\begin{Rem}\label{graded-HH-one}
Note that if the characteristic of the field $k$ is equal to 0, then $L_{-1}=0$ since there exists for every loop $a\|e\in Q_1\|Q_0$ a relation $r=a^m \in Z$ for some $m\geq 2$ such that $\delta^1(a\|e)$ has a summand $m r\|a^{m-1}$ which can not be cancelled because $\delta^1(a\|e)\neq 0$. It follows that $\bigoplus_{i\geq 1}L_i$ is a solvable Lie ideal of $\mathrm{HH^1}(\Lambda)=\bigoplus_{i\geq 0}L_i$ since $\mathrm{HH^1}(\Lambda)$ is finite dimensional. It is also obvious that $L_0$ is a Lie subalgebra of $\mathrm{HH^1}(\Lambda)$.
\end{Rem}

For the quiver $Q$, the parallelism is an equivalence relation on the set of arrows $Q_1$; for $\alpha\in Q_1$, $[\alpha]$ denotes the equivalence class of $\alpha$. We denote $\bar{Q}_1$ the set of equivalence classes of parallel arrows. The quiver which has $Q_0$ as vertices and
$\bar{Q}_1$ as set of arrows, will be denoted by $\bar{Q}$. We denote by $\chi(\bar{Q})$ the first Betti number of $\bar{Q}$ (see Section \ref{fundsec}). In order to ensure each $L_0^{[\alpha]}$ (in the Lie algebra decomposition $(\dag)$ of $L_0$ below) to be a Lie ideal, we need to use the following variation of \cite[Proposition 4.7]{STR}).

\begin{Prop}\label{basis-L_0} The basis $\mathcal{B}_{L_0}$ of $L_0$ is given by the union of the following sets:
	\begin{itemize}
		\item [$(i)$] all the elements $a\|b \in L_0$ such that $a\neq b$;
		\item [$(ii)$] for every class of parallel arrows $[\alpha]=\{\alpha_1,\alpha_2,\cdots,\alpha_n \} \in \bar{Q}_1$, all the elements $\alpha_i \| \alpha_i-\alpha_n\| \alpha_n \in L_0$ such that $i< n$;
		\item [$(iii)$] for each (oriented or undirected) cycle in $\bar{Q}$, choose one class of parallel arrows $[\alpha]=\{\alpha_1,\alpha_2,\cdots,\alpha_n \}$ in this cycle and take $\alpha_n\| \alpha_n$ (there are $\chi(\bar{Q})$ linearly independent such elements).
		\end{itemize}
\end{Prop}

For each class of parallel arrows $[\alpha] \in \bar{Q}_1$ we denote by $L_0^{[\alpha]}$ the Lie ideal of $L_0$ generated by the elements $\alpha_i\|\alpha_j\in \mathcal{B}_{L_0}$ and $\alpha_i \| \alpha_i-\alpha_n\| \alpha_n\in \mathcal{B}_{L_0}$ where $[\alpha]=\{\alpha_1,\alpha_2,\cdots,\alpha_n\}$. Obviously the Lie algebra $L_0$ is the product of these Lie algebras: $$L_0=\prod_{[\alpha] \in \bar{Q}_1} L_0^{[\alpha]},\quad \quad (\dag)$$
where the above decomposition depends on the choice of the basis $\mathcal{B}_{L_0}$.

To describe the generators of the center $Z(L_0)$ of $L_0$, we adopt the definitions introduced by Strametz in \cite{STR}: for every class of parallel arrows $[\alpha]$ of $\bar{Q}_1$ we call a set $C \subset [\alpha]$ connected, if for every two arrows $\alpha_1$ and $\alpha_m$ of $C$ there exist arrows $\alpha_2,\cdots,\alpha_{m-1} \in C$ such that we have $r^{\alpha_{i+1}\|\alpha_i}=0$ or $r^{\alpha_i\|\alpha_{i+1}}=0$ for all $1\le i\le m-1$ and all $r\in Z$. A connected set $C \subset [\alpha]$ is called a connected component of $[\alpha]$ if it is maximal for the connection, that is, for every arrow $\beta\in [\alpha]\backslash C$ there is no arrow $\alpha \in C$ such that $r^{\beta\|\alpha}=0$ or $r^{\alpha\|\beta}=0$ for all $r\in Z$.

\begin{Rem}
In fact, it is straightforward to check that $r^{\alpha\|\beta}=0$ for all $r\in Z$ is equivalent to $\alpha\|\beta \in \mathrm{Ker}(\delta^1)$, where $\alpha\|\beta \in Q_1\| Q_1$.
\end{Rem}

\begin{Prop}\label{center-L_0} {\rm(\cite[Lemma 4.20]{STR})}
	Let $Q$ be a connected quiver and $\Lambda=kQ/\langle Z \rangle$ a finite dimensional monomial algebra.
	\begin{itemize}
		\item [$(i)$] The center $Z(L_0)$ of the Lie algebra $L_0$ is generated by the elements $\sum_{a\in C}a\|a$, where $C$ denotes a connected component of a class of parallel arrows of $Q$.
		\item [$(ii)$] If the field $k$ has characteristic 0, then the center $Z(\mathrm{HH^1}(\Lambda))$ of the Lie algebra $\mathrm{HH^1}(\Lambda)$ is contained in the center $Z(L_0)$ of $L_0$.
\end{itemize}
\end{Prop}

Since $L_0=\prod\limits_{[\alpha] \in \bar{Q}_1} L_0^{[\alpha]}$ as Lie algebras, we have $Z(L_0)=\prod\limits_{[\alpha] \in \bar{Q}_1} Z(L_0^{[\alpha]})$. And we have the parallel conclusion for $Z(L_0^{[\alpha]})$, in fact, $Z(L_0^{[\alpha]})=\bigoplus\limits_C \langle \sum\limits_{a\in C} a\|a \ |\  \sum\limits_{a\in C} a\|a \in L_0^{[\alpha]}\rangle$ as abelian Lie algebras for each $[\alpha]\in \bar{Q}_1$, where $C$ takes over all the connected components of $[\alpha]$.

Let $A$ be a finite dimensional algebra. By the Wedderburn--Malcev theorem we have $A=E\oplus rad(A)$. Another complex which computes Hochschild cohomology and which very helpful for calculations is the reduced bar complex:
\[ 0 \rightarrow A^{E}\rightarrow \mathrm{Hom}_{E^e}(r,A)\rightarrow \dots \rightarrow \mathrm{Hom}_{E^e}(r^{\otimes^{n}_E}, A)\rightarrow \mathrm{Hom}_{E^e}(r^{\otimes^{n+1}_E}, A)\rightarrow \dots
\]
where $A^E=\{a\in A | ae=ea \textrm{ for all } e\in E\}$,  $r:=rad(A)$ and $r^{\otimes^{n}_E}$ denotes the $n$-th fold tensor product over $E$. The differential is described in \cite[Proposition 2.2]{C}.

For radical square zero algebras, Cibils \cite{C} provides  an isomorphic complex to the reduced bar complex
which can be described entirely  in terms of the combinatorics of the quiver.

We recall  very briefly some constructions from \cite{C}: The author denotes by $
(kQ)_2$ the quotient of the path algebra of $Q$ by the two-sided ideal generated by paths of length $2$. From
Gabriel's theorem it follows that every radical square zero algebra $\Lambda$ over an algebraically closed field $k$ is
Morita equivalent to an algebra $(kQ)_2$. The algebra  $\Lambda$ is assumed to be
indecomposable, therefore the Gabriel quiver $Q$ of $\Lambda$ is connected and finite. By Lemma 2.1 in \cite{C} we have that $r^{\otimes_E^n}$ has
basis given by $Q_n$. By Lemma 2.2 in \cite{C}  the vector space $\mathrm{Hom}_{E^{e}}(r^{\otimes_E^n}, \Lambda)$ is isomorphic to $k (Q_n\| Q_0 ) \oplus k(Q_n\| Q_1 )$. The differential is defined as follows (cf. \cite[Proposition 2.4]{C2}):
$$k(Q_n\| Q_0)\oplus k(Q_n\| Q_1) \xrightarrow{\delta^n} k(Q_{n+1}\| Q_0)\oplus k(Q_{n+1}\| Q_1),$$
where $\delta^n$ is the $2\times2$-matrix that has  $D_{n}$ in position $(2,1)$ and $0$ elsewhere, where $D_n$ is  defined as:
$$D_{n}(\gamma\| e)=\sum_{s(a)=e ,a\in (Q_A)_1}
 a\gamma \| a+(-1)^{n+1}\sum_{t(b)=e,b\in (Q_A)_1} \gamma b\| b.$$

{\bf Fundamental groups of a bound quiver.}
The elements of $I$  of an ideal of the path algebra $kQ$ are usually called relations. A relation $r$ is minimal
in $I$ if it is a nonzero relation $r = \sum_{i=1}^s a_ip_i$,
where the $p_i$ are distinct parallel paths in $Q$ and $a_i \in k\setminus \{0\}$, such that there is no proper nonempty subset $T \subset \{1, \dots , s\}$ for which $\sum_{i\in T}  a_i p_i \in I $.
In order to construct the fundamental group of a bound quiver, the first step is to consider for each arrow $\alpha \in Q_1$ the formal inverse $\alpha^{-1}$ which is an arrow such that $s(\alpha^{-1})=t(\alpha)$ and
$t(\alpha^{-1})=s(\alpha)$. Then we consider the double quiver $\bar{\bar{Q}}$ where $\bar{\bar{Q}}_0=Q_0$ and
$\bar{\bar{Q}}_1=Q_1 \cup \{\alpha^{-1} \ | \alpha\in Q_1\}$. A walk is an oriented path in $\bar{\bar{Q}}$. As for the classical fundamental group, one needs to define an homotopy relation $\sim_I$ which is an equivalence relation on the set of walks in $\bar{\bar{Q}}$ generated by:

\begin{itemize}
\item $\alpha\alpha^{-1}\sim_I t(\alpha)$,  $\alpha^{-1}\alpha\sim_I s(\alpha)$,
\item if $v \sim_I  v^{'}$, then $wvu \sim_I wv^{'}u$,
\item  if $u$ and $v$  are paths which occur with a nonzero coefficient in the
same minimal relation, then $u \sim_I v$.
\end{itemize}

It is worth noting that the definition of the fundamental group does depend on the relation, that is, two different presentations $I$ and $J$ of a finite dimensional algebra $A$ have different fundamental groups, see for example \cite[Example 1.2]{LM}.
Let $e_i\in Q_0$, then we denote by $\pi(Q,I, e_i)$ the set of equivalence classes of
walks having source and target $e_i$. This set is endowed with a group structure given by
concatenation of walks. The unit is the trival walk $e_i$. The group $\pi(Q,I, e_i)$ is called the fundamental group of the bound quiver $(Q,I)$ based at $e_i$.

\section{First Hochschild cohomology}
\label{HH1sec}

We assume that $A$ is a finite dimensional algebra which is isomorphic to $kQ_A/I$, where $k$ is a field with characteristic $\geq 0$, $Q_A$ is a finite quiver (with vertices $e_1,\cdots,e_n$) and $I$ is an admissible ideal in the path algebra $kQ_A$. Throughout we assume that $e_1\neq e_n$ and we exclude the case that $e_1$ or $e_n$ is an isolated vertex.

Having in mind the cochain complex $\mathcal{C}_{mon}$ in the previous section, in order to understand the behaviour of the first Hochschild cohomology under  idempotent gluings, we should start by considering how idempotent gluings behave with respect to parallelism of arrows and paths.

\begin{Lem} \label{parallel-arrows}
Let $B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_{1}$ and $e_{n}$ of $A$. Let $\alpha,\beta \in (Q_{A})_{1}$. If $\alpha \| \beta$, then $\alpha^{*} \| \beta^{*}$, where $\varphi:Q_{A} \to Q_{B}$ sends $\alpha$ to $\alpha^{*}$ (cf. Notations in Section \ref{pre}).
\end{Lem}

\begin{Proof}
Recall that we denote the vertices of $Q_{B}$ by $f_1,\cdots,f_{n-1}$, where $f_1$ is obtained by gluing $e_1$ and $e_n$. Let $\alpha,\beta \in (Q_{A})_{1}$ and $\alpha \| \beta$. We may assume that $s(\alpha)=s(\beta)=e_{i}$ and $t(\alpha)=t(\beta)=e_{j}$. It is clear that	
	$$s(\alpha^{*})=\left\{
	\begin{array}{rcl}
	f_{i} &  & {for\ 2\le i \le n-1} \\
	f_{1} &  & {for\ i=1 \  or \  n}
	\end{array},
	\right.$$	
	and	
	$$t(\alpha^{*})=\left\{
	\begin{array}{rcl}
	f_{j} &  & {for\ 2\le j \le n-1} \\
	f_{1} &  & {for\ j=1\  or\  n}
	\end{array}.
	\right.$$
	Similarly, for $\beta^{*}$, we have	
	$$s(\beta^{*})=\left\{
	\begin{array}{rcl}
	f_{i} &  & {for\  2\le i \le n-1} \\
	f_{1} &  & {for\  i=1 \  or \  n}
	\end{array},
	\right.$$
	and
	$$t(\beta^{*})=\left\{
	\begin{array}{rcl}
	f_{j} &  & {for\  2\le j \le n-1} \\
	f_{1} &  & {for\  j=1 \  or \  n}
	\end{array}.
	\right.$$
	Hence, if $\alpha \| \beta$, then $s(\alpha^{*})=s(\beta^{*})$ and $t(\alpha^{*})=t(\beta^{*})$, that is to say $\alpha^{*} \| \beta^{*}$.
\end{Proof}

\begin{Lem}\label{bijection-parallel-arrows}
Let $B$ be a radical embedding of $A$ obtained by gluing a source and a sink of $A$. Let $\alpha,\beta \in (Q_{A})_{1}$. Then $\alpha \| \beta$ if and only if $\alpha^{*} \| \beta^{*}$.
\end{Lem}

\begin{Proof}
	The sufficiency is obvious by Lemma \ref{parallel-arrows}, it suffices to show the necessity. If $\alpha^{*} \| \beta^{*}$, then to show $\alpha \| \beta$ we need to use the assumption that we are gluing a source, say $e_{1}$, and a sink, say $e_{n}$. We show that if $\alpha \nparallel \beta$, then $\alpha^{*} \nparallel \beta^{*}$. If $\alpha \nparallel \beta$, then either $s(\alpha) \neq s(\beta)$ or $t(\alpha) \neq t(\beta)$. Assume $s(\alpha)=e_{i} \neq e_{j}=s(\beta)$ where $i\neq j$. We consider three cases:
	
	a) If $2\le i \le n-1,1 \le j \le n$ and $i \ne j$, then
	$$s(\alpha^{*})=f_{i} \ne s(\beta^{*})=\left\{
	\begin{array}{rcl}
	f_{j} &  & {for\ 2\le j \le n-1} \\
	f_{1} &  & {for\  j=1\  or\  n}
	\end{array},
	\right.$$
	which means $\alpha^{*} \nparallel \beta^{*}$.
	
	b) If $i=1,1 \le j \le n$ and $i\ne j$, then $s(\alpha^{*})=f_{1}$, $s(\beta^{*})=\left\{
	\begin{array}{rcl}
	f_{j} &  & {for\ 2\le j \le n-1} \\
	f_{1} &  & {for\ j=1 \  or \  n}
	\end{array}
	\right.$.
	Since $i \ne j, s(\beta^{*})=\left\{
	\begin{array}{rcl}
	f_{j} &  & {for\ 2\le j \le n-1} \\
	f_{1} &  & {for\ j=n}
	\end{array}
	\right.$. Only when $j=n$, we have $s(\beta^{*})=f_{1}=s(\alpha^{*})$. That is to say $s(\beta)=e_{n}$. But this is not possible since $e_{n}$ is a sink, so there no exists $\beta$ such that $s(\beta)=e_{n}$. Hence $s(\alpha^{*}) \ne s(\beta^{*})$, which means $\alpha^{*} \nparallel \beta^{*}$.

    c) We can deduce the same for $i=n$, $1 \le j \le n$ and $i \ne j$.
	
	Similar arguments apply if we assume $t(\alpha) \ne t(\beta)$.
\end{Proof}

We now extend partly the above results to parallel paths for monomial algebras. Recall that for a monomial algebra $A=kQ_A/I_A$, there is a $k$-basis $\mathcal{B}_A$ of $A$ consisting of paths of $Q_A$ which do not contain any element of $Z_A$ as a subpath, where $Z_A$ is a minimal generating set of the ideal $I_A$.

\begin{Prop}\label{parallel-paths-in-monomial-algebras}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents of $A$. Let $\varphi:Q_{A} \to Q_{B}$ be the quiver morphism defined as in Section \ref{pre}. Then we have the following.

$(1)$ $\varphi: Q_{A} \to Q_{B}$ induces a surjective map $\widetilde{\varphi}: \mathcal{B}_A\rightarrow \mathcal{B}_B$ such that $\widetilde{\varphi}^{-1}(p^*)=\{p\}$ for $p^*\neq f_1$ and $\widetilde{\varphi}^{-1}(f_1)=\{e_1,e_n\}$, where we denote $\widetilde{\varphi}(p)$ by $p^*$ for $p\in \mathcal{B}_A$.

$(2)$ Let $p,q \in \mathcal{B}_A$. Then $p \| q$ in $Q_A$ implies $p^{*} \| q^{*}$ in $Q_B$.

$(3)$ $\widetilde{\varphi}: \mathcal{B}_A\rightarrow \mathcal{B}_B$ induces $k$-linear maps $\psi_0: k((Q_A)_0\| \mathcal{B}_A) \to k((Q_B)_0\| \mathcal{B}_B)$, $\psi_1: k((Q_A)_1\| \mathcal{B}_A) \to k((Q_B)_1\| \mathcal{B}_B)$, $\psi_2: k(Z_A\| \mathcal{B}_A) \to k(Z_B\| \mathcal{B}_B)$.
\end{Prop}

\begin{Proof} As before, suppose that the vertices of $Q_{A}$ are given by $e_1,\cdots,e_n$ and that the vertices of $Q_{B}$ are given by $f_1,\cdots,f_{n-1}$, where $f_1$ is obtained by gluing $e_1$ and $e_n$; for each arrow $\alpha$ in $Q_{A}$, we denote the corresponding arrow in $Q_{B}$ by $\alpha^{*}$. Then the quiver morphism $\varphi:Q_{A} \to Q_{B}$ is given by the following formula: $\varphi(e_{i})=f_{i}$ for $2\le i \le n-1$, $\varphi(e_{1})=\varphi(e_{n})=f_{1}$, and $\varphi(\alpha)=\alpha^{*}$ for each arrow $\alpha$. Notice that since $A$ is monomial, $B$ is also monomial.

The quiver morphism $\varphi: Q_{A} \to Q_{B}$ extends to a $k$-linear map $kQ_{A} \to kQ_{B}$ between path algebras by sending a path $p=a_1\cdots a_m$ ($a_i\in (Q_A)_1$ for $1\leq i\leq m$) in $Q_A$ to a path $p^*:=a_1^*\cdots a_m^*$ in $Q_B$. Since the newly formed relations in $I_B$ are of the forms $\cdot\rightarrow f_1\rightarrow\cdot$, $p\in \mathcal{B}_A$ implies $p^*\in \mathcal{B}_B$. Now the statement (1) follows from the fact that $\mathrm{dim}_kB=\mathrm{dim}_kA-1$, and the statements (2) and (3) follow from Lemma \ref{parallel-arrows}.
\end{Proof}


From now on, we fix $A=kQ_A/I_A$ and $B=kQ_B/I_B$ to be the monomial algebras as in Proposition \ref{parallel-paths-in-monomial-algebras}. Then we have the following diagram:

\begin{center}
		\begin{tikzcd}
		0 \arrow[r] & k((Q_A)_{0}\| \mathcal{B}_A) \arrow[d, "\psi_{0}"] \arrow[r, "\delta_A^{0}"] & k((Q_A)_{1}\| \mathcal{B}_A) \arrow[d, "\psi_{1}"] \arrow[r, "\delta_A^{1}"] & k(Z_A\| \mathcal{B}_A) \arrow[d, "\psi_{2}"]\\
        0 \arrow[r] & k((Q_B)_{0}\| \mathcal{B}_B) \arrow[r, "\delta_B^{0}"] & k((Q_B)_{1}\| \mathcal{B}_B) \arrow[r, "\delta_B^{1}"] & k(Z_B\| \mathcal{B}_B)~,
		\end{tikzcd}	
	    \quad \quad $(*)$
\end{center}
where $\psi_0: k((Q_A)_0\| \mathcal{B}_A) \to k((Q_B)_0\| \mathcal{B}_B)$, $\psi_1: k((Q_A)_1\| \mathcal{B}_A) \to k((Q_B)_1\| \mathcal{B}_B)$, $\psi_2: k(Z_A\| \mathcal{B}_A) \to k(Z_B\| \mathcal{B}_B)$ are the induced $k$-linear maps from the quiver morphism $\varphi: Q_{A} \to Q_{B}$ as mentioned in Proposition \ref{parallel-paths-in-monomial-algebras}. Note that the top and the bottom  complexes are truncations of the complexes $\mathcal{C}_{mon}$ of $A$ and of $B$, respectively.
Both squares in the diagram $(*)$ are not commutative in general, however, there are  close connections between the coboundary elements (resp. the cocycle elements) of the top complex and the coboundary elements (respectively the cocycle elements) of the bottom complex in the diagram $(*)$.

We briefly outline the structure of the rest of this section.  Firstly, we will compare $\mathrm{Im}(\delta^0_A)$ and $\mathrm{Im}(\delta^0_B)$. Then we will study the Lie algebra structure of $\mathrm{Ker}(\delta^1_A)$ and $\mathrm{Ker}(\delta^1_B)$. Lastly, we will compare the dimensions and the Lie structures of $\mathrm{HH}^1(A)$ and $\mathrm{HH}^1(B)$.
In order to express these connections more precisely, we need some definitions and a lemma.

\begin{Def}
\label{kerdeltaA0}
With Proposition \ref{straCmon} in mind, we define $\delta^0_{(A)_0}$ to be the map $\delta^0_A$ restricted to the subspace $k((Q_A)_0||(Q_A)_0)$. Then $\mathrm{Im}$$(\delta^0_{(A)_0})$ is the $k$-vector space generated by the image of $\delta^0_A$ on $e_i\| e_i$, where $e_i$ ($1\leq i\leq n$) are idempotents corresponding to vertices of $Q_A$. Similarly, we define $\mathrm{Ker}(\delta^0_{(A)_0})$ to be the kernel of the map $\delta^0_{(A)_0}$.
\end {Def}

\begin{Lem}\label{Image-zero-part} Let $A$ be a monomial algebra. Then
$\mathrm{dim}_k\mathrm{Im}(\delta^0_{(A)_0})=n_A-c_A$, where $n_A$ is the number of vertices of $Q_A$ and $c_A$ is the number of connected components
of $Q_A$.
\end{Lem}

\begin{Proof}
Note that it is enough to assume that $A$ is indecomposable. In fact, if it holds for each block $A_i$ of $A$, then
\begin {equation*}
\mathrm{dim}_k (\mathrm{Im}(\delta^0_{(A)_0}))= \sum_{A_i}(n_{A_i}-c_{A_i})=n_A-c_A.
\end {equation*}
Hence assume $A$ is indecomposable.
Note that:
\[\mathrm{dim}_k(k((Q_A)_0||(Q_A)_0)=n_A= \mathrm{dim}_k (\mathrm{Im}(\delta^0_{(A)_0}))+\mathrm{dim}_k (\mathrm{Ker}(\delta^0_{(A)_0})).\]
Consequently it is enough to show that $\mathrm{dim}_k (\mathrm{Ker}(\delta^0_{(A)_0}))=1$. It is straightforward to check that $\sum_{i=1}^{n_A} e_i||e_i$ is in $\mathrm{Ker}(\delta^0_{(A)_0})$. Consequently $\mathrm{Ker}(\delta^0_{(A)_0})$ has dimension at least one. We will prove by contradiction that the dimension of $\mathrm{Ker}(\delta^0_{(A)_0})$ is exactly $1$.

Assume the dimension of $\mathrm{Ker}(\delta^0_{(A)_0})$ is greater than $1$. Then we can assume without loss of generality that there exists $T\subsetneqq \{1, \dots, n_A\}$ such that $\sum_{i\in T} \lambda_i e_i||e_i$ is an element of  $\mathrm{Ker}(\delta^0_{(A)_0})$ where $\lambda_i$ are non-zero scalars. Indeed, if there exists an element $\sum_{i=1}^{n_A} \lambda_i e_i||e_i$ in $\mathrm{Ker}(\delta^0_{(A)_0})$, then by taking a linear combination with $\sum_{i=1}^{n_A} e_i||e_i$ we can always find such $T$. Consider the full subquiver $\overline{Q}$ having the vertices indexed by $T$. Since $Q_A$ is connected and since $T\subsetneqq \{1, \dots, n_A\}$, then
$\delta_A^0(\sum_{i\in T} \lambda_i e_i||e_i)$ has one summand of the form $c||c$ where $c$ is an arrow such that $s(c)\in \overline{Q}_0$ and  $t(c)\notin \overline{Q}_0$ (or $s(c)\notin \overline{Q}_0$ and  $t(c)\in \overline{Q}_0$). Since $c||c$ cannot be written  as a linear combination of other elements of $k((Q_A)_1||\mathcal{B}_A)$ and since $\lambda_i$ are non-zero, then $\sum_{i\in T} \lambda_i e_i||e_i$ is not in $\mathrm{Ker}(\delta^0)$. The statement follows.
\end{Proof}

Let $p$ be a path either from $e_1$ to $e_n$ or from $e_n$ to $e_1$ in $Q_A$. Then $p^*$ is an oriented cycle at $f_1$ in $Q_B$. If $p$ is a path from $e_1$ to $e_n$, then we have $$\delta_{B}^{0}(f_1 \| p^*)=\sum_{s(a)=e_n,a\in (Q_{A})_1,ap\in \mathcal{B}_A}a^*\| a^*p^*-\sum_{t(b)=e_1,b\in (Q_{A})_1,pb\in \mathcal{B}_A}b^*\| p^*b^*.$$
Note that we have omitted some zero terms in the above sum, for example, if $d\in (Q_{A})_1$ is an arrow staring at $e_1$, then $d^*\| d^*p^*$ appears as a term in the above sum, however, it is zero since $d^*p^*$ lies in $I_B$.
If $p$ is a path from $e_n$ to $e_1$, then we have $$\delta_{B}^{0}(f_1 \| p^*)=\sum_{s(a)=e_1,a\in (Q_{A})_1,ap\in \mathcal{B}_A}a^*\| a^*p^*-\sum_{t(b)=e_n,b\in (Q_{A})_1,pb\in \mathcal{B}_A}b^*\| p^*b^*.$$
As in the previous case, we have omitted some zero terms in the above sum. Moreover, in both cases, $\delta_{B}^{0}(f_1 \| p^*)$ is zero if and only if $ap,pa\in I_A$ for all $a\in (Q_{A})_1$. This observation leads to the following definition:

\begin{Def}\label{special-path}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$ and let $p$ be a path either from $e_1$ to $e_n$ or from $e_n$ to $e_1$ in $Q_A$. We call $p$ is a special path between $e_1$ and $e_n$ in $Q_A$ if $\delta_{B}^{0}(f_1 \| p^*)\neq 0$, or equivalently, if there exists some $a\in (Q_{A})_1$ such that $ap\notin I_A$ or $pa\notin I_A$.

We denote by $\mathrm{Sp}(1,n)$ the set of special paths between $e_1$ and $e_n$ in $Q_A$, and by $\mathrm{sp}(1,n)$ the number of these special paths. Furthermore, we denote by $Z_{sp}$ the $k$-subspace of $\mathrm{Im}(\delta^0_B)$ generated by the elements $\delta_B^0(f_1 \| p^*)$, where $p \in \mathrm{Sp}(1,n)$.
\end{Def}

\begin{Lem}\label{summand-disjoint}
The summands of $\delta_B^0(f_1\|p^*)$ and $\delta_B^0(f_i\|q^*)$ $(1\leq i\leq n-1)$ are disjoint, where $p$ is a path either from $e_1$ to $e_n$ or from $e_n$ to $e_1$ in $\mathcal{B}_A$, $q$ is a path in $\mathcal{B}_A$ but different from $p$.
\end{Lem}

\begin{Proof}
Without loss of generality, we assume that $p$ is a path from $e_1$ to $e_n$. Then $$\delta_{B}^{0}(f_1 \| p^*)=\sum_{s(\alpha)=e_n,\alpha\in (Q_{A})_1,\alpha p\in \mathcal{B}_A}\alpha^*\| \alpha^*p^*-\sum_{t(\beta)=e_1,\beta\in (Q_{A})_1,p\beta\in \mathcal{B}_A}\beta^*\| p^*\beta^*,$$ $$\delta_{B}^{0}(f_i \| q^*)=\sum_{s(a^*)=f_i,a^*\in (Q_{B})_1,a^*q^*\in \mathcal{B}_B}a^*\| a^*q^*-\sum_{t(b^*)=f_i,b^*\in (Q_{B})_1,q^*b^*\in \mathcal{B}_B}b^*\| q^*b^*.$$ Note that $\alpha^*\| \alpha^*p^*\neq a^*\| a^*q^*$ since
$p\neq q$. Also $\alpha^*\| \alpha^*p^*\neq b^*\| q^*b^*$, otherwise $\alpha=b$ and $\alpha p=qb$ which cause $p$ to be of the form $x_m\cdots x_2\alpha$ where $x_2, \cdots, x_m\in (Q_{A})_1$. It follows that $s(p)=s(\alpha)=e_n$, which is a contradiction. We can similarly show that $\beta^*\| p^*\beta^*\neq a^*\| a^*q^*$ and $\beta^*\| p^*\beta^*\neq b^*\| q^*b^*$.
\end{Proof}

\begin{Rem} \label{dimension-zsp}
\begin{itemize}
\item[$(1)$] The dimension of $Z_{sp}$ is equal to $\mathrm{sp}(1,n)$. This follows from the fact that the summands of $\delta_B^0(f_1\|p^*)$ and $\delta_B^0(f_1\|q^*)$ are disjoint for $p,q\in \mathrm{Sp}(1,n) \ (p\neq q)$ by Lemma \ref{summand-disjoint}.
\item[$(2)$] If $e_1$ and $e_n$ belong to different blocks of $A$ or $A$ is a radical square zero algebra, then $\mathrm{sp}(1,n)=0$.
\item[$(3)$]In general, the number $\mathrm{sp}(1,n)$ could be arbitrarily large, see Example \ref{ex-image-kernel}.
\end{itemize}
\end{Rem}

We can now compare the dimensions of $\mathrm{Im}(\delta^0_A)$ and $\mathrm{Im}(\delta^0_B)$:

\begin{Prop}\label{Iamge-delta-zero}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$. Let $\mathrm{sp}(1,n)$ be the number of special paths between $e_1$ and $e_n$ in $Q_A$. Then $$\mathrm{dim}_k\mathrm{Im}(\delta^0_A) = \mathrm{dim}_k\mathrm{Im}(\delta^0_B)+1+c_B-c_A-\mathrm{sp}(1,n),$$ where $c_A$ and $c_B$ are the number of connected components of $Q_A$ and $Q_B$ respectively. In particular, if we glue $e_1$ and $e_n$ from the same block of $A$, then $$\mathrm{dim}_k\mathrm{Im}(\delta^0_A)=\mathrm{dim}_k\mathrm{Im}(\delta^0_B)+1-\mathrm{sp}(1,n);$$ if we glue $e_1$ and $e_n$ from different blocks of $A$, then $$\mathrm{dim}_k\mathrm{Im}(\delta^0_A)=\mathrm{dim}_k\mathrm{Im}(\delta^0_B).$$
\end{Prop}

\begin{Proof} We keep the notations as in the proof of Proposition \ref{parallel-paths-in-monomial-algebras}, that is, the vertices of $Q_{A}$ are given by $e_1,\cdots,e_n$ and the vertices of $Q_{B}$ are given by $f_1,\cdots,f_{n-1}$, where $f_1$ is obtained by gluing $e_1$ and $e_n$. The quiver morphism $\varphi:Q_{A} \to Q_{B}$ is given by the following formula: $\varphi(e_{i})=f_{i}$ for $2\le i \le n-1$, $\varphi(e_{1})=\varphi(e_{n})=f_{1}$, and $\varphi(\alpha)=\alpha^{*}$ for each arrow $\alpha$. We begin with describing the basis elements in $\mathrm{Im}(\delta_{A}^{0})$ and in $\mathrm{Im}(\delta_{B}^{0})$.

Let $e_i \| p \in k((Q_A)_{0}\| \mathcal{B}_A)$. We consider two cases depending if $p=e_i$ or $p\neq e_i$.

($a1$) $p=e_i$ ($1\leq i\leq n$). We have $$\delta_{A}^{0}(e_i \| e_i)=\sum_{s(a)=e_i,a\in (Q_{A})_1}a\| a-\sum_{t(b)=e_i,b\in (Q_{A})_1}b\| b.$$
By Lemma \ref{Image-zero-part}, the subspace $\mathrm{Im}$($\delta^0_{(A)_0}$) of $\mathrm{Im}\delta^0_A$ generated by the elements of the form $\delta^0_A(e_i\| e_i)$ has dimension $n_A-c_A$.

($a2$) $p\neq e_i$. Then $p$ is an oriented cycle at $e_i$ and $$\delta_{A}^{0}(e_i \| p)=\sum_{s(a)=e_i,a\in (Q_{A})_1,ap\in \mathcal{B}_A}a\| ap-\sum_{t(b)=e_i,b\in (Q_{A})_1,pb\in \mathcal{B}_A}b\| pb.$$
Let denote $\mathrm{Im}$($\delta^0_{(A)_{\geq 1}}$) the $k$-vector space generated by the image of $\delta^0_A$ on $e_i\| p$ ($1\leq i\leq n$), where $p\in \mathcal{B}_A$ and $p\neq e_i$. It is clear that $$\mathrm{Im}(\delta^0_A) = \mathrm{Im}(\delta^0_{(A)_0})\bigoplus \mathrm{Im}(\delta^0_{(A)_{\geq 1}}).$$

Similarly, we let $f_i \| q \in k((Q_B)_{0}\| \mathcal{B}_B)$ and consider four cases.

($b_1$) $q=f_i$ ($1\leq i\leq n-1$). We have $$\delta_{B}^{0}(f_i \| f_i)=\sum_{s(a^*)=f_i,a^*\in (Q_{B})_1}a^*\| a^*-\sum_{t(b^*)=f_i,b^*\in (Q_{B})_1}b^*\| b^*.$$
By Lemma \ref{Image-zero-part}, the subspace $\mathrm{Im}$($\delta^0_{(B)_0}$) of $\mathrm{Im}\delta^0_B$ generated by the elements of the form $\delta^0_B(f_i\| f_i)$ has dimension $n_B-c_B$.

($b_2$) $q$ is an oriented cycle at $f_i$ and $i\neq 1$. By Proposition \ref{parallel-paths-in-monomial-algebras}, $q=p^*$ for some oriented cycle $p\in \mathcal{B}_A$ at $e_i$ ($2\leq i\leq n-1$). We have $$\delta_{B}^{0}(f_i \| p^*)=\sum_{s(a^*)=f_i,a^*\in (Q_{B})_1}a^*\| a^*p^*-\sum_{t(b^*)=f_i,b^*\in (Q_{B})_1}b^*\| p^*b^*=\psi_1(\delta_{A}^{0}(e_i \| p)).$$

($b_3$) $q$ is an oriented cycle at $f_1$ such that $q=p^*$ for some oriented cycle $p\in \mathcal{B}_A$ at $e_1$ or $e_n$. If $p$ is an oriented cycle at $e_1$, then we have $$\delta_{B}^{0}(f_1 \| p^*)=\sum_{s(a^*)=f_1,a^*\in (Q_{B})_1,a^*p^*\in \mathcal{B}_B}a^*\| a^*p^*-\sum_{t(b^*)=f_1,b^*\in (Q_{B})_1,p^*b^*\in \mathcal{B}_B}b^*\| p^*b^*=\psi_1(\delta_{A}^{0}(e_1 \| p)).$$
If $p$ is an oriented cycle at $e_n$, then we have $$\delta_{B}^{0}(f_1 \| p^*)=\sum_{s(a^*)=f_1,a^*\in (Q_{B})_1,a^*p^*\in \mathcal{B}_B}a^*\| a^*p^*-\sum_{t(b^*)=f_1,b^*\in (Q_{B})_1,p^*b^*\in \mathcal{B}_B}b^*\| p^*b^*=\psi_1(\delta_{A}^{0}(e_n \| p)).$$

($b_4$) $q$ is an oriented cycle at $f_1$ with $q=p^*$ for some $p\in \mathcal{B}_A$, where $p$ is a path either from $e_1$ to $e_n$ or from $e_n$ to $e_1$. We may assume $p$ is a special path since otherwise $\delta_{B}^{0}(f_1 \| p^*)$ is zero. Note that $q$ is of the form $f_1\stackrel{a^*}{\rightarrow}\cdots\stackrel{b^*}{\rightarrow}f_1$ and might be a loop at $f_1$. If $p$ is a path from $e_1$ to $e_n$, then we have $$\delta_{B}^{0}(f_1 \| p^*)=\sum_{s(a)=e_n,a\in (Q_{A})_1,ap\in \mathcal{B}_A}a^*\| a^*p^*-\sum_{t(b)=e_1,b\in (Q_{A})_1,pb\in \mathcal{B}_A}b^*\| p^*b^*.$$ If $p$ is a path from $e_n$ to $e_1$, then we have $$\delta_{B}^{0}(f_1 \| p^*)=\sum_{s(a)=e_1,a\in (Q_{A})_1,ap\in \mathcal{B}_A}a^*\| a^*p^*-\sum_{t(b)=e_n,b\in (Q_{A})_1,pb\in \mathcal{B}_A}b^*\| p^*b^*.$$In both cases, $\delta_{B}^{0}(f_1 \| p^*)$ is nonzero since $p$ is a special path.

Let denote $\mathrm{Im}$($\delta^0_{(B)_{\geq 1}}$) the $k$-vector space generated by the image of $\delta^0_B$ on $f_i\| q$ ($1\leq i\leq n-1$), where $q\in \mathcal{B}_B$ and $q\neq f_i$. Then we have
$$\mathrm{Im}(\delta^0_B) = \mathrm{Im}(\delta^0_{(B)_0})\oplus \mathrm{Im}(\delta^0_{(B)_{\geq 1}}).$$
We now claim that
  $$\mathrm{Im}(\delta^0_{(B)_{\geq 1}})=\psi_1({\mathrm{Im}(\delta^0_{(A)_{\geq 1}})})\oplus Z_{sp}.$$
It suffices to show that the summands of $\delta_B^0(f_1\|p^*)$ and the summands of elements in $\psi_1({\mathrm{Im}(\delta^0_{(A)_{\geq 1}})})$ are disjoint for $p\in \mathrm{Sp}(1,n)$. Since the element in $\psi_1({\mathrm{Im}(\delta^0_{(A)_{\geq 1}})})$ is of the form $\psi_1(\delta^0_A(e_i\ \|\ q))=\delta_B^0(f_i\ \|\ q^*)$, where $q$ is an oriented cycle at $e_i \ (1\leq i\leq n)$ (here we identify $f_n$ with $f_1$), the statement follows from Lemma \ref{summand-disjoint}.

From the above claim we see that there is a bijection between the basis elements of $\mathrm{Im}$($\delta^0_{(A)_{\geq 1}}$) and the basis elements of $\mathrm{Im}$($\delta^0_{(B)_{\geq 1}}$) that are not appeared in the above case ($b_4$). According to Remark \ref{dimension-zsp} (1), the dimension of $\mathrm{Im}$($\delta^0_{(B)_{\geq 1}}$) is equal to the dimension of $\mathrm{Im}$($\delta^0_{(A)_{\geq 1}}$) plus $\mathrm{sp}(1,n)$. Combining this and ($a_1$), ($b_1$) and the fact that $n_A=n_B+1$,  we get that the dimension of $\mathrm{Im}(\delta_{(A)_0}^0)$ is equal to the dimension of $\mathrm{Im}(\delta_{(B)_0}^0)$ plus $1+c_B-c_A$. Therefore we have $$\mathrm{dim}_k\mathrm{Im}(\delta^0_A)=\mathrm{dim}_k\mathrm{Im}(\delta^0_B)+1+c_B-c_A-\mathrm{sp}(1,n).$$

In particular, if we glue $e_1$ and $e_n$ from the same block of $A$, then we have $c_B=c_A$. However, if $e_1$ and $e_n$ are from two different blocks of $A$, then we have $c_B=c_A-1$ and $sp(1,n)=0$. We are done.
\end{Proof}

We obtain an important corollary that it will be useful for stable equivalences induced by idempotent gluings.

\begin{Cor}\label{Iamge-delta-zero-source-and-sink}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$. Then we have $$\mathrm{dim}_k\mathrm{Im}(\delta^0_A)=\mathrm{dim}_k\mathrm{Im}(\delta^0_B)+1+c_B-c_A$$
 under each of the following two conditions:
\begin{itemize}
		\item[$(i)$] $e_1$ is a source and $e_n$ is a sink;
		\item[$(ii)$] $A$ is a radical square zero algebra.
\end{itemize}
\noindent In particular, if we glue $e_1$ and $e_n$ from the same block of $A$, then $$\mathrm{dim}_k\mathrm{Im}(\delta^0_A)=\mathrm{dim}_k\mathrm{Im}(\delta^0_B)+1;$$ if $e_1$ and $e_n$ are from two different blocks of $A$, then we have $$\mathrm{dim}_k\mathrm{Im}(\delta^0_A)=\mathrm{dim}_k\mathrm{Im}(\delta^0_B).$$
\end{Cor}

\begin{Proof} It is clear that under the condition $(i)$ or $(ii)$ there is no special path between $e_1$ and $e_n$, so we have $\mathrm{sp}(1,n)=0$. If we glue $e_1$ and $e_n$ from the same block of $A$, then $c_B=c_A$; if $e_1$ and $e_n$ are from two different blocks of $A$, then $c_B=c_A-1$. Hence the result follows from Proposition \ref{Iamge-delta-zero}.
\end{Proof}

From now on, we often use the following assumption on the characteristic of the filed $k$.

\begin{Assumption}\label{assump}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$. For each loop $\alpha$ at $e_1$ or at $e_n$ with $\alpha^m\in Z_A$ for some $m\geq 2$, char$(k)\nmid m$.
\end{Assumption}

Clearly Assumption \ref{assump} holds when the characteristic of the field $k$ is zero or big enough. We now proceed to compare the Lie structures of $\mathrm{Ker}(\delta^1_A)$ and $\mathrm{Ker}(\delta^1_B)$:

\begin{Prop}\label{Kernel-delta-one-injective}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$. If char$(k)$ satisfies Assumption \ref{assump}, then there exists an injective (restricted) Lie algebra homomorphism $\mathrm{Ker}(\delta^1_A)\hookrightarrow$ $\mathrm{Ker}(\delta^1_B)$ induced from $\psi_1: k((Q_A)_1\| \mathcal{B}_A) \to k((Q_B)_1\| \mathcal{B}_B)$, which we still denote by $\psi_1$.
\end{Prop}

\begin{Proof} First we notice that $I_A=\mathrm{Span}(Z_A)$ and $I_B=\mathrm{Span}(Z_B)$, and by obvious identification we can write $Z_B=Z_A\cup Z_{new}$, where $Z_{new}=\{b^*c^*\mid b^*,c^*\in (Q_B)_1, t(c^*)=f_1=s(b^*), bc\notin \mathcal{B}_A\}$.
Having the diagram  $(*)$  in mind,
let $\alpha \| p \in k((Q_A)_{1}\| \mathcal{B}_A)$ and let $\psi_1(\alpha \| p)=\alpha^{*} \| p^{*}$ be the corresponding element in $k((Q_B)_1\| \mathcal{B}_B)$. On the one hand, we have
$$\psi_2(\delta_{A}^{1}(\alpha \| p))=\psi_2(\sum_{r \in Z_A}r\| r^{\alpha \| p})=\sum_{r \in Z_A}r\| r^{\alpha^* \| p^*};$$
On the other hand, we have
$$\delta_{B}^{1}(\psi_1(\alpha \| p))=\delta_{B}^{1}(\alpha^* \| p^*)=\sum_{r \in Z_A}r\| r^{\alpha^* \| p^*}+\sum_{r' \in Z_{new}}r'\| {r'}^{\alpha^* \| p^*}.$$

We consider four cases.

(c1) $\alpha$ is a loop at $e_i$ ($2\leq i\leq n-1$), $p=e_i$ or $p$ is an oriented cycle at $e_i$. Then $\sum_{r' \in Z_{new}}r'\| {r'}^{\alpha^* \| p^*}=0$ since $\alpha^*$ does not appear in any $r'\in Z_{new}$. Therefore $\psi_2(\delta_{A}^{1}(\alpha \| e_i))=\delta_{B}^{1}(\psi_1(\alpha \| e_i))$.

(c2) $\alpha$ is a loop at $e_1$ (respectively $e_n$) and $p=e_1$ (resp. $p=e_n$). In case $p=e_1$, since $A$ is finite dimensional, $Z_A$ contains an element $r=\alpha^m$ for some $m\geq 2$, $\delta_{A}^{1}(\alpha\| e_1)$ contains a summand $m r \| \alpha^{m-1}$, which can not be cancelled in $\mathrm{Im}(\delta_{A}^{1})$ unless char$(k)\mid m$. That is, if char$(k) \nmid m$, then $\alpha \| e_1 \notin \mathrm{Ker}(\delta_A^1)$. On the other hand, $\alpha^*$ must appears in some $r'\in Z_{new}$ and therefore $\delta_{B}^{1}(\psi_1(\alpha \| e_1))=\delta_{B}^{1}(\alpha^* \| f_1)\neq 0$. Therefore, if char$(k) \nmid m$, then both $\alpha\| e_1\notin \mathrm{Ker}(\delta_{A}^{1})$ and $\psi_1(\alpha \| e_1)\notin \mathrm{Ker}(\delta_{B}^{1})$ hold. The similar result holds in case $p=e_n$.

(c3) $\alpha$ is a loop at $e_1$ (resp. $e_n$) and $p$ is an oriented cycle at $e_1$ (resp. $e_n$). Since once we replace $\alpha^*$ in any $r'\in Z_{new}$ by $p^*$, $r'$ becomes a path in $Q_B$ that still contains some relation in $Z_{new}$, we have $\sum_{r' \in Z_{new}}r'\| {r'}^{\alpha^* \| p^*}=0$. Therefore $\psi_2(\delta_{A}^{1}(\alpha \| p))=\delta_{B}^{1}(\psi_1(\alpha \| p))$.

(c4) $\alpha$ is a non-loop arrow and $p\in \mathcal{B}_A$ is a parallel path to $\alpha$. In this case it is easy to see that once $\alpha^*$ appears in some $r'\in Z_{new}$ (note that $r'\notin \mathcal{B}_B$), the element obtained from $r'$ by replacing $\alpha^*$ by $p^*$ is again not in $\mathcal{B}_B$. We still have $\sum_{r' \in Z_{new}}r'\| {r'}^{\alpha^* \| p^*}=0$. Therefore $\psi_2(\delta_{A}^{1}(\alpha \| p))=\delta_{B}^{1}(\psi_1(\alpha \| p))$.

The above discussion shows that, if char$(k)$ satisfies Assumption \ref{assump}, then there is a $k$-linear map $\psi_{1}: \mathrm{Ker}(\delta_{A}^{1})\longrightarrow \mathrm{Ker}(\delta_{B}^{1})$ induced from the following mapping: $\alpha\| e_i\mapsto \alpha^*\| f_i$ ($i\neq 1,n$), $\alpha\| p\mapsto \alpha^*\| p^*$ ($p\in \mathcal{B}_A$ has length $\geq 1$). It is also clear that $\psi_{1}: \mathrm{Ker}(\delta_{A}^{1})\longrightarrow \mathrm{Ker}(\delta_{B}^{1})$ is injective and preserves the Lie bracket, since $\psi_1$ preserves the parallel paths.
\end{Proof}

\begin{Rem}\label{no-loop-at-e1-or-en-Ker}
Since the characteristic condition only makes sense in the case (c2), if there is no loop both at $e_1$ and at $e_n$, then we do not need Assumption \ref{assump} in Proposition \ref{Kernel-delta-one-injective}. In particular, we do not need Assumption \ref{assump} in Proposition \ref{Kernel-delta-one-injective} when we glue a source vertex and a sink vertex.
\end{Rem}

\begin{Rem}\label{radical-square-zero}
$(1)$ If $A$ (hence also $B$) is a radical square zero algebra, then Assumption \ref{assump} is equivalent to $\mathrm{char}(k)\neq 2$ and we do not need this assumption in Proposition \ref{Kernel-delta-one-injective} when we glue $e_1$ and $e_n$ from the same block of $A$ or when we glue $e_1$ and $e_n$ from different blocks (say $A_1$ and $A_2$) of $A$ such that both $A_1$ and $A_2$ are not isomorphic to $k[x]/(x^2)$. In fact, the characteristic condition only makes sense in the case (c2), however, in each of the above two cases, the loop $\alpha$ must appears in a relation $\alpha\beta$ (where $\beta$ is an arrow different from $\alpha$) or in a relation $\gamma\alpha$ (where $\gamma$ is an arrow different from $\alpha$), and so both $\alpha\| e_1\notin \mathrm{Ker}(\delta_{A}^{1})$ and $\psi_1(\alpha \| e_1)\notin \mathrm{Ker}(\delta_{B}^{1})$ hold.

$(2)$ It is easy to see that $\mathrm{Ker}(\delta_{A}^{1})=k((Q_A)_1\|(Q_A)_1)$ when $A$ is a radical square zero algebra, except in the case that $\mathrm{char}(k)=2$ and $A\simeq k[x]/(x^{2})$ (for this case, see Remark \ref{exceptional-case}).
\end{Rem}

In order to describe the elements in $\mathrm{Ker}(\delta_{B}^{1})$ which are in the complement of the subspace $\psi_1(\mathrm{Ker}(\delta_{A}^{1}))$, we introduce some further notation.

\begin{Def}\label{elements-in-Z-spp}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$. Let $\alpha$ be an arrow and $p$ be a path in $\mathcal{B}_A$. We call $(\alpha,p)$ is a special pair with respect to the gluing of $e_1$ and $e_n$ if the following three conditions are satisfied:

$(1)$ $\alpha$ is starting from $e_1$, or ending at $e_1$, or starting from $e_n$, or ending at $e_n$;

$(2)$ $\alpha^* \| p^*$ in $Q_B$;

$(3)$ $\alpha \nparallel p$ in $Q_A$.

We denote by $\mathrm{Spp}(1,n)$ the set of all special pairs with respect to the gluing of $e_1$ and $e_n$ and by $\langle\mathrm{Spp}(1,n)\rangle$ the $k$-subspace of $k((Q_B)_1\| \mathcal{B}_B)$ generated by the elements $\alpha^* \| p^*$, where $(\alpha,p) \in \mathrm{Spp}(1,n)$. Furthermore, we denote by $Z_{spp}$ the intersection of $\langle\mathrm{Spp}(1,n)\rangle$ and $\mathrm{Ker}(\delta^1_B)$, and by $\mathrm{kspp}(1,n)$ the dimension of the $k$-subspace $Z_{spp}$ of $\mathrm{Ker}(\delta^1_B)$.
\end{Def}

Note that every nonzero element of $Z_{spp}$ is a linear combination of the parallel paths corresponding to special pairs and it gives a $E^e$-derivation in $\mathrm{Ker}(\delta^1_B)$ which lies in the complement of the subspace $\psi_1(\mathrm{Ker}(\delta_{A}^{1}))$. Moreover, the condition (1) can be derived from conditions (2) and (3) in Definition \ref{elements-in-Z-spp}, but for  convenience we  include condition (1). Note also that there are usually many types of special pairs, see Example \ref{example-type-special-pair}.

\begin{Rem}\label{radical-square-zero-of-spp}
Note that although in radical square zero case there is no special path in $Q_A$, there may exists special pair when we glue $e_1$ and $e_n$ whether from the same block or from two different blocks. In fact, if we glue two idempotents $e_1$ and $e_n$ from the same block and there is an arrow $\alpha$ connects $e_1$ and $e_n$, then both $(\alpha,e_1)$ and $(\alpha,e_n)$ are special pairs. More precisely, in this case the special pairs $(\alpha,p)$ have the following $5$ types (cf. Example \ref{example-type-special-pair}): \begin{itemize}
		\item[$(i)$] $\alpha$ is a loop at $e_1$ or $e_n$, $p$ is an arrow between $e_1$ and $e_n$;
		\item[$(ii)$] $\alpha$ is an arrow between $e_1$ and $e_n$, $p=e_1$ (or $e_n$) or $p$ is a loop at $e_1$ (or $e_n$);
		\item[$(iii)$] $s(\alpha)=s(p)=e_i$ for $i\neq 1,n$, and $t(\alpha),t(p)$ are both in $\{e_1,e_n\}$ but $t(\alpha)\neq t(p)$, also the dual case;
		\item[$(iv)$] $\alpha$ is an arrow from $e_1$ to $e_n$, and $p$ is an arrow from $e_n$ to $e_1$, also the dual case;
		\item[$(v)$] $\alpha$ is a loop at $e_1$, and $p=e_n$ or $p$ is a loop at $e_n$, also the dual case.
\end{itemize}
\noindent For the gluing different blocks case, $\mathrm{Spp}(1,n)$ consists of all special pairs of the form $(\alpha,e_n),(\alpha,\beta),(\beta,e_1)$ and $(\beta,\alpha)$, where $\alpha$ is a loop at $e_1$ and $\beta$ is a loop at $e_n$. Then under the condition $\mathrm{char}(k)\neq 2$, neither $\alpha^*\|f_1$ nor $\beta^*\|f_1$ belongs to $\mathrm{Ker}(\delta_B^1)$, hence $Z_{spp}$ must be generated by all special pairs $\alpha^*\| p^*$ and $\beta^*\| q^*$. However, once we exclude the case that there are loops at $e_1$ and $e_n$ simultaneously, these generators vanish and $Z_{spp}$ is zero.
\end{Rem}

\begin{Prop}\label{Kernel-delta-one-difference}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$. If $\mathrm{char}(k)$ satisfies Assumption \ref{assump}, then we have $\mathrm{dim}_k\mathrm{Ker}(\delta^1_B)$ = $\mathrm{dim}_k\mathrm{Ker}(\delta^1_A)+$ $\mathrm{kspp}(1,n)$.
\end{Prop}

\begin{Proof}
By Proposition \ref{Kernel-delta-one-injective}, we only need to describe the elements $\theta$ in $\mathrm{Ker}(\delta_{B}^{1})$ which are in the complement of the subspace $\psi_1(\mathrm{Ker}(\delta_{A}^{1}))$, under Assumption \ref{assump}. According to the proof of Proposition \ref{Kernel-delta-one-injective}, we may assume that $\theta$ is a linear combination of the elements of the form $\alpha^* \| p^*$ such that $(\alpha,p)$ is a special pair with respect to the gluing of $e_1$ and $e_n$. Clearly in this case $\theta\in Z_{spp}$, where $Z_{spp}$ is the subspace of $\mathrm{Ker}(\delta_{B}^{1})$ defined in Definition \ref{elements-in-Z-spp}. Therefore, we have the following decomposition: $\mathrm{Ker}(\delta_B^1)=\psi_1(\mathrm{Ker}(\delta_{A}^{1}))\oplus Z_{spp}$. Hence the dimension formula follows.
\end{Proof}

\begin{Cor}\label{Kernel-delta-one-source-and-sink}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing a source vertex $e_1$ and a sink vertex $e_n$ of $A$. Then we have $\mathrm{Ker}(\delta^1_A) \simeq \mathrm{Ker}(\delta^1_B)$ as Lie algebras, except in the case that $B\simeq k[x]/(x^{2})$ in $\mathrm{char}(k)=2$.
\end{Cor}

\begin{Proof} By Lemma \ref{bijection-parallel-arrows}, in this case the only possible special pair with respect to the gluing of $e_1$ and $e_n$ has the form $(\alpha,e_1)$ or $(\alpha,e_n)$ such that $\alpha^*$ is a loop at $f_1$ but $\alpha$ is neither a loop at $e_1$ nor a loop at $e_n$.  Therefore $\langle\mathrm{Spp}(1,n)\rangle$ is generated by the elements of the form $\alpha^*||f_1$. Suppose now that $\alpha^*||f_1 \in\mathrm{Ker}(\delta_{B}^{1})$. Then we consider two cases. If $Q_B$ is a single loop $\alpha^*$ so that $B\simeq k[x]/(x^{2})$, then $\delta_{B}^{1}(\alpha^* \| f_1)=2r'\| \alpha^*=0$ (where $r'=\alpha^*\alpha^*$) implies that char$(k)=2$. If $Q_B$ is not a single loop, then either there is an arrow $\beta^*\neq \alpha^*$ starting from $f_1$ or there is an arrow $\gamma^*\neq \alpha^*$ ending at $f_1$ in $Q_B$. Therefore $\delta_{B}^{1}(\alpha^* \| f_1)$ will contain a summand $\alpha^*\beta^*\| \beta^*$ or a summand $\gamma^*\alpha^*\| \gamma^*$, which clearly can not be cancelled in $\mathrm{Im}\delta_{B}^{1}$, so $\alpha^* \| f_1\notin \mathrm{Ker}(\delta_{B}^{1})$. It follows that $\alpha^*\| f_1\in \mathrm{Ker}(\delta_{B}^{1})$ if and only if $B\simeq k[x]/(x^{2})$ and char$(k)=2$. Summarizing the above discussion we get $\mathrm{kspp}(1,n)=0$ when gluing a source and a sink and excluding the case $B\simeq k[x]/(x^{2})$ in $\mathrm{char}(k)=2$. Now the result follows from Proposition \ref{Kernel-delta-one-difference}, Proposition \ref{Kernel-delta-one-injective} and Remark \ref{no-loop-at-e1-or-en-Ker}.	
\end{Proof}

\begin{Rem}\label{exceptional-case}
In case that $\mathrm{char}(k)=2$ and $B\simeq k[x]/(x^{2})$, $A=kQ_A$ where $Q_A$ is given by the quiver $1\stackrel{\alpha}\longrightarrow 2$. By a direct computation, we have the following: $\mathrm{Im}(\delta^0_A)=\mathrm{Ker}\delta^1_A$ is 1-dimensional with $k$-basis $\{\alpha\|\alpha\}$, $\mathrm{Im}(\delta^0_B)=0$ and $\mathrm{Ker}(\delta^1_B)$ is 2-dimensional with $k$-basis $\{\alpha^*\|f_1, \alpha^*\|\alpha^*\}$. Here $\mathrm{Spp}(1,2)=\{(\alpha,e_1),(\alpha,e_2)\}$ and $Z_{spp}=\langle \alpha^*\| f_1 \rangle $, hence the formula $\mathrm{dim}_k\mathrm{Ker}(\delta^1_B)$ = $\mathrm{dim}_k\mathrm{Ker}(\delta^1_A)+$ $\mathrm{kspp}(1,2)$ still holds in this case.
\end{Rem}

We can finally compare the dimensions of $\mathrm{HH}^1(A)$ and of $\mathrm{HH}^1(B)$.

\begin{Thm}\label{hh1-quotient-isomorphism-for-monomial-algebras}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$. If char$(k)$ satisfies Assumption \ref{assump}, then we have $$\mathrm{dim}_k\mathrm{HH}^{1}(A)=\mathrm{dim}_k\mathrm{HH}^{1}(B)-1-\mathrm{kspp}(1,n)+\mathrm{sp}(1,n) +c_A-c_B.$$
In particular, if we glue $e_1$ and $e_n$ from the same block of $A$, then $$\mathrm{dim}_k\mathrm{HH}^{1}(A)=\mathrm{dim}_k\mathrm{HH}^{1}(B)-1-\mathrm{kspp}(1,n)+\mathrm{sp}(1,n);$$ if $e_1$ and $e_n$ are from two different blocks of $A$, then $\mathrm{HH}^{1}(A)$ is a Lie subalgebra of $\mathrm{HH}^{1}(B)$ and $$\mathrm{dim}_k\mathrm{HH}^{1}(A)=\mathrm{dim}_k\mathrm{HH}^{1}(B)-\mathrm{kspp}(1,n).$$
\end{Thm}

\begin{Proof} Since $\mathrm{HH}^{1}(A) \simeq \mathrm{Ker}(\delta_A^1) / \mathrm{Im}(\delta_A^0)$, this is a direct consequence of Proposition \ref{Iamge-delta-zero}, Proposition \ref{Kernel-delta-one-injective} and Proposition \ref{Kernel-delta-one-difference}.
\end{Proof}

Concerning the relation between Lie structures of $\mathrm{HH}^{1}(A)$ and $\mathrm{HH}^{1}(B)$, we have the following strengthened form of Theorem \ref{hh1-quotient-isomorphism-for-monomial-algebras}.

\begin{Thm}\label{Lie-strucutre-of-hh1} Under the conditions of Theorem \ref{hh1-quotient-isomorphism-for-monomial-algebras}, we have the following exact commutative diagram:
\begin{center}
	\begin{tikzcd}
	& 0 & 0 & 0 \\
	0 & {Z_{sp}} & {Z_{spp}} & {\mathrm{Coker}(\varphi)} & 0 \\
	0 & Y & {\mathrm{Ker}(\delta_B^1)} & {\frac{\mathrm{Ker}(\delta_B^1)}{Y}} & 0 \\
	0 & {\mathrm{Im}(\delta_A^0)} & {\mathrm{Ker}(\delta_A^1)} & {\mathrm{HH}^1(A)} & 0, \\
	& 0 & 0 & 0
	\arrow[from=2-1, to=2-2]
	\arrow["{\tilde{\iota}_B}", from=2-2, to=2-3]
	\arrow[from=2-3, to=2-4]
	\arrow[from=2-4, to=2-5]
	\arrow[from=3-1, to=3-2]
	\arrow["{\iota_B}", from=3-2, to=3-3]
	\arrow[from=3-3, to=3-4]
	\arrow[from=3-4, to=3-5]
	\arrow[from=4-1, to=4-2]
	\arrow["{\iota_A}", from=4-2, to=4-3]
	\arrow[from=4-3, to=4-4]
	\arrow[from=4-4, to=4-5]
	\arrow["{\psi_1|_{\mathrm{Im}(\delta_A^0)}}", from=4-2, to=3-2]
	\arrow["{\pi^0}", from=3-2, to=2-2]
	\arrow["{\pi^1}", from=3-3, to=2-3]
	\arrow["{\psi_1}", from=4-3, to=3-3]
	\arrow["\pi", from=3-4, to=2-4]
	\arrow["\varphi", from=4-4, to=3-4]
	\arrow[from=5-2, to=4-2]
	\arrow[from=5-3, to=4-3]
	\arrow[from=2-2, to=1-2]
	\arrow[from=2-3, to=1-3]
	\arrow[from=2-4, to=1-4]
	\arrow[from=5-4, to=4-4]
	\end{tikzcd}
\quad \quad $(**)$
\end{center}
where $\pi^0,\pi^1$ are canonical projections, $\iota_A$ and $\iota_B$ are canonical injections, $\varphi$ is an
injective map induced from $\psi_1$ and $\pi$ is a surjective map induced from $\pi^1$, and where $Y=\psi_1(\mathrm{Im}(\delta_A^0))\oplus Z_{sp}$ is a subspace of $\mathrm{Ker}
(\delta_B^1)$ which is equal to $\mathrm{Im}(\delta_B^0)$ in case that $e_1$ and $e_n$ are from two
different blocks of $A$ and which contains $\mathrm{Im}(\delta_B^0)$ as a codimension $1$ subspace
in case that $e_1$ and $e_n$ are from the same block of $A$. In particular, if $Z_{spp}=Z_{sp}$, then $Y$ is a Lie
ideal of $\mathrm{Ker}(\delta_B^1)$ and there is a Lie algebra epimorphism from $\mathrm{HH}^1(B)$ to $
\mathrm{Ker}(\delta_B^1)/Y\simeq \mathrm{HH}^1(A)$ with kernel $I:=Y/\mathrm{Im}(\delta_B^0)$, where $I$
is zero if $e_1$ and $e_n$ are from two different blocks of $A$ and dim$_kI=1$ if $e_1$ and $e_n$ are
from the same block of $A$.
\end{Thm}

\begin{Proof} By Proposition \ref{Kernel-delta-one-injective}, there exists an injective Lie algebra homomorphism $\psi_1: \mathrm{Ker}(\delta^1_A)\hookrightarrow \mathrm{Ker}(\delta^1_B)$, which is induced from the canonical map $\psi_1: k((Q_A)_1\| \mathcal{B}_A) \to k((Q_B)_1\| \mathcal{B}_B)$ ($\alpha\| p\mapsto \alpha^*\| p^*$). Moreover, by Proposition \ref{Kernel-delta-one-difference}, we have the decomposition $\mathrm{Ker}(\delta_B^1)=\psi_1(\mathrm{Ker}(\delta_{A}^{1}))\oplus Z_{spp}$, where $Z_{spp}$ denotes the intersection of $\langle\mathrm{Spp}(1,n)\rangle$ and $\mathrm{Ker}(\delta^1_B)$ (cf. Definition \ref{elements-in-Z-spp}).

Combining with Proposition \ref{Iamge-delta-zero} and noting the fact $\delta_B^0(f_1\|f_1)=\psi_1(\delta_A^0(e_1\|e_1))+\psi_1(\delta_A^0(e_n\|e_n))$, we have that $\psi_1: \mathrm{Ker}(\delta^1_A)\hookrightarrow \mathrm{Ker}(\delta^1_B)$ restricts to an injective map $$\psi_1|_{\mathrm{Im}(\delta_A^0)}:\mathrm{Im}(\delta_A^0)=\mathrm{Im}(\delta_{(A)_0}^0)\oplus \mathrm{Im}(\delta_{(A)_{\ge 1}}^0) \hookrightarrow X\oplus \mathrm{Im}(\delta_{(B)_{\ge 1}}^0)\subseteq \mathrm{Ker}(\delta_B^1),$$ where $X$ is the subspace of $\mathrm{Ker}(\delta_B^1)$ generated by the elements $\psi_1(\delta_A^0(e_1\|e_1))$, $\psi_1(\delta_A^0(e_n\|e_n))$ and $\delta_B^0(f_i\|f_i)$ $(2\le i\le n-1)$. If we denote $X\oplus \mathrm{\mathrm{Im}}(\delta_{(B)_{\ge 1}}^0)$ by $Y$, then we have the decomposition $Y=\psi_1(\mathrm{Im}(\delta_A^0))\oplus Z_{sp}$, where $Z_{sp}$ (cf. Definition \ref{special-path}) is the subspace of $\mathrm{Im}(\delta_{(B)_{\ge1}}^0)$ generated by the elements $\delta_{B}^{0}(f_1 \| p^*)$ for $p \in \mathrm{Sp}(1,n)$.

Note that the dimension of $X$ is equal to $\mathrm{dim}_k\mathrm{Im}(\delta_{(A)_0}^0)$, $X=\mathrm{Im}(\delta_{(B)_0}^0)$ if $e_1$ and $e_n$ are from two
different blocks of $A$, and $\mathrm{Im}(\delta_{(B)_0}^0)\subseteq X$ has codimension $1$ in $X$ if $e_1$ and $e_n$ are from the same block of $A$ (cf. Lemma \ref{Image-zero-part}). It follows that $\mathrm{Im}$($\delta_B^0$)=$\mathrm{Im}$($\delta_{(B)_0}^0$)$\oplus \mathrm{Im}$($\delta_{(B)_{\ge1}}^0$) is equal to $Y$ if $e_1$ and $e_n$ are from two
different blocks of $A$ and has codimension $1$ in $Y$ if $e_1$ and $e_n$ are from the same block of $A$. However, when $Y\supsetneqq \mathrm{Im}(\delta_B^0)$, $Y$ is usually not a Lie ideal of $\mathrm{Ker}(\delta_B^1)$ since in general $[Y,Z_{spp}]$ is not contained in $Y$ (In fact, $Y$ is a Lie ideal of $\mathrm{Ker}(\delta_B^1)$ if and only if $[\psi_1(\mathrm{Im}(\delta_A^0)),Z_{spp}] \subset Y$). It is clear that if $p$ is a special path from $e_1$ to $e_n$, then each summand $a^*\|a^*p^*$ (or $b^*\|p^*b^*$) of $\delta_{B}^{0}(f_1 \| p^*)$, where $a$ is an arrow starting from $e_n$ such that $ap\in \mathcal{B}_A$ (or where $b$ is an arrow ending at $e_1$ such that $pb\in \mathcal{B}_A$), is induced from a special pair $(a,ap)$ (or $(b,pb)$). In the case that $p$ is a special path from $e_n$ to $e_1$, we have the similar conclusion. Therefore the canonical injective map $Y\hookrightarrow \mathrm{Ker}(\delta^1_B)$ restricts to an injective map $Z_{sp}\hookrightarrow Z_{spp}$.

Summarizing the above discussion we obtain the exact commutative diagram $(**)$. From this we know that the conclusions under the condition $Z_{spp}=Z_{sp}$ are also clear.
\end{Proof}

\begin{Rem}\label{minimal-generator-of-ideal-I}
	Note that the one-dimensional ideal $I:=Y/\mathrm{Im}(\delta_B^0)$ of $\mathrm{HH}^1(B)$ in Theorem \ref{Lie-strucutre-of-hh1} is generated by $\psi_1(\delta_A^0(e_1\|e_1))=\psi_1(\sum_{[\alpha]\in (\bar{Q}_A)_1e_1}I_{[\alpha]}-\sum_{[\alpha]\in e_1(\bar{Q}_A)_1}I_{[\alpha]})$, where $I_{[\alpha]}:=\sum_{i=1}^{m}\alpha_i\|\alpha_i$ for $[\alpha]=\{\alpha_1,\alpha_2,\cdots,\alpha_m \}$. If we glue a source $e_1$ and a sink $e_n$ from the same block of $A$ to get $B$, then we can rewrite the generator $\psi_1(\delta_A^0(e_1\|e_1))=\psi_1(\sum_{[\alpha]\in (\bar{Q}_A)_1e_1}I_{[\alpha]})$ of $I$. Specifically, $\psi_1(\delta_A^0(e_1\|e_1))=\psi_1(\sum_{[\alpha]\in \Delta}I_{[\alpha]})$ (modulo an element in $\mathrm{Im}(\delta_B^0)$), where $\Delta$ is a subset of $(\bar{Q}_A)_1e_1$ consisting of some equivalence classes of parallel arrows $[\alpha]$ starting from $e_1$ such that $\alpha$ satisfies one of the following two conditions:
		\begin{itemize}
			\item [$(i)$] $t(\alpha)=e_n$;
			\item [$(ii)$] $\alpha$ lies in an undirected cycle in the quiver $Q_A$ with at least one arrow in this cycle ending at $e_n$.
		\end{itemize}
For a concrete example for $\Delta$, see Example \ref{eg-I-summand-of-HH}.
\end{Rem}

In order to define $\Delta$ we could proceed as follows. We define two sub-quivers of $Q_A$: $Q^c_A$ and $Q^d_A$. The arrows of $Q^c_A$ satisfy one of the following two conditions:
		\begin{itemize}
			\item [$(i)$] $t(\alpha)=e_n$;
			\item [$(ii)$] $\alpha$ lies in an undirected cycle in the quiver $Q_A$ with at least one arrow in this cycle ending at $e_n$.
		\end{itemize}
The arrows of $Q^d_A$ are the arrows of $Q_A$ which are not in $Q^c_A$. Note that by construction the sets of arrows of $Q^c_A$ and of $Q^d_A$ are disjoint and that both quivers only share the vertex $e_1$. Then we define $\Delta$ to be a set $(\bar{Q}^c_A)_1e_1$. We also define the corresponding sub-quivers $Q^c_B$ and $Q^d_B$ via the map $\varphi$ in Section \ref{pre}.
In order to show that $\psi_1(\delta_A^0(e_1\|e_1))$ is equal to $\psi_1(\sum_{[\alpha]\in \Delta}I_{[\alpha]})$ (modulo an element in $\mathrm{Im}(\delta_B^0)$) we proceed as follows.
We have
	\begin{equation*}
		\begin{split}
\psi_1(\delta_A^0(e_1\|e_1))&=\psi_1(\sum_{[\alpha]\in (\bar{Q}_A)_1e_1}I_{[\alpha]})= \psi_1(\sum_{[\alpha]\in (\bar{Q}^c_A)_1e_1}I_{[\alpha]})+  \psi_1(\sum_{[\alpha]\in (\bar{Q}^d_A)_1e_1}I_{[\alpha]})\\
&=  \psi_1(\sum_{[\alpha]\in\Delta}I_{[\alpha]})+  \psi_1(\sum_{[\alpha]\in (\bar{Q}^d_A)_1e_1}I_{[\alpha]})
\end{split}
	\end{equation*}
Now we add $\psi_1(\delta_A^0(e_1\|e_1))$ and an element in $\mathrm{Im}(\delta_B^0)$ to obtain $\psi_1(\sum_{[\alpha]\in\Delta}I_{[\alpha]})$. More precisely, we pick the element  $\sum_{f_i \in (Q_B^d)_0, f_i\neq f_1} \delta^0_B(f_i || f_i)$ which is equal to $-\psi_1(\sum_{[\alpha]\in (\bar{Q}^d_A)_1e_1}I_{[\alpha]})$. In fact,  each summand $I_{[\alpha^*]}$ in
$\delta_B^0(f_i\|f_i)$ is cancelled by the summand $-I_{[\alpha^*]}$  obtained from
$\delta_A^0(f_j\|f_j)$ where $f_j$ is equal to the source (or the target) of $\alpha^*$. Therefore the only non-zero summands are of the form $-\sum_{[\alpha^*]\in f_i(\bar{Q}^d_B)_1}I_{[\alpha^*]}$,
where the source of $\alpha^*$ is equal to $f_1$. Note that the sum of all non-zero summands of $\sum_{f_i \in (Q_B^d)_0, f_i\neq f_1} \delta^0_B(f_i || f_i)$ is equal to
$-\sum_{[\alpha^*]\in (\bar{Q}^d_B)_1f_1}I_{[\alpha^*]}$. The statement follows by observing that $$\sum_{[\alpha^*]\in (\bar{Q}^d_B)_1f_1}I_{[\alpha^*]}=\psi_1(\sum_{[\alpha]\in (\bar{Q}^d_A)_1e_1}I_{[\alpha]})$$
since there is a bijection between the set of arrows in $Q^d_A$ having source $e_1$ and the set of arrows in $Q^d_B$ having source $f_1$.

\begin{Cor}\label{relationship-between-spp-and-sp}
Under the conditions of Theorem \ref{hh1-quotient-isomorphism-for-monomial-algebras}, we have that $\mathrm{kspp}(1,n)\geq \mathrm{sp}(1,n)$. In particular, the number of special pairs is bigger than or equal to the number of special paths.
\end{Cor}

\begin{Cor}\label{quotient-lie-algebra-source-and-sink}
	Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing a source vertex $e_1$ and a sink vertex $e_n$ of $A$. Then we have $$\mathrm{dim}_k\mathrm{HH}^{1}(A)=\mathrm{dim}_k\mathrm{HH}^{1}(B)+c_A-c_B-1,$$ except in the case that $B\simeq k[x]/(x^{2})$ in $\mathrm{char}(k)=2$. In particular, if we glue $e_1$ and $e_n$ from two different blocks of $A$, then there is a (restricted) Lie algebra isomorphism $\mathrm{HH}^{1}(A)\simeq \mathrm{HH}^{1}(B)$; if $e_1$ and $e_n$ are from the same block of $A$, then $\mathrm{HH}^{1}(A)\simeq \mathrm{HH}^{1}(B)/I$ as (restricted) Lie algebras, where $I$ is a one-dimensional (restricted) Lie ideal of $\mathrm{HH}^{1}(B)$. Moreover, if $\mathrm{char}(k)=0$, then the one-dimensional Lie ideal $I$ lies in the center of $\mathrm{HH}^{1}(B)$ and $$\mathrm{HH}^{1}(B)\simeq \mathrm{HH}^{1}(A) \times I \simeq \mathrm{HH}^{1}(A) \times k$$ as Lie algebras.
\end{Cor}

\begin{Proof} First we notice that by Remark \ref{no-loop-at-e1-or-en-Ker}, we do not need Assumption \ref{assump} here. By Corollary \ref{Kernel-delta-one-source-and-sink}, we have $Z_{spp}=0$ since we glue a source and a sink. Now our conclusions except for the last one follow directly from Theorem \ref{hh1-quotient-isomorphism-for-monomial-algebras} and Theorem \ref{Lie-strucutre-of-hh1}.
	
It suffices to show the last statement of the Corollary. From now on, we fix the `minimal' generator $\psi_1(\sum_{[\alpha]\in \Delta}I_{[\alpha]})$ of the Lie ideal $I$ by Remark \ref{minimal-generator-of-ideal-I}. We claim that $I$ is contained in the center $Z(L_0)$ of $L_0$ and $I$ is a direct summand of $L_0$ as Lie algebras. Then the statement follows. Indeed, if we assume that $L_0=I\oplus G$ as Lie algebras, then by the graduation of $\mathrm{HH}^{1}(B)$ we have $$\mathrm{HH}^{1}(B)=L_0\oplus \bigoplus_{i\geq 1}L_i=(I\oplus G)\oplus \bigoplus_{i\geq 1}L_i=I\oplus(G\oplus \bigoplus_{i\geq 1}L_i)=:I\oplus L,$$ where $L$ is a Lie ideal of $\mathrm{HH}^{1}(B)$. Indeed,  $[I,L]=[I,G]+[I,\bigoplus_{i\geq 1}L_i]=[I,\bigoplus_{i\geq 1}L_i]\subset \bigoplus_{i\geq 1}L_i$ by Remark \ref{graded-HH-one} and the above claim. Consequently, to show $\mathrm{HH}^{1}(B)=I\oplus L$ as Lie algebras is equivalent to show $0=[I,L]=[I,\bigoplus_{i\geq 1}L_i]$. By Remark \ref{graded-HH-one} we know that both $\bigoplus_{i\geq 1}L_i$ and $I$ are Lie ideals of $\mathrm{HH}^{1}(B)$, hence $[I,\bigoplus_{i\geq 1}L_i]=0$. Therefore, $\mathrm{HH}^{1}(B)=I\oplus L$ as Lie algebras. Since $\mathrm{HH}^{1}(A)\simeq \mathrm{HH}^{1}(B)/I$ as Lie algebras, then $L\simeq \mathrm{HH}^{1}(A)$ as Lie algebras.
	
Now we prove the above claim. Firstly, we show that $I$ is contained in the center $Z(L_0)$ of $L_0$. We also adopt symbol $[\alpha]$ rather than $[\alpha^*]$ in $(\bar{Q}_B)_1$ for every $[\alpha]$ in $(\bar{Q}_A)_1$. According to Proposition \ref{basis-L_0}, without loss of generality, we can always take $L_0^{[\alpha]}=\langle \alpha_i^*\|\alpha_j^*,\alpha_l^*\|\alpha_l^* \ |\ 1\leq l\leq m,1\leq i\neq j\leq m,\alpha_i^*\|\alpha_j^* \in L_0 \rangle$ for each $[\alpha]=\{\alpha_1,\cdots,\alpha_m \}$ in $\Delta$. Then by Proposition \ref{center-L_0} we have $Z(L_0^{[\alpha]})=\oplus_C \langle \sum_{\alpha_i\in C}\alpha_i^*\|\alpha_i^* \rangle$, where $C$ traverses connected components of $[\alpha]$ and $[\alpha] \in \Delta$. And it is easy to see that $I=\langle \psi_1(\sum_{[\alpha]\in \Delta}I_{[\alpha]}) \rangle=\langle \sum_{[\alpha] \in \Delta}\sum_{i=1}^{m}\alpha_i^*\|\alpha_i^* \rangle \subset \oplus_{[\alpha] \in \Delta}(\oplus_C \langle \sum_{\alpha_i \in C}\alpha_i^*\|\alpha_i^* \rangle)=\oplus_{[\alpha] \in \Delta}Z(L_0^{[\alpha]})\subset Z(L_0)$.
	
Secondly, we show that $I$ is a direct summand of $L_0$ as Lie algebras. We sketch the proof in case that $\Delta$ only contains two equivalence classes of parallel arrows (cf. Remark \ref{minimal-generator-of-ideal-I}), namely $\Delta=\{[\alpha],[\beta] \}$, where $[\alpha]=\{\alpha_1,\cdots,\alpha_m \}$ and $[\beta]=\{\beta_1,\cdots,\beta_t \}$. Then $I=\langle \psi_1(I_{[\alpha]}+I_{[\beta]}) \rangle=\langle \sum_{i=1}^{m}\alpha_i^*\| \alpha_i^*+\sum_{j=1}^{t}\beta_j^*\| \beta_j^* \rangle$. Since $L_0^{[\alpha]}=\langle \alpha_i^*\|\alpha_j^*,\alpha_l^*\|\alpha_l^* \ |\ 1\leq l\leq m,1\leq i\neq j\leq m,\alpha_i\|\alpha_j \in \mathrm{Ker}(\delta_A^1) \rangle$ and $L_0^{[\beta]}=\langle \beta_i^*\|\beta_j^*,\beta_l^*\|\beta_l^* \ |\ 1\leq l\leq t,1\leq i\neq j\leq t,\beta_i\|\beta_j \in \mathrm{Ker}(\delta_A^1) \rangle$, it is easy to see that $I$ is a summand of $\oplus_{[\alpha] \in \Delta}L_0^{[\alpha]}=L_0^{[\alpha]}\oplus L_0^{[\beta]}$, hence it is a summand of $L_0$. In fact, we have Lie algebra decompositions:
	\begin{equation*}
		\begin{split}
		    L_0^{[\alpha]} & =\langle \sum_{i=1}^{m}\alpha_i^*\| \alpha_i^*\rangle \oplus \langle \alpha_i^*\|\alpha_j^*,\alpha_l^*\|\alpha_l^*-\alpha_m^*\|\alpha_m^* \ |\ 1\leq l\leq m-1,1\leq i\neq j\leq m,\alpha_i\|\alpha_j \in \mathrm{Ker}(\delta_A^1) \rangle \\ & =:\langle \psi_1(I_{[\alpha]}) \rangle \oplus J_1,
		\end{split}
	\end{equation*}	
	$$L_0^{[\beta]}=\langle \sum_{i=1}^{t}\beta_i^*\| \beta_i^*\rangle \oplus \langle \beta_i^*\|\beta_j^*,\beta_l^*\|\beta_l^*-\beta_t^*\|\beta_t^* \ |\ 1\leq l\leq t-1,1\leq i\neq j\leq t,\beta_i\|\beta_j \in \mathrm{Ker}(\delta_A^1) \rangle=:\langle \psi_1(I_{[\beta]}) \rangle \oplus J_2.$$ As a consequence, we can easily check that there are Lie algebra decompositions
    \begin{equation*}
    	\begin{split}
    	L_0^{[\alpha]}\oplus L_0^{[\beta]} & =(\langle \psi_1(I_{[\alpha]}) \rangle \oplus \langle \psi_1(I_{[\beta]}) \rangle) \oplus J_1 \oplus J_2= (I\oplus \langle \psi_1(I_{[\alpha]}) \rangle) \oplus J_1 \oplus J_2 \\ & = I\oplus (\langle \psi_1I_{[\alpha]}) \rangle \oplus J_1 \oplus J_2)=:I\oplus J.
    	\end{split}
    \end{equation*}
    It follows that $L_0=\oplus_{[\alpha] \in (\bar{Q}_B)_1}L_0^{[\alpha]}=I\oplus G$ as Lie algebras, where $G:=J\oplus \bigoplus_{\alpha\in(\bar{Q}_B)_1\backslash\Delta}L_0^{[\alpha]}$. We are done.
\end{Proof}

\begin{Rem}
\begin{itemize}
\item[$(1)$] From Corollary \ref{quotient-lie-algebra-source-and-sink} we deduce that although the (restricted) Lie algebra structure of $\mathrm{HH}^1$ is an invariant under stable equivalences of Morita type, $\mathrm{HH}^1$ is not invariant under stable equivalences obtained by gluing a source and a sink. However, there is still a close relation between these two Lie algebras, namely, $\mathrm{HH}^1(A)$ is a quotient of $\mathrm{HH}^1(B)$ by an (often splitting) one-dimensional Lie ideal.
\item[$(2)$] It would interesting to know when there is a Lie algebra decomposition $$\mathrm{HH}^1(B)=\mathrm{HH}^1(A)\times I\simeq \mathrm{HH}^1(A)\times k$$ if we glue two idempotents from the same block of $A$ such that $Z_{spp}=Z_{sp}$. When $\mathrm{char}(k)\neq 0$, we can  easily find examples in which $I$ is not a Lie algebra direct summand of $\mathrm{HH}^1(B)$ and therefore $\mathrm{HH}^1(B)\ncong \mathrm{HH}^1(A)\times k$. When $\mathrm{char}(k)=0$, Corollary \ref{quotient-lie-algebra-source-and-sink} shows that $\mathrm{HH}^1(B)\simeq \mathrm{HH}^1(A)\times k$ when we glue a source and a sink. Moreover, at least for radical square zero algebras over a field of characteristic zero, when we glue two idempotents from the same block such that $Z_{spp}=0$, we have $\mathrm{HH}^1(B)\simeq \mathrm{HH}^1(A)\times k$ as Lie algebras (see Corollary \ref{hh1-radical-square-zero-case-structure} and Remark \ref{reduce-cycle}).
\end{itemize}
  \end{Rem}

\begin{Cor}\label{hh1-radical-square-zero-case}
Let $A=kQ_A/I_A$ be a radical square zero algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$. If $\mathrm{char}(k)\neq 2$, then we have $$\mathrm{dim}_k\mathrm{HH}^{1}(A)=\mathrm{dim}_k\mathrm{HH}^{1}(B)-1-\mathrm{kspp}(1,n)-c_B+c_A.$$
In particular, if we glue $e_1$ and $e_n$ from the same block of $A$, then
	$$\mathrm{dim}_k\mathrm{HH}^{1}(A)=\mathrm{dim}_k\mathrm{HH}^{1}(B)-1-\mathrm{kspp}(1,n);$$ if we glue $e_1$ and $e_n$ from two different blocks of $A$, then
$$\mathrm{dim}_k\mathrm{HH}^{1}(A)=\mathrm{dim}_k\mathrm{HH}^{1}(B)-\mathrm{kspp}(1,n)$$ and then $\mathrm{HH}^{1}(A)\simeq \mathrm{HH}^{1}(B)$ as Lie algebras if we exclude the case that there are loops both at $e_1$ and $e_n$. Moreover, if one of the following conditions holds, then the above results still hold in the case $\mathrm{char}(k)=2$:
\begin{itemize}
  \item[$(i)$] glue $e_1$ and $e_n$ from the same block of $A$;
  \item[$(ii)$] glue $e_1\in A_1$ and $e_n\in A_2$ from the different blocks of $A$ such that both $A_1$ and $A_2$ are not isomorphic to $k[x]/(x^2)$.
\end{itemize}
\end{Cor}

\begin{Proof} In the radical square zero case, the number of special paths is zero, and by Remark \ref{radical-square-zero}, Assumption \ref{assump} is equivalent to $\mathrm{char}(k)\neq 2$. Thus the dimension formulas follow immediately from Remark \ref{radical-square-zero-of-spp}, Theorem \ref{hh1-quotient-isomorphism-for-monomial-algebras} and Theorem \ref{Lie-strucutre-of-hh1}. Moreover, if we glue $e_1$ and $e_n$ from two different blocks of $A$ and exclude the case that there are loops at $e_1$ and $e_n$ simultaneously, then $Z_{spp}=0$ and $\mathrm{HH}^{1}(A)\simeq \mathrm{HH}^{1}(B)$ as Lie algebras again by Theorem \ref{Lie-strucutre-of-hh1}. Finally, it is also easy to see that we do not need the assumption $\mathrm{char}(k)\neq 2$ under the condition (i) or (ii).
\end{Proof}

\begin{Rem} \label{exceptional-case-different-blocks} Let $B$ be a radical embedding of a radical square zero algebra $A$ obtained by gluing two idempotents $e_1\in A_1$ and $e_n\in A_2$ from the different blocks $A_1, A_2$.
\begin{itemize}
  \item[$(1)$] If there are loops at $e_1$ or at $e_n$, then in general $\mathrm{HH}^{1}(A)$ is not isomorphic to $\mathrm{HH}^{1}(B)$ and the difference between the dimensions of $\HH^1(A)$ and $\HH^1(B)$ can be arbitrary large, see Example \ref{diffblockarblarg}.
  \item[$(2)$] If $\mathrm{char}(k)=2$ and exactly one of $A_1, A_2$ is isomorphic to $k[x]/(x^2)$, then by a direct computation the dimension formula should be changed as $$\mathrm{dim}_k\mathrm{HH}^{1}(A)=\mathrm{dim}_k\mathrm{HH}^{1}(B)+1.$$
\end{itemize}
\end{Rem}

\begin{Cor}\label{hh1-radical-square-zero-case-structure}
Let $A=kQ_A/I_A$ be a radical square zero algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ from the same block of $A$. If $Z_{spp}=0$ and $\mathrm{char}(k)=0$, then we have a Lie algebra decomposition $$\mathrm{Ker}(\delta_A^1)=\mathrm{Im}(\delta_A^0)\oplus G,$$ where $G$ is some Lie ideal of $\mathrm{Ker}(\delta_A^1)$. As a result, we have a Lie algebra isomorphism $$\mathrm{HH}^1(B)\simeq \mathrm{HH}^1(A)\times k.$$
\end{Cor}

\begin{Proof} We use the notations in Theorem \ref{Lie-strucutre-of-hh1}. Since $Z_{spp}=0$, we have a Lie algebra epimorphism from $\mathrm{HH}^1(B)$ to $\mathrm{Ker}(\delta_B^1)/Y\simeq \mathrm{HH}^1(A)$ with one-dimensional kernel $I:=Y/\mathrm{Im}(\delta_B^0)$, where $\mathrm{Ker}(\delta_B^1)=\psi_1(\mathrm{Ker}(\delta_A^1))$ and $Y$ is a Lie ideal of $\mathrm{Ker}(\delta_B^1)$. Also note that this epimorphism and equality do not depend on the Assumption \ref{assump} since we glue $e_1$ and $e_n$ from the same block (cf. Remark \ref{radical-square-zero}). Furthermore, since $A$ is radical square zero, we have $\mathrm{Ker}(\delta_A^1)=k((Q_A)_1\|(Q_A)_1)$  (cf. Remark \ref{radical-square-zero}), $\mathrm{Im}(\delta_A^0)=\mathrm{Im}(\delta_{(A)_0}^0)$ and $Z_{sp}=0$. Therefore $Y=\psi_1(\mathrm{Im}(\delta_A^0))\oplus Z_{sp}=\psi_1(\mathrm{Im}(\delta_A^0)).$ Similarly we have $\mathrm{Ker}(\delta_B^1)=k((Q_B)_1\|(Q_B)_1)$ and $\mathrm{Im}(\delta_B^0)=\mathrm{Im}(\delta_{(B)_0}^0)$.
	
We claim that there is a Lie algebra decomposition $\mathrm{Ker}(\delta_A^1)=\mathrm{Im}(\delta_A^0)\oplus G$ where $G$ is some Lie ideal of $\mathrm{Ker}(\delta_A^1)$. From this we deduce that $\mathrm{Ker}(\delta_B^1)=Y\oplus L$ as Lie algebras, where $L=\psi_1(G)$. Since $\mathrm{Im}(\delta_B^0) \subset Y$, we get the following Lie algebra isomorphism $$\mathrm{HH}^1(B)=\frac{\mathrm{Ker}(\delta_B^1)}{\mathrm{Im}(\delta_B^0)}=\frac{Y\oplus L}{\mathrm{Im}(\delta_B^0)}\simeq\frac{Y}{\mathrm{Im}(\delta_B^0)}\times L=I\times L.$$ Combining the facts that $\mathrm{HH}^1(B)/I\simeq \mathrm{HH}^1(A)$ and $I$ is one-dimensional we know that $$\mathrm{HH}^1(B) \simeq I \times \mathrm{HH}^1(A)\simeq \mathrm{HH}^1(A)\times k$$ as Lie algebras.
	
	We now prove the above claim. Note that by assumption $Q_A$ contains at least two idempotents. Denote by $S$ a complete set of representatives of non-trivial classes of $(Q_A)_1$. Then $t:=|(\bar{Q}_A)_1|-|S|$ is the number of single arrows in $(Q_A)_1$, where $(\bar{Q}_A)_1$ denotes the set of equivalence classes of parallel arrows (see also Subsection \ref{rszssec} for notations $S$ and $(\bar{Q}_A)_1$). We assume that $|(Q_A)_1|=m$ and $|(Q_A)_0|=n$, so $m=\sum_{\alpha \in S}|\alpha|+t$. Write the equivalence class $[\alpha]=\{\alpha_1,\alpha_2,\dots,\alpha_{|\alpha|}\}$ for $\alpha \in S$ and let $\beta_1,\beta_2,\dots,\beta_t$ be all the single arrows in $(Q_A)_1$. Then the Lie algebra $$\mathrm{Ker}(\delta_A^1)=\langle \alpha_i\|\alpha_j,\beta_s\|\beta_s\ |\ 1\leq i,j\leq |\alpha|,\alpha \in S,1\leq s\leq t \rangle$$ has dimension $\sum_{\alpha \in S}|\alpha|^2+t$ and its Lie ideal $\mathrm{Im}(\delta_A^0)=\mathrm{Im}(\delta_{(A)_0}^0)$ has dimension $n-1$.
 On the other hand, according to \cite[Lemma 2.1]{S-F}, $\mathrm{Ker}(\delta_A^1)\simeq\bigoplus_{\alpha\in(\bar{Q}_A)_1}\mathfrak{gl}_{|\alpha|}(k)$ as Lie algebras, and therefore $\mathrm{rad}(\mathrm{Ker}(\delta_A^1))\simeq \bigoplus_{\alpha\in(\bar{Q}_A)_1}k$
 is an abelian Lie ideal of $\mathrm{Ker}(\delta_A^1)$ with dimension $|(\bar{Q}_A)_1|=|S|+t$. Since $\mathrm{char}(k)=0$, $\mathfrak{gl}_{|\alpha|}(k) \simeq \sl_{|\alpha|}(k)\times k$ as Lie algebras. It follows that we have a decomposition $$\mathrm{Ker}(\delta_A^1)=\mathrm{rad}(\mathrm{Ker}(\delta_A^1))\oplus L_1,$$
 where $L_1$ is a Lie ideal of $\mathrm{Ker}(\delta_A^1)$.
 Observe that $\mathrm{Im}(\delta_A^0)=\mathrm{Im}(\delta_{(A)_0}^0)$ is again an abelian Lie ideal of $\mathrm{Ker}(\delta_A^1)$, $\mathrm{Im}(\delta_A^0)$ must be contained in $\mathrm{rad}(\mathrm{Ker}(\delta_A^1))$. Therefore there exists some abelian ideal $L_2$ of $\mathrm{rad}\mathrm{Ker}(\delta_A^1)$ such that $\mathrm{rad}(\mathrm{Ker}(\delta_A^1))=\mathrm{Im}(\delta_A^0) \oplus L_2$. Let $G=L_1\oplus L_2$, then $G$ is a Lie ideal of $\mathrm{Ker}(\delta_A^1)$ and $\mathrm{Ker}(\delta_A^1)=\mathrm{Im}(\delta_A^0)\oplus G$ as Lie algebras. We are done.
\end{Proof}

\begin{Rem} \label{Levi-decomposition} Note that $\mathrm{Ker}(\delta_A^1)=\mathrm{rad}(\mathrm{Ker}(\delta_A^1))\oplus L_1$ gives a Levi decomposition of the Lie algebra $\mathrm{Ker}(\delta_A^1)$ and it induces a Levi decomposition $\mathrm{HH}^1(A)\simeq L_2\oplus L_1$ of the Lie algebra $\mathrm{HH}^1(A)$, where $L_2\simeq \mathrm{rad}(\mathrm{HH}^1(A))$.  It is also interesting to note that the dimension of $L_2$ is equal to $|S|+t-n+1$, the first Betti number $\chi(\bar{Q}_A)$ of $\bar{Q}_A$.
\end{Rem}

\begin{Rem}\label{reduce-cycle}
	Let $A=kQ_A/I_A$ be a radical square zero algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ from the same block of $A$. If we exclude the case that $\mathrm{char}(k)=2$ and $B\simeq k[x]/(x^{2})$, $A=kQ_A$ where $Q_A$ is given by the quiver $1\stackrel{\alpha}\longrightarrow 2$, then it is straightforward to check that $Z_{spp}=0$ under each of the following conditions (cf. Remark \ref{radical-square-zero-of-spp}):
		\begin{itemize}
			\item[$(i)$] $e_1$ is a source and $e_n$ is a sink;
			\item[$(ii)$] Both $e_1$ and $e_n$ are sinks such that $$\{s(\alpha)\ |\ t(\alpha)=e_1,\alpha\in (Q_A)_1\} \cap \{s(\beta)\ |\ t(\beta)=e_n,\beta\in (Q_A)_1\}=\emptyset;$$
			\item[$(iii)$] Both $e_1$ and $e_n$ are sources such that $$\{t(\alpha)\ |\ s(\alpha)=e_1,\alpha\in (Q_A)_1\} \cap \{t(\beta)\ |\ s(\beta)=e_n,\beta\in (Q_A)_1\}=\emptyset.$$
	\end{itemize}
\end{Rem}


\begin{Rem}\label{reduced-case}
Let $A$ be a radical square zero algebra defined by some quiver $Q$. By direct computations, we can determine the Lie algebra structure of $\HH^1(A)$ in the following special situations.

$(1)$ $\HH^1(A)\simeq \mathfrak{gl}_n(k)$ if $Q$ is the $n$-multiple loop quiver (the quiver with one vertex and $n$ loops) , except in the case $n=1$ and $\mathrm{char}(k)=2$ (for this exceptional case, see Remark \ref{exceptional-case}); The isomorphism sends $\alpha_i||\alpha_j$ to $E_{ji}$ where $E_{ij}$ is the matrix that has  $1$ in position $(i,j)$ and $0$ elsewhere. Note that if the characteristic of the field $k$ does not divide $n$, then $\mathfrak{gl}_n(k) \simeq \sl_n(k)\times k$ as Lie algebras.

$(2)$ $\HH^1(A)\simeq \mathfrak{pgl}_m(k)$ if $Q$ is the $m$-Kronecker quiver (the quiver with $m$ parallel arrows), with the convention that $1$-Kronecker quiver is the Dnykin quiver $A_2$ where $\mathfrak{pgl}_m(k)$ is defined as $\mathfrak{gl}_m(k)/ k\cdot Id$. In other words, $\mathfrak{pgl}_m(k)$ is the quotient $\mathfrak{gl}_m(k)$ by its center. Let $e$ be the source vertex of the $m$-Kronecker quiver. Then the above isomorphism can be obtained by observing that $\mathfrak{gl}_m(k)\simeq \mathrm{Ker}(\delta^1_A)$. The isomorphism sends $\alpha_i||\alpha_j$ to $E_{ji}$ where $E_{ij}$ is the matrix that has  $1$ in position $(i,j)$ and $0$ elsewhere. In addition, $\mathrm{Im}(\delta^0_A)$ is one dimensional and it is generated by the element $\sum_{s(\alpha_i)=e} \alpha_i || \alpha_i$ which is sent to $Id$ via the isomorphism above. If the characteristic of the field $k$  does not divide $m$, then $\mathfrak{pgl}_m(k) \simeq \sl_m(k)$.
\end{Rem}

\subsection{An interpretation of the Lie algebra structure of $\mathrm{HH}^1$ for radical square zero algebras by inverse gluing operations}
\label{rszssec}

Given a quiver $Q$ we have that the parallelism is an equivalence relation on the set of arrows $Q_1$. The set $S$ denotes  a complete set of representatives of the non-trivial classes, that is, equivalence classes having at least two arrows, and for $\alpha\in S$, $|\alpha|$ denotes the number of arrows in the equivalence class $[\alpha]$ of $\alpha$. We denote $\bar{Q}_1$ the set of equivalence classes of parallel arrows. The quiver which has $Q_0$ as vertices and
$\bar{Q}_1$ as set of arrows, will be denoted by $\bar{Q}$. We denote by $\chi(\bar{Q})$ the first Betti number of $\bar{Q}$ (see also Section \ref{fundsec}), which is equal to $|\bar{Q}_1|-|\bar{Q}_0|+c_{\bar{Q}}$, where $c_{\bar{Q}}$ is the number of connected components of $\bar{Q}$. Using these notations S\'{a}nchez-Flores' description of the first Hochschild cohomology for radical square zero algebras can be stated as follows.

\begin{Thm} {\rm(\cite[Theorem 2.9]{S-F})} \label{S-F-1}
Let $A$ be an indecomposable radical square
zero algebra defined by some quiver $Q$ over a field $k$ of characteristic zero. Then (as Lie algebras)
\[
    \HH^1(A)\simeq \prod_{\alpha \in S} \sl_{|\alpha|}(k) \times k^{\chi(\bar{Q})}.
\]
\end{Thm}

Note that intuitively we can say that $\chi(\bar{Q})$ counts the number of holes in $\bar{Q}$. From this point of view we could give an interpretation of the above result by inverse gluing operations. To be more intuitive we will demonstrate our method by a typical example. Note also that the characteristic zero condition in the above result is necessary since the proof uses the Lie algebra decomposition $\mathfrak{gl}_{|\alpha|}(k) \simeq \sl_{|\alpha|}(k)\times k$ when $\mathrm{char}(k)=0$.

\begin{Ex1}\label{typical-example}
	Let $A$ be a radical square	zero algebra defined by the following quiver $Q_A$ over a field $k$ of characteristic zero. Note that here $\chi(\bar{Q}_A)=4$ and $S=\{[\alpha_1],[\beta_1]\}$.
	\begin{center}
		\begin{tikzcd}
		& j\bullet \arrow["\alpha_2", loop, distance=2em, in=55, out=125] \arrow["\alpha_1"', loop, distance=2em, in=215, out=145] \arrow[r, "\eta_1"] & \bullet i \arrow[r, "\xi_4"] \arrow[d, "\xi_1"]                                                                              & \bullet h \arrow[r, "\eta_2"]                   & \bullet g                     \\
		Q_A: & a\bullet \arrow[d, "\gamma_2", shift left=2]                                                                                                 & \bullet d \arrow[l, "\beta_1" description, shift right=2] \arrow[l, "\beta_2" description, shift left=2] \arrow[ld, "\beta_3"] & \bullet e \arrow[u, "\xi_3"] \arrow[l, "\xi_2"] & \bullet f \arrow[l, "\eta_3"] \\
		& b\bullet \arrow[u, "\gamma_1", shift left]                                                                                                   &                                                                                                                              &                                                 &
		\end{tikzcd}
	\end{center}

\textbf{Step 1 (reduce loops):} We separate loops at vertex $j$ of $Q_A$ to get $Q_B$. The new algebra $B$ has two blocks, say $B_1$ and $B_2$.
\begin{center}
	\begin{tikzcd}
	& j_1\bullet \arrow[r, "\eta_1"]               & \bullet i \arrow[r, "\xi_4"] \arrow[d, "\xi_1"]                                                                              & \bullet h \arrow[r, "\eta_2"]                   & \bullet g                     &  &                                                                                                                           \\
	Q_B: & a\bullet \arrow[d, "\gamma_2", shift left=2] & \bullet d \arrow[l, "\beta_1" description, shift right=2] \arrow[l, "\beta_2" description, shift left=2] \arrow[ld, "\beta_3"] & \bullet e \arrow[u, "\xi_3"] \arrow[l, "\xi_2"] & \bullet f \arrow[l, "\eta_3"] &  & \bullet j_2 \arrow["\alpha_1"', loop, distance=2em, in=215, out=145] \arrow["\alpha_2", loop, distance=2em, in=55, out=125] \\
	& b\bullet \arrow[u, "\gamma_1", shift left]   &                                                                                                                              & B_1                                             &                               &  & B_2
	\end{tikzcd}
\end{center}

The inverse operation is given by gluing two vertices (one of which has no loops) from two different blocks, that is, we glue $j_1\in Q_{B_1}$ and $j_2\in Q_{B_2}$. By Corollary \ref{hh1-radical-square-zero-case}, this operation does not change the dimension and the Lie structure of $\mathrm{HH}^1(A)$, that is, $$\mathrm{HH}^1(A)\simeq \mathrm{HH}^1(B) \simeq \mathrm{HH^1}(B_1)\times \mathrm{HH^1}(B_2).$$ Combining Remark \ref{reduced-case} (1) we obtain $\mathrm{HH}^1(B_2) \simeq \mathfrak{gl}_2(k)\simeq sl_2\times k$, where the summand $k$ contributes $1$ for the number $\chi(\bar{Q}_A)$. We have reduced $Q_A$ to the no loop quiver $Q_{B_1}$.

\textbf{Step 2 (reduce oriented $l$-cycles ($l\geq 2$)):} We reduce the oriented cycle $p:=\gamma_2\gamma_1$ in $Q_{B_1}$. Choose the vertex $b$ in $p$ and split it into a source vertex $b_1$ and a sink vertex $b_2$:
\begin{center}
	\begin{tikzcd}
	&                                   & j_1\bullet \arrow[r, "\eta_1"]             & \bullet i \arrow[r, "\xi_4"] \arrow[d, "\xi_1"]                                                                              & \bullet h \arrow[r, "\eta_2"]                   & \bullet g                     \\
	Q_C: &                                   & a\bullet \arrow[d, "\gamma_2", shift left] & \bullet d \arrow[l, "\beta_1" description, shift right=2] \arrow[l, "\beta_2" description, shift left=2] \arrow[ld, "\beta_3"] & \bullet e \arrow[u, "\xi_3"] \arrow[l, "\xi_2"] & \bullet f \arrow[l, "\eta_3"] \\
	& b_1\bullet \arrow[ru, "\gamma_1"] & b_2\bullet                                 &                                                                                                                              &                                                 &
	\end{tikzcd}
\end{center}

The inverse operation is given by gluing $b_1$ and $b_2$ from the same block. By Remark \ref{reduce-cycle} (i), reducing $p$ from $Q_{B_1}$ we get one summand isomorphic to $k$ (cf. Corollary \ref{hh1-radical-square-zero-case-structure}), which contributes $1$ for the number $\chi(\bar{Q}_B)=\chi(\bar{Q}_A)$. So $$\mathrm{HH}^1(B_1)\simeq \mathrm{HH}^1(C) \times k$$ and we have reduced $Q_{B_1}$ to the no oriented cycle quiver $Q_C$.

\textbf{Step 3 (reduce undirected $l$-cycles ($l\geq 3$)):} We first deal with the undirected $3$-cycle $q_1:=\beta_3-\gamma_1-\beta_1$ in $Q_C$. Actually, we can split $b_2$ into two sinks, say $b_3$ and $b_4$, denote the corresponding quiver and algebra by $Q_D$ and $D$, respectively.
\begin{center}
	\begin{tikzcd}
	&                                   & j_1\bullet \arrow[r, "\eta_1"]             & \bullet i \arrow[r, "\xi_4"] \arrow[d, "\xi_1"]                                                                                          & \bullet h \arrow[r, "\eta_2"]                   & \bullet g                     \\
	Q_D: &                                   & a\bullet \arrow[d, "\gamma_2", shift left] & \bullet d \arrow[l, "\beta_1" description, shift right=2] \arrow[l, "\beta_2" description, shift left=2] \arrow[d, "\beta_3", shift right] & \bullet e \arrow[u, "\xi_3"] \arrow[l, "\xi_2"] & \bullet f \arrow[l, "\eta_3"] \\
	& b_1\bullet \arrow[ru, "\gamma_1"] & b_3\bullet                                 & \bullet b_4                                                                                                                              &                                                 &
	\end{tikzcd}
\end{center}

The inverse operation is given by gluing $b_3$ and $b_4$ from the same block. By Corollary \ref{hh1-radical-square-zero-case-structure} and Remark \ref{reduce-cycle}, reducing $q_1$ from $Q_C$ we get a summand isomorphic to $k$, which again contributes $1$ for the number $\chi(\bar{Q}_A)$. Therefore $$\mathrm{HH}^1(C)\simeq \mathrm{HH}^1(D) \times k.$$

After this, we reduce another undirected cycle $q_2:=\xi_4-\xi_3-\xi_2-\xi_1$. Choose the vertex $i$ in $q_2$ and split $i$ into a sink vertex $i_1$ and a source vertex $i_2$ to get $Q_E$, denote the corresponding algebra by $E$.
\begin{center}
	\begin{tikzcd}
	& j\bullet \arrow[r, "\eta_1"]      & \bullet i_1                                & i_2\bullet \arrow[r, "\xi_4"] \arrow[d, "\xi_1"]                                                                                         & \bullet h \arrow[r, "\eta_2"]                   & \bullet g                     \\
	Q_E: &                                   & a\bullet \arrow[d, "\gamma_2", shift left] & \bullet d \arrow[l, "\beta_1" description, shift right=2] \arrow[l, "\beta_2" description, shift left=2] \arrow[d, "\beta_3", shift right] & \bullet e \arrow[u, "\xi_3"] \arrow[l, "\xi_2"] & \bullet f \arrow[l, "\eta_3"] \\
	& b_1\bullet \arrow[ru, "\gamma_1"] & b_3\bullet                                 & \bullet b_4                                                                                                                              &                                                 &
	\end{tikzcd}
\end{center}

The inverse operation is given by gluing $i_1$ and $i_2$ from two different blocks. By Corollary \ref{hh1-radical-square-zero-case}, this operation does not change the dimension and the Lie structure of $\mathrm{HH}^1(D)$, that is, $$\mathrm{HH}^1(D) \simeq \mathrm{HH}^1(E).$$

Note that the above reduction produces a new undirected cycle $q_2':=\xi_4-\xi_3-\xi_2-\xi_1$ in $Q_E$. However, we can reduce $q_2'$ in $Q_E$ by splitting $i_2$ into two sources, say $i_3$ and $i_4$ (the corresponding quiver is $Q_F$).
\begin{center}
	\begin{tikzcd}
	& i_1\bullet                                 & i_3\bullet \arrow[rd, "\xi_1"]             & i_4\bullet \arrow[r, "\xi_4"]                                                                                                            & \bullet h \arrow[r, "\eta_2"]                   & \bullet g                     \\
	Q_F: & j\bullet \arrow[u, "\eta_1"', shift right] & a\bullet \arrow[d, "\gamma_2", shift left] & \bullet d \arrow[l, "\beta_1" description, shift right=2] \arrow[l, "\beta_2" description, shift left=2] \arrow[d, "\beta_3", shift right] & \bullet e \arrow[u, "\xi_3"] \arrow[l, "\xi_2"] & \bullet f \arrow[l, "\eta_3"] \\
	& b_1\bullet \arrow[ru, "\gamma_1"]          & b_3\bullet                                 & \bullet b_4                                                                                                                              &                                                 &
	\end{tikzcd}
\end{center}

The inverse operation is given by gluing two sources from the same block. Again by Corollary \ref{hh1-radical-square-zero-case-structure} and Remark \ref{reduce-cycle}, we get that $$\mathrm{HH}^1(E)\simeq \mathrm{HH}^1(F) \times k,$$ where the summand $k$ also contributes $1$ for the number $\chi(\bar{Q}_A)$. Now we have reduced to a quiver $Q_F$ that has neither oriented cycles nor undirected cycles.

\textbf{Step 4 (Split into several $m$-Kronecker quivers):} Since $Q_F$ contains no cycles (whether oriented or undirected), we can do the last step to split $Q_F$ into several quivers, each of these quivers is a $m$-Kronecker quiver for some $m\geq 1$.
\begin{center}
	\begin{tikzcd}
	&                                                &                                                & i_4\bullet \arrow[r, "\xi_4"']                                                    & \bullet h_3                     &                                 \\
	& i_1\bullet                                     & i_3\bullet \arrow[r, "\xi_1"']                 & \bullet d_1                                                                       & \bullet h_2 \arrow[r, "\eta_2"] & \bullet g                       \\
	Q_G: & j\bullet \arrow[u, "\eta_1"', shift right]     & a\bullet                                       & \bullet d_2 \arrow[l, "\beta_1"', shift right] \arrow[l, "\beta_2", shift left=2] & h_1\bullet                      & \bullet e_3 \arrow[l, "\xi_3"]  \\
	& a_1\bullet                                     & \bullet a_2 \arrow[d, "\gamma_2", shift right] & \bullet d_3 \arrow[d, "\beta_3", shift right]                                     & f\bullet                        & \bullet e_2 \arrow[l, "\eta_3"] \\
	& b_1\bullet \arrow[u, "\gamma_1"', shift right] & \bullet b_3                                    & \bullet b_4                                                                       & d_4\bullet                      & \bullet e_1 \arrow[l, "\xi_2"']
	\end{tikzcd}
\end{center}

The inverse of the above operations are given by repeatedly applying three kinds of operations: gluing a source and a sink from different blocks, gluing two sources from different blocks, gluing two sinks from different blocks. By Corollary \ref{hh1-radical-square-zero-case}, these operations do not change the dimension and the Lie structure of $\mathrm{HH}^1(F)$. Therefore, $$\mathrm{HH}^1(F)\simeq \mathrm{HH}^1(G).$$
Since the $\mathrm{HH}^1$ of a $m$-Kronecker algebra is $\sl_m(k)$ by Remark \ref{reduced-case} (2), $\mathrm{HH}^1(G) \simeq \sl_2(k)$.

We conclude that $\mathrm{HH}^1(B_1) \simeq \mathrm{HH}^1(G) \times k^3$, therefore $$\mathrm{HH}^1(A)\simeq \mathrm{HH}^1(B_1) \times \mathrm{HH^1}(B_2) \simeq {\sl_2(k)}^2\times k^4.$$ We are done.
\end{Ex1}


\section{Examples}
\label{eg}

The first two examples show that the characteristic condition in Proposition \ref{Kernel-delta-one-injective} is necessary.

\begin{Ex1}\label{example-characteristic-condition} Here we assume that char$(k)=2$, and that $B$ is obtained from $A$ by gluing $e_{1}$ and $e_{3}$:
	\begin{center}
		\begin{tikzcd}
		Q_A:\hspace{-2em} & e_2\bullet \arrow[r, "\beta"] & e_1\bullet \arrow[r, "\gamma"] \arrow["\alpha", loop, distance=2em, in=55, out=125] & \bullet e_3 &  & Q_B:\hspace{-2em} & f_2\bullet \arrow[r, "\beta^*"] & \bullet f_1 \arrow["\gamma^*", loop, distance=2em, in=325, out=35] \arrow["\alpha^*", loop, distance=2em, in=55, out=125]
		\end{tikzcd}
	\end{center}
Where $Z_A=\{r=\alpha^2\}$, $Z_{new}=\{r_1=(\gamma^*)^2,r_2=\alpha^*\gamma^*\}$ and $Z_B=Z_A\cup Z_{new}$. Then $\delta_A^1(\alpha\|e_1)=r\|r^{\alpha\|e_1}=2r\|\alpha=0$ since char$(k)=2$, but $\delta_B^1(\alpha^*\|f_1)=2r^*\|\alpha^*+r_2\|\gamma^*=r_2\|\gamma^*\ne0$,  which means that although $\alpha\|e_1\in \mathrm{Ker}(\delta_A^1)$, $\psi_1(\alpha\|e_1)=\alpha^*\|f_1\notin \mathrm{Ker}(\delta_B^1)$, hence $\psi_1$ does not induce an injective $k$-linear map $\mathrm{Ker}(\delta_A^1)\hookrightarrow \mathrm{Ker}(\delta_B^1)$.
\end{Ex1}


\begin{Ex1}
\label{Diffblockinc}
Let $A$ be given by two blocks $A_1$ and $A_2$ such that $A_1$ is isomorphic to $k[x]/(x^2)$ and $A_2$ is isomorphic to $k[y]/(y^2)$. Let $B$ be obtained by gluing the units of $A_1$ and $A_2$. Then $\mathrm{Ker}(\delta_B^1)=\HH^1(B)\simeq \mathfrak{gl}_2(k)$ and has $k$-basis
given by $\{ x \| x,\ x\| y,\ y\| x,\  y\| y\}$. However, there are two cases for $A$.

$(1)$ If $\mathrm{char}(k)\neq 2$, then $\mathrm{Ker}(\delta_A^1)=\HH^1(A)\simeq k\times k$ has
$k$-basis given by $x\| x$ and $y\| y$, and there is an injective Lie algebra homomorphism $\mathrm{Ker}(\delta_A^1)\hookrightarrow \mathrm{Ker}(\delta_B^1)$.

$(2)$ If $\mathrm{char}(k)=2$, then $\mathrm{Ker}(\delta_A^1)=\HH^1(A)$ has
$k$-basis given by $\{ x \| x,\ x\| e_1,\ y\| y,\  y\| e_2\}$. Clearly in this case we can not get an injective Lie algebra homomorphism from $\mathrm{Ker}(\delta_A^1)$ to $\mathrm{Ker}(\delta_B^1)$.
\end{Ex1}


We verify the special paths and the k-space $Z_{sp}$ (resp. the special pairs and the k-space $Z_{spp}$) appeared in Definition \ref{special-path} and Proposition \ref{Iamge-delta-zero} (resp. in Definition \ref{elements-in-Z-spp} and Proposition \ref{Kernel-delta-one-difference}) by the following example.

\begin{Ex1}\label{ex-image-kernel}
$B$ is obtained from $A$ by gluing $e_{1}$ and $e_{4}$:	
	\begin{center}
		\begin{tikzcd}
		&                             &                                                                                                                   &                             &              &        &  f_{3} \bullet & {} \arrow[rd, "\ddots", phantom, shift right=-1]                                                                                                                                              &    \\
		Q_{A}:\hspace{-2em} & e_{2}\bullet \arrow[r, "a"] & e_{1}\bullet \arrow[r, "\alpha_{1}", bend left] \arrow[r, "\alpha_{n}"', bend right] \arrow[r, "\vdots", phantom] & e_{4}\bullet \arrow[r, "b"] & e_{3}\bullet & Q_{B}:\hspace{-2em} & f_{2}\bullet \arrow[r, "a^{*}"] & \bullet f_{1} \arrow["\alpha_{1}^{*}"', loop, distance=2em, in=125, out=55] \arrow["\alpha_{n}^{*}"', loop, distance=2em, in=35, out=325] \arrow[lu, "b^{*}"', shift left] & {}
		\end{tikzcd}  	
	\end{center}
Where $Z_A=\emptyset$, $Z_{B}=Z_{new}=\{\alpha_{i}^{*}\alpha_{j}^{*}\ |\ 1\le i,j \le n\}$. Since $\alpha_ia \notin I_A$ for $1\le i \le n$, $\alpha_i$ is a special path from $e_1$ to $e_4$ for $1\le i \le n$, we have $\mathrm{Sp}(1,4)=\{\alpha_i\ |\ 1\le i \le n \}$ and
\begin{equation*}
\begin{split}
 Z_{sp} & =\langle \delta_B^0(f_1 \| \alpha_i^*)\ |\ 1\le i\le n \rangle \\ & =\langle b^*\|b^*\alpha_i^*-a^*\|\alpha_i^*a^*\ |\ 1\le i \le n \rangle.
\end{split}
\end{equation*}
Hence $\mathrm{sp}(1,4)=n=\mathrm{dim}_kZ_{sp}$. Since $a^{*}\|\alpha_{i}^{*}a^{*}, b^{*}\|b^{*}\alpha_{i}^{*},\alpha_i^*\|f_1$, $a\nparallel\alpha_{i}a, b\nparallel b\alpha_{i},\alpha_i\nparallel e_1,\alpha_i\nparallel e_4$, we know that $(a,\alpha_{i}a), (b,b\alpha_{i}), (\alpha_i,e_1), (\alpha_i,e_n)$ are special pairs with respect to the gluing of $e_{1}$ and $e_{4}$ for $1\le i \le n$, and $\mathrm{Spp}(1,4)=\{(a,\alpha_i a), (b,b\alpha_i), (\alpha_i,e_1), (\alpha_i,e_n)\ |\ 1\le i\le n \}$. As a result we get $$\langle \mathrm{Spp}(1,4) \rangle=\langle a^*\| \alpha_i^*a^*,b^*\| b^*\alpha_i^*, \alpha_i^*\|f_1\ |\ 1\le i\le n \rangle,$$
\begin{equation*}
	\begin{split}
	Z_{spp} & = \langle \mathrm{Spp}(1,4) \rangle \cap \mathrm{Ker}(\delta_B^1) \\ & =\langle a^*\| \alpha_i^*a^*,b^*\| b^*\alpha_i^*\ |\ 1\le i\le n\rangle.
	\end{split}
\end{equation*}
Hence $\mathrm{kspp}(1,4)=\mathrm{dim}_kZ_{spp}=2n$. A direct computation shows that $\mathrm{Im}(\delta_{A}^{0}),\mathrm{Im}(\delta_{B}^{0})$ are 3-dimensional and $(n+2)$-dimensional, respectively, since $$\mathrm{Im}(\delta_{A}^{0})=\langle a\|a, b\|b,\sum\limits_{i=1}^n\alpha_{i}\|\alpha_{i}\rangle,$$
$$\mathrm{Im}(\delta_{B}^{0})=\langle a^{*}\|a^{*}, b^{*}\| b^{*},b^{*}\| b^{*}\alpha_{i}^{*}-a^{*}\| \alpha_{i}^{*}a^{*}\ |\ 1\le i\le n\rangle.$$  Therefore $$\mathrm{dim}_{k}\mathrm{Im}(\delta_{A}^{0})=\mathrm{dim}_{k}\mathrm{Im}(\delta_{B}^{0})+1-\mathrm{sp}(1,4).$$
$$\mathrm{Ker}(\delta_{A}^{1})=\langle a\| a, b\| b, \alpha_{i}\| \alpha_{j}\ |\ 1\le i, j \le n\rangle$$ is $(n^{2}+2)$-dimensional and
$$\mathrm{Ker}(\delta_{B}^{1})=\langle a^{*}\| a^{*}, b^{*}\| b^{*}, \alpha_{i}^{*}\| \alpha_{j}^{*}, b^{*}\| b^{*}\alpha_{i}^{*}, a^{*}\| \alpha_{i}^{*}a^{*}\ |\ 1\le i, j \le n\rangle$$ is $(n^{2}+2n+2)$-dimensional, hence $$\mathrm{dim}_{k}\mathrm{Ker}(\delta_{B}^{1})=\mathrm{dim}_{k}\mathrm{Ker}(\delta_{A}^{1})+\mathrm{kspp}(1,4).$$
One can verify that $\mathrm{HH}^1(A)$ is isomorphic to $\mathfrak{pgl}_n(k)$ and $\mathrm{HH}^1(B)$ contains a subalgebra isomorphic to $\mathfrak{gl}_n(k)$. Note also that by the notations in the proof of Theorem \ref{Lie-strucutre-of-hh1}, in this example the subspace $Y$ of $\mathrm{Ker}(\delta_{B}^{1})$ is equal to $\mathrm{Im}(\delta_{B}^{0})\oplus \langle \sum\limits_{i=1}^n\alpha_{i}^*\|\alpha_{i}^*\rangle$ and $Y$ is not a Lie ideal of $\mathrm{Ker}(\delta_{B}^{1})$.		
\end{Ex1}

By Corollary \ref{Kernel-delta-one-source-and-sink}, if $B$ is a radical embedding of $A$ obtained by gluing a source vertex $e_1$ and a sink vertex $e_n$ of $A$, then $\mathrm{Ker}(\delta_B^1)\simeq \mathrm{Ker}(\delta_A^1)$. However, the converse of Corollary \ref{Kernel-delta-one-source-and-sink} is not true in general which reveals by the following example.

\begin{Ex1}\label{ex-converse-coro}
$B$ is obtained from $A$ by gluing $e_{1}$ and $e_{4}$:
\begin{center}
			\begin{tikzcd}
			&                             &                                                                                                             &            &                            &  &      & f_3 \bullet \arrow[rd, "b^*", shift right] & {} \arrow[rd, "\ddots", phantom, shift right=-1]                                                                               &    \\
			Q_A: \hspace{-2em}& e_{2}\bullet \arrow[r, "a"] & e_1\bullet \arrow[r, "\alpha_n"', bend right] \arrow[r, "\alpha_1", bend left] \arrow[r, "\vdots", phantom] & e_4\bullet & e_3\bullet \arrow[l, "b"'] &  & Q_B:\hspace{-2em} & f_2\bullet \arrow[r, "a^*"]                & \bullet f_1 \arrow["\alpha_n^*", loop, distance=2em, in=325, out=35] \arrow["\alpha_1^*", loop, distance=2em, in=55, out=125] & {}
			\end{tikzcd}	
\end{center}
Where $Z_A=\{\alpha_ia\ |\ 1\le i\le n\}$, $Z_{new}=\{\alpha_i^*b^*, \alpha_{i}^{*}\alpha_{j}^{*}\ |\ 1\le i, j \le n\}$ and $Z_B=Z_A\cup Z_{new}$. Note that although $\mathrm{Spp(1,4)}=\{(\alpha_i,e_1),(\alpha_i,e_4)\ |\ 1\le i\le n\}$, we have $Z_{spp}=\langle \mathrm{Spp}(1,4) \rangle \cap \mathrm{Ker}(\delta_B^1)=\langle \alpha_i^*\|f_1 \rangle \cap \mathrm{Ker}(\delta_B^1)=0$.
 We infer from Proposition \ref{Kernel-delta-one-difference} that we have $\mathrm{dim}_k\mathrm{Ker}(\delta_B^1)=\mathrm{dim}_k\mathrm{Ker}(\delta_A^1)$. In fact, a direct computation shows that both
$$\mathrm{Ker}(\delta_A^1)=\langle a\|a, b\|b, \alpha_i\|\alpha_j\ |\ 1\le i,j\le n\rangle$$
and
$$\mathrm{Ker}(\delta_B^1)=\langle a^*\|a^*, b^*\|b^*, \alpha_i^*\|\alpha_j^*\ |\ 1\le i,j\le n\rangle$$
are $(n^2+2)$-dimensional. Hence although we do not glue a source and a sink, we have $\mathrm{Ker}(\delta_B^1)\simeq \mathrm{Ker}(\delta_A^1)$.	
\end{Ex1}


In order to have an intuitive feeling, we give various types of special pairs in the following example.

\begin{Ex1}\label{example-type-special-pair} In this example we always assume that $B$ is obtained from $A$ by gluing $e_1$ and $e_n$, and that $\alpha$ is an arrow in $Q_A$ and $p$ is a path in $\mathcal{B}_A$. It can be proved that the special pairs ($\alpha$,$p$) are exclusively from the following 7 typical cases and their dual cases:
	
	$(\romannumeral1):$ $\alpha$ is a loop at $e_1$ or $e_n$, assume that \begin{tikzcd}
	e_1\bullet \arrow["\alpha", loop, distance=2em, in=55, out=125].
	\end{tikzcd} (The case that \begin{tikzcd}
	e_n\bullet \arrow["\alpha", loop, distance=2em, in=55, out=125]
	\end{tikzcd} is dual.)
	
	Case 1: $p=a_n\cdots a_1$ is an oriented cycle at $e_n$ or $p=e_n$, such as:
\begin{center}
	\begin{tikzcd}
	& \arrow[r, "\cdots", phantom]     & \arrow[d, "a_n"]      \\
	e_1\bullet \arrow["\alpha", loop, distance=2em, in=55, out=125] \arrow[r] & \bullet\quad\cdots\quad\bullet & \arrow[l]\bullet e_n \arrow[lu, "a_1"']
	\end{tikzcd} ;
\end{center}

	Case 2: $p=a_n\cdots a_1$ is a path between $e_1$ and $e_n$, such as:
\begin{center}
	\begin{tikzcd}
	e_1\bullet \arrow["\alpha", loop, distance=2em, in=55, out=125] \arrow[r, "a_1"] & \bullet\quad\cdots\quad\bullet \arrow[r, "a_n"] & \bullet e_n
	\end{tikzcd} ;
\end{center}

	
	$(\romannumeral2):$ $\alpha$ is an arrow between $e_1$ and $e_n$, assume that \begin{tikzcd}
	e_1\bullet \arrow[r, "\alpha"] & \bullet e_n.
	\end{tikzcd} (The case that \begin{tikzcd}
	e_n\bullet \arrow[r, "\alpha"] & \bullet e_1
	\end{tikzcd} is dual.)
	
	Case 3: $p=a_n\cdots a_1$ is an oriented cycle at $e_1$ or $e_n$ or $p=e_1$ or $e_n$, such as:
\begin{center}
	\begin{tikzcd}
	{} \arrow[r, "\cdots", phantom, shift right=3]                      & {} \arrow[ld, "a_n"] \\
	e_1\bullet \arrow[r, "\alpha"] \arrow[u] \arrow[u, "a_1"] \arrow[r] & \bullet e_n
	\end{tikzcd} ;
\end{center}

	Case 4: $p=a_n\cdots a_1$ is a path from $e_n$ and $e_1$, such as:
\begin{center}
	\begin{tikzcd}
	{e_1\bullet} & {\bullet e_n} \\
	{} & {}
	\arrow["\alpha", from=1-1, to=1-2]
	\arrow[""{name=0, anchor=center, inner sep=0}, "\cdots"{description}, shift right=4, draw=none, from=2-2, to=2-1]
	\arrow["{a_1}"{pos=0.3}, shift left=1, shorten >=4pt, from=1-2, to=0]
	\arrow["{a_n}"{pos=0.7}, shift left=1, shorten <=4pt, from=0, to=1-1]
	\end{tikzcd} ;
\end{center}

	
	$(\romannumeral3):$ Exactly one of the vertex of $\alpha$ is $e_1$ or $e_n$, assume that \begin{tikzcd}
	e_1\bullet \arrow[r, "\alpha"] \arrow[r] & \bullet .
	\end{tikzcd} (The other cases are dual.)
	
	Case 5: $p=a_n\cdots a_1$ is a path from $e_n$ to $t(\alpha)$, such as:
\begin{center}
	\begin{tikzcd}
	{} & {e_1\bullet} & \bullet & \bullet \cdots \bullet & {\bullet e_n} \\
	&&& \cdots
	\arrow["\alpha", from=1-2, to=1-3]
	\arrow[from=1-3, to=1-4]
	\arrow[from=1-5, to=1-4]
	\arrow["{a_1}", from=1-5, to=2-4]
	\arrow["{a_n}", from=2-4, to=1-3]
	\end{tikzcd} ;
\end{center}

	Case 6: $p=\alpha p_1$, where $p_1=a_n \cdots a_1$ is a path from $e_n$ to $e_1$, such as:
\begin{center}
	\begin{tikzcd}
	{e_1\bullet} & {\bullet \cdots \bullet} & {\bullet e_n} \\
	& \cdots
	\arrow["\alpha", from=1-1, to=1-2]
	\arrow[from=1-3, to=1-2]
	\arrow["{a_1}", from=1-3, to=2-2]
	\arrow["{a_n}", from=2-2, to=1-1]
	\end{tikzcd};
\end{center}

	Case 7: $p=p_2\alpha p_1$, where $p_1=a_n \cdots a_1$ is a path from $e_n$ to $e_1$ and $p_2=b_m\cdots b_1$ is a cycle at $t(\alpha)$, such as:
\begin{center}
	\begin{tikzcd}
	{} & {} & {} \\
	{e_1\bullet} & {\bullet } & \bullet & {\bullet e_n} \\
	& \cdots
	\arrow["\alpha", from=2-1, to=2-2]
	\arrow["{a_n}", from=3-2, to=2-1]
	\arrow["\cdots"{description, pos=0.7}, shift right=2, draw=none, from=1-1, to=1-3]
	\arrow["\cdots"{description}, draw=none, from=2-2, to=2-3]
	\arrow[from=2-4, to=2-3]
	\arrow["{b_1}", from=2-2, to=1-2]
	\arrow["{b_m}", shorten <=4pt, from=1-3, to=2-2]
	\arrow["{a_1}", from=2-4, to=3-2]
	\end{tikzcd}.
\end{center}

	
	After giving relations in specific examples, we can show that the special pair $(\alpha,p)$ in each of the above cases can appear. Indeed, the following example covers all the above 7 cases:

	\begin{center}
		\begin{tikzcd}
		Q_A: & e_2\bullet \arrow["d"', loop, distance=2em, in=215, out=145] & e_1\bullet \arrow[r, "\beta", shift left] \arrow["\alpha"', loop, distance=2em, in=125, out=55] \arrow[l, "\gamma"'] & \bullet e_3 \arrow[l, "a"', shift left=2] \arrow[ll, "c", bend left] & Q_B: & f_2\bullet \arrow["d^*"', loop, distance=2em, in=215, out=145] & \bullet f_1 \arrow["\alpha^*"', loop, distance=2em, in=125, out=55] \arrow["\beta^*"', loop, distance=2em, in=35, out=325] \arrow["a^*"', loop, distance=2em, in=305, out=235] \arrow[l, "\gamma^*"', shift right] \arrow[l, "c^*", shift left]
		\end{tikzcd}
	\end{center}
Where $Z_A$ consists of all paths in $Q_A$ of length $3$ except $d\gamma a$, $Z_{new}=\{a^*\alpha^*, c^*\alpha^*,\alpha^*\beta^*,(\beta^*)^2,\gamma^*\beta^*,(a^*)^2,c^*a^*\}$ and $Z_B=Z_A\cup Z_{new}$. We list all special pairs ($\alpha,p$) for each case as follows:

Case 1: ($\alpha,\beta a$), ($\alpha,e_3$);

Case 2: ($\alpha,a$), ($\alpha,\beta$), ($\alpha,\beta\alpha$), ($\alpha,\alpha a$);

Case 3: ($\beta,\alpha$), ($\beta,a\beta$), ($\beta,\beta a$), ($\beta,e_1$), ($\beta,e_3$) ($a,\alpha$), ($a,a\beta$), ($a,\beta a$), ($a,e_1$), ($a, e_3$);

Case 4: ($\beta,a$), ($a,\beta$);

Case 5: ($\gamma,c$), ($\gamma,dc$), ($c,\gamma \alpha$), ($c,d\gamma$);

Case 6: ($\gamma,\gamma a$), ($c,c\beta$);

Case 7: ($\gamma,d\gamma a$).

By check one by one, we have $\mathrm{Spp}(1,3)$ is the set consisting of these 25 special pairs and $\langle \mathrm{Spp}(1,3) \rangle = \langle \alpha^* \| p^* \ |\ (\alpha,p) \in \mathrm{Spp}(1,3) \rangle $ and therefore
\begin{equation*}
	\begin{split}
		Z_{spp} & =\langle \mathrm{Spp}(1,3) \rangle \cap \mathrm{Ker}(\delta_B^1) \\ & =\langle a^*\| \beta^*a^*, \alpha^*\| \beta^*\alpha^*, \alpha^*\|\alpha^*a^*, \beta^*\| a^*\beta^*, \beta^*\| \beta^* a^*, a^*\|a^*\beta^*, \\ & a^*\| \beta^*a^*,\gamma^* \| d^*c^*, c^*\| \gamma^*\alpha^*, c^*\| d^*\gamma^*, \gamma^*\|\gamma^*a^*, c^*\|c^*\beta^*, \gamma^*\|d^*\gamma^*a^* \rangle.
	\end{split}
\end{equation*}
Hence $\mathrm{kspp}(1,3)=13$. Note also that the special paths in this example are $\beta$ and $a$, so $\mathrm{sp}(1,3)=2$.
\end{Ex1}

It deserves to mention that although the $k$-space $\langle \mathrm{Spp}(1,n) \rangle$ is generated by the elements of the form $\alpha^* \| p^*$ (where $\alpha$ is an arrow and $p$ is a path), an element in $Z_{spp}$ is usually a $k$-linear combination of such elements.

\begin{Ex1}\label{linear-combination-zspp} $B$ is obtained from $A$ by gluing $e_1$ and $e_5$:
	\begin{center}
		\begin{tikzcd}
		&&&&&&&&& {\bullet f_4} \\
		{Q_A:} \hspace{-2em} & {e_2\bullet} & {e_1\bullet} & {e_3 \bullet} & {e_5 \bullet} & {\bullet e_4} && {Q_B:} \hspace{-2em} & {f_2\bullet} & {\bullet f_1} & {\bullet f_3}
		\arrow["b", from=2-2, to=2-3]
		\arrow["c", from=2-3, to=2-4]
		\arrow["d", from=2-4, to=2-5]
		\arrow["a", from=2-5, to=2-6]
		\arrow["{b^*}", from=2-9, to=2-10]
		\arrow["{a^*}", shift left=1, from=2-10, to=1-10]
		\arrow["{c^*}", shift left=1, from=2-10, to=2-11]
		\arrow["{d^*}", shift left=1, from=2-11, to=2-10]
		\end{tikzcd}
	\end{center}

  Where $Z_A=\emptyset$ and $Z_B=Z_{new}=\{ a^*b^*,c^*d^* \}$. It follows from a direct calculation that $$\mathrm{Im}(\delta_A^0)=\langle a\|a,b\|b,c\|c,d\|d \rangle = \mathrm{Ker}(\delta_A^1),$$ hence $\mathrm{HH}^1(A)=0$. Similarly we have $$\mathrm{Im}(\delta_B^0)=\langle a^*\|a^*,b^*\|b^*,d^*\|d^*-c^*\|c^*,a^*\|a^*d^*c^*-b^*\|d^*c^*b^* \rangle,$$ $$\mathrm{Ker}(\delta_B^1)=\langle a^*\|a^*,b^*\|b^*,c^*\|c^*,d^*\|d^*,a^*\|a^*d^*c^*-b^*\|d^*c^*b^* \rangle,$$ hence $\mathrm{HH}^1(B)\simeq \langle c^*\|c^* \rangle$. Using the notation in Theorem \ref{Lie-strucutre-of-hh1}, we get the ideal $I \simeq \langle \psi_1(\delta_A^0(e_1\|e_1)) \rangle=\langle c^*\|c^*-b^*\|b^* \rangle$ and $\mathrm{HH}^1(A)\simeq \mathrm{HH}^1(B)/I$. It is clear that $\mathrm{Spp}(1,5)=\{ (a,adc),(b,dcb) \}$, therefore $\langle \mathrm{Spp}(1,5) \rangle =\langle a^*\|a^*d^*c^*,b^*\|d^*c^*b^* \rangle$ and
  \begin{equation*}
  \begin{split}
    Z_{spp} & = \langle \mathrm{Spp}(1,5) \rangle \cap \mathrm{Ker}(\delta_B^1) \\ & = \langle a^*\|a^*d^*c^*-b^*\|d^*c^*b^* \rangle.
  \end{split}
  \end{equation*}
\end{Ex1}

The following example shows that the difference between the dimensions of $\HH^1(A)$ and $\HH^1(B)$ can be arbitrarily large.

\begin{Ex1}
\label{diffblockarblarg}
Let $A$ be given by two blocks $A_1$ and $A_2$ such that $A_1$ and $A_2$ are radical square zero local algebras having $m$-loops and $n$-loops respectively. If we exclude the case that $m=1$ and $n=1$ in $\mathrm{char}(k)=2$ (for this case, see Example \ref{Diffblockinc}), then the dimension of $\HH^1(A)$ is the sum of the dimensions of $\HH^1(A_1)\simeq \mathfrak{gl}_{m}(k)$ and $\HH^1(A_2)\simeq \mathfrak{gl}_{n}(k)$, that is, $m^2+n^2$. Let $B$ be obtained by gluing the units of $A_1$ and $A_2$. Then $\HH^1(B)\simeq \mathfrak{gl}_{m+n}(k)$ and consequently has dimension $(m+n)^2$.
\end{Ex1}

We use the following example to demonstrate the last statement of Corollary \ref{quotient-lie-algebra-source-and-sink}. This example also shows that in general the center of $\mathrm{HH}^1(B)$ is properly contained in the center of $L_0$ (cf. Proposition \ref{center-L_0}).

\begin{Ex1}\label{eg-I-summand-of-HH}
Suppose $\mathrm{char}(k)=0$, $B$ is obtained from $A$ by gluing $e_1$ and $e_4$:
    \begin{center}
    	\begin{tikzcd}
    	&            & e_2\bullet \arrow[rd, "\beta", shift left]                                                                    &             &  &      &            &                                                                                                                                                                    &                                                \\
    	Q_A:\hspace{-2em} & e_3\bullet & e_1\bullet \arrow[r, "\gamma"'] \arrow[u, "\alpha_1"] \arrow[u, "\alpha_2"', shift right=3] \arrow[l, "\eta"] & \bullet e_4 &  & Q_B:\hspace{-2em} & f_3\bullet &  f_1 \bullet \arrow["\gamma^*"', loop, distance=2em, in=125, out=55] \arrow[r, "\alpha_1^*", shift left=4] \arrow[r, "\alpha_2^*" description] \arrow[l, "\eta^*"] & \bullet f_2 \arrow[l, "\beta^*", shift left=4]
    	\end{tikzcd}
    \end{center}
Where $Z_A=\{\beta\alpha_1\}$ and $Z_{new}=\{(\gamma^*)^2,\alpha_i^*\gamma^*,\alpha_i^*\beta^*,\gamma^*\beta^*,\eta^*\gamma^*,\eta^*\beta^*\ |\ i=1,2 \}$. From a straightforward computation we have $$\mathrm{Im}(\delta_A^0)=\langle \alpha_1\|\alpha_1+\alpha_2\|\alpha_2+\gamma\|\gamma,\beta\|\beta+\gamma\|\gamma,\eta\|\eta \rangle,$$ $$\mathrm{Im}(\delta_B^0)=\langle \alpha_1^*\|\alpha_1^*+\alpha_2^*\|\alpha_2^*-\beta^*\|\beta^*,\eta^*\|\eta^* \rangle,$$ $$\mathrm{Ker}(\delta_A^1)=\langle \alpha_2\|\alpha_1,\alpha_1\|\alpha_1,\alpha_2\|\alpha_2,\beta\|\beta,\gamma\|\gamma,\gamma\|\beta\alpha_2,\eta\|\eta \rangle.$$ Since we glue a source and a sink, Corollary \ref{Kernel-delta-one-source-and-sink} shows that $\mathrm{Ker}(\delta_B^1)\simeq\mathrm{Ker}(\delta_A^1)$. As a consequence, $$\mathrm{HH}^1(A)\simeq \langle \alpha_2\|\alpha_1,\alpha_1\|\alpha_1,\alpha_2\|\alpha_2,\gamma\|\beta\alpha_2\rangle,$$ $$\mathrm{HH}^1(B)\simeq \langle \alpha_2^*\|\alpha_1^*,\alpha_1^*\|\alpha_1^*,\alpha_2^*\|\alpha_2^*,\gamma^*\|\gamma^*,\gamma^*\|\beta^*\alpha_2^*\rangle.$$ Using the notation in Theorem \ref{Lie-strucutre-of-hh1}, we get the ideal $I=\langle \psi_1(\delta_A^0(e_1\|e_1)) \rangle=\langle \alpha_1^*\|\alpha_1^*+\alpha_2^*\|\alpha_2^*+\gamma^*\|\gamma^*+\eta^*\|\eta^* \rangle$ and $\mathrm{HH}^1(A)\simeq \mathrm{HH}^1(B)/I$. Note that the symbol $\Delta$ in Remark \ref{minimal-generator-of-ideal-I} is exactly equal to $\{[\alpha],[\gamma] \}$, where $[\alpha]=\{\alpha_1,\alpha_2 \}$ and $[\gamma]=\{\gamma \}$, we can rewrite the generator of $I$ as $\psi_1(I_{[\alpha]}+I_{[\gamma]})= \alpha_1^*\|\alpha_1^*+\alpha_2^*\|\alpha_2^*+\gamma^*\|\gamma^*$ since $\eta^*\|\eta^*\in \mathrm{Im}(\delta_B^0)$. Also $L_0^{[\alpha^*]}=\langle \alpha_2^*\|\alpha_1^*,\alpha_1^*\|\alpha_1^*,\alpha_2^*\|\alpha_2^* \rangle$, $L_0^{[\gamma^*]}=\langle \gamma^*\|\gamma^* \rangle$, hence $$L_0=L_0^{[\alpha^*]}\oplus L_0^{[\gamma^*]}=\langle \alpha_2^*\|\alpha_1^*,\alpha_1^*\|\alpha_1^*,\alpha_2^*\|\alpha_2^* \rangle \oplus \langle \alpha_1^*\|\alpha_1^*+\alpha_2^*\|\alpha_2^*+\gamma^*\|\gamma^*\rangle= L_0^{[\alpha^*]} \oplus I$$ as Lie algebras, and therefore $Z(L_0)=Z(L_0^{[\alpha^*]})\oplus Z(L_0^{[\gamma^*]})=\langle \alpha_1^*\|\alpha_1^*+\alpha_2^*\|\alpha_2^* \rangle \oplus \langle \gamma^*\|\gamma^* \rangle$. Clearly $Z(\mathrm{HH}^1(B))=I\subsetneq Z(L_0)$. Since $L_1=\langle \gamma^*\|\beta^*\alpha_2^* \rangle$, $$\mathrm{HH}^1(B)=L_0\oplus L_1=(L_0^{[\alpha^*]} \oplus I)\oplus L_1=(L_0^{[\alpha^*]} \oplus L_1)\oplus I\simeq \mathrm{HH}^1(A)\times I\simeq \mathrm{HH}^1(A)\times k$$ as Lie algebras.
\end{Ex1}

\section{Fundamental group}
\label{fundsec}
Let $\pi_1(Q, I)$ be a fundamental group of a bound quiver $(Q,I)$.
Suppose that a quiver $Q$ has $n$ vertices and $m$ edges and $c$ connected components. We adopt the notation that the first Betti number of $Q$, denoted by $\chi(Q)$, equals $m - n + c$. Note that the first Betti number is equal to the dimension of the first cohomology group of the underlying graph of $Q$, see for example  \cite[Lemma 8.2]{D}. Intuitively we can say that $\chi(Q)$ counts the number of holes in $Q$.

Recall from \cite[Lemma 1.7]{BR} that for a bound quiver $(Q,I)$ we have  $\dim_k(\Hom(\pi_1(Q, I), k^{+}) \leq \chi(Q)$. Equality holds if I is a monomial ideal, and more generally if $I$ is semimonomial  \cite[Section 1]{GAS} and in positive characteristic if $I$ is $p$-semimonomial; see after Remark 1.8 in \cite{BR}. Therefore by Theorem C in \cite{BR} we have that \[\pi_1\!\text{-}\!\rank(A):=\max\{\dim_k\pi_1(Q,I)^\vee ~:~ A\simeq kQ/I, \textrm{ $I$ is an admissible ideal} \}\] is equal to $\chi(Q_A)$.

Note that the maximum should be over the minimal presentations, however, since for monomial algebras the maximum is obtained for an admissible ideal it is enough to restrict to this subset. The $\pi_1\!\text{-}\!\rank(A)$ is a derived invariant and an invariant under stable equivalences of Morita type for selfinjective algebras, see \cite[Theorem B]{BR}. However, the next lemma shows that it is not an invariant under stable equivalences induced by gluing idempotents.

\begin{Lem} \label{pi-one-rank}
Let $A=kQ_A/I_A$ be a finite dimensional monomial (or semimonomial) algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents of $A$. Then $$\pi_1\!\text{-}\!\rank(A)=\pi_1\!\text{-}\!\rank(B)+c_A-c_B-1.$$ In particular, if we glue two idempotents from different blocks, then $$\pi_1\!\text{-}\!\rank(A)=\pi_1\!\text{-}\!\rank(B);$$ and if we glue two idempotents from the same block, then $$\pi_1\!\text{-}\!\rank(A)=\pi_1\!\text{-}\!\rank(B)-1.$$
\end{Lem}

\begin{Proof} Since $A$ and $B$ are monomial algebras we have by Theorem C in \cite{BR} that $\pi_1\!\text{-}\!\rank(A)=\chi(Q_A)$ and $\pi_1\!\text{-}\!\rank(B)=\chi(Q_B)$. The statement follows from the fact the number of arrows of $Q_A$ and $Q_B$ is the same, that is, $m_A=m_B$, and from the observation that $n_A=n_B+1$. Also note that the number of connected components have the relation $c_A=c_B+1$ when we glue two idempotents from different blocks but $c_A=c_B$ when we glue two idempotents from the same block. The same argument applies if $A$ and $B$ are semimonomial algebras.
\end{Proof}

\begin{Rem}
When the characteristic of the field is positive, Lemma \ref{pi-one-rank} holds also for $p$-semimonomial algebras since the $\pi_1\!\text{-}\!\rank$ coincides with the first Betti number.
\end{Rem}

\begin{Rem}
Note that Lemma \ref{pi-one-rank} and Lemma \ref{Image-zero-part} are intimately related. In fact $\pi_1\!\text{-}\!\rank(A)=m_A-\mathrm{dim}_k (\mathrm{Im}(\delta^0_{(A)_0}))$.
\end{Rem}

\begin{Rem}
It is worth noting that if we glue a source and a sink from the same block we have that  $\pi_1\!\text{-}\!\rank(B)- \pi_1\!\text{-}\!\rank(A)=1$ and this quantity coincides with the difference $\mathrm{dim}_k\mathrm{HH}^1(B)-\mathrm{dim}_k\mathrm{HH}^1(A)=1$. The reason why we obtain such equality is because if we glue a source and a sink,
 then  $\mathrm{HH}^1(A)$ is isomorphic to the  quotient of $\mathrm{HH}^1(B)$. More precisely, the ideal  of this quotient is a $1$-dimensional torus  $\psi_1(\mathrm{Im}(\delta^0_A))/\mathrm{Im}(\delta^0_B)$ where $\psi_1: \mathrm{Ker}(\delta^1_A) \to \mathrm{Ker}(\delta^1_B)$.  Therefore, we can say that the $\pi_1\!\text{-}\!\rank$ controls the change of the dimensions between $\mathrm{HH}^1(B)$ and $\mathrm{HH}^1(A)$. This process also changes the Lie algebra structure. For example, in characteristic different from $2$, from a reductive Lie algebra $\mathfrak{gl}_2(k)$ we pass to a simple Lie algebra $\mathfrak{sl}_2(k)$, or for example from a one dimensional abelian Lie algebra to the trivial Lie algebra. It follows from Theorem \ref{hh1-quotient-isomorphism-for-monomial-algebras} that for a general gluing we do not obtain the equality above.
\end{Rem}




\section{Centers and Hochschild cohomology groups in higher degrees}
\label{HHsec}


\subsection{Center}

In this subsection, we study the behaviour of the centers of finite dimensional monomial algebras under gluing idempotents. Throughout we will denote by $Z(A)$ the center of an algebra $A$.

\begin{Def}\label{non-special-path}
	Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$ and let $p$ be a path either from $e_1$ to $e_n$ or from $e_n$ to $e_1$ in $Q_A$. We call that $p$ is a non-special path between $e_1$ and $e_n$ in $Q_A$ if $\delta_{B}^{0}(f_1 \| p^*)= 0$, or equivalently, if $ap\in I_A$ and $pb\in I_A$ for arbitrary $a,b \in (Q_{A})_1$.
	
	We denote by $\mathrm{NSp}(1,n)$ the set of non-special paths between $e_1$ and $e_n$ in $Q_A$, and by $\mathrm{nsp}(1,n)$ the number of these non-special paths. Furthermore, we denote by $Z_{nsp}$ the $k$-subspace of $k((Q_B)_0\| \mathcal{B}_B)$ generated by the elements $f_1 \| p^*$, where $p \in \mathrm{NSp}(1,n)$.
\end{Def}

Actually, as the name shows, non-special path is exactly the opposite notion of special path, and it is clear that there is no non-special path between $e_1$ and $e_n$ when we glue these two idempotents from different blocks.

Similarly to Definition \ref{kerdeltaA0}, let $\delta^0_{(A)_{\geq 1}}$ denote the map $\delta^0_A$ restricted to the subspace $k((Q_A)_0||(\mathcal{B}_A)_{\geq 1})$ and define $\mathrm{Ker}(\delta^0_{(A)_{\geq 1}})$ to be the kernel of the map $\delta^0_{(A)_{\geq 1}}$.

\begin{Lem}\label{kernel-delta}
Let $A=kQ_A/I_A$ be a monomial algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents $e_1$ and $e_n$ of $A$. Then there is a decomposition as $k$-vector spaces $$\mathrm{Ker}(\delta^0_{(B)_{\geq1}})=\psi_0(\mathrm{Ker}(\delta^0_{(A)_{\geq1}})) \oplus Z_{nsp}.$$ In particular, if we glue $e_1$ and $e_n$ from the same block, then $\mathrm{dim}_k \mathrm{Ker}(\delta^0_{(B)_{\geq1}})=\mathrm{dim}_k \mathrm{Ker}(\delta^0_{(A)_{\geq1}})+nsp(1,n)$; if we glue $e_1$ and $e_n$ from different blocks, then $\mathrm{dim}_k \mathrm{Ker}(\delta^0_{(B)_{\geq1}})=\mathrm{dim}_k \mathrm{Ker}(\delta^0_{(A)_{\geq1}}).$
\end{Lem}

\begin{Proof}
Recall from Proposition \ref{parallel-paths-in-monomial-algebras} that there is a $k$-linear map $\psi_0: k((Q_A)_0\| \mathcal{B}_A) \to k((Q_B)_0\| \mathcal{B}_B)$. A direct computation shows that $\delta_B^0(\psi_0(e_i \| p))=\psi_1(\delta_A^0(e_i \| p))$ for $1\leq i\leq n$ and $p\in \mathcal{B}_A\backslash \{e_1,e_n\}$. It follows that $\psi_0$ induces an injective $k$-linear map from $\mathrm{Ker}(\delta_{(A)_{\geq 1}}^{0})$ to $\mathrm{Ker}(\delta_{(B)_{\geq 1}}^{0})$.

Let $\theta\in\mathrm{Ker}(\delta_{(B)_{\geq 1}}^{0})$ lies in the complement of the subspace $\psi_0(\mathrm{Ker}(\delta_{(A)_{\geq 1}}^{0}))$. Then we may assume that $\theta$ is a linear combination of the elements of the form $f_1\| p^*$ such that $p$ is a path between $e_1$ and $e_n$. If $p$ is a non-special path, then $f_1\| p^*\in Z_{nsp}\subseteq \mathrm{Ker}(\delta^0_{(B)_{\geq1}})$. Moreover, by Lemma \ref{summand-disjoint}, every element $\sum_{p\in \mathrm{Sp}(1,n)} \lambda_p f_1 \| p^*$ does not belong to $\mathrm{Ker}(\delta^0_{(B)_{\geq1}})$. It follows easily that  $\mathrm{Ker}(\delta^0_{(B)_{\geq1}})=\psi_0(\mathrm{Ker}(\delta^0_{(A)_{\geq1}})) \oplus Z_{nsp}$.
\end{Proof}

First, we deal with the case that the algebra $A$ is indecomposable.

\begin{Prop}\label{center-from-the-same-block}
Let $A$ be an indecomposable finite
	dimensional monomial $k$-algebra and let $B$ be a radical embedding of
	$A$ obtained by gluing two idempotents $e_{1}$ and $e_{n}$
	of $A$. Then there is an algebra monomorphism $Z(A)\hookrightarrow Z(B)$. Moreover, we have the formula $$\mathrm{dim}_kZ(B)=\mathrm{dim}_kZ(A)+nsp(1,n).$$
\end{Prop}

\begin{Proof} We keep the notations as in Proposition \ref{parallel-paths-in-monomial-algebras} and identify the centers $Z(A), Z(B)$ as $\mathrm{Ker}(\delta^0_A), \mathrm{Ker}(\delta^0_B)$ respectively. Also notice that $\mathrm{Ker}(\delta_{A}^{0})=\mathrm{Ker}(\delta^0_{(A)_0})\oplus \mathrm{Ker}(\delta_{(A)_{\geq 1}}^{0})$ as $k$-vector spaces and the similar decomposition applies for $\mathrm{Ker}(\delta_{B}^{0})$.
	
By Lemma \ref{kernel-delta} we know that $\psi_0$ induces an injective $k$-linear map from $\mathrm{Ker}(\delta_{(A)_{\geq 1}}^{0})$ to $\mathrm{Ker}(\delta_{(B)_{\geq 1}}^{0})$, and $\mathrm{dim}_k \mathrm{Ker}(\delta^0_{(B)_{\geq1}})=\mathrm{dim}_k \mathrm{Ker}(\delta^0_{(A)_{\geq1}})+nsp(1,n)$. Combine the fact (cf. Proof of Lemma \ref{Image-zero-part}) that $\mathrm{Ker}(\delta^0_{(A)_0})=\langle \sum_{1\leq i\leq n}e_i \| e_i \rangle$ and $\mathrm{Ker}(\delta^0_{(B)_0})=\langle \sum_{1\leq i\leq n-1}f_i \| f_i \rangle$, we deduce that $\mathrm{dim}_k \mathrm{Ker}(\delta^0_{(B)_0})=\mathrm{dim}_k \mathrm{Ker}(\delta^0_{(A)_0})$, hence the second statement follows. Moreover, there is an injective $k$-linear map $\psi_0:\mathrm{Ker}(\delta_{A}^{0})\to \mathrm{Ker}(\delta_{B}^{0})$. Note that we can identify $\mathrm{Ker}(\delta_{A}^{0})$ with $Z(A)$ by $\sum e_i \| p \mapsto \sum p$ and $\sum_{i=1}^{n}e_i \| e_i \mapsto 1_A$, so does for $\mathrm{Ker}(\delta_{B}^{0})$ and $Z(B)$. Then, by the fact that $p^*q^*=(pq)^*$ for $p,q \in (\mathcal{B}_A \backslash \{e_1,\cdots,e_n\})$, $\psi_{0}$ gives an algebra monomorphism, and the first statement follows.
\end{Proof}

\begin{Cor}\label{center-source-sink-same-block}
	Let $A$ be an indecomposable finite
	dimensional monomial $k$-algebra and let $B$ be a radical embedding of
	$A$ obtained by gluing a source vertex $e_{1}$ and a sink vertex $e_{n}$. Then $\psi_0:\mathrm{Ker}(\delta^0_A) \hookrightarrow
	\mathrm{Ker}(\delta^0_B)$ is an isomorphism if and only if there is no path from $e_1$ to $e_n$.
\end{Cor}

\begin{Proof} Note that in this case, $p$ is a non-special path between $e_1$ and $e_n$ if and only if $p$ is a path from $e_1$ to $e_n$. Thus the result follows from Proposition \ref{center-from-the-same-block}.
\end{Proof}

\begin{Cor}\label{center-radical-square-zero-same-block}
	Let $A$ be a radical square zero indecomposable finite
	dimensional algebra and let $B$ be a radical embedding of
	$A$ obtained by gluing two idempotents $e_{1}$ and $e_{n}$
	of $A$. Then $\psi_0:\mathrm{Ker}(\delta^0_A) \hookrightarrow
	\mathrm{Ker}(\delta^0_B)$ is isomorphism if and only if there are no arrows between $e_1$ and $e_n$ in $Q_A$.
\end{Cor}

\begin{Proof}This is because in radical square zero case, the set $\mathrm{NSp}(1,n)$ consists of all arrows between $e_1$ and $e_n$ in $Q_A$.
\end{Proof}

Note that Cibils has shown in \cite{C} that the dimension of the center
of an indecomposable radical square zero algebra is given by $|Q_1\| Q_0|+1$. Indeed, by the proof of Proposition \ref{center-from-the-same-block}, we know that the basis of the center of an indecomposable radical square zero algebra is provided by the set of loops together with the unit element of the algebra.

\begin{Cor}
	Let $A$ be a radical square zero indecomposable finite
	dimensional algebra and let $B$ be a radical embedding of
	$A$ obtained by gluing two idempotents $e_{1}$ and $e_{n}$
	of $A$. Then we have $$|(Q_B)_1\| (Q_B)_0|=|(Q_A)_1\| (Q_A)_0|+nsp(1,n),$$ where $nsp(1,n)$ is equal to the number of arrows between $e_1$ and $e_n$.
\end{Cor}

\begin{Proof} This is obvious by Corollary \ref{center-radical-square-zero-same-block} and by Cibils' dimension formula as mentioned above.
\end{Proof}

Next we deal with the case that the algebra $A$ is not indecomposable. Without loss of generality we assume that $A$ has two blocks, say $A_1$ and $A_2$, and assume that $B$ is an algebra obtained from $A$ by gluing $e_1 \in A_1$ and $e_n \in A_2$.

\begin{Prop}\label{center-from-different-block}
Let $A$ be a finite dimensional monomial algebra with two blocks $A_1$ and $A_2$. Let $B$ be a radical embedding of $A$ obtained by gluing idempotents $e_1 \in A_1$ and $e_n \in A_2$. Then the radical embedding $B \to A$ restricts to a radical embedding $Z(B)\to Z(A)$. In particular,  $\mathrm{dim}_kZ(A)=\mathrm{dim}_kZ(B)+1$.
\end{Prop}

\begin{Proof} Let $\mathcal{B}_A=\{e_1,\cdots,e_n,p_1,\cdots,p_u\ |\ \mbox{the length of each }p_i\mbox{ is }\geq 1 \}$ denotes the properly chosen $k$-basis of the monomial algebra $A$ (cf. Section \ref{pre}). Then the subalgebra $B$ of $A$ has a $k$-basis $\mathcal{B}_B=\{e_1+e_n,e_2,\cdots,e_{n-1},p_1,\cdots,p_u\}$. We identify the centers $Z(A), Z(B)$ as $\mathrm{Ker}(\delta^0_A), \mathrm{Ker}(\delta^0_B)$ respectively. Let $Z(A)=Z(A)_0\oplus Z(A)_{\geq 1}$ be the decomposition corresponding to $\mathrm{Ker}(\delta_{A}^{0})=\mathrm{Ker}(\delta^0_{(A)_0})\oplus \mathrm{Ker}(\delta_{(A)_{\geq 1}}^{0})$ as $k$-vector spaces, so does for $Z(B)$.

By Lemma \ref{kernel-delta}, we obtain that $\mathrm{Ker}(\delta^0_{(B)_{\geq1}})\simeq \mathrm{Ker}(\delta^0_{(A)_{\geq1}})$, hence $Z(A)_{\geq 1}=\langle \sum p \ | \ p\ is\ a\ cycle\ in\ \mathcal{B}_A \rangle = Z(B)_{\geq 1}$. Note that $Z(A)_0=\langle 1_{A_1},1_{A_2} \rangle$, where $1_{A_j}$ denotes the unit element in $A_j$ for $j=1,2$, and $Z(B)_0=\langle 1_B=1_{A_1}+1_{A_2} \rangle$. Therefore there is an embedding from $Z(B)$ to $Z(A)$ which sends $1_B$ to $1_{A_1}+1_{A_2}$ and let the elements in $Z(B)_{\geq 1}$ one-corresponding-one to $Z(A)_{\geq 1}$.	

It is clear that this embedding from $Z(B)$ to $Z(A)$ is an injection of algebras and preserves the radical, hence we get a radical embedding from $Z(B)$ to $Z(A)$ by gluing $e_1\in A_1$ and $e_n\in A_2$. In particular, we have $\mathrm{dim}_kZ(A)=\mathrm{dim}_kZ(B)+1$.
\end{Proof}

\subsection{Higher degrees}

In this subsection, we assume that all algebras considered are indecomposable and radical square zero.

\begin{Def}[\cite{C}]
A $n$-crown is a quiver with $n$ vertices
cyclically labeled by the cyclic group of
order $n$, and $n$ arrows $a_0, \dots ,
a_{n-1}$ such that $s(a_i)=i$ and
$t(a_i)=i+1$. A $1$-crown is a loop, and a
$2$-crown is an oriented 2-cycle.
\end{Def}

Theorem 2.1 in \cite{C} provides the dimension of Hochschild cohomology: Let $Q$ be a connected quiver which is not a crown.
The dimension of Hochschild cohomology is:
$$\mathrm{dim}_k \HH^n(A)=|Q_n\| Q_1|-|Q_{n-1}\| Q_0|$$

For the $n$-crown case, see Proposition 2.3 in \cite{C}. Note that there is a typo in \cite{C}  since the formula above holds for $n>1$ and not for $n>0$. On page 24 of S\'anchez Flores' PhD thesis \cite{S1}  this is corrected.

\begin{Lem} \label{parallel-path}
Let $B$ be a radical embedding of $A$ obtained by gluing two idempotents and let $n\geq 2$. Let $\alpha \in (Q_{A})_{1}$ and $p \in (Q_{A})_{n}$. If $p \| \alpha$, then $p^{*} \| \alpha^{*}$, where  $\varphi_n:(Q_{A})_n \to (Q_{B})_n$  sends $p$ to $p^*$  (cf. Notations in Section \ref{pre}). In addition, $\varphi_n$ is injective.
\end{Lem}

\begin{Proof}
The first statement is easy. The map $\varphi_n$ is injective because the map $\varphi:(Q_A)_1 \rightarrow (Q_B)_1$ sending $\alpha$ to $\alpha^*$ is injective.
\end{Proof}


\begin{Lem}\label{parallel-paths-in-radicalsquarezero-algebras}
Let $A=kQ_A/I_A$ be an indecomposable radical square zero algebra and let $B=kQ_B/I_B$ be a radical embedding of $A$ obtained by gluing two idempotents of $A$. Then $\varphi_n:(Q_{A})_n \to (Q_{B})_n$ induces $k$-linear maps $\psi_{n,0}:  k( (Q_A)_n \| (Q_A)_0) \to k((Q_B)_n\|(Q_B)_0)$ and $\psi_{n,1}: k( (Q_A)_n \| (Q_A)_1) \to k( (Q_B)_n\| (Q_B)_1)$ for $n\geq 2$. In addition, $\psi_{n,1}$ in injective.
\end{Lem}

\begin{Proof}
Follow the same arguments of Proposition \ref{parallel-paths-in-monomial-algebras} (3). The injectivity of $\psi_{n,1}$ follows from the injectivity of $\varphi_n$.
\end{Proof}

\begin{Prop}\label{highhochgluing}
Let $A$ be an indecomposable radical square zero  algebra and let $B$ be a radical embedding of
	$A$ obtained by gluing a source  and a sink
	of $A$. Then there is an injective map $\psi_n:\mathrm{Ker}(\delta^n_A) \hookrightarrow \mathrm{Ker}(\delta^n_B)$ which restricts to  $\mathrm{Im}(\delta^{n-1}_A) \hookrightarrow \mathrm{Im}(\delta^{n-1}_B)$ for $n\geq 2$ (cf. Notation in Section 2). In addition, $\mathrm{dim}_k\HH^n(B)-\mathrm{dim}_k\HH^n(A)\geq 0$.
\end{Prop}

\begin{Proof}
Note that $\mathrm{Ker}(\delta^n_A)=k( (Q_A)_n \| (Q_A)_1) \oplus \mathrm{Ker}(D_n)$, by the proof of Theorem 2.1 in \cite{C} we know that $D_n$ is injective for $n\geq 2$ since the Gabriel quiver of $A$ is not a $n$-crown, hence $\mathrm{Ker}(\delta^n_A)=  k( (Q_A)_n \| (Q_A)_1)$, and the map $\psi_n:\mathrm{Ker}(\delta^n_A) \rightarrow \mathrm{Ker}(\delta^n_B)$ is well defined by Lemma
\ref{parallel-path} and Lemma \ref{parallel-paths-in-radicalsquarezero-algebras}. The injectivity of $\psi_n$
follows from Lemma \ref{parallel-paths-in-radicalsquarezero-algebras}. In order to check that $\psi_n$ restricts
to  $\mathrm{Im}(\delta^{n-1}_A) \rightarrow \mathrm{Im}(\delta^{n-1}_B)$, it is enough to check that
$\psi_{n,1}\circ D_{n-1}=D_{n-1}\circ\psi_{n-1,0}$ (cf. Notations in Section \ref{pre}). Let $e$ be a vertex of $Q_A$ and let $\gamma\in (Q_A)_{n-1}$ such that $\gamma$ is parallel to $e$.  On the one hand
 $$\psi_{n,1}\circ D_{n-1}(\gamma\| e)=\sum_{s(a)=e ,a\in (Q_A)_1}
 a^*\gamma^*\| a^*+(-1)^{n}\sum_{t(b)=e,b\in (Q_A)_1} \gamma^*b^*\| b^*.$$
On the other hand
$$D_{n-1}\circ\psi_{n-1,0}(\gamma\| e)= \sum_{s(a^*)=e,a^*\in (Q_B)_1}
 a^*\gamma^*\| a^*+(-1)^{n}\sum_{t(b^*)=e,b^*\in (Q_B)_1} \gamma^*b^*\| b^*.$$
Note that the vertex $e$ cannot be a source or a sink. Since for the rest of the vertices there is a bijection between the number of incoming (respectively outcoming) arrows of $Q_A$ and $Q_B$, then $\psi_{n,1}\circ D_{n-1}=D_{n-1}\circ\psi_{n-1,0}$ for every $n\geq2$.

Assume that the Gabriel quiver of $B$ is not a $n$-crown. Then by \cite[Theorem 2.1]{C} the expected inequality can be written as:
$$|(Q_B)_n\| (Q_B)_1|-|(Q_A)_n\| (Q_A)_1|\geq
|(Q_B)_{n-1}\| (Q_B)_0|
-|(Q_A)_{n-1}\| (Q_A)_0|.$$

Let $q||f$ be an element of $k((Q_B)_{n-1}\| (Q_B)_0)$ which is not in $\mathrm{Im}(\psi_{n-1,0})$ restricted to $k((Q_A)_{n-1}\| (Q_A)_0$). This means that either $p=0$ or $p$ is not an oriented cycle, that is, $s(p)\neq  t(p)$ where $p^*=q\in (Q_B)_{n-1}$. Consider now $a^{*}_1q||a^{*}_1$ where $q=a^*_n\dots a^*_1$. Then $a^*_1q||a^*_1\in k((Q_B)_n\| (Q_B)_1)$ but it is not an element of  $\mathrm{Im}(\psi_{n,1})$.  In fact, if $p=0$, then $a_1p=0$. If $s(a_1)=s(p)\neq t(p)$, then $a_1p=0$. This proves the above inequality.

Assume that the Gabriel quiver of $B$ is a $n$-crown for $n\geq 1$. Then $A$ is an $A_{n+1}$-quiver.  For $A_{n+1}$, the dimensions of Hochschild cohomology groups is zero since $A_{n+1}$ is hereditary. By Proposition 2.3  and  Proposition 2.4 in \cite{C} the statement follows.
\end{Proof}

Assume that we glue a source and a sink from the same block. The following two examples show that the difference of dimensions of higher Hochschild cohomology groups is not always one. This is very different from the case $n=1$ where $\mathrm{dim}_k\mathrm{HH}^1(B)-\mathrm{dim}_k\mathrm{HH}^1(A)=1$, see Remark \ref{quotient-lie-algebra-source-and-sink}.

\begin{Ex1}
\label{exeq}
Let $A$ be radical square zero algebra with Gabriel quiver given by a zig-zag type $A_n$ quiver such that  $e_1$ is a source vertex and $e_{2n}$ is a sink vertex.
\[\begin{tikzcd}
	{e_1} & {e_2} & {e_3} & \dots & {e_{2n-1}} & {e_{2n}}
	\arrow[from=1-1, to=1-2]
	\arrow[from=1-3, to=1-2]
	\arrow[from=1-3, to=1-4]
	\arrow[from=1-5, to=1-4]
	\arrow[from=1-5, to=1-6]
\end{tikzcd}\]
Let $B$ be the radical embedding obtained by gluing $e_1$ and $e_{2n}$. Then $\HH^n(A)$ and  $\HH^n(B)$ are zero for $n>1$ since there are no elements of the form $|Q_n\| Q_1|$.
\end{Ex1}

\begin{Ex1}
\label{Kro}
Let $m$ be a positive integer greater than $1$. Let $Q_A$ be the $m$-Kronecker quiver
and $Q_B$ be the $m$-multiple loop quiver. For $n>1$ the dimension of each
$\HH^n$ of the $m$-multiple loop quiver is $|Q_n\| Q_1|-|Q_{n-1}\| Q_0|=m^{n+1}-m^{n-1}$
whilst for $m$-Kronecker quiver the Hochschild cohomology groups are zero.
\end{Ex1}

Example \ref{Kro} shows that any $n>1$ although we glue a source and a sink vertex from the same block, the difference between the
dimensions of $\HH^n(B)$ and $\HH^n(A)$
can be arbitrarily large. In other words, for any  $n>1$ and $M>0$ we can always find a radical embedding obtained by gluing a source and a sink vertex from the same block such that $\mathrm{dim}_k\mathrm{HH}^n(B)-\mathrm{dim}_k\mathrm{HH}^n(A)>M$.

\medskip
{\bf Acknowledgements.} This first author was partially supported by NSFC (No. 12031014). The second author has been partially supported by the project PRIN 2017 - Categories, Algebras: Ring-Theoretical and Homological Approaches and  by the project REDCOM: Reducing complexity in algebra, logic,
combinatorics, financed by the programme Ricerca Scientifica di Eccellenza 2018 of the Fondazione Cariverona. The second author is also grateful to Beijing Normal University  for its hospitality during the research for this paper.

