\section{Introduction}
Higher Auslander-Reiten theory was introduced by Iyama in \cite{Iya07a} and further developed in \cite{Iya07,Iya11}. It has several connections to other areas, for example non-commutative algebraic geometry \cite{HI11a,HIMO14,Iya07}, combinatorics \cite{OT12}, higher category theory \cite{DJW19}, and symplectic geometry \cite{DJL19}. One of the main objects of study are $d$-cluster tilting subcategories of abelian, exact, and triangulated categories. The study of their intrinsic properties, called higher homological algebra, is an active area of research, see for example \cite{Fed19,Fed20,HJV17,JJ17,JJ20,J16,Rei20}. This approach was catalysed by the papers \cite{GKO13} and \cite{Jas16}, where they introduced $d$-abelian, $d$-exact, and $(d+2)$-angulated categories as an axiomatization of $d$-cluster tilting subcategories. In particular, they showed that $d$-cluster tilting subcategories of abelian, exact or triangulated categories are $d$-abelian, $d$-exact, or $(d+2)$-angulated, respectively, and that any projectively generated $d$-abelian category is equivalent to a $d$-cluster tilting subcategory of an abelian category. 

Axiomatizing subcategories of abelian categories is closely related to  characterizing $\Lambda$-modules $M$, where $\Lambda$ is an Artin algebra, in terms of properties of the endomorphism algebra $\Gamma:=\operatorname{End}_{\Lambda}(M)$. In an unpublished manuscript \cite{MT} Morita and Tachikawa showed that $M\mapsto \Gamma$ gives a correspondence between generating and cogenerating modules $M$ and algebras $\Gamma$ with dominant dimension $\geq 2$. Auslander \cite{Aus71} showed that this specializes to the case where $M$ is an additive generator of a module category and $\Gamma$ is an algebra with dominant dimension $\geq 2$ and global dimension $\leq 2$. This is typically called the Auslander-correspondence. It was later extended by Iyama \cite{Iya07} to a bijection between $d$-cluster tilting modules $M$ and algebras $\Gamma$ with dominant dimension $\geq d+1$ and global dimension $\leq d+1$. Recently Iyama and Solberg \cite{IS18} introduced $d$-precluster tilting modules $M$ and showed that the assignment $M\mapsto \Gamma$ gives a bijection to algebras of dominant dimension $\geq d+1$ and selfinjective dimension $\leq d+1$. In all of these cases one characterizes the module $M$ in terms of properties of the category of finitely presented $\Gamma$-modules. One can interpret axiomatization similarly, but where the characterization is in terms of the category  of finitely generated projective $\Gamma$-modules. 

In this paper we continue the idea of axiomatizing  subcategories of abelian categories and study their properties. The following definition clarifies what we mean:

\begin{Definition}\label{Axiomatizing subcategories}
Let $\mathtt{P}$ be a set of axioms of additive categories, and let $\mathtt{S}$ be a class of subcategories of abelian categories. We say that  $\mathtt{P}$ \emphbf{axiomatizes} subcategories in $\mathtt{S}$ if the following hold:
\begin{enumerate}
\item\label{Axiomatizing subcategories:1} If ${\mathcal X}$ is in $\mathtt{S}$, then ${\mathcal X}$ satisfies $\mathtt{P}$ as an additive category;
\item\label{Axiomatizing subcategories:2} If ${\mathcal X}$ satisfies $\mathtt{P}$, then there exists an abelian category ${\mathcal A}$ and a fully faithful functor ${\mathcal X}\to {\mathcal A}$ such that its essential image is in $\mathtt{S}$;
\item\label{Axiomatizing subcategories:3} If ${\mathcal A}$ and ${\mathcal A}'$ are abelian categories and $\Phi\colon {\mathcal X}\to {\mathcal A}$ and $\Phi'\colon {\mathcal X}\to {\mathcal A}'$ are two fully faithful functors such that the essential images of $\Phi$ and $\Phi'$ are in $\mathtt{S}$, then there exists an equivalence $\Psi\colon {\mathcal A}\xrightarrow{\cong} {\mathcal A}'$ and a natural isomorphism $\Psi \circ \Phi\cong \Phi'$.
\end{enumerate}
\end{Definition}

Part \ref{Axiomatizing subcategories:1} and \ref{Axiomatizing subcategories:2} tells us that an additive category ${\mathcal X}$ satisfies $\mathtt{P}$ if and only if it is equivalent to a subcategory in $\mathtt{S}$. Part \ref{Axiomatizing subcategories:3} tells us that the ambient abelian category of a subcategory in $\mathtt{S}$ must be unique up to equivalence. 

We prove the following theorem, which gives examples of axioms $\mathtt{P}$ and classes of subcategories $\mathtt{S}$. See the end of the introduction for the list of axioms.
\begin{Theorem}\label{Main theorem}
Let $\mathtt{P}$ be a set of axioms for additive categories, and let $\mathtt{S}$ be a class of subcategories of abelian categories. We have that $\mathtt{P}$ axiomatizes subcategories in $\mathtt{S}$ in the following cases:
\begin{enumerate}
\item\label{Main theorem:1} $\mathtt{P}$ consists of the axioms (A1), (A1)\ensuremath{^{\mathrm{op}}}, (A2), (A2)\ensuremath{^{\mathrm{op}}}, (A3), and (A3)\ensuremath{^{\mathrm{op}}} , and $\mathtt{S}$ is the class of generating cogenerating functorially finite subcategories;
\item\label{Main theorem:2} $\mathtt{P}$ consists of the axioms (A1), (A1)\ensuremath{^{\mathrm{op}}}, (A2), (A2)\ensuremath{^{\mathrm{op}}}, (A3), and (A3)\ensuremath{^{\mathrm{op}}}, and ($d$-Rigid) and $\mathtt{S}$ is the class of generating cogenerating functorially finite subcategories ${\mathcal X}$ satisfying
\[
\operatorname{Ext}^i_{{\mathcal A}}(X,X')=0 \text{ for all }X,X'\in {\mathcal X} \text{ and }0<i<d
\]
where ${\mathcal A}$ is the ambient abelian category;
\item\label{Main theorem:3} $\mathtt{P}$ consists of the axioms (A0), (A1), (A1)\ensuremath{^{\mathrm{op}}}, (A2), (A2)\ensuremath{^{\mathrm{op}}}, (A3), and (A3)\ensuremath{^{\mathrm{op}}}, ($d$-Rigid), (A4.$d$), and (A4.$d$)\ensuremath{^{\mathrm{op}}} , and $\mathtt{S}$ is the class of $d$-precluster tilting subcategories;
\item\label{Main theorem:4} $\mathtt{P}$ consists of the axioms (A0), (A1), (A1)\ensuremath{^{\mathrm{op}}}, (A2), (A2)\ensuremath{^{\mathrm{op}}}, (A3), and (A3)\ensuremath{^{\mathrm{op}}}, ($d$-Rigid), ($d$-Ker), and ($d$-Coker) and $\mathtt{S}$ is the class of $d$-cluster tilting subcategories.
\end{enumerate} 
\end{Theorem}
We even find a class of subcategories axiomatized by (A1) and (A2), see Theorem \ref{Embedding Theorem}. Note that the definition of $d$-precluster tilting subcategories in \cite{IS18} can be reformulated in a way that makes sense for any abelian category, see Theorem \ref{reformulation precluster tilting}, and this reformulated definition is what we use in Theorem \ref{Main theorem} \ref{Main theorem:3}. 

As a corollary of Theorem \ref{Main theorem} \ref{Main theorem:4} we show that any $d$-abelian category is equivalent to a $d$-cluster tilting subcategory, without the assumption of being projectively generated. Since for any $d\geq 1$ there exist examples of $d$-cluster tilting subcategories without any non-zero projective or injective objects, see \cite{JK19}, the result is necessary to complete the axiomatization of $d$-cluster tilting subcategories in terms of $d$-abelian categories.

\begin{Corollary}\label{d-abelian is d-cluster tilting main}
Let ${\mathcal X}$ be an additive category. The following hold:
\begin{enumerate}
\item\label{d-abelian is d-cluster tilting main:1} ${\mathcal X}$ is $d$-abelian if and only if it satisfies (A0), (A1), (A1)\ensuremath{^{\mathrm{op}}}, (A2), (A2)\ensuremath{^{\mathrm{op}}}, (A3), and (A3)\ensuremath{^{\mathrm{op}}}, (d-Rigid), (d-Ker), and (d-Coker);
\item\label{d-abelian is d-cluster tilting main:2} If ${\mathcal X}$ is $d$-abelian, then there exists an abelian category ${\mathcal A}$ and a fully faithful functor ${\mathcal X}\to {\mathcal A}$ such that its essential image is $d$-cluster tilting.\footnote{The author gave a talk about this result in the LMS Northern Regional Meeting and Workshop on Higher Homological Algebra in 2019}
\end{enumerate}
\end{Corollary}

We end the introduction by giving the list of axioms we use:
\begin{enumerate}
\item[(A0)] ${\mathcal X}$ is idempotent complete;

\item[(A1)] ${\mathcal X}$ has weak kernels;

\item[(A1)\ensuremath{^{\mathrm{op}}}] ${\mathcal X}$ has weak cokernels;

\item[(A2)] Any epimorphism in ${\mathcal X}$ is a weak cokernel;

\item[(A2)\ensuremath{^{\mathrm{op}}}] Any monomorphism in ${\mathcal X}$ is a weak kernel;

\item[(A3)] Consider the following diagram
\begin{equation*}
\begin{tikzcd}
X_2 \arrow{rr}{f} \arrow{rd}{l} & & X_1 \arrow{r}{g} & X_0 \\
 & \arrow{ru}{h}X_2' & 
\end{tikzcd}
\end{equation*}
where $f$ is an arbitrary morphism in ${\mathcal X}$, where $g$ is a weak cokernel of $f$, where $h$ is a weak kernel of $g$, and where $l$ is an induced map satisfying $h\circ l=f$. Then for any weak kernel $k\colon X_3'\to X_2'$ of $h$ the map $\begin{bmatrix}l & k\end{bmatrix}\colon X_2\oplus X_3'\to X_2'$ is an epimorphism;

\item[(A3)\ensuremath{^{\mathrm{op}}}] Consider the following diagram
\begin{equation*}
\begin{tikzcd}
X_0 \arrow{r}{g} & X_1 \arrow{rr}{f} \arrow{rd}{h} & & X_2 \\
& & \arrow{ru}{l}X_2' & 
\end{tikzcd}
\end{equation*}
where $f$ is an arbitrary morphism in ${\mathcal X}$, where $g$ is a weak kernel of $f$, where $h$ is a weak cokernel of $g$, and where $l$ is a map satisfying $l\circ h=f$. Then for any weak cokernel $k\colon X_2'\to X_3'$ of $h$ the map $\begin{bmatrix}l \\ k\end{bmatrix}\colon X_2'\to X_2\oplus X_3'$ is a monomorphism;

\item[(A4.$d$)] Let
\[
X_{d+1}\xrightarrow{f_{d+1}}X_d\xrightarrow{f_d}\cdots \xrightarrow{f_3}X_2\xrightarrow{f_2}X_1\xrightarrow{f_1}X_0\xrightarrow{f_0}X_{-1}
\]
be a sequence with $f_{i+1}$ a weak kernel of $f_i$ for all $0\leq i\leq d$. Then $f_{d+1}$ is a weak cokernel;

\item[(A4.$d$)\ensuremath{^{\mathrm{op}}}] Let
\[
X_{d+1}\xrightarrow{f_{d+1}}X_d\xrightarrow{f_d}\cdots \xrightarrow{f_3}X_2\xrightarrow{f_2}X_1\xrightarrow{f_1}X_0\xrightarrow{f_0}X_{-1}
\]
be a sequence with $f_{i}$ a weak cokernel of $f_{i+1}$ for all $0\leq i\leq d$. Then $f_{0}$ is a weak kernel;

\item[($d$-Rigid)] For all epimorphism $f_1\colon X_1\to X_0$ in ${\mathcal X}$ there exists a sequence
\[
X_{d+1}\xrightarrow{f_{d+1}}X_d\xrightarrow{f_d}\cdots \xrightarrow{f_3}X_2\xrightarrow{f_2}X_1\xrightarrow{f_1}X_0
\]
with $f_{i+1}$ a weak kernel of $f_i$ and $f_i$ a weak cokernel of $f_{i+1}$ for all $1\leq i\leq d$;

\item[($d$-Ker)] Any map in ${\mathcal X}$ has a $d$-kernel;

\item[($d$-Coker)] Any map in ${\mathcal X}$ has a $d$-cokernel.
\end{enumerate}

Note that the axiom ($d$-Rigid) is self-dual under the assumption of axioms (A1), (A1)\ensuremath{^{\mathrm{op}}}, (A2), (A2)\ensuremath{^{\mathrm{op}}}, (A3), and (A3)\ensuremath{^{\mathrm{op}}}, since these axioms together axiomatizes generating cogenerating functorially finite $d$-rigid subcategories by Theorem \ref{Main theorem} \ref{Main theorem:2}, and the definition of such subcategories are self-dual. 

\subsection{Conventions}
All categories are assumed to be additive, i.e. enriched over abelian groups and admitting finite direct sums. For an additive category ${\mathcal X}$ we let ${\mathcal X}(X,X')$ denote the set of morphism between two objects $X,X'\in {\mathcal X}$ and $\operatorname{Hom}_{{\mathcal X}}(F_1,F_2)$ the set of natural transformations between two additive functors $F_1,F_2\colon {\mathcal X}\ensuremath{^{\mathrm{op}}}\to \operatorname{Ab}$. A subcategory ${\mathcal X}$ of an abelian category ${\mathcal A}$ is called generating (resp cogenerating) if for any object $A\in {\mathcal A}$ there exists an epimorphism $X\to A$ (resp a monomorphism $A\to X$) with $X\in {\mathcal X}$.

\subsection{Acknowledgement}
Corollary \ref{d-abelian is d-cluster tilting main} \ref{d-abelian is d-cluster tilting main:2} is proved independently by Ramin Ebrahimi and Alireza Nasr-Isfahani in \cite{RA20}. The author would like to thank the anonymous referee for helpful suggestions which has improved the readability of the paper.

\section{Serre subcategories}
In this section we recall the localization of an abelian category by a Serre subcategory. Let ${\mathcal A}$ be an abelian category. A subcategory ${\mathcal S}$ of ${\mathcal A}$ is called a \emph{Serre subcategory} if for any exact sequence in ${\mathcal A}$
\[
0\to A_1\to A_2\to A_3\to 0
\]
we have that $A_1\in {\mathcal S}$ and $A_3\in {\mathcal S}$ if and only if $A_2\in {\mathcal S}$. If ${\mathcal S}$ be a Serre subcategory of ${\mathcal A}$, category ${\mathcal A}/{\mathcal S}$ is defined to be the localization of ${\mathcal A}$ by the class of morphisms $f\colon X\to X'$ satisfying 
\[
\operatorname{Ker} f\in {\mathcal S}\quad \text{and} \quad \operatorname{Coker} f \in {\mathcal S}.
\]
Note that the objects in ${\mathcal A}/{\mathcal S}$ are the same as the objects in ${\mathcal A}$. We need the following results for this localization, which follows from  Proposition 1 and Lemma 2 in Chapter 3 in \cite{Gab62}.

\begin{Theorem}\label{Serre subcategory}
Let ${\mathcal S}$ be a Serre subcategory of ${\mathcal A}$, and let $q\colon {\mathcal A}\to {\mathcal A}/{\mathcal S}$ denote the canonical functor to the localization. The following hold:
\begin{enumerate}
\item  ${\mathcal A}/{\mathcal S}$ is an abelian category;
\item  $q$ is an exact functor;
\item  For a morphism $f\colon A_1\to A_2$ in ${\mathcal A}$, we have that $q(f)=0$ if and only if $\operatorname{im} f \in {\mathcal S}$. 
\end{enumerate} 
\end{Theorem}

\begin{Remark}
Note that the class of morphisms $f$ with $\operatorname{Ker} f\in {\mathcal S}$ and $\operatorname{Coker} f\in {\mathcal S}$ forms a multiplicative system in ${\mathcal A}$, see \cite[Exercise 10.3.2 (1)]{Wei94}. In particular, the morphisms in ${\mathcal A}/{\mathcal S}$ can be described using a calculus of fractions, see \cite[Chapter 10.3]{Wei94}.
\end{Remark}

\section{Weak kernels and weak cokernels}
In this section we recall the definition of weak kernels and cokernels and their basic properties. Let ${\mathcal X}$ be an additive category.  We denote the category of additive functors from ${\mathcal X}\ensuremath{^{\mathrm{op}}}$ to $\operatorname{Ab}$ by $\operatorname{Mod} {\mathcal X}$. Note that the Yoneda embedding gives a fully faithful functor
\begin{align*}
& {\mathcal X}\to \operatorname{Mod} {\mathcal X} \quad \quad X\mapsto {\mathcal X}(-,X) 
\end{align*}
A functor $F\colon {\mathcal X}\ensuremath{^{\mathrm{op}}}\to \operatorname{Ab}$ is called \emph{finitely presented} if there exists an exact sequence
\[
{\mathcal X}(-,X_1)\to {\mathcal X}(-,X_0)\to F\to 0
\]
in $\operatorname{Mod} {\mathcal X}$, and the subcategory of finitely presented functors is denoted by $\operatorname{mod} {\mathcal X}$. We have that $\operatorname{mod} {\mathcal X}$ is closed under cokernels in $\operatorname{Mod} {\mathcal X}$. Let $g\colon X''\to X'$ and $f\colon X'\to X$  be two composable morphisms in ${\mathcal X}$. We say that $g$ is a \emph{weak kernel} of $f$ if 
\[
{\mathcal X}(Y,X'')\xrightarrow{g\circ -}{\mathcal X}(Y,X')\xrightarrow{f\circ -}{\mathcal X}(Y,X)
\]
is an exact sequence of abelian groups for all $Y\in {\mathcal X}$. Dually, we say that $f$ is a \emph{weak cokernel} of $g$ if 
\[
{\mathcal X}(X,Y)\xrightarrow{-\circ f}{\mathcal X}(X',Y)\xrightarrow{-\circ g}{\mathcal X}(X'',Y)
\]
is an exact sequence of abelian groups for all $Y\in {\mathcal X}$. The category ${\mathcal X}$ has \emph{weak kernels} or \emph{weak cokernels} if any morphism in ${\mathcal X}$ has a weak kernel or weak cokernel, respectively. In the following theorem we relate these notions to $\operatorname{mod} {\mathcal X}$.

\begin{Theorem}[Theorem 1.4 in \cite{Fre66}]\label{Freyd}
Let ${\mathcal X}$ be an additive category. Then ${\mathcal X}$ has weak kernels if and only if $\operatorname{mod} {\mathcal X}$ is abelian. 
\end{Theorem}

We call a morphism $f$ a weak kernel or weak cokernel if there exists a morphism $g$ such that $f$ is the weak kernel or weak cokernel of $g$, respectively. The following result is known to the experts. We give a proof for the readers convenience.

\begin{Lemma}\label{Weak kernels and cokernels}
Let $f\colon X\to X'$ be a morphism in ${\mathcal X}$. The following hold:
\begin{enumerate}
\item\label{Weak kernels and cokernels:1} If $f$ is a weak kernel and admits a weak cokernel, then it is a weak kernel of its weak cokernel;
\item\label{Weak kernels and cokernels:2} If $f$ is a weak cokernel and admits a weak kernel, then it is a weak cokernel of its weak kernel.
\end{enumerate}
\end{Lemma}

\begin{proof}
We prove \ref{Weak kernels and cokernels:1}, \ref{Weak kernels and cokernels:2} is proved dually. Assume $f$ is a weak kernel of $g\colon X'\to X''$, and let $h\colon X'\to \tilde{X}$ be a weak cokernel of $f$. Since $g\circ f=0$ and $h$ is a weak cokernel of $f$, there exists a morphism $k\colon \tilde{X}\to X''$ such that $k\circ h=g$. Hence, if a morphism $l\colon \tilde{X'}\to X'$ satisfies $h\circ l=0$, then $g\circ l=k\circ h\circ l=0$, so $l$ factors through $f$ since $f$ is a weak kernel of $g$. This implies that $f$ is a weak kernel of $h$.
\end{proof}

Finally, we recall the definition of contravariantly and covariantly finite subcategories. Assume ${\mathcal X}$ is an additive subcategory of an abelian category ${\mathcal A}$. A morphism $X\xrightarrow{f} A$ in ${\mathcal A}$ with $X\in {\mathcal X}$ is called a \emph{right} ${\mathcal X}$-\emph{approximation} of $A$ if any map $X'\to A$ with $X'\in {\mathcal X}$ factors through $f$. Dually, a morphism $g\colon A\to X$ with $X\in {\mathcal X}$ is a \emph{left} ${\mathcal X}$-\emph{approximation} of $A$ if it is a right ${\mathcal X}\ensuremath{^{\mathrm{op}}}$-approximation of $A$ in ${\mathcal A}\ensuremath{^{\mathrm{op}}}$.  We say that ${\mathcal X}$ is \emph{contravariantly finite} (resp \emph{covariantly finite}) if any object $A$ in ${\mathcal A}$ admits a right (resp left) ${\mathcal X}$-approximation. We say that ${\mathcal X}$ is \emph{functorially finite} if it is both contravariantly finite and covariantly finite. If ${\mathcal X}$ is a contravariantly finite subcategory, then it has weak kernels. In fact, if $f\colon X'\to X$ is a morphism in ${\mathcal X}$ and  $X''\to \operatorname{Ker} f$ is a right ${\mathcal X}$-approximation, then the composite $X''\to \operatorname{Ker} f\to X'$ is a weak kernel of $f$. Similarly, any covariantly finite subcategory has weak cokernels, which are constructed in the dual way.



\section{Embeddings into abelian categories}
In this section we compare intrinsic axioms of additive categories with properties of subcategories of abelian categories. For an additive category ${\mathcal X}$, the intrinsic axioms we consider are:

\begin{enumerate}[label=(A\arabic*)]
\item\label{A1} ${\mathcal X}$ has weak kernels;
\item\label{A2} Any epimorphism in ${\mathcal X}$ is a weak cokernel.
\end{enumerate}

For an abelian category ${\mathcal A}$ and a full subcategory ${\mathcal X}$ of ${\mathcal A}$, the properties we consider are:
\begin{enumerate}[label=(B\arabic*)]
\item\label{B1} ${\mathcal X}$ is a generating subcategory of ${\mathcal A}$;
\item\label{B2} If $A\in {\mathcal A}$ satisfies ${\mathcal A}(A,X)=0$ for all $X\in {\mathcal X}$, then $A=0$;
\item\label{B3} Any $A\in \Omega^2_{{\mathcal X}}({\mathcal A})$ admits a right ${\mathcal X}$-approximation.
\end{enumerate}
Here $\Omega^n_{{\mathcal X}}({\mathcal A})$ denotes the subcategory of ${\mathcal A}$ consisting of all objects $A$ for which there exists an exact sequence
\[
0\to A\to X_1\to \cdots \to X_n
\]
where $X_i\in {\mathcal X}$ for all $1\leq i\leq n$. Our main goal is to prove the following theorem:

\begin{Theorem}\label{Embedding Theorem}
Assume $\mathsf{P}$ consists of the axioms \ref{A1} and \ref{A2}. Then $\mathsf{P}$ axiomatizes subcategories of abelian categories satisfying \ref{B1}, \ref{B2} and \ref{B3}. 
\end{Theorem}

We can show one part of the theorem immediately.

\begin{Lemma}\label{B1,B2,B3 implies A1,A2}
Assume ${\mathcal X}\subseteq {\mathcal A}$ is a full subcategory of an abelian category ${\mathcal A}$ satisfying \ref{B1}, \ref{B2} and \ref{B3}. The following hold: 
\begin{enumerate}
\item\label{B1,B2,B3 implies A1,A2:1} The inclusion functor ${\mathcal X}\to {\mathcal A}$ preserves epimorphisms and sends a sequence $X_2\xrightarrow{f}X_1\xrightarrow{g}X_0$ in ${\mathcal X}$ with $f$ a weak kernel of $g$ to an exact sequence in ${\mathcal A}$;
\item\label{B1,B2,B3 implies A1,A2:2} ${\mathcal X}$ satisfies \ref{A1} and \ref{A2} as an additive category.
\end{enumerate}
\end{Lemma}

\begin{proof}
Assume ${\mathcal X}\subseteq {\mathcal A}$ satisfies \ref{B1}, \ref{B2} and \ref{B3}.  Let $X_1\xrightarrow{g}X_0$ be an arbitrary morphism in ${\mathcal X}$. Then $\operatorname{Ker} g\in \Omega^2_{{\mathcal X}}({\mathcal A})$, and hence there exists a right ${\mathcal X}$-approximation $X_2\to \operatorname{Ker} g$, which is surjective since ${\mathcal X}$ is a generating subcategory of ${\mathcal A}$. It follows that the composite $X_2\to \operatorname{Ker} g\to X_1$ is a weak kernel of $g$. This proves \ref{A1} and shows that there exists a weak kernel $X_2\to X_1$ of $g$ such that $X_2\to X_1\xrightarrow{g}X_0$ is exact in $\mathcal{A}$. But then this property must hold for any weak kernel of $g$, since $X_2\to X_1$ must factor through any such map. 

Now assume $g$ is an epimorphism, and let $\operatorname{Coker} g$ be the cokernel of $g$ in ${\mathcal A}$. Applying ${\mathcal A}(-,X)$ with $X\in {\mathcal X}$ to the exact sequence
\[
X_1\xrightarrow{g}X_0\to \operatorname{Coker} g\to 0
\]
gives an exact sequence 
\[
0\to {\mathcal A}(\operatorname{Coker} g,X)\to {\mathcal A}(X_0,X)\xrightarrow{-\circ g} {\mathcal A}(X_1,X)
\]
If $X\in {\mathcal X}$, then since $g$ is an epimorphism in ${\mathcal X}$ it follows that $-\circ g$ is a monomorphism, and hence ${\mathcal A}(\operatorname{Coker} g,X)=0$. By \ref{B2} it follows that $\operatorname{Coker} g=0$, so $g$ is an epimorphism in ${\mathcal A}$. Hence the inclusion ${\mathcal X}\to {\mathcal A}$ preserves epimorphisms. Finally, since $X_2\to X_1\xrightarrow{g}X_0\to 0$ is exact, $g$ is a cokernel of $X_2\to X_1$, which proves \ref{A2}.  
\end{proof}

Now assume ${\mathcal X}$ is an additive category satisfying \ref{A1} and \ref{A2}. Since ${\mathcal X}$ has weak kernels, the category of finitely presented functors $\operatorname{mod}{\mathcal X}$ is abelian by Theorem \ref{Freyd}. Let $\operatorname{eff}{\mathcal X}$ denote the subcategory of $\operatorname{mod}{\mathcal X}$ consisting of all functors $F$ for which there exists an exact sequence
\[
{\mathcal X}(-,X_1)\xrightarrow{f\circ -}{\mathcal X}(-,X_0)\to F\to 0
\]
where $f\colon X_1\to X_0$ is an epimorphism in ${\mathcal X}$. We show that $\operatorname{eff} {\mathcal X}$ is a Serre subcategory of $\operatorname{mod} {\mathcal X}$.

\begin{Proposition}\label{effaceable functors}
Let ${\mathcal X}$ be an additive category satisfying \ref{A1} and \ref{A2}. The following hold:
\begin{enumerate}
\item\label{effaceable functors:1} If 
\[
{\mathcal X}(-,X'_1)\xrightarrow{f'\circ -}{\mathcal X}(-,X'_0)\to F\to 0
\]
is exact with $F\in \operatorname{eff} {\mathcal X}$, then $f'\colon X_1'\to X_0'$ is an epimorphism in ${\mathcal X}$;
\item\label{effaceable functors:2} $\operatorname{eff} {\mathcal X}=\{F\in \operatorname{mod}{\mathcal X}\mid \operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))=0 \text{ for all } X\in {\mathcal X}\}$;
\item\label{effaceable functors:3} If $F\in \operatorname{eff} {\mathcal X}$ then $\operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F,{\mathcal X}(-,X))=0$ for all $X\in {\mathcal X}$;
\item\label{effaceable functors:4} $\operatorname{eff} {\mathcal X}$ is a Serre subcategory of $\operatorname{mod} {\mathcal X}$.
\end{enumerate}
\end{Proposition}

\begin{proof}
If $F\in \operatorname{eff} {\mathcal X}$, then there exists an exact sequence 
\[
{\mathcal X}(-,X_1)\xrightarrow{f\circ -}{\mathcal X}(-,X_0)\to F\to 0
\]
where $f\colon X_1\to X_0$ is an epimorphism in ${\mathcal X}$. Applying $\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))$ gives the exact sequence
\[
0\to \operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))\to {\mathcal X}(X_0,X)\xrightarrow{-\circ f}{\mathcal X}(X_1,X)
\]
Since $f$ is an epimorphism in ${\mathcal X}$, the map $-\circ f$ is a monomorphism, and hence 
\[
\operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))=0.
\]
Conversely, assume $\operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))=0$ for all $X\in {\mathcal X}$. Applying $\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))$ to an exact sequence of the form 
\[
{\mathcal X}(-,X'_1)\xrightarrow{f'\circ -}{\mathcal X}(-,X'_0)\to F\to 0
\]
gives an exact sequence
\[
0\to \operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))\to {\mathcal X}(X'_0,X)\xrightarrow{-\circ f'}{\mathcal X}(X'_1,X)
\]
Since $\operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))=0$, the map $-\circ f'$ is a monomorphism, and hence $f'\colon X_1'\to X_0$ is an epimorphism. This proves \ref{effaceable functors:1} and \ref{effaceable functors:2}. For \ref{effaceable functors:3}, assume again we have an exact sequence ${\mathcal X}(-,X_1)\xrightarrow{f\circ -}{\mathcal X}(-,X_0)\to F\to 0$ where $f\colon X_1\to X_0$ is an epimorphism in ${\mathcal X}$. Let $g\colon X_2\to X_1$ be a weak kernel of $f$. Then 
\[
{\mathcal X}(-,X_2)\xrightarrow{g\circ -}{\mathcal X}(-,X_1)\xrightarrow{f\circ -}{\mathcal X}(-,X_0)\to F\to 0
\]
is exact. Applying $\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))$ with $X\in {\mathcal X}$ gives a complex
\[
0\to \operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))\to {\mathcal X}(X_0,X)\xrightarrow{-\circ f}{\mathcal X}(X_1,X)\xrightarrow{-\circ g}{\mathcal X}(X_2,X)
\]
Since $f$ is an epimorphism in ${\mathcal X}$, it is a weak cokernel by \ref{A2}. By Lemma \ref{Weak kernels and cokernels} \ref{Weak kernels and cokernels:2} we know that $f$ must be a weak cokernel of $g$. Therefore the sequence 
\[
{\mathcal X}(X_0,X)\xrightarrow{-\circ f}{\mathcal X}(X_1,X)\xrightarrow{-\circ g}{\mathcal X}(X_2,X)
\]
 must be exact. This shows that $\operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F,{\mathcal X}(-,X))=0$.

Finally, we show that $\operatorname{eff} {\mathcal X}$ is a Serre subcategory. Let 
\[
0\to F_1\to F_2\to F_3\to 0
\]
be an exact sequence in $\operatorname{mod} {\mathcal X}$. Applying $\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))$, we get an exact sequence
\begin{multline*}
0\to \operatorname{Hom}_{{\mathcal X}}(F_3,{\mathcal X}(-,X))\to \operatorname{Hom}_{{\mathcal X}}(F_2,{\mathcal X}(-,X))\to \operatorname{Hom}_{{\mathcal X}}(F_1,{\mathcal X}(-,X)) \\
\to \operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F_3,{\mathcal X}(-,X))\to \operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F_2,{\mathcal X}(-,X))
\end{multline*}
Now if $F_1\in\operatorname{eff}{\mathcal X}$ and $F_3\in \operatorname{eff} {\mathcal X}$, then $\operatorname{Hom}_{{\mathcal X}}(F_1,{\mathcal X}(-,X))=0$ and $\operatorname{Hom}_{{\mathcal X}}(F_3,{\mathcal X}(-,X))=0$ for all $X\in {\mathcal X}$ by part \ref{effaceable functors:2} of the theorem. Therefore, $\operatorname{Hom}_{{\mathcal X}}(F_2,{\mathcal X}(-,X))=0$ for all $X\in {\mathcal X}$, so $F_2\in \operatorname{eff} {\mathcal X}$. Conversely, assume $F_2\in \operatorname{eff} {\mathcal X}$. Since $\operatorname{Hom}_{{\mathcal X}}(F_2,{\mathcal X}(-,X))=0$ for all $X\in {\mathcal X}$ by \ref{effaceable functors:2} and $\operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F_2,{\mathcal X}(-,X)=0$ by \ref{effaceable functors:3}, the exact sequence above correspondence to the exact sequence
\[
0\to \operatorname{Hom}_{{\mathcal X}}(F_3,{\mathcal X}(-,X))\to 0 \to \operatorname{Hom}_{{\mathcal X}}(F_1,{\mathcal X}(-,X))\to \operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F_3,{\mathcal X}(-,X))\to 0
\]
Hence $\operatorname{Hom}_{{\mathcal X}}(F_3,{\mathcal X}(-,X))=0$ and therefore $\operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F_3,{\mathcal X}(-,X))=0$ by \ref{effaceable functors:3}. This implies that $\operatorname{Hom}_{{\mathcal X}}(F_1,{\mathcal X}(-,X))=0$ for all $X\in {\mathcal X}$, and so $F_1\in \operatorname{eff}{\mathcal X}$ and $F_3\in \operatorname{eff} {\mathcal X}$.
\end{proof}

Since $\operatorname{eff} {\mathcal X}$ is a Serre subcategory of $\operatorname{mod} {\mathcal X}$, the localization $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ is an abelian category, see Theorem \ref{Serre subcategory}.  We let $q\colon \operatorname{mod} {\mathcal X}\to \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ denote the localization functor. By abuse of notation, we denote an object in $\operatorname{mod} {\mathcal X}$ and its image in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ by the same letter. Consider the functor $\Phi\colon {\mathcal X}\to \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ given by the composite
\[
 {\mathcal X} \to \operatorname{mod} {\mathcal X} \xrightarrow{q} \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X} 
\]
where ${\mathcal X}\to \operatorname{mod} {\mathcal X}$ is the Yoneda functor. It plays the role of the fully faithful functor in Definition \ref{Axiomatizing subcategories} \ref{Axiomatizing subcategories:2}. To show this, we need the following lemma.

\begin{Lemma}\label{Hom sending maps to isos}
Let ${\mathcal X}$ be an additive category satisfying \ref{A1} and \ref{A2}. Let $\phi\colon F_0\to F_1$ be a morphism in $\operatorname{mod} {\mathcal X}$ with $\operatorname{Ker} \phi\in \operatorname{eff} {\mathcal X}$ and $\operatorname{Coker} \phi \in \operatorname{eff} {\mathcal X}$. Then 
\[
\operatorname{Hom}_{{\mathcal X}}(F_1,{\mathcal X}(-,X))\xrightarrow{-\circ \phi}\operatorname{Hom}_{{\mathcal X}}(F_0,{\mathcal X}(-,X))
\]
is an isomorphism for all $X\in {\mathcal X}$.
\end{Lemma}

\begin{proof}
Applying $\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))$ to the exact sequence
\[
0\to \operatorname{im} \phi \to F_1\to \operatorname{Coker} \phi \to 0
\]
gives an exact sequence
\begin{multline*}
0\to \operatorname{Hom}_{{\mathcal X}}(\operatorname{Coker} \phi,{\mathcal X}(-,X))\to \operatorname{Hom}_{{\mathcal X}}(F_1,{\mathcal X}(-,X)) \\
\to \operatorname{Hom}_{{\mathcal X}}(\operatorname{im} \phi,{\mathcal X}(-,X))\to \operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(\operatorname{Coker} \phi,{\mathcal X}(-,X))
\end{multline*}
Since $\operatorname{Coker} \phi\in \operatorname{eff} {\mathcal X}$, it follows that 
\[
\operatorname{Hom}_{{\mathcal X}}(\operatorname{Coker} \phi,{\mathcal X}(-,X))=0 \quad \text{and} \quad \operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(\operatorname{Coker} \phi,{\mathcal X}(-,X))=0
\]
by Theorem \ref{effaceable functors}. Hence the map 
\[
\operatorname{Hom}_{{\mathcal X}}(F_1,{\mathcal X}(-,X)) \to \operatorname{Hom}_{{\mathcal X}}(\operatorname{im} \phi,{\mathcal X}(-,X))
\]
 is an isomorphism. Similarly, applying $\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))$ to the exact sequene
\[
0\to \operatorname{Ker} \phi \to F_0\to \operatorname{im} \phi\to 0
\]
and using that $\operatorname{Hom}_{{\mathcal X}}(\operatorname{Ker} \phi,{\mathcal X}(-,X))=0$, it follows that the map 
\[
\operatorname{Hom}_{{\mathcal X}}(\operatorname{im} \phi,{\mathcal X}(-,X)) \to \operatorname{Hom}_{{\mathcal X}}(F_0,{\mathcal X}(-,X))
\]
is an isomorphism. Since $\operatorname{Hom}_{{\mathcal X}}(F_1,{\mathcal X}(-,X))\xrightarrow{-\circ \phi}\operatorname{Hom}_{{\mathcal X}}(F_0,{\mathcal X}(-,X))$ is equal to the composite
\[
\operatorname{Hom}_{{\mathcal X}}(F_1,{\mathcal X}(-,X)) \xrightarrow{\cong} \operatorname{Hom}_{{\mathcal X}}(\operatorname{im} \phi,{\mathcal X}(-,X)) \xrightarrow{\cong} \operatorname{Hom}_{{\mathcal X}}(F_0,{\mathcal X}(-,X))
\]
of two isomorphisms, the claim follows.
\end{proof}

To simplify notation we let ${\mathcal A}:= \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$. The localization functor $q\colon \operatorname{mod} {\mathcal X}\to {\mathcal A}$ induces a map
\[
q\colon \operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))\to {\mathcal A}(F,{\mathcal X}(-,X)) \quad f\mapsto q(f)
\]
for all $X\in {\mathcal X}$ and $F\in \operatorname{mod} {\mathcal X}$, which we also denote by $q$ by abuse of notation. We use the previous lemma to show that it is an isomorphism.

\begin{Lemma}\label{Fully faithful}
Let ${\mathcal X}$ be an additive category satisfying \ref{A1} and \ref{A2}. The map
\[
q\colon \operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))\to {\mathcal A}(F,{\mathcal X}(-,X)) \quad f\mapsto q(f)
\]
is an isomorphism for all $X\in {\mathcal X}$ and $F\in \operatorname{mod} {\mathcal X}$.
\end{Lemma}

\begin{proof}
The functor
\[
\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))\colon (\operatorname{mod} {\mathcal X})\ensuremath{^{\mathrm{op}}}\to \operatorname{Ab}
\]
induces a well-defined functor
\[
\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))\colon {\mathcal A}\ensuremath{^{\mathrm{op}}}\to \operatorname{Ab}
\]
by Lemma \ref{Hom sending maps to isos} and the universal property of the localization. Furthermore, $q$ induces a natural transformation $q\colon  \operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))\to {\mathcal A}(-,{\mathcal X}(-,X))$ of functors ${\mathcal A}\ensuremath{^{\mathrm{op}}}\to \operatorname{Ab}$. Also, by Yoneda's Lemma the element $1_{{\mathcal X}(-,X)}\in \operatorname{Hom}_{{\mathcal X}}({\mathcal X}(-,X),{\mathcal X}(-,X))$ corresponds to a natural transformation
\[
\mu\colon {\mathcal A}(-,{\mathcal X}(-,X))\to \operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))
\]
of functors ${\mathcal A}\ensuremath{^{\mathrm{op}}}\to \operatorname{Ab}$. Since the composite
\[
{\mathcal A}(-,{\mathcal X}(-,X))\xrightarrow{\mu} \operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))\xrightarrow{q} {\mathcal A}(-,{\mathcal X}(-,X))
\]
sends $1_{{\mathcal X}(-,X)}\in {\mathcal A}({\mathcal X}(-,X),{\mathcal X}(-,X))$ to itself, it must be the identity map. Furthermore, the composite 
\[
\mu\circ q\colon \operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))\to \operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))
\]
must also be a natural transformation when we consider $\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X}(-,X))$ as a functor $(\operatorname{mod} {\mathcal X})\ensuremath{^{\mathrm{op}}}\to \operatorname{Ab}$. Since it sends $1_{{\mathcal X}(-,X)}\in \operatorname{Hom}_{{\mathcal X}}({\mathcal X}(-,X),{\mathcal X}(-,X))$ to itself, it must be the identity map. This shows that $q$ is an isomorphism.
\end{proof}

Now we are ready to show that any additive category satisfying \ref{A1} and \ref{A2} is equivalent to a subcategory satisfying \ref{B1}, \ref{B2} and \ref{B3}. Recall that $\Phi\colon {\mathcal X}\to \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ is the functor given by the composite of the Yoneda embedding ${\mathcal X}\to \operatorname{mod} {\mathcal X}$ and the localization functor $q\colon \operatorname{mod} {\mathcal X}\to \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$. For simplicity, we also denote the essential image of $\Phi$ by ${\mathcal X}$ in the proof. 

\begin{Proposition}\label{essential image satisfy B1-B3}
Let ${\mathcal X}$ be an additive category satisfying \ref{A1} and \ref{A2}. Then the functor $\Phi\colon {\mathcal X}\to \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ is fully faithful, and its essential image satisfies \ref{B1}, \ref{B2} and \ref{B3}.
\end{Proposition}

\begin{proof}
By Lemma \ref{Fully faithful} the canonical map
\[
\operatorname{Hom}_{{\mathcal X}}({\mathcal X}(-,X'),{\mathcal X}(-,X))\to {\mathcal A}({\mathcal X}(-,X'),{\mathcal X}(-,X))
\]
is an isomorphism for all $X,X'\in {\mathcal X}$, where ${\mathcal A}=\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$. Since the Yoneda embedding is fully faithful, the functor $\Phi$ must therefore be fully faithful. Now let $F\in \operatorname{mod} {\mathcal X}$ be arbitrary, and choose an epimorphism $p\colon{\mathcal X}(-,X)\to F$ in $\operatorname{mod} {\mathcal X}$. Since $q$ is exact, $q(p)$ is an epimorphism in ${\mathcal A}$. This shows \ref{B1}, i.e. that ${\mathcal X}$ is generating. Next, assume ${\mathcal A}(F,{\mathcal X}(-,X))=0$ for all $X\in {\mathcal X}$. Then $\operatorname{Hom}_{{\mathcal X}}(F,{\mathcal X}(-,X))=0$ for all $X\in {\mathcal X}$ by Lemma \ref{Fully faithful}, and therefore $F\in \operatorname{eff} {\mathcal X}$ by Lemma \ref{effaceable functors}. It follows that $F\cong 0$ considered as an object in ${\mathcal A}$, which shows \ref{B2}. Finally, to prove \ref{B3}, assume $F\in \Omega^2_{{\mathcal X}}({\mathcal A})$. Choose an exact sequence
\[
0\to F\xrightarrow{\phi} {\mathcal X}(-,X_1)\xrightarrow{\psi}{\mathcal X}(-,X_0)
\]
in ${\mathcal A}$. By Lemma \ref{Fully faithful} the maps $\phi$ and $\psi$ can be lifted to morphisms in $\operatorname{mod} {\mathcal X}$ (which we denote by the same name). In particular, we have a morphism $g\colon X_1\to X_0$ in ${\mathcal X}$ such that $\psi=g\circ-$. Let $f$ be the weak kernel of $g$ in ${\mathcal X}$. Then the sequence of functors
\[
{\mathcal X}(-,X_2)\xrightarrow{f\circ -}{\mathcal X}(-,X_1)\xrightarrow{g\circ -}{\mathcal X}(-,X_0)
\]
is exact in $\operatorname{mod} {\mathcal X}$. Since $q$ is an exact functor, the sequence is also exact in ${\mathcal A}$. Furthermore, since $F$ is the kernel of $g\circ -$ in ${\mathcal A}$, there exists an epimorphism $\xi\colon {\mathcal X}(-,X_2)\to F$ in ${\mathcal A}$ such that $f\circ -=\phi\circ \xi$. Now let $\kappa\colon {\mathcal X}(-,X)\to F$ be an arbitrary morphism in ${\mathcal A}$ with $X\in {\mathcal X}$. Then the composite $\phi\circ \kappa\colon {\mathcal X}(-,X)\to {\mathcal X}(-,X_1)$ can be written as $h\circ -\colon {\mathcal X}(-,X)\to {\mathcal X}(-,X_1)$ for some morphism $h\colon X\to X_1$ in ${\mathcal X}$, by Lemma \ref{Fully faithful}. Since $g\circ h=0$, and $f$ is a weak kernel of $g$, it follows that $h$ factors through $f$. Therefore the map $\phi \circ \kappa$ factors through $f\circ -\colon {\mathcal X}(-,X_2)\to {\mathcal X}(-,X_1)$. Since $\phi$ is a monomorphism, it follows that $\kappa$ factors through $\xi$. Hence $\xi$ is a right ${\mathcal X}$-approximation, which proves the claim.
\end{proof}

Our next goal is to prove the uniqueness of the ambient abelian category as required in Definition \ref{Axiomatizing subcategories} \ref{Axiomatizing subcategories:3}. To this end, we need the following results describing a universal property of $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$.

\begin{Lemma}\label{universal property quotient}
Let ${\mathcal X}$ be an additive category satisfying \ref{A1} and \ref{A2}. The following hold:
\begin{enumerate}
\item\label{universal property quotient:1} The functor $\Phi\colon {\mathcal X}\to \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ preserves epimorphisms and sends a sequence $X_2\xrightarrow{f}X_1\xrightarrow{g}X_0$ in ${\mathcal X}$ with $f$ a weak kernel of $g$ to an exact sequence
\[
\Phi(X_2)\xrightarrow{\Phi(f)}\Phi(X_1)\xrightarrow{\Phi(g)}\Phi(X_0)
\]
in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$;
\item\label{universal property quotient:2} Let ${\mathcal B}$ be an abelian category and let $\Psi\colon {\mathcal X}\to {\mathcal B}$ be an additive functor which preserves epimorphisms and sends a sequence $X_2\xrightarrow{f}X_1\xrightarrow{g}X_0$ in ${\mathcal X}$ with $f$ a weak kernel of $g$ to an exact sequence
\[
\Psi(X_2)\xrightarrow{\Psi(f)}\Psi(X_1)\xrightarrow{\Psi(g)}\Psi(X_0)
\]
in ${\mathcal B}$. Then there exists an exact functor $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}\to {\mathcal B}$ extending $\Psi$, which is unique up to natural isomorphism.
\end{enumerate}
\end{Lemma}

\begin{proof}
Since the essential image of $\Phi$ satisfies \ref{B1}, \ref{B2} and \ref{B3} by Proposition \ref{essential image satisfy B1-B3}, part \ref{universal property quotient:1} follows from Lemma \ref{B1,B2,B3 implies A1,A2}. For part \ref{universal property quotient:2}, let $\Psi$ be a functor as in the lemma. Then there exists a right exact functor $\tilde{\Psi}\colon \operatorname{mod}{\mathcal X}\to {\mathcal B}$ extending $\Psi$, see Property 2.1 in \cite{Kra98}. Since $\Psi$ sends weak kernels to exact sequences, it follows that $\tilde{\Psi}$ is an exact functor by Lemma 2.5 in \cite{Kra98}. Now let $F\in \operatorname{mod} {\mathcal X}$ be arbitrary, and choose an exact sequence ${\mathcal X}(-,X_1)\xrightarrow{f\circ -}{\mathcal X}(-,X_0)\to F\to 0$. Applying $\tilde{\Psi}$ gives an exact sequence
\[
\Psi(X_1)\xrightarrow{\Psi(f)}\Psi(X_0)\to \tilde{\Psi}(F)\to 0
\]
in ${\mathcal B}$. If $F\in \operatorname{eff} {\mathcal X}$, then $f\colon X_1\to X_0$ is an epimorphism in ${\mathcal X}$ by Proposition \ref{effaceable functors} \ref{effaceable functors:1}. Therefore $\Psi(f)$ is an epimorphism in ${\mathcal B}$, which implies that $\tilde{\Psi}(F)=0$. This shows that $\operatorname{eff} {\mathcal X} \subseteq \operatorname{Ker} \tilde{\Psi}$. Therefore, by \cite[Corollaire 2 and 3 on page 368-369]{Gab62} there exists an exact functor
\[
\overline{\Psi}\colon \operatorname{mod}{\mathcal X}/\operatorname{eff} {\mathcal X}\to {\mathcal B}
\]
satisfying $\overline{\Psi}\circ q=\tilde{\Psi}$. The fact that $\tilde{\Psi}$ is unique follows readily from the fact that ${\mathcal X}$ is generating in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$.
\end{proof}


Now we can show the uniqueness of the ambient abelian category.

\begin{Proposition}\label{uniqueness ambient abelian}
Let ${\mathcal A}'$ be an abelian category and ${\mathcal X}$ an additive subcategory of ${\mathcal A}'$ which satisfies \ref{B1}, \ref{B2} and \ref{B3}. Then there exists an equivalence $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}\xrightarrow{\cong} {\mathcal A}'$ unique up to natural isomorphism which makes the diagram
\begin{equation*}
\begin{tikzcd}
\operatorname{mod}{\mathcal X}/\operatorname{eff} {\mathcal X} \arrow{rr}{\cong} & & {\mathcal A}' \\
 & \arrow{lu}{\Phi} \arrow{ru}[swap]{\operatorname{inclusion}}{\mathcal X} & 
\end{tikzcd}
\end{equation*}
commute.
\end{Proposition}

\begin{proof}
By Lemma \ref{B1,B2,B3 implies A1,A2} and Lemma \ref{universal property quotient} \ref{universal property quotient:2} we have exact functors 
\[
\tilde{\Psi}\colon \operatorname{mod} {\mathcal X}\to {\mathcal A}' \quad \text{and} \quad \overline{\Psi}\colon \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}\to {\mathcal A}'
\]
extending the inclusion ${\mathcal X}\subseteq {\mathcal A}'$. Our goal is to show that $\overline{\Psi}$ is an equivalence. Since ${\mathcal X}$ is a generating subcategory of ${\mathcal A}'$, it follows immediately that $\overline{\Psi}$ is dense. We show that $\overline{\Psi}$ is faithful. Let $F\in \operatorname{mod} {\mathcal X}$ be arbitrary and let ${\mathcal X}(-,X_1)\xrightarrow{f\circ -}{\mathcal X}(-,X_0)\to F\to 0$ be an exact sequence in $\operatorname{mod} {\mathcal X}$. Applying $\tilde{\Psi}$, we get an exact sequence
\[
X_1\xrightarrow{f}X_0\to \tilde{\Psi}(F)\to 0
\]
in ${\mathcal A}'$. Hence, if $\tilde{\Psi}(F)\cong 0$, then  $X_1\xrightarrow{f}X_0$ is surjective in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$. Therefore, $f$ must be surjective in ${\mathcal X}$, so $F\in \operatorname{eff} {\mathcal X}$. This shows that $\operatorname{Ker} \tilde{\Psi}=\operatorname{eff} {\mathcal X}$. Now if $\phi$ is a morphism in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$, then $\overline{\Psi}(\phi)=0$ if and only $\overline{\Psi}(\operatorname{im} \phi)=0$ since $\overline{\Psi}$ is exact. Also, since $\operatorname{Ker} \tilde{\Psi}=\operatorname{eff} {\mathcal X}$, it follows that $\operatorname{im} \phi \in \operatorname{eff} {\mathcal X}$, which implies that $\phi=0$ in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$. This shows that $\overline{\Psi}$ is faithful.

 It now only remains to show that $\overline{\Psi}$ is full. Let $F,F'\in \operatorname{mod} {\mathcal X}$, and let 
\[
{\mathcal X}(-,X_1)\xrightarrow{f\circ -}{\mathcal X}(-,X_0)\xrightarrow{\pi}F\to 0 \quad \text{and} \quad  {\mathcal X}(-,X'_1)\xrightarrow{f'\circ -}{\mathcal X}(-,X'_0)\xrightarrow{\pi'}F'\to 0 
\] 
be exact sequences in $\operatorname{mod} {\mathcal X}$. Let $\phi\colon \overline{\Psi}(F)\to \overline{\Psi}(F')$ be a morphism in ${\mathcal A}'$. We then get a commutative diagram
\begin{equation*}
\begin{tikzcd}[row sep =5ex, column sep =11ex, ampersand replacement =\&]
X_1 \arrow{r}{f} \arrow{d}{} \& X_0 \arrow{r}{\overline{\Psi}(\pi)} \arrow{d}{\begin{bmatrix}
1 \\ 0\end{bmatrix}} \& \overline{\Psi}(F) \arrow{r}{} \arrow{d}{\phi} \& 0 \\
K \arrow{r}{} \& X_0\oplus X_0'\arrow{r}{\begin{bmatrix}\phi\circ \overline{\Psi}(\pi)& \overline{\Psi}(\pi')\end{bmatrix}} \& \overline{\Psi}(F') \arrow{r}{} \& 0 \\
X_1'\arrow{r}{f'} \arrow{u}{} \& X_0'\arrow{r}{\overline{\Psi}(\pi')} \arrow{u}{\begin{bmatrix}0 \\ 1\end{bmatrix}} \& \overline{\Psi}(F') \arrow{r}{} \arrow{u}{1} \& 0
\end{tikzcd}
\end{equation*}
where the rows are exact and $K=\operatorname{Ker} \begin{bmatrix}\phi\circ \overline{\Psi}(\pi)& \overline{\Psi}(\pi')\end{bmatrix}$. Choose an epimorphism $\tilde{X_1}\to K$ in ${\mathcal A}'$ with $\tilde{X_1}\in {\mathcal X}$ and set $X_1''=X_1\oplus X_1'\oplus \tilde{X}_1$. Then we can factorize the map $X_1\oplus X_1'\to K$ as a composition $X_1\oplus X_1'\to X_1''\to K$ where $X_1''\to K$ is an epimorphism. Let $\begin{bmatrix}g \\ h\end{bmatrix}\colon X_1''\to X_0\oplus X_0'$ denote the map obtained by composing this epimorphism $X_1''\to K$ with the inclusion $K\to X_0\oplus X_0'$, let 
\[
F''=\operatorname{Coker} ({\mathcal X}(-,X_1'')\xrightarrow{\begin{bmatrix}g \\ h\end{bmatrix}\circ -}{\mathcal X}(-,X_0\oplus X_0'))
\]
and let $\pi''\colon {\mathcal X}(-,X_0\oplus X_0')\to F''$ denote the projection. Then we get morphisms $\phi'\colon F\to F''$ and $\phi''\colon F'\to F''$ making the diagram
\begin{equation*}
\begin{tikzcd}[row sep =5ex, column sep =11ex, ampersand replacement =\&]
{\mathcal X}(-,X_1) \arrow{r}{f\circ -} \arrow{d}{} \& {\mathcal X}(-,X_0) \arrow{r}{\pi} \arrow{d}{\begin{bmatrix}
1 \\ 0\end{bmatrix}\circ -} \& F \arrow{r}{} \arrow{d}{\phi'} \& 0 \\
{\mathcal X}(-,X_1'') \arrow{r}{\begin{bmatrix}g \\ h\end{bmatrix}\circ -} \& {\mathcal X}(-,X_0\oplus X_0')\arrow{r}{\pi''} \& F'' \arrow{r}{} \& 0 \\
{\mathcal X}(-,X_1')\arrow{r}{f'\circ -} \arrow{u}{} \& {\mathcal X}(-,X_0')\arrow{r}{\pi'} \arrow{u}{\begin{bmatrix}0 \\ 1\end{bmatrix}\circ -} \&F' \arrow{r}{} \arrow{u}{\phi''} \& 0
\end{tikzcd}
\end{equation*}
commutative. Note furthermore that $\tilde{\Psi}(\phi'')$ is an isomorphism, hence 
\[
\tilde{\Psi}(\operatorname{Ker} \phi'')\cong 0\cong \tilde{\Psi}(\operatorname{Coker} \phi'')
\]
since $\tilde{\Psi}$ is exact. This implies that $\operatorname{Ker} \phi''\in \operatorname{eff} {\mathcal X}$ and $\operatorname{Coker} \phi''\in \operatorname{eff} {\mathcal X}$ since $\operatorname{Ker} \tilde{\Psi}=\operatorname{eff} {\mathcal X}$. Therefore, $\phi''$ is an isomorphism in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$, and hence admits an inverse $\phi''^{-1}$. It is now clear that 
\[
\phi=\overline{\Psi}(\phi''^{-1})\circ \overline{\Psi}(\phi')=\overline{\Psi}(\phi''^{-1}\circ \phi')
\]
which shows that $\overline{\Psi}$ is full. This proves the claim.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Embedding Theorem}]
This follows from Lemma \ref{B1,B2,B3 implies A1,A2}, Proposition \ref{essential image satisfy B1-B3}, and Proposition \ref{uniqueness ambient abelian}.
\end{proof}
As a consequence we get that the ambient abelian category also satisfies a universal property.

\begin{Corollary}\label{universal property ambient}
Let ${\mathcal A}'$ an abelian category and let ${\mathcal X}$ an additive subcategory of ${\mathcal A}'$ which satisfies \ref{B1}, \ref{B2} and \ref{B3}. The following hold:
\begin{enumerate}
\item\label{universal property ambient:1} The inclusion functor ${\mathcal X}\to {\mathcal A}'$ preserves epimorphisms and sends a sequence $X_2\xrightarrow{f}X_1\xrightarrow{g}X_0$ in ${\mathcal X}$ with $f$ a weak kernel of $g$ to an exact sequence in ${\mathcal A}'$;
\item\label{universal property ambient:2} Let ${\mathcal B}$ be an abelian category and let $\Psi\colon {\mathcal X}\to {\mathcal B}$ be an additive functor which preserves epimorphisms and sends a sequence $X_2\xrightarrow{f}X_1\xrightarrow{g}X_0$ in ${\mathcal X}$ with $f$ a weak kernel of $g$ to an exact sequence
\[
\Psi(X_2)\xrightarrow{\Psi(f)}\Psi(X_1)\xrightarrow{\Psi(g)}\Psi(X_0)
\]
in ${\mathcal B}$. Then there exists an exact functor ${\mathcal A}'\to {\mathcal B}$ extending $\Psi$, which is unique up to natural isomorphism.
\end{enumerate}
\end{Corollary}

\begin{proof}
This follows immediately from Lemma \ref{universal property quotient} and Proposition \ref{uniqueness ambient abelian}.
\end{proof}

\section{Functorially finite generating cogenerating subcategories}

Let ${\mathcal X}$ be an additive category. We would like to find intrinsic axioms on ${\mathcal X}$ which axiomatizes  functorially finite generating and cogenerating subcategories of abelian categories, as in Definition \ref{Axiomatizing subcategories}. To this end, by Theorem \ref{Embedding Theorem} we know that ${\mathcal X}$ must satisfy \ref{A1} and \ref{A2} and their duals
\begin{enumerate}[label=(A\arabic*)$^{\operatorname{op}}$]
\item\label{A1op} ${\mathcal X}$ has weak cokernels;
\item\label{A2op} Any monomorphism in ${\mathcal X}$ is a weak kernel.
\end{enumerate}
To continue, we investigate the duality functor on $\operatorname{mod} {\mathcal X}$. For each $F\in \operatorname{mod} {\mathcal X}$ choose a projective presentation 
\[
{\mathcal X}(-,X_1)\xrightarrow{f\circ -}{\mathcal X}(-,X_0)\to F\to 0
\]
in $\operatorname{mod} {\mathcal X}$. Applying the contravariant Yoneda functor to $f\colon X_1\to X_0$ gives a map \\$-\circ f\colon {\mathcal X}(X_0,-)\to {\mathcal X}(X_1,-)$ in $\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}$. We define $F^*=\operatorname{Ker}(-\circ f)$ and $\operatorname{Tr} F=\operatorname{Coker} (-\circ f)$, so that we have an exact sequence
\[
0\to F^*\to {\mathcal X}(X_0,-)\xrightarrow{-\circ f}{\mathcal X}(X_1,-)\to \operatorname{Tr} F\to 0.
\]
in $\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}$. Dually, for $F'\in \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}$ we choose a projective presentation 
\[
{\mathcal X}(X_0',-)\xrightarrow{-\circ f'}{\mathcal X}(X_1',-)\to F'\to 0
\]
and define $\operatorname{Tr} F'$ and $F'^*$ by the exact sequence
\[
0\to F'^*\to {\mathcal X}(-,X_1')\xrightarrow{f'\circ -}{\mathcal X}(-,X_0')\to \operatorname{Tr} F'\to 0.
\]
Note that we have natural isomorphisms
\[
(-)^*\cong\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X})\colon \operatorname{mod} {\mathcal X} \to \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}  
\]
and 
\[
(-)^*\cong\operatorname{Hom}_{{\mathcal X}\ensuremath{^{\mathrm{op}}}}(-,{\mathcal X}\ensuremath{^{\mathrm{op}}})\colon \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}} \to \operatorname{mod} {\mathcal X}  
\]
which we use to identify these functors, where $\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X})$ and $\operatorname{Hom}_{{\mathcal X}\ensuremath{^{\mathrm{op}}}}(-,{\mathcal X}\ensuremath{^{\mathrm{op}}})$ denote the functors given by $\operatorname{Hom}_{{\mathcal X}}(-,{\mathcal X})(F)=\operatorname{Hom}_{{\mathcal X}}(F,-)|_{{\mathcal X}}$ and $\operatorname{Hom}_{{\mathcal X}\ensuremath{^{\mathrm{op}}}}(-,{\mathcal X}\ensuremath{^{\mathrm{op}}})(G)=\operatorname{Hom}_{{\mathcal X}\ensuremath{^{\mathrm{op}}}}(G,-)|_{{\mathcal X}\ensuremath{^{\mathrm{op}}}}$ for $F\in \operatorname{mod}{\mathcal X}$ and $G\in \operatorname{mod}{\mathcal X}\ensuremath{^{\mathrm{op}}}$, respectively. It follows that the functors $(-)^*\colon \operatorname{mod} {\mathcal X} \to \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}$ and $(-)^*\colon \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}} \to \operatorname{mod} {\mathcal X}$ form an adjoint pair. The unit and counit are part of exact sequences 
\begin{align}\label{unit of (-)^*}
& 0\to \operatorname{Ext}^2_{\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}}(\operatorname{Tr}F,{\mathcal X}\ensuremath{^{\mathrm{op}}})\to F\to F^{**}\to \operatorname{Ext}^1_{\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}}(\operatorname{Tr}F,{\mathcal X}\ensuremath{^{\mathrm{op}}})\to 0 
\end{align}
\begin{align}\label{counit of (-)^*}
 0\to \operatorname{Ext}^2_{\operatorname{mod} {\mathcal X}}(\operatorname{Tr}F,{\mathcal X})\to F\to F^{**}\to \operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(\operatorname{Tr}F,{\mathcal X})\to 0
\end{align}
see for example Proposition 6.3 in \cite{Aus66} in the case ${\mathcal X}$ is a ring. Now by Lemma \ref{Hom sending maps to isos} the functor $(-)^*$ sends morphisms in $\operatorname{mod} {\mathcal X}$ with kernel and cokernel in $\operatorname{eff} {\mathcal X}$ to isomorphisms in $\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}$. Together with the dual statement this implies that there are induced contravariant functors
\[
\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}\to \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}} \quad \text{and} \quad \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}}\to \operatorname{mod} {\mathcal X}
\]
making the following diagrams commute
\begin{equation*}
\begin{tikzcd}
\operatorname{mod}{\mathcal X} \arrow{r}{(-)^*} \arrow{d}{q} & \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}} \\
 \arrow{ru}{} \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X} & 
\end{tikzcd} \quad 
\begin{tikzcd}
\operatorname{mod}{\mathcal X}\ensuremath{^{\mathrm{op}}} \arrow{r}{(-)^*} \arrow{d}{q} & \operatorname{mod} {\mathcal X} \\
 \arrow{ru}{} \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}} & 
\end{tikzcd}
\end{equation*}
Composing with $q$ gives contravariant functors
\begin{align*}
&(-)^*\colon \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}\to \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}} \\
& (-)^*\colon \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}}\to \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}
\end{align*}
which we denote by the same symbol. The natural transformations $\operatorname{Id} \to (-)^{**}$ satisfy the triangular identities in $\operatorname{mod} {\mathcal X}$ and $\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}$, and hence also in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ and $\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}}$. This implies that the functors $(-)^*$ still form an adjoint pair as functors between $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ and $\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}}$.

Next we want to find conditions on ${\mathcal X}$ which ensures that $(-)^*$ induces an equivalence
\[
\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}\xrightarrow{\cong}(\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}})\ensuremath{^{\mathrm{op}}}.
\] 
Note first that for $F\in \operatorname{mod}{\mathcal X}$ we have an isomorphism
\[
\operatorname{Ext}^2_{\operatorname{mod} {\mathcal X}}(F,G)\cong \operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F',G)
\]
where $F'$ is a syzygy of $F$, i.e. fits in an exact sequence
\[
0\to F'\to {\mathcal X}(-,X)\to F\to 0
\] 
Using this, we see that the unit and counit given by \eqref{unit of (-)^*} and \eqref{counit of (-)^*} becomes isomorphisms in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ and $\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}}$ if
\begin{align*}
& \operatorname{Ext}^1_{\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}}(F,{\mathcal X}\ensuremath{^{\mathrm{op}}})\in \operatorname{eff} {\mathcal X} \quad \text{for all } F\in \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}} \\
& \operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F,{\mathcal X})\in \operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}} \quad \text{for all } F\in \operatorname{mod} {\mathcal X}
\end{align*}
which ensures that $(-)^*\colon \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}\xrightarrow{}\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}}$ is an equivalence. To capture this requirement, we introduce the following axiom:
\begin{enumerate}[label=(A\arabic*)]
\setcounter{enumi}{2}
\item\label{A3} Consider the following diagram
\begin{equation*}
\begin{tikzcd}
X_2 \arrow{rr}{f} \arrow{rd}{l} & & X_1 \arrow{r}{g} & X_0 \\
 & \arrow{ru}{h}X_2' & 
\end{tikzcd}
\end{equation*}
where $f$ is an arbitrary morphism in ${\mathcal X}$, where $g$ is a weak cokernel of $f$, where $h$ is a weak kernel of $g$, and where $l$ is an induced map satisfying $h\circ l=f$ (which exists since $g\circ f=0$ and $h$ is a weak kernel of $g$). Then for any weak kernel $k\colon X_3'\to X_2'$ of $h$ the map $\begin{bmatrix}l & k\end{bmatrix}\colon X_2\oplus X_3'\to X_2'$ is an epimorphism. 
\end{enumerate}

\begin{Proposition}\label{Equivalent criteria for A3}
Assume ${\mathcal X}$ is an additive category satsifying \ref{A1}, \ref{A1op}, \ref{A2}, \ref{A2op}. Then the following statements hold:
\begin{enumerate}
\item\label{Equivalent criteria for A3:1} ${\mathcal X}$ satisfies \ref{A3} if and only if $\operatorname{Ext}^1_{\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}}(F,{\mathcal X})\in \operatorname{eff} {\mathcal X}$ for all $F\in \operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}$;
\item\label{Equivalent criteria for A3:2} ${\mathcal X}$ satisfies \ref{A3}$\ensuremath{^{\mathrm{op}}}$ if and only if $\operatorname{Ext}^1_{\operatorname{mod}{\mathcal X}}(F,{\mathcal X})\in \operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}}$ for all $F\in \operatorname{mod} {\mathcal X}$.
\end{enumerate}
\end{Proposition}
\begin{proof}
We prove \ref{Equivalent criteria for A3:1}, \ref{Equivalent criteria for A3:2} is proved dually.  Let $f\colon X_2\to X_1$ be arbitrary, and choose $g$, $h$, $k$, $l$ as in \ref{A3}. Let $F$ be the cokernel of $-\circ f\colon {\mathcal X}(X_1,-)\to {\mathcal X}(X_2,-)$. Applying $(-)^*$ to the exact sequence
\[
{\mathcal X}(X_0,-)\xrightarrow{-\circ g}{\mathcal X}(X_1,-)\xrightarrow{-\circ f}{\mathcal X}(X_2,-)\to F\to 0
\] 
we get a complex 
\begin{equation}\label{complex for A3}
F^*\to {\mathcal X}(-,X_2)\xrightarrow{f\circ -}{\mathcal X}(-,X_1)\xrightarrow{g\circ -}{\mathcal X}(-,X_0)
\end{equation}
Let $K$ be the kernel of $g\circ -$. Since $g\circ h=0$, it follows that the map $h\circ -\colon {\mathcal X}(-,X_2')\to {\mathcal X}(-,X_1)$ factors through $K$ via a morphism $p\colon {\mathcal X}(-,X_2')\to K$. Since $h$ is a weak kernel of $g$, it follows that any map ${\mathcal X}(-,X)\to K$ with $X\in {\mathcal X}$ must factor through $p$. Hence, $p$ is an epimorphism. Similarly, if we let $K'=\operatorname{Ker} ({\mathcal X}(-,X_2')\xrightarrow{h\circ -}{\mathcal X}(-,X_1))$, then since $k\colon X_3'\to X_2'$ is a weak kernel of $h$, it follows that ${\mathcal X}(-,X_3')\xrightarrow{k\circ -}{\mathcal X}(-,X_2')$ factors through $K'$ via an epimorphism ${\mathcal X}(-,X'_3)\xrightarrow{q} K'$. Hence, we get a commutative diagram 
\begin{equation*}
\begin{tikzcd}[ampersand replacement =\&]
0\arrow{r}{} \& {\mathcal X}(-,X_3') \arrow{r}{\begin{bmatrix}0 \\ 1\end{bmatrix}} \arrow{d}{q} \& {\mathcal X}(-,X_2)\oplus {\mathcal X}(-,X_3') \arrow{r}{\begin{bmatrix}1 & 0\end{bmatrix}} \arrow{d}{r} \& {\mathcal X}(-,X_2) \arrow{r}{} \arrow{d}{} \& 0 \\
0\arrow{r} \& K' \arrow{r}{} \& {\mathcal X}(-,X_2')\arrow{r}{p} \& K \arrow{r}{} \& 0 
\end{tikzcd}
\end{equation*}
with exact rows, where $r=\begin{bmatrix}
l\circ - & k\circ -\end{bmatrix}$. Note that the cokernel of ${\mathcal X}(-,X_2)\to K$ is $\operatorname{Ext}^1_{\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}}(F,{\mathcal X})$. By the snake lemma, it follows that the cokernel of 
\[
r\colon {\mathcal X}(-,X_2)\oplus {\mathcal X}(-,X_3')\to {\mathcal X}(-,X_2')
\]
is also $\operatorname{Ext}^1_{\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}}(F,{\mathcal X})$, since $q$ is an epimorphism. Hence, $\operatorname{Ext}^1_{\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}}(F,{\mathcal X})\in \operatorname{eff} {\mathcal X}$ if and only if $r$ is an epimorphism in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$. Since a map in ${\mathcal X}$ is an epimorphism in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ if and only if it is an epimorphism in ${\mathcal X}$, it follows that $r$ is an epimorphism in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$ if and only if $\begin{bmatrix}l & k\end{bmatrix}\colon X_2\oplus X_3'\to X_2'$ is an epimorphism in ${\mathcal X}$. This proves the claim.
\end{proof}

\begin{Remark}\label{Remark A3 and A3'}
Note that the "only if" direction of the proof of Proposition \ref{Equivalent criteria for A3} uses the following alternative version of \ref{A3}, which therefore must be equivalent to \ref{A3} under the assumption of \ref{A1}, \ref{A1op}, \ref{A2}, \ref{A2op}:
\begin{enumerate}[label=(A\arabic*')]
\setcounter{enumi}{2}
\item\label{A3'} Let $f\colon X_2\to X_1$ be a morphism in ${\mathcal X}$. Then there exists a diagram 
\begin{equation*}
\begin{tikzcd}
X_2 \arrow{rr}{f} \arrow{rd}{l} & & X_1 \arrow{r}{g} & X_0 \\
X_3'\arrow{r}{k} & \arrow{ru}{h}X_2' & 
\end{tikzcd}
\end{equation*}
where $g$ is a weak cokernel of $f$, where $h$ is a weak kernel of $g$, where $k$ is a weak kernel of $h$, where $l$ is a map satisfying $h\circ l=f$, and where the map $\begin{bmatrix}l & k\end{bmatrix}\colon X_2\oplus X_3'\to X_2'$ is an epimorphism. 
\end{enumerate}
\end{Remark}

We finish by giving a characterization of generating cogenerating functorially finite subcategories in terms of intrinsic axioms.

\begin{Theorem}\label{Embedding Theorem generating cogenerating functorially finite}
Let $\mathsf{P}$ be the axioms  \ref{A1}, \ref{A1op}, \ref{A2}, \ref{A2op}, \ref{A3} and \ref{A3}\ensuremath{^{\mathrm{op}}}. Then $\mathsf{P}$ axiomatizes generating, cogenerating functorially finite subcategories of abelian categories.
\end{Theorem}

\begin{proof}
If ${\mathcal X}$ satisfies \ref{A1}, \ref{A1op}, \ref{A2}, \ref{A2op}, \ref{A3} and \ref{A3}\ensuremath{^{\mathrm{op}}}, then by Proposition \ref{Equivalent criteria for A3} and the exact sequences \eqref{unit of (-)^*} and \eqref{counit of (-)^*} it follows that $(-)^*$ induces an equivalence
\[
\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}\cong (\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}})\ensuremath{^{\mathrm{op}}}
\]
which commutes with the natural inclusions 
\[
{\mathcal X}\to \operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}\quad \text{and} \quad {\mathcal X}\to (\operatorname{mod} {\mathcal X}\ensuremath{^{\mathrm{op}}}/\operatorname{eff} {\mathcal X}\ensuremath{^{\mathrm{op}}})\ensuremath{^{\mathrm{op}}}.
\]
By Proposition \ref{essential image satisfy B1-B3} and its dual we get that ${\mathcal X}$ is (equivalent to) a generating cogenerating subcategory of $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$. Therefore $\Omega^2_{{\mathcal X}}(\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X})=\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$, and since ${\mathcal X}$ satisfies \ref{B3} it follows that ${\mathcal X}$ is contravariantly finite in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$. Similarly, ${\mathcal X}$ is also covariantly finite in $\operatorname{mod} {\mathcal X}/\operatorname{eff} {\mathcal X}$. This shows that ${\mathcal X}$ is a generating cogenerating functorially finite subcategory. 

Now assume ${\mathcal X}\subseteq {\mathcal A}$ is a generating cogenerating functorially finite subcategory of an abelian category ${\mathcal A}$. By Theorem \ref{Embedding Theorem} and its dual it follows that ${\mathcal X}$ satisfies \ref{A1}, \ref{A1op}, \ref{A2} and \ref{A2op}. We only need to show that ${\mathcal X}$ satisfies \ref{A3}, since the proof for \ref{A3}\ensuremath{^{\mathrm{op}}}{ } is dual. Note first that a morphism $g\colon X_1\to X_0$ in ${\mathcal X}$ is a weak cokernel of $f\colon X_2\to X_1$ if and only if the induced map $\operatorname{Coker} f\to X_0$ is a left ${\mathcal X}$-approximation (and therefore also a monomorphism). The dual statement holds for weak kernels. Now assume we are given $l$,$f$,$g$,$h$,$k$ as in \ref{A3}. Since $\operatorname{Ker} g \cong \operatorname{Ker} (X_1\to \operatorname{Coker} f)$, it follows that $f$ factors through $\operatorname{Ker} g$ via an epimorphism $p\colon X_2\to \operatorname{Ker} g$. Also, since $h$ is a weak kernel of $g$ and $k$ is a weak kernel of $h$, the map $h$ factors through $\operatorname{Ker} g$ via an epimorphism $q\colon X_2'\to \operatorname{Ker} g$ and the map $k$ factors through $\operatorname{Ker} h$ via an epimorphism $p'\colon X_3'\to \operatorname{Ker} h$. Hence we get a commutative diagram with exact rows
\begin{equation*}
\begin{tikzcd}[ampersand replacement =\&]
0\arrow{r}{} \& X_3' \arrow{r}{\begin{bmatrix}0 \\ 1\end{bmatrix}} \arrow{d}{p'} \& X_2\oplus X_3' \arrow{r}{\begin{bmatrix}1 & 0\end{bmatrix}} \arrow{d}{\begin{bmatrix}l & k\end{bmatrix}} \& X_2 \arrow{r}{} \arrow{d}{p} \& 0 \\
0\arrow{r} \& \operatorname{Ker} h \arrow{r}{} \& X_2'\arrow{r}{q} \& \operatorname{Ker} g \arrow{r}{} \& 0 
\end{tikzcd}.
\end{equation*}
Since the leftmost and rightmost vertical map are epimorphism, the middle map must be an epimorphism. Therefore, ${\mathcal X}$ satisfies \ref{A3}.
\end{proof}

\section{Rigid subcategories}

In this section we assume ${\mathcal X}$ is a generating cogenerating functorially finite subcategory of an abelian category ${\mathcal A}$. By Theorem \ref{Embedding Theorem generating cogenerating functorially finite} we know that this is equivalent to \ref{A1}, \ref{A1op}, \ref{A2}, \ref{A2op}, \ref{A3}, \ref{A3}\ensuremath{^{\mathrm{op}}}. Now we want to determine the intrinsic axiom needed to capture the property
\begin{equation}\label{Ext vanishing}
\operatorname{Ext}^i_{{\mathcal A}}(X,X')=0 \text{ for } 0< i< d \text{ and } X,X'\in {\mathcal X}.
\end{equation}
We consider the following:
\begin{enumerate}[label=($d$-Rigid)]
\item\label{d-Rigid} For all epimorphism $f_1\colon X_1\to X_0$ in ${\mathcal X}$ there exists a sequence
\[
X_{d+1}\xrightarrow{f_{d+1}}X_d\xrightarrow{f_d}\cdots \xrightarrow{f_3}X_2\xrightarrow{f_2}X_1\xrightarrow{f_1}X_0
\]
with $f_{i+1}$ a weak kernel of $f_i$ and $f_i$ a weak cokernel of $f_{i+1}$ for all $1\leq i\leq d$.
\end{enumerate}

The following theorem relates these two notions:

\begin{Theorem}\label{d-Rigid and Ext vanishing}
Let ${\mathcal X}$ be a generating cogenerating functorially finite subcategory of an abelian category ${\mathcal A}$. Then ${\mathcal X}$ satisfies \ref{d-Rigid} if and only if
\[
\operatorname{Ext}^i_{{\mathcal A}}(X,X')=0 
\] 
for all $0< i< d$ and all $ X,X'\in {\mathcal X}$.
\end{Theorem}

\begin{Remark}
By Lemma \ref{Weak kernels and cokernels} it follows that \ref{d-Rigid} with $d=1$ is equivalent to \ref{A2} (under the assumption that weak kernels and cokernels exist). Hence, it holds automatically for a generating cogenerating functorially finite subcategory. This is reflected by the fact that condition \eqref{Ext vanishing} is empty for $d=1$.
\end{Remark}

\begin{proof}[Proof of "if" part of Theorem \ref{d-Rigid and Ext vanishing}]
Assume $f_1\colon X_1\to X_0$ is an epimorphism in ${\mathcal X}$. Then $f_1$ must also be an epimorphism in ${\mathcal A}$. Choose a right ${\mathcal X}$-approximation $X_2\to \operatorname{Ker} f_1$, which must be an epimorphism since ${\mathcal X}$ is generating. Let $f_2$ denote the composite $X_2\to \operatorname{Ker} f\to X_1$. We continue this construction iteratively for $1\leq i\leq d$, i.e. we choose a right ${\mathcal X}$-approximation $X_{i+1}\to \operatorname{Ker} f_i$ and we let $f_{i+1}$ denote the composite $X_{i+1}\to \operatorname{Ker} f_i\to X_i$. Then we get an exact sequence
\[
X_{d+1}\xrightarrow{f_{d+1}}X_d\xrightarrow{f_d}\cdots \xrightarrow{f_3}X_2\xrightarrow{f_2}X_1\xrightarrow{f_1}X_0\to 0
\]
in ${\mathcal A}$ where $f_{i+1}$ is a weak kernel of $f_i$ for all $1\leq i\leq d$. Applying ${\mathcal A}(-,X')$ with $X'\in {\mathcal X}$ and using that $\operatorname{Ext}^j_{{\mathcal A}}(X_i,X')=0$ for all $0<j<d$  and $0\leq i\leq d+1$, we get an exact sequence
\[
0\to {\mathcal A}(X_0,X')\xrightarrow{-\circ f_1}{\mathcal A}(X_1,X')\xrightarrow{-\circ f_2}\cdots \xrightarrow{-\circ f_d}{\mathcal A}(X_d,X')\xrightarrow{-\circ f_{d+1}}{\mathcal A}(X_{d+1},X')
\] 
In particular, since the sequences ${\mathcal A}(X_{i-1},X')\xrightarrow{-\circ f_i}{\mathcal A}(X_i,X')\xrightarrow{-\circ f_{i+1}}{\mathcal A}(X_{i+1},X')$ are exact for all $1\leq i\leq d$ and all $X'\in {\mathcal X}$, it follows that $f_{i+1}$ is a weak cokernel of $f_i$ for all $1\leq i\leq d$. This proves the claim.
\end{proof}

The goal in the remaining part of the section is prove the converse.

\begin{Lemma}\label{1-Rigid}
Let ${\mathcal X}$ be a generating cogenerating functorially finite subcategory of an abelian category ${\mathcal A}$. Assume ${\mathcal X}$ satisfies \ref{d-Rigid} for some $d>1$. Then $\operatorname{Ext}^1_{{\mathcal A}}(X,X')=0$ for all $X,X'\in {\mathcal X}$. 
\end{Lemma}
 
\begin{proof}
Let $0\to X'\xrightarrow{g}A\xrightarrow{f}X\to 0$ be an exact sequence in ${\mathcal A}$ with $X,X'\in {\mathcal X}$. Choose a right ${\mathcal X}$-approximation $p\colon X_1\to A$. Then $g$ factors through $p$ via a monomorphism $i\colon X'\to X_1$. We therefore get a commutative diagram
\begin{equation*}
\begin{tikzcd}
0\arrow{r}{} & X' \arrow{r}{i} \arrow{d}{1} & X_1 \arrow{r}{} \arrow{d}{p} & \operatorname{Coker} i \arrow{r}{} \arrow{d}{} & 0\\
0\arrow{r} & X' \arrow{r}{g} & A \arrow{r}{f} & X \arrow{r}{} & 0 
\end{tikzcd}.
\end{equation*}
with exact rows. This gives an exact sequence
\[
0\to X'\oplus \operatorname{Ker} p\xrightarrow{\begin{bmatrix}i & j\end{bmatrix}} X_1\xrightarrow{f\circ p} X\to 0
\]
where $j\colon \operatorname{Ker} p\to X_1$ is the canonical monomorphism. By \ref{d-Rigid}, there exists a sequence
\[
X_3\xrightarrow{l}X_2\xrightarrow{k}X_1\xrightarrow{f\circ p}X\to 0
\] 
where $l$ is a weak kernel of $k$ and $k$ is a weak kernel of $f\circ p$, and where $k$ is weak cokernel of $l$ and $f\circ p$ is a weak cokernel of $k$, and where $X_3,X_2\in {\mathcal X}$. Then 
\[
\operatorname{Coker} l\cong\operatorname{im} k\cong\operatorname{Ker} (f\circ p)=X'\oplus \operatorname{Ker} p.
\]
Since $k$ is a weak cokernel of $l$ it follows that the induced map
\[
\operatorname{Coker} l\cong X'\oplus \operatorname{Ker} p \xrightarrow{\begin{bmatrix}i & j\end{bmatrix}} X_1
\] 
is a left ${\mathcal X}$-approximation. Hence $X'\oplus \operatorname{Ker} p \xrightarrow{\begin{bmatrix}1 & 0\end{bmatrix}}X'$ factors through $X_1$. This means that there exists a map $s\colon X_1\to X'$ such that $s\circ i=1_{X'}$ and $s\circ j=0$. Since $A\cong \operatorname{Coker} j$, we get an induced map $t\colon A\to X'$ satisfying $t\circ g=1_{X'}$. Therefore the sequence $0\to X'\xrightarrow{g}A\xrightarrow{f}X\to 0$ is split. Since the elements in $\operatorname{Ext}^1_{{\mathcal A}}(X,X')$ can be described in terms of Yoneda extensions, this proves the claim.
\end{proof}

\begin{Lemma}\label{1 rigid not just in X}
Let ${\mathcal X}$ be a generating cogenerating functorially finite subcategory of an abelian category ${\mathcal A}$. Assume ${\mathcal X}$ satisfies \ref{d-Rigid} for some $d>1$, and let $f\colon X_1\to X_0$ be a weak cokernel in ${\mathcal X}$. Then $\operatorname{Ext}^1_{{\mathcal A}}(\operatorname{Coker} f,X)=0$ for all $X\in {\mathcal X}$. 
\end{Lemma}

\begin{proof}
Applying ${\mathcal A}(-,X)$ for $X\in {\mathcal X}$ to the exact sequence $0\to \operatorname{im} f\to X_0\to \operatorname{Coker} f\to 0$ gives an exact sequence
\[
0\to {\mathcal A}(\operatorname{Coker} f,X)\to {\mathcal A}(X_0,X)\to {\mathcal A}(\operatorname{im} f, X)\to \operatorname{Ext}^1_{{\mathcal A}}(\operatorname{Coker} f,X) \to \operatorname{Ext}^1_{{\mathcal A}}(X_0,X)
\]
Now $\operatorname{Ext}^1_{{\mathcal A}}(X_0,X)=0$ by Lemma \ref{1-Rigid}. Also, since $f$ is a weak cokernel, the map $\operatorname{im} f \to X_0$ is a left ${\mathcal X}$-approximation. Therefore ${\mathcal A}(X_0,X)\to {\mathcal A}(\operatorname{im} f,X)$ is surjective. By considering the exact sequence above it follows that $\operatorname{Ext}^1_{{\mathcal A}}(\operatorname{Coker} f,X)=0$. This proves the claim.
\end{proof}

\begin{proof}[Proof of "only if" part of Theorem \ref{d-Rigid and Ext vanishing}]
We prove that $\operatorname{Ext}^i_{{\mathcal A}}(X',X)=0$ for all $X,X'\in {\mathcal X}$ and $0<i<d$ by induction on $i$. For $i=1$ this follows from Lemma \ref{1-Rigid}. Assume $\operatorname{Ext}^i_{{\mathcal A}}(X',X)=0$ for all $X,X'\in {\mathcal X}$ and all $0<i\leq j$ with $j<d-1$. We prove that $\operatorname{Ext}^{j+1}_{{\mathcal A}}(X',X)=0$ for all $X,X'\in {\mathcal X}$. Let
\[
0\to X\xrightarrow{f_{j+2}}A_{j+1}\xrightarrow{f_{j+1}}\cdots \xrightarrow{f_2}A_1\xrightarrow{f_1}X'\to 0
\]
be an exact sequence with $X,X'\in {\mathcal X}$. Choose an epimorphism $g\colon X_1\to A_1$ with $X_1\in {\mathcal X}$, and let $g_1=f_1\circ g$. Next take the pullback square
\begin{equation*}
\begin{tikzcd}
 A_2' \arrow{r}{} \arrow{d}{} & \operatorname{Ker} g_1 \arrow{d}{} \\
 A_2 \arrow{r}{} & \operatorname{Ker} f_1  
\end{tikzcd}.
\end{equation*}
and let $g_2$ be the composite $A_2'\to \operatorname{Ker} g_1\to X_1$. Now constrict iteratively $A_k'$ and $g_k\colon A_k'\to A_{k-1}'$ for $k\leq j+1$ such that
\begin{equation*}
\begin{tikzcd}
 A_k' \arrow{r}{} \arrow{d}{} & \operatorname{Ker} g_{k-1} \arrow{d}{} \\
 A_k \arrow{r}{} & \operatorname{Ker} f_{k-1} 
\end{tikzcd}.
\end{equation*} 
is a pullback square and $g_k$ is the composite $A_k'\to \operatorname{Ker} g_{k-1}\to A_{k-1}'$ (note that $\operatorname{Ker} g_k\cong \operatorname{Ker} f_k$ for $k\geq 2$ and $A_k'\cong A_k$ for $k\geq 3$). Then we get a commutative diagram with exact rows
\begin{equation*}
\begin{tikzcd}
0\arrow{r}{} & X \arrow{r}{g_{j+2}} \arrow{d}{1} & A'_{j+1} \arrow{r}{g_{j+1}} \arrow{d}{} & \cdots \arrow{r}{g_3} & A_2' \arrow{r}{g_2} \arrow{d}{} & X_1 \arrow{r}{g_1} \arrow{d}{} & X'\arrow{r}{} \arrow{d}{1} & 0\\
0\arrow{r}{} & X \arrow{r}{f_{j+2}} & A_{j+1} \arrow{r}{f_{j+1}} & \cdots \arrow{r}{f_3} & A_2 \arrow{r}{f_2} & A_1 \arrow{r}{f_1} & X'\arrow{r}{}  & 0 
\end{tikzcd}.
\end{equation*}
Hence, both exact sequences represents the same element in the Yoneda Ext-group \\ $\operatorname{Ext}^{j+1}_{{\mathcal A}}(X',X)$. Therefore, it is sufficient to show that the upper exact sequence is $0$ as an element in  $\operatorname{Ext}^{j+1}_{{\mathcal A}}(X',X)$. For this, it suffices to show that $\operatorname{Ext}^j_{{\mathcal A}}(\operatorname{Ker} g_1,X)=0$. Now by axiom \ref{d-Rigid} there exists an exact sequence
\[
X_{j+3}\xrightarrow{h_{j+3}}\cdots \xrightarrow{h_4}X_3\xrightarrow{h_3}X_2\xrightarrow{h_2}X_1\xrightarrow{h_1}X'\to 0
\]
where $g_1=h_1$, and where $h_{i+1}$ is a weak kernel of $h_i$ and $h_i$ is a weak cokernel of $h_{i+1}$ for $1\leq i\leq j+2$, and where $X_i\in {\mathcal X}$ for $1\leq i\leq j+3$. Now consider the exact sequence
\[
0\to \operatorname{Ker} h_{i+1}\to X_{i+1}\to \operatorname{Ker} h_i\to 0
\]
where $1\leq i \leq j-1$. Applying ${\mathcal A}(-,X)$, we get exact sequences
\begin{multline*}
0\to {\mathcal A}(\operatorname{Ker} h_i,X)\to {\mathcal A}(X_{i+1},X)\to {\mathcal A}(\operatorname{Ker} h_{i+1},X)\to \cdots \to \operatorname{Ext}^{j-i}_{{\mathcal A}}(X_{i+1},X) \\
 \to \operatorname{Ext}^{j-i}_{{\mathcal A}}(\operatorname{Ker} h_{i+1},X)\to \operatorname{Ext}^{1+j-i}_{{\mathcal A}}(\operatorname{Ker} h_i,X)\to \operatorname{Ext}^{1+j-i}_{{\mathcal A}}(X_{i+1},X)\to \cdots
\end{multline*}
Since $\operatorname{Ext}^{j-i}_{{\mathcal A}}(X_{i+1},X)=0=\operatorname{Ext}^{1+j-i}_{{\mathcal A}}(X_{i+1},X)$ for $1\leq i\leq j-1$ by the induction hypothesis, we get that 
\[
\operatorname{Ext}^{1+j-i}_{{\mathcal A}}(\operatorname{Ker} h_i,X)\cong \operatorname{Ext}^{j-i}_{{\mathcal A}}(\operatorname{Ker} h_{i+1},X)
\]
Hence
\[
\operatorname{Ext}^j_{{\mathcal A}}(\operatorname{Ker} h_1,X)\cong \operatorname{Ext}^{j-1}_{{\mathcal A}}(\operatorname{Ker} h_2,X)\cong \cdots \cong \operatorname{Ext}^1_{{\mathcal A}}(\operatorname{Ker} h_{j},X)
\]
Since $\operatorname{Ker} h_j\cong \operatorname{Coker} h_{j+2}$ and $h_{j+2}$ is a weak cokernel, it follows that 
\[
\operatorname{Ext}^1_{{\mathcal A}}(\operatorname{Coker} h_{j+2},X)=0
\] by Lemma \ref{1 rigid not just in X}.
Hence $\operatorname{Ext}^j_{{\mathcal A}}(\operatorname{Ker} h_1,X)=0$, which proves the claim.
\end{proof}

\section{d-abelian categories are d-cluster tilting}

In this section we show that any $d$-abelian category is equivalent to a $d$-cluster tilting subcategory of an abelian category. More precisely, we show that being a $d$-abelian is equivalent to having $d$-kernels and $d$-cokernels and satisfying axioms \ref{A1}, \ref{A1op}, \ref{A2}, \ref{A2op}, \ref{A3}, \ref{A3}\ensuremath{^{\mathrm{op}}}, \ref{d-Rigid}, and we show that such categories axiomatizes $d$-cluster tilting subcategories.  

We first recall the definition of a $d$-cluster tilting subcategory:

\begin{Definition}\label{Definition cluster tilting subcategory}
Let ${\mathcal X}$ be a full subcategory of an abelian category ${\mathcal A}$, and let $d>0$ be a positive integer. We say that ${\mathcal X}$ is $d$\emphbf{-cluster tilting} in ${\mathcal A}$ if the following hold:
\begin{enumerate}
\item\label{Definition cluster tilting subcategory:1} ${\mathcal X}$ is a generating cogenerating functorially finite subcategory of ${\mathcal A}$;
\item\label{Definition cluster tilting subcategory:2} We have
\begin{align*}
{\mathcal X} & = \{A\in {\mathcal A} \mid \operatorname{Ext}^i_{{\mathcal A}}(A,X)=0 \text{ for }1\leq i\leq d-1 \text{ and }X\in {\mathcal X}\} \\ 
& = \{A\in {\mathcal A} \mid \operatorname{Ext}^i_{{\mathcal A}}(X,A)=0 \text{ for }1\leq i\leq d-1 \text{ and }X\in {\mathcal X}\};
\end{align*}
\end{enumerate}
\end{Definition}

We need the following result on $d$-cluster tilting subcategories

\begin{Lemma}\label{result on d-cluster tilting}
Let ${\mathcal X}$ be a generating cogenerating functorially finite subcategory of an abelian category ${\mathcal A}$. Assume $\operatorname{Ext}^i_{{\mathcal A}}(X,X')=0$ for $1\leq i\leq d-1$ and $X,X'\in {\mathcal X}$. The following are equivalent:
\begin{enumerate}
\item\label{result on d-cluster tilting:1} ${\mathcal X}$ is $d$-cluster tilting in ${\mathcal A}$;
\item\label{result on d-cluster tilting:2} ${\mathcal X}$ is closed under direct summands, and for any $A\in {\mathcal A}$ there exists exact sequences
\[
0\to A\to X_{-1}\to \cdots \to X_{-d}\to 0 \quad \text{and} \quad 0\to X_d'\to \cdots \to X'_1\to A\to 0
\]
where $X_i,X_i'\in {\mathcal X}$ for $1\leq i\leq d$.
\end{enumerate}
\end{Lemma}

\begin{proof}
This follows from \cite[Proposition 2.2.2]{Iya07a}.
\end{proof}

Let ${\mathcal X}$ be an additive category, and let $f\colon X_1\to X_0$ be a morphism in ${\mathcal X}$. Following \cite{Jas16}, we say that a sequence
\[
X_{d+1}\to X_d \to \cdots \to X_1
\]
in ${\mathcal X}$ is a $d$\emphbf{-kernel} of $f$ if the sequence of abelian groups
\[
0\to {\mathcal X}(X,X_{d+1})\to {\mathcal X}(X,X_d)\to \cdots \to {\mathcal X}(X,X_1)\xrightarrow{f\circ -} {\mathcal X}(X,X_0)
\]
is exact for all $X\in {\mathcal X}$. Dually, the sequence
\[
X_0\to X_{-1}\to \cdots \to X_{-d}
\]
is a $d$\emphbf{-cokernel} of $f$ if the sequence of abelian groups
\[
0\to {\mathcal X}(X_{-d},X)\xrightarrow{} {\mathcal X}(X_{-d+1},X)\to \cdots \to {\mathcal X}(X_{0},X)\xrightarrow{-\circ f} {\mathcal X}(X_1,X)
\]
is exact for all $X\in {\mathcal X}$. 

\begin{Theorem}\label{axioms implying d-cluster tilting}
Let ${\mathcal X}$ be an idempotent complete additive category satisfying \ref{A1}, \ref{A1op}, \ref{A2}, \ref{A2op}, \ref{A3}, \ref{A3}\ensuremath{^{\mathrm{op}}}, and \ref{d-Rigid}. Assume furthermore that every morphism in ${\mathcal X}$ has a $d$-kernel and a $d$-cokernel. Then ${\mathcal X}$ is equivalent to a $d$-cluster tilting subcategory of an abelian category.
\end{Theorem}

\begin{proof}
By Theorem \ref{Embedding Theorem generating cogenerating functorially finite} and Theorem \ref{d-Rigid and Ext vanishing} we can assume ${\mathcal X}$ is a generating cogenerating functorially finite subcategory of an abelian category ${\mathcal A}$ satisfying $\operatorname{Ext}^i_{{\mathcal A}}(X,X')=0$ for all $0<i<d$ and $X,X'\in {\mathcal X}$. In particular, since ${\mathcal X}$ is generating and cogenerating, for any object $A\in {\mathcal A}$ there exists morphisms $f\colon X_{1}\to X_0$ and $g\colon X_0'\to X_{-1}'$ in ${\mathcal X}$ with $\operatorname{Coker} f\cong A$ and $\operatorname{Ker} g\cong A$. Since ${\mathcal X}$ is idempotent complete, taking the $d$-kernel of $f$ and the $d$-cokernel of $g$, we see that condition \ref{result on d-cluster tilting:2} in Lemma \ref{result on d-cluster tilting} holds. Hence ${\mathcal X}$ must be $d$-cluster tilting in ${\mathcal A}$.
\end{proof}

Next we recall the definition of $d$-abelian categories. Following \cite{Jas16}, we say that a complex
\[
X_{d+1}\xrightarrow{f_{d+1}} \cdots \to X_1\xrightarrow{f_1} X_0
\] 
is $d$\emphbf{-exact} if $X_{d+1}\xrightarrow{f_{d+1}} \cdots \xrightarrow{f_2} X_1$ is a $d$-kernel of $f_1$ and $X_{d}\xrightarrow{f_{d}} \cdots \to X_1\xrightarrow{f_1} X_0$ is a $d$-cokernel of $f_{d+1}$.

\begin{Definition}[Definition 3.1 in \cite{Jas16}]\label{d-abelian categories}
Let $d$ be a positive integer, and let ${\mathcal X}$ be an additive category. We say that ${\mathcal X}$ is a $d$-abelian category if it satisfies the following
\begin{enumerate}
\item\label{d-abelian categories:1} ${\mathcal X}$ is idempotent complete;
\item\label{d-abelian categories:2} Every morphism in ${\mathcal X}$ has a $d$-kernel and a $d$-cokernel;
\item\label{d-abelian categories:3} Any complex 
\[
X_{d+1}\xrightarrow{f_{d+1}} \cdots \to X_1\xrightarrow{f_1} X_0
\] 
where $f_1$ is an epimorphism and $X_{d+1}\xrightarrow{f_{d+1}} \cdots \xrightarrow{f_2} X_1$ is a $d$-kernel of $f$ must be $d$-exact;
\item\label{d-abelian categories:4} Any complex 
\[
X_{d+1}\xrightarrow{f_{d+1}} \cdots \to X_1\xrightarrow{f_1} X_0
\] 
where $f_{d+1}$ is a monomorphism and $X_{d}\xrightarrow{f_{d}} \cdots \to X_1\xrightarrow{f_1} X_0$ is a $d$-cokernel of $f_{d+1}$ must be $d$-exact.
\end{enumerate}
\end{Definition}

\begin{Proposition}\label{d-abelian satisfies axioms}
Let ${\mathcal X}$ be an additive category.  Then ${\mathcal X}$ is $d$-abelian if and only if it is idempotent complete, has $d$-kernels and cokernels, and satisfies \ref{A1}, \ref{A1op}, \ref{A2}, \ref{A2op}, \ref{A3}, \ref{A3}\ensuremath{^{\mathrm{op}}} and \ref{d-Rigid}.
\end{Proposition}

\begin{proof}
The "if" direction follows from Theorem \ref{axioms implying d-cluster tilting} and the fact that any $d$-cluster tilting subcategory is $d$-abelian by \cite[Theorem 3.16]{Jas16}. Conversely, assume ${\mathcal X}$ is a $d$-abelian category. Since ${\mathcal X}$ has $d$-kernels and $d$-cokernels, axioms \ref{A1}, \ref{A1op} hold automatically. Also by Definition \ref{d-abelian categories} \ref{d-abelian categories:3} and the fact that ${\mathcal X}$ has $d$-kernels it follows that \ref{A2} and \ref{d-Rigid} hold, and dually by Definition \ref{d-abelian categories} \ref{d-abelian categories:4} and the fact that ${\mathcal X}$ has $d$-cokernels it follows that \ref{A2}\ensuremath{^{\mathrm{op}}} \text{ }holds. We show that ${\mathcal X}$ satisfies \ref{A3}. Let $f^0\colon X^0\to X^1$ be a morphism, and let $f^1\colon X^1\to X^2$ be a weak cokernel of $f^0$. Then by \cite[Proposition 3.13]{Jas16} there exists objects $Y^1_1$ and $Y^2_1$ in ${\mathcal X}$ and morphisms $g^1_1\colon Y^1_1\to X^1$ and $g^2_1\colon Y^2_1\to Y^1_1$ and $p^0_0\colon X^0\to Y^1_1$ such that
\begin{enumerate}
\item $g^1_1$ is a weak kernel of $f^1$, $g_1^2$ is a weak kernel of $g^1_1$;
\item $g^1_1\circ p^0_0=f^0$;
\item The map $\begin{bmatrix}p^0_0 & g^2_1\end{bmatrix}\colon X^0\oplus Y^2_1\to Y^1_1$ is an epimorphism.
\end{enumerate}
This shows that ${\mathcal X}$ satisfies axiom \ref{A3'}, which by Remark \ref{Remark A3 and A3'} is equivalent to \ref{A3}. Axiom \ref{A3}\ensuremath{^{\mathrm{op}}}  \text{ }is proved dually.
\end{proof}

\section{Precluster tilting subcategories}
Let ${\mathcal X}$ be an additive category satisfying \ref{A1}, \ref{A1op}, \ref{A2}, \ref{A2op}, \ref{A3}, \ref{A3}\ensuremath{^{\mathrm{op}}} \text{ }and \ref{d-Rigid}. The goal in this section is to find additional axioms on ${\mathcal X}$ so that it gives an axiomatization of $d$-precluster tilting subcategories as introduced in \cite{IS18}.  In order to do this, we need to reformulate the definition of precluster tilting subcategories so that it makes sense for any abelian category.

 In the following we fix a commutative artinian ring $R$, an Artin $R$-algebra $\Lambda$, and we let $\operatorname{mod} \Lambda$ be the category of finitely generated (right) $\Lambda$-modules. We denote by $\underline{\operatorname{mod}}\Lambda$ and $\overline{\operatorname{mod}} \Lambda$ the quotients of $\operatorname{mod} \Lambda$ by the ideals of morphisms factoring through a projective or injective object, respectively, and $\Omega\colon \underline{\operatorname{mod}}\Lambda\to \underline{\operatorname{mod}}\Lambda$ and $\Omega^-\colon \overline{\operatorname{mod}}\Lambda\to \overline{\operatorname{mod}}\Lambda$ the syzygy and cosyzygy functor,  respectively. The objects in $\operatorname{mod} \Lambda$ and their image in $\underline{\operatorname{mod}}\Lambda$ and $\overline{\operatorname{mod}} \Lambda$ will be denoted by the same letter. For a subcategory ${\mathcal X}$ of $\operatorname{mod} \Lambda$, we let $\underline{{\mathcal X}}$ and $\overline{{\mathcal X}}$ denote the smallest subcategories of  $\underline{\operatorname{mod}}\Lambda$ and $\overline{\operatorname{mod}} \Lambda$ which are closed under isomorphisms and contain all $X\in {\mathcal X}$.
 
To define $d$-precluster tilting subcategories, we consider the $d$-Auslander-Reiten translations defined by
\[
\tau_d := \tau \circ \Omega^{d-1}\colon \underline{\operatorname{mod}}\Lambda \to \overline{\operatorname{mod}} \Lambda \quad \text{and} \quad \tau^-_d := \tau^- \circ \Omega^{-(d-1)}\colon \overline{\operatorname{mod}}\Lambda \to \underline{\operatorname{mod}} \Lambda
\] 
where $\tau\colon \underline{\operatorname{mod}}\Lambda \to \overline{\operatorname{mod}}\Lambda$ denotes the classical Auslander-Reiten translation with quasi-inverse $\tau^-\colon \overline{\operatorname{mod}}\Lambda \to \underline{\operatorname{mod}}\Lambda$. 

\begin{Definition}[Definition 3.2 in \cite{IS18}]\label{precluster tilting}
Let ${\mathcal X}$ be an additive subcategory of $\operatorname{mod} \Lambda$. Assume ${\mathcal X}$ is closed under direct summands. We say that ${\mathcal X}$ is a $d$\emphbf{-precluster tilting subcategory} if it satisfies the following:
\begin{enumerate}
\item\label{precluster tilting:1} ${\mathcal X}$ is a generating cogenerating subcategory of $\operatorname{mod} \Lambda$;
\item\label{precluster tilting:2} $\tau_d(X)\in \overline{{\mathcal X}}$ and $\tau^-_d(X)\in \underline{{\mathcal X}}$ for all $X\in {\mathcal X}$;
\item\label{precluster tilting:3} $\operatorname{Ext}^i_{\Lambda}(X,X')=0$ for all $X,X'\in {\mathcal X}$ and $0<i<d$;
\item\label{precluster tilting:4} ${\mathcal X}$ is a functorially finite subcategory of $\operatorname{mod} \Lambda$.
\end{enumerate}
\end{Definition}

The appearance of $\tau_d$ and $\tau^-_d$ makes precluster tilting subcategories difficult to axiomatize. Luckily, criterion \ref{precluster tilting:2} in Definition \ref{precluster tilting} can be reformulated in homological terms. Our first goal is to do this. For $d>1$ such a reformulation is already known, as the following result shows. For simplicity we set
\[
{}^{\perp_d}{\mathcal X}:=\{M\in \operatorname{mod} \Lambda\mid \operatorname{Ext}^i_{\Lambda}(M,X)=0 \text{ for all } 0<i<d \text{ and }X\in {\mathcal X}\}
\]
and 
\[
{\mathcal X}^{\perp_d}:=\{M\in \operatorname{mod} \Lambda\mid \operatorname{Ext}^i_{\Lambda}(X,M)=0 \text{ for all } 0<i<d \text{ and }X\in {\mathcal X}\}.
\]



\begin{Lemma}[Proposition 3.8 part b) in \cite{IS18}]\label{reformulation of precluster tilting d>1}
Let $d>1$ be an integer. Assume ${\mathcal X}$ is an  additive subcategory of $\operatorname{mod} \Lambda$ closed under direct summands and satisfying \ref{precluster tilting:1}, \ref{precluster tilting:3}, \ref{precluster tilting:4} in Definition \ref{precluster tilting}. Then the following are equivalent:
\begin{enumerate}
\item ${\mathcal X}$ is $d$-precluster tilting;
\item ${}^{\perp_d}{\mathcal X}={\mathcal X}^{\perp_d}$.
\end{enumerate}
\end{Lemma}

We also need the following lemma which gives a simpler criterion for when ${}^{\perp_d}{\mathcal X}={\mathcal X}^{\perp_d}$.

\begin{Lemma}\label{useful lemma precluster tilting}
Let $d>1$ be a positive integer, and assume ${\mathcal X}$ is an additive subcategory of $\operatorname{mod} \Lambda$ closed under direct summands and satisfying \ref{precluster tilting:1}, \ref{precluster tilting:3}, \ref{precluster tilting:4} in Definition \ref{precluster tilting}. Assume furthermore that ${{\mathcal X}}^{\perp_d}\subseteq {}^{\perp_2}{\mathcal X}$ and ${}^{\perp_d}{\mathcal X}\subseteq{\mathcal X}^{\perp_2}$. Then ${}^{\perp_d}{\mathcal X}={\mathcal X}^{\perp_d}$.
\end{Lemma}

\begin{proof}
We prove by induction on $2\leq i\leq d$ that ${\mathcal X}^{\perp_d}\subseteq {}^{\perp_i}{\mathcal X}$. For $i=2$ this follows by assumption. Assume the claim holds for $2\leq i<d$, and we want to show that it holds for $i+1$. Let $M\in {\mathcal X}^{\perp_d}$, choose a right ${\mathcal X}$-approximation $f\colon X\to M$, and let $M'=\operatorname{Ker} f$. Applying $\operatorname{Hom}_{\Lambda}(X',-)$ with $X'\in {\mathcal X}$ to the exact sequence 
\begin{equation}\label{equation reference}
0\to M'\to X\xrightarrow{f}M\to 0
\end{equation} gives a long exact sequence
\begin{multline*}
0\to \operatorname{Hom}_{\Lambda}(X',M')\to \operatorname{Hom}_{\Lambda}(X',X)\xrightarrow{f\circ -} \operatorname{Hom}_{\Lambda}(X',M)\to \operatorname{Ext}^1_{\Lambda}(X',M') \\
 \to \operatorname{Ext}^1_{\Lambda}(X',X) \to \cdots \to \operatorname{Ext}^{j-1}_{\Lambda}(X',M)\to \operatorname{Ext}^j_{\Lambda}(X',M') 
 \to \operatorname{Ext}^j_{\Lambda}(X',X)\to \cdots
\end{multline*}
Since $f$ is a right ${\mathcal X}$-approximation, it follows that the map 
\[
\operatorname{Hom}_{\Lambda}(X',X)\xrightarrow{f\circ -} \operatorname{Hom}_{\Lambda}(X',M)
\]
is an epimorphism. Also, since $\operatorname{Ext}^j_{\Lambda}(X',X)=0$ for $0<j<d$ by Definition \ref{precluster tilting} \ref{precluster tilting:3} and $\operatorname{Ext}^j_{\Lambda}(X',M)=0$ for $0<j<d$ by assumption, it follows that $\operatorname{Ext}^j_{\Lambda}(X',M')=0$ for $0<j<d$. Hence $M'\in {\mathcal X}^{\perp_d}$, and therefore $M'\in {}^{\perp_i}{\mathcal X}$ by induction hypothesis. Now applying $\operatorname{Hom}_{\Lambda}(-,X')$ to \eqref{equation reference} and considering the long exact sequence we get
\[
\operatorname{Ext}^{i}_{\Lambda}(M,X')\cong \operatorname{Ext}^{i-1}_{\Lambda}(M',X')=0
\]
since $\operatorname{Ext}^i_{\Lambda}(X,X')=0=\operatorname{Ext}^{i-1}_{\Lambda}(X,X')$. This shows that $M\in {}^{\perp_{i+1}}{\mathcal X}$. Therefore, by induction we get that ${\mathcal X}^{\perp_d}\subseteq {}^{\perp_d}{\mathcal X}$. The inclusion ${}^{\perp_d}{\mathcal X}\subseteq{\mathcal X}^{\perp_d}$ is proved dually. Combining the inclusions, we get that ${}^{\perp_d}{\mathcal X}={\mathcal X}^{\perp_d}$, which proves the claim.
\end{proof}

Note that Lemma \ref{reformulation of precluster tilting d>1} only holds when $d>1$, so we still need a homological reformulation of  Definition \ref{precluster tilting} \ref{precluster tilting:2} when $d=1$. This is done by the following result.

\begin{Theorem}\label{reformulation precluster tilting}
Let $d$ be a positive integer, and assume ${\mathcal X}$ is an additive subcategory of $\operatorname{mod} \Lambda$ closed under direct summands and satisfying \ref{precluster tilting:1}, \ref{precluster tilting:3} and \ref{precluster tilting:4} in Definition \ref{precluster tilting}. The following are equivalent:
\begin{enumerate}
\item\label{reformulation precluster tilting:1} ${\mathcal X}$ is a $d$-precluster tilting subcategory;
\item\label{reformulation precluster tilting:2} For any exact sequence in $\operatorname{mod} \Lambda$
\[
0\to M'\xrightarrow{f_{d+1}} X_d\xrightarrow{f_d} \cdots \xrightarrow{f_3} X_2\xrightarrow{f_2} X_1\xrightarrow{f_1} M\to 0
\]
with $X_i\in {\mathcal X}$ for $1\leq i\leq d$, the following hold:
\begin{enumerate}
\item If the induced map $X_i\to \operatorname{im} f_i$ is a right ${\mathcal X}$-approximation for all $1\leq i\leq d$, then $f_{d+1}\colon M'\to X_d$ is a left ${\mathcal X}$-approximation;
\item If the induced map $\operatorname{im} f_i\to X_{i-1}$ is a left ${\mathcal X}$-approximation for all $2\leq i\leq d+1$, then $f_1\colon X_1\to M$ is a right ${\mathcal X}$-approximation.
\end{enumerate}
\end{enumerate}
\end{Theorem} 

\begin{proof}
We prove the cases $d>1$ and $d=1$ separately. First assume $d>1$ and that ${\mathcal X}$ is a $d$-precluster tilting subcategory. Let
\[
0\to M'\xrightarrow{f_{d+1}} X_d\xrightarrow{f_d} \cdots \xrightarrow{f_3} X_2\xrightarrow{f_2} X_1\xrightarrow{f_1} M\to 0
\]
be an exact sequences such that $X_i\in {\mathcal X}$ and $X_i\to \operatorname{im} f_i$ is a right ${\mathcal X}$-approximation for all $1\leq i\leq d$. Applying $\operatorname{Hom}_{\Lambda}(X,-)$ with $X\in {\mathcal X}$ to the exact sequence
\[
0\to \operatorname{im} f_{i+1}\to X_{i}\to \operatorname{im} f_{i}\to 0
\] 
and using that $\operatorname{Ext}^j_{\Lambda}(X,X_{i})=0$ for $1\leq j<d$, we get that 
\[
\operatorname{Ext}^1_{\Lambda}(X,\operatorname{im} f_{i+1})=0 \quad \text{and} \quad \operatorname{Ext}^{j}_{\Lambda}(X,\operatorname{im} f_{i+1})\cong \operatorname{Ext}^{j-1}_{\Lambda}(X,\operatorname{im} f_{i})
\]
for $1\leq i\leq d$ and $2\leq j<d$. Hence, we have that
\[
\operatorname{Ext}^j_{\Lambda}(X,\operatorname{im} f_d)\cong \operatorname{Ext}^{j-1}_{\Lambda}(X,\operatorname{im} f_{d-1})\cong \cdots \cong \operatorname{Ext}^1_{\Lambda}(X,\operatorname{im} f_{d-j+1})=0
\]
for $0<j<d$. This shows that $\operatorname{im} f_d\in {\mathcal X}^{\perp_d}$, so $\operatorname{im} f_{d}\in {}^{\perp_2}{\mathcal X}$ by Lemma \ref{reformulation of precluster tilting d>1}, and therefore $f_{d+1}\colon M'\to X_d$ is a left ${\mathcal X}$-approximation. This together with the dual argument shows the implication \ref{reformulation precluster tilting:1} $\implies$ \ref{reformulation precluster tilting:2} for $d>1$.

Conversely, assume $d>1$ and that ${\mathcal X}$ satisfies  part \ref{reformulation precluster tilting:2} of the theorem. Let $M'\in {\mathcal X}^{\perp_d}$, and choose a right ${\mathcal X}$-approximation $X_d\to M'$ and an exact sequence
\[
0\to M'\xrightarrow{} X_{d-1}\xrightarrow{f_{d-1}}X_{d-2}\xrightarrow{f_{d-2}}\cdots \xrightarrow{f_2}X_1\xrightarrow{f_1}M\to 0
\]
Since $\operatorname{Ext}^i_{\Lambda}(X,M')=0$ and $\operatorname{Ext}^i_{\Lambda}(X,X')=0$ for $0<i<d$ and $X,X'\in {\mathcal X}$, it follows that the sequence
\[
\operatorname{Hom}_{\Lambda}(X,X_{d-1})\xrightarrow{f_{d-1}\circ -} \cdots \xrightarrow{f_2\circ -} \operatorname{Hom}_{\Lambda}(X,X_1) \xrightarrow{f_1\circ -} \operatorname{Hom}_{\Lambda}(X,M)\to 0
\]
is exact for $X\in {\mathcal X}$. Hence the canonical map $X_i\to \operatorname{im} f_i$ is a right ${\mathcal X}$-approximation for $1\leq i\leq d-1$. Therefore, by assumption we get that the map $\operatorname{Ker} f_d\to X_d$ is a left ${\mathcal X}$-approximation, where $f_d$ is the composite $X_d\to M'\to X_{d-1}$. Hence, applying $\operatorname{Hom}_{\Lambda}(-,X)$ with $X\in {\mathcal X}$ to the exact sequence
\[
0\to \operatorname{Ker} f_d\to X_d\to M'\to 0
\]
we get that $\operatorname{Ext}^1_{\Lambda}(M',X)=0$ so $M'\in {}^{\perp_2}{\mathcal X}$. Since $M'\in {\mathcal X}^{\perp_d}$ was arbitrary, this shows that ${\mathcal X}^{\perp_d}\subseteq {}^{\perp_2}{\mathcal X}$. The inclusion ${}^{\perp_d}{\mathcal X}\subseteq {\mathcal X}^{\perp_2}$ is proved dually, and the fact that ${\mathcal X}$ is $d$-precluster tilting follows from Lemma \ref{reformulation of precluster tilting d>1} and Lemma \ref{useful lemma precluster tilting}.

Now we assume $d=1$ and ${\mathcal X}$ is a $1$-precluster tilting subcategory. Let 
\[
0\to M\xrightarrow{f}X\xrightarrow{g}M'\to 0
\]
 be an exact sequence with $f$ a left ${\mathcal X}$-approximation. Then since $\tau(X')\in \overline{{\mathcal X}}$ for all $X'\in \underline{{\mathcal X}}$, all morphisms $M\to \tau(X')$ will factor through $f$. Hence by \cite[Chapter IV, Corollary 4.4]{ARS95} all morphisms $X'\to M'$ with $X'\in {\mathcal X}$ will factor through $g$. Therefore $g$ is a right ${\mathcal X}$-approximation. Together with the dual argument this shows that $1$-precluster tilting implies condition \ref{reformulation precluster tilting:2} in the theorem. 
 
Finally, assume $d=1$ and that condition \ref{reformulation precluster tilting:2} holds for ${\mathcal X}$. Assume furthermore that there exists an indecomposable module $X\in {\mathcal X}$ for which $Y=\tau(X)\notin \overline{{\mathcal X}}$. By abuse of notation we let $\tau(X)$ denote the indecomposable module in $\operatorname{mod} \Lambda$ corresponding $Y$ in $\overline{\operatorname{mod}}\Lambda$. Consider the commutative diagram with exact rows
\begin{equation*}
\begin{tikzcd}
0\arrow{r}{} & \tau(X) \arrow{r}{} \arrow{d}{1} & E \arrow{r}{} \arrow[dashed]{d}{} & X \arrow{r}{} \arrow[dashed]{d}{} & 0\\
0\arrow{r} & \tau(X) \arrow{r}{} & X'\arrow{r}{} & M \arrow{r}{} & 0 
\end{tikzcd}
\end{equation*}
where the top row is an almost split sequence, and where $\tau(X)\to X'$ is a left ${\mathcal X}$-approximation with cokernel $M$. Since $\tau(X)\notin {\mathcal X}$, the morphism $\tau(X)\to X'$ is not a split monomorphism, and therefore it factors through $\tau(X)\to E$. Hence we obtain vertical maps $E\to X'$ and $X\to M$ making the diagram commute. By assumption we have that $X'\to M$ is a right ${\mathcal X}$-approximation, and hence $X\to M$ factors through $X'\to M$. Since the rightmost square is a pushout square, it follows that $E\to X$ is a split epimorphism. Therefore the sequence $0\to \tau(X)\to E\to X\to 0$ must be split, which is a contradiction. This shows that $\tau(X)\in {\mathcal X}$. The implication $X\in \overline{{\mathcal X}}\implies \tau^-(X)\in \underline{{\mathcal X}}$ is proved dually.
\end{proof}

Motivated by this, we define $d$-precluster tilting subcategories for arbitrary abelian categories. By Theorem \ref{reformulation precluster tilting} it coincides with the classical definition for ${\mathcal A}=\operatorname{mod} \Lambda$.

\begin{Definition}\label{precluster tilting abelian}
Let ${\mathcal X}$ be an additive subcategory of an abelian category ${\mathcal A}$. Assume ${\mathcal X}$ is closed under direct summands. We say that ${\mathcal X}$ is a $d$\emphbf{-precluster tilting subcategory} if it satisfies the following:
\begin{enumerate}
\item\label{precluster tilting abelian:1} ${\mathcal X}$ is a generating cogenerating subcategory of ${\mathcal A}$;
\item\label{precluster tilting abelian:2} For any exact sequence in ${\mathcal A}$
\[
0\to A'\xrightarrow{f_{d+1}} X_d\xrightarrow{f_d} \cdots \xrightarrow{f_3} X_2\xrightarrow{f_2} X_1\xrightarrow{f_1} A\to 0
\]
with $X_i\in {\mathcal X}$ for $1\leq i\leq d$, the following hold:
\begin{enumerate}
\item If the induced map $X_i\to \operatorname{im} f_i$ is a right ${\mathcal X}$-approximation for all $1\leq i\leq d$, then $f_{d+1}\colon A'\to X_d$ is a left ${\mathcal X}$-approximation;
\item If the induced map $\operatorname{im} f_i\to X_{i-1}$ is a left ${\mathcal X}$-approximation for all $2\leq i\leq d+1$, then $f_1\colon X_1\to A$ is a right ${\mathcal X}$-approximation.
\end{enumerate}
\item\label{precluster tilting abelian:3} $\operatorname{Ext}^i_{{\mathcal A}}(X,X')=0$ for all $X,X'\in {\mathcal X}$ and $0<i<d$;
\item\label{precluster tilting abelian:4} ${\mathcal X}$ is a functorially finite subcategory of ${\mathcal A}$.
\end{enumerate}
\end{Definition}

We now introduce the necessary axiom to capture Definition \ref{precluster tilting abelian} \ref{precluster tilting abelian:2}.

\begin{enumerate}[label=(A4.$d$)]
\item\label{A4} Consider a sequence
\[
X_{d+1}\xrightarrow{f_{d+1}}X_d\xrightarrow{f_d}\cdots \xrightarrow{f_3}X_2\xrightarrow{f_2}X_1\xrightarrow{f_1}X_0\xrightarrow{f_0}X_{-1}
\]
with $f_{i+1}$ a weak kernel of $f_i$ for all $0\leq i\leq d$. Then $f_{d+1}$ is a weak cokernel.
\end{enumerate}

\begin{Theorem}\label{axiomatization precluster tilting}
Let ${\mathcal X}$ be a generating cogenerating functorially finite subcategory of an abelian category ${\mathcal A}$ satisfying $\operatorname{Ext}^i_{{\mathcal A}}(X,X')=0$ for all $X,X'\in {\mathcal X}$ and $0<i<d$. Then ${\mathcal X}$ satisfies \ref{A4} and \ref{A4}\ensuremath{^{\mathrm{op}}}\text{ } if and only if it is $d$-precluster tilting.
\end{Theorem}

\begin{proof}
This follows immediately from the fact that a  map $X\xrightarrow{f}X'$ is a weak kernel or weak cokernel if and only if the projection $X\to \operatorname{im} f$ is a right ${\mathcal X}$-approximation or the inclusion $\operatorname{im} f\to X'$ is a left ${\mathcal X}$-approximation, respectively.
\end{proof}


