\section{Introduction}
	
	In his highly influential work \cite{tarski}, Tarski proved that the complete theory $\Trcf$ of the real closed field $\R$ in the language of ordered rings $\Lor=\{+,-,\cdot,0,1,<\}$ is decidable, by presenting an explicit quantifier-elimination algorithm for $\Trcf$. As a result, he noted that the $\Lor$-structure $(\R,+,-,\cdot,0,\allowbreak1,\allowbreak<)$ is -- in modern terminology -- o-minimal, i.e.\ any unary definable subset of this structure is a finite union of points and open intervals. In the same work, Tarski asked whether decidability can also be obtained for the complete theory $\Texp$ of the real exponential field $\Rexp = (\R,+,-,\cdot,0,1,\allowbreak <,\exp)$, where $\exp$ denotes the standard exponential function $x\mapsto \mathrm{e}^x$ on $\R$. While this question is open to the date, considerable progress has been made since the 1990s: In \cite{wilkie}, Wilkie proved that $\Rexp$ is model complete, and thus this structure is o-minimal (cf.\ also van den Dries, Macintyre and Marker~\cite{dries2}). Bulding on this result, Macintyre and Wilkie \cite{macintyre} showed that under the assumption of the real version of Schanuel's Conjecture below, $\Texp$ is decidable.
	\begin{schanuel}\thlabel{schnauel}
		Let $n\in\N$ and let $\alpha_1,\ldots,\alpha_n\in \R$ be $\Q$-linearly independent. Then the transcendence degree of 		$\Q(\alpha_1,\ldots,\alpha_n,\mathrm{e}^{\alpha_1},\ldots,\mathrm{e}^{\alpha_n})$ over $\Q$ is at least $n$.
	\end{schanuel}
	In general, an ordered exponential field $\KK=(K,+,-,\cdot,0,1,<,E)$ is an expansion of an ordered field $(K,+,-,\cdot,0,1,<)$ by an exponential $E$, i.e.\ an order-preserving isomorphism from the ordered additive group $(K,+,0,<)$ to the ordered multiplicative group $(K^{>0},\cdot,1,<)$. We denote the corresponding first-order language $\Lor\cup \{E\}$, where $E$ is a unary function symbol, by $\Lexp$. Following the terminology of Krapp~\cite{krappla}, we call an ordered exponential field $\KK$ an $\EXP$-field if its exponential satisfies the first-order $\Lexp$-sentence expressing the differential equation $E'=E$. While $\Rexp$ and, more generally, any model of $\Texp$ is an o-minimal $\EXP$-field, it is not known whether \emph{every} o-minimal $\EXP$-field is already a model of $\Texp$. In \cite[page~153~f.]{dries4}, van den Dries stated a transfer conjecture for o-minimal structures, which specialises to $\EXP$-fields as follows (cf.\ Krapp~\cite[Observation~4.54]{krappdiss} for details):
	\begin{transfer}\thlabel{transfer}
		Any o-minimal $\EXP$-field is elementarily equivalent to $\Rexp$.
	\end{transfer}
	Berarducci and Servi~\cite{berarducci} showed that the Transfer Conjecture would imply that $\Texp$ is decidable. This result motivated the study of o-minimal $\EXP$-fields in connection to the Transfer Conjecture.
	
	One approach towards proving the Transfer Conjecture is to show that the (unique) prime model of real exponentiation, i.e.\ the prime model of $\Texp$, elementarily embeds into any o-minimal $\EXP$-field, which is also the motivating question for this line of research.
	
	\begin{question}[Main question]\thlabel{qu:main}
		Does the prime model of $\Texp$ elementarily embed into any o-minimal $\EXP$-field?
	\end{question}
	
	Note that \Autoref{qu:main} can be answered positively if and only if the Transfer Conjecture holds: given an o-minimal $\EXP$-field $\mathcal{K}$, if $\KK$ is a model of $\Texp$, then the prime model $\mathcal{P}$ of $\Texp$ elementarily embeds into $\KK$, and if $\mathcal{P}$ elementarily embeds into $\KK$, then $\KK$ is already a model of $\Texp$.
	It is thus natural to attempt answering \Autoref{qu:main} affirmatively only under the assumption of Schanuel's Conjecture, as a positive answer would imply the decidability of $\Texp$ via the Transfer Conjecture. 
	
	While this note does not provide a complete answer to \Autoref{qu:main}, we show in \Autoref{cor:scprimeemb} that under the assumption of Schanuel's Conjecture, the prime model of $\Texp$ embeds into any o-minimal $\EXP$-field (however, this embedding might not necessarily be elementary). This result is deduced from the technical heart of this paper, \Autoref{thm:eclembeddable}, which is a more general \emph{unconditional} embeddability result for exponential algebraic closures of $\Z$ within one o-minimal $\EXP$-field into another. Finally, we complete this note by discussing in \Autoref{sec:residue} how \Autoref{qu:main} relates to 
	another approach towards the Transfer Conjecture via elementary embeddability of the residue exponential field (cf.\ Krapp~\cite[Proposition~3.6]{krappla}).
	
	\subsection{Notation and terminology}

		More background on the model theoretic notation and terminology we use can be found in Marker~\cite{marker}. If it is clear from the context, then the $\Lor$-structure of an ordered field $(K,+,-,\cdot,0,1,<)$ is simply denoted by $K$ and the $\Lexp$-structure of an ordered exponential field $(K,+,-,\cdot,0,1,<,E)$ simply by $(K,E)$.
		Given an ordered exponential field $(K,E)$, we say that a subfield $F\subseteq K$ is \textbf{exponentially closed} in $(K,E)$ if the restriction $E|_{F}$ is an exponential on $F$. In this case, if no confusion is likely to arise, then we also denote $E|_F$ simply by $E$ and write $(F,E)\subseteq (K,E)$ for the corresponding embedding of ordered exponential fields.  
		For any structure $\mathcal{M}=(M,\ldots)$, its complete theory is denoted by $\Th(\MM)$ and its existential theory by $\Th_\exists(\MM)$. For instance, $\Texp = \Th(\Rexp)$ denotes the theory of real exponentiation and $\Th_\exists(\Rexp)$ denotes the existential theory of real exponentiation. 
		The definable closure of a set $C$ in $\MM$ is denoted by $\dcl(C;\MM)$. We say that a set is definable if it is definable \emph{with} parameters, and we say that it is $A$-definable if we wish to specify the set of possible parameters $A$. 
		Variable tuples are denoted by $\ul{x}$, and we only specify their length if it is of importance. 
		If $\ul{x}=(x_1,\ldots,x_n)$, we write $E(\ul{x})$ for the tuple $(E(x_1),\ldots,E(x_n))$. 
		We denote by $\N$ the set of natural numbers \emph{without} $0$.
		\textbf{Throughout the rest of this note, let $(K,E)$ denote an o-minimal $\EXP$-field.}
		
	\begin{remark}
		The Transfer Conjecture motivates us to entirely focus on o-minimal ordered exponential fields $(K,E)$ whose exponential satisfies the differential equation $E'=E$.
		For a general o-minimal ordered exponential field $(F,e)$, it is ensured by o-minimality that $e$ is differentiable in all but finitely many points in $F$ (cf.\ van den Dries~\cite[Chapter~7, Proposition~2.5]{dries}). Due to a general property of exponentials, this implies that $e$ satisfies the differential equation $e'=ae$ for some positive $a\in F$ (cf.~Krapp~\cite[Corollary~2.14]{krappdiss}). We can thus change the base of exponentiation, i.e.\ set $E\colon F\to F^{>0}, x\mapsto e(a^{-1}x)$, to obtain an exponential $E$ on $F$ with $E'=E$, that is, $(F,E)$ becomes an o-minimal $\EXP$-field. By this base change argument, most results in this note, in particular \Autoref{thm:eclembeddable}, can also be generalised to \emph{any} o-minimal ordered exponential field, e.g.\ o-minimal exponential fields $(F,e)$ with the property $e(1)=2$ (base-$2$ exponentiation as done in Carl and Krapp~\cite[Section~4.2]{carlkrapp}).
	\end{remark}

	\section{Exponential algebraic closures}
	
	In the following, we mostly follow the terminology of Macintyre~\cite{macintyre2} and Kirby~\cite{kirby} adjusted to our context. 
	For a subring $A\subseteq K$, we 
	say that $A$ is an \textbf{$E$-ring} if it is closed under $E$, i.e.\ for any $a\in A$ also $E(a)\in A$. 
	The smallest $E$-ring in $K$ is denoted by $\Z^{E}$. 
	If $A$ is an $E$-ring, then we	
	denote by $A[x_1,\ldots,x_n]^{E}$ the ring of \textbf{exponential polynomials} in $n$ variables, that is, the smallest ring containing the polynomial ring $A[x_1,\ldots,x_n]$ such that for any $y\in A[x_1,\ldots,x_n]^{E}$ also $E(y)\in A[x_1,\ldots,x_n]^{E}$. (We refer the reader to \cite[Section~1.7]{macintyre2} for an explicit construction of $A[x_1,\ldots,x_n]^{E}$.)
	
	
	\begin{definition}
		\begin{enumerate}[label = (\roman*)]
			\item Let $n \in \N$ and let $A \subseteq K$ be an $E$-ring. 
			Moreover, let $f_1,\ldots,f_n \in A[x_1,\ldots,x_n]^{E}$ and let  $|J_{f_1,\ldots,f_n}(\ul{x})|$ denote the determinant of the Jacobian matrix $$J_{f_1,\ldots,f_n}(\ul{x}) = \brackets{\begin{array}{ccc}
					\frac{\partial f_1}{\partial x_1}(\ul{x})& \ldots & \frac{\partial f_1}{\partial x_n}(\ul{x})\\
					\vdots & & \vdots\\
					\frac{\partial f_n}{\partial x_1}(\ul{x})& \ldots & \frac{\partial f_n}{\partial x_n}(\ul{x})
			\end{array}}\!\!.$$
			Then the \textbf{Khovanskii system $S(\ul{x})$} of $f_1,\ldots,f_n$ over $A$ is the system of equations and inequations 
			$$f_1(\ul{x}) = \ldots = f_n(\ul{x}) = 0 \text{ and }|J_{f_1,\ldots,f_n}(\ul{x})| \neq 0.$$
		
			\item Let $B$ be a subset of $K$ and let $A\subseteq K$ be the smallest $E$-ring containing $B$. 
			An element $a_1 \in K$ is said to be \textbf{exponentially algebraic} over $B$ if for some $n \in \N$ there exist a Khovanskii system $S(x_1,\ldots,x_n)$ over $A$ and $a_2,\ldots,a_n \in K$ such that $(a_1,\ldots,a_n)$ solves this system. The set of all elements of $K$ that are exponentially algebraic over $B$ is called the \textbf{exponential algebraic closure} of $B$ in $(K,E)$ and is denoted by $\ecl^{E}_K(B)$.
		\end{enumerate}
	\end{definition}
	
	Note that the condition on the determinant of the Jacobian to be non-zero implies that Khovanskii systems only have isolated solutions. This is a consequence of the Implicit Function Theorem in o-minimal structures (cf.~van den Dries~\cite[page~113]{dries}).
	
	\begin{lemma}\thlabel{lem:expclosed}
		Let $B$ be a subset of $K$. Then $\ecl^{E}_K(B)$ is a subfield of $K$ that is exponentially closed in $(K,E)$.
	\end{lemma}

	\begin{proof}
		Let $F=\ecl^{E}_K(B)$ and let $A\subseteq K$ be the smallest $E$-ring containing $B$.
		By Kirby~\cite[Lemma~3.3]{kirby} and Macintyre~\cite[Lemma~21]{macintyre2}, $F$ is an $E$-ring and a field. Thus, $E|_F$ defines an order-preserving homomorphism from $(F,+,0,<)$ to $(F^{>0},\cdot,1,<)$. Since $E$ is injective, so is its restriction to $F$. To verify that $E|_F$ is also surjective, let $a_1\in F^{>0}$ and fix $n \in \N$, $f_1(\ul{x}),\ldots, f_n(\ul{x})\in A[x_1,\ldots,x_n]^{E}$ and $a_2,\ldots,a_n\in K$ such that $\ul{a}=(a_1,\ldots,a_n)$ solves the Khovanskii system $S(\ul{x})$ of $f_1,\ldots,f_n$. Set $f_0(y,x_1)=E(y)-x_1$. Then there is some $d\in K$ with $E(d)=a_1$, i.e.\ $f_0(d,a_1)=0$. Note that
		$|J_{f_0,\ldots,f_n}(y,\ul{x})|=E(y)|J_{f_1,\ldots,f_n}(\ul{x})|$. Hence, $(d,a_1,\ldots,a_n)$ solves the Khovanskii system $\widetilde{S}(y,\allowbreak \ul{x})$ of $f_0,\ldots,f_n$. This yields $y\in F$, as required.
	\end{proof}
	
	We now investigate the connection between the definable closure and the exponential algebraic closure in o-minimal $\EXP$-fields. In particular, we show that for any model of real exponentiation $(K,E)$ we have that $(\ecl^{E}_K(\emptyset),E)$ is the prime model of $\Texp$ (see \Autoref{prop:primekhovanskii}).
	
	\begin{lemma}\thlabel{lem:khovanskiifinite}
		Let $A\subseteq K$ be an $E$-ring and let $S(\ul{x})$ be a Khovanskii system over $A$. Then $S$ only has finitely many solutions in $K$.
	\end{lemma}
	
	\begin{proof}
		Since $(K,E)$ is o-minimal, $S(\ul{x})= S(x_1,\ldots,x_n)$ defines finitely many connected components in $K^n$ (cf.~van den Dries~\cite[\S~3.2]{dries}). By the condition that the Jacobian matrix of the system of exponential polynomial equations in $S$ has to be non-singular, all solutions of $S(\ul{x})$ in $K$ are isolated. Thus, $S(\ul{x})$ defines a set consisting of finitely many isolated points in $K^n$. In particular, $S(\ul{x})$ only has finitely many solutions.
	\end{proof}
	
	\begin{lemma}\thlabel{prop:dclexpecl2}
		Let $B$ be a subset of $K$. Then $\ecl^{E}_K(B) \subseteq \dcl(B;\allowbreak\Kexp)$.
	\end{lemma}
	
	\begin{proof}
		Let $A\subseteq K$ be the smallest $E$-ring containing $B$, 
		let $a_1 \in \ecl^{E}_K(A)$ and let $S(\ul{x})$ be a Khovanskii system over $A$ such that for some $a_2,\ldots,a_n \in K$ the tuple $(a_1,\ldots,a_n)$ solves $S$. Let $C\subseteq K^n$ be the $A$-definable set of all $\ul{c}\in K^n$ with $\Kexp \models S(\ul{c})$. \Autoref{lem:khovanskiifinite} shows that $C$ is finite.
		
		Let $C' \subseteq K$ be the projection of $C$ onto the first coordinate. Then $C'$ is $A$-definable as well as finite and $a_1 \in C'$. Suppose that $a_1$ is the $k$-th element of $C'$ in the ordering on $C'$ induced by the ordering on $K$. Then there exists an $\Lexp$-formula $\varphi(x)$ with parameters in $A$ expressing that $x$ is the $k$-th element of $C'$, and $\Kexp \models \varphi(a_1) \wedge \exists! x \ \varphi(x)$. Hence, $a_1 \in \dcl(A;\allowbreak\Kexp)$. It remains to note that any element of $A$ is $B$-definable in $(K,E)$ and thus $\dcl(A;\allowbreak\Kexp)=\dcl(B;\allowbreak\Kexp)$.
	\end{proof}
	
	Recall that due to definable Skolem functions in o-minimal expansions of ordered groups, for any subset $B$ of $K$ we have that  $(\dcl(B;\Kexp),E)\preceq (K,E)$, i.e.\ $\dcl(B;\Kexp)$ is the domain of an elementary substructure of $(K,E)$. Hence, $(\dcl(\emptyset;\Kexp),E)$ is the unique prime model of $\Th(K,E)$ (cf.\ Krapp~\cite[Proposition~4.75]{krappdiss} for further details).
	
	If $(K,E)$ is not only assumed to be an o-minimal $\EXP$-field but already a model of real exponentiation, then we can strengthen the conclusion of \Autoref{prop:dclexpecl2} to $\ecl^{E}_K(B) = \dcl(B;\allowbreak\Kexp)$. 
	The following result is mentioned by Macintyre in \cite[Theorem~22]{macintyre2}, who attributes it to Wilkie~\cite{wilkie}. For the convenience of the reader, we give a proof based on Macintyre and Wilkie~\cite[Theorem~2.1]{macintyre}.
	
	
	\begin{proposition}\thlabel{prop:dclexpecl}\thlabel{prop:dclexpecl3}\thlabel{prop:primekhovanskii}
		Let $(K,E) \models \Texp$ and let $B$ be a subset of $K$. Then $$\dcl(B;(K, E)) = \ecl_K^{E}(B).$$ 
		In particular, $(\ecl_K^{E}(B),E)\preceq (K,E)$ and  $(\ecl_K^{E}(\emptyset),E)$ is the prime model of $\Texp$.
	\end{proposition}
	
	\begin{proof}
		By \Autoref{prop:dclexpecl2}, we only have to show $\dcl(B;(K, E)) \subseteq \ecl^{E}_K(B)$.
		Let $A\subseteq K$ be the smallest $E$-ring containing $B$, and let $b \in \dcl(A;\allowbreak (K, E)) =\dcl(B;(K, E))$. Then there exists a function $g$ that is $\emptyset$-definable in $(K,E)$ such that $g(\ul{a}) = b$ for some $\ul{a} \in A$. Since $(K,E)$ is model complete, there is some atomic $\Lexp$-formula $\varphi(\ul{v},\ul{w})$ such that $\exists \ul{w}\ \varphi(\ul{v},\ul{w})$ defines $g$.
		After replacing the variables, this defining formula is equivalent to one of the form
		$$\exists \ul{z} \ p(\ul{x},{y},\ul{z}) = 0,$$
		where $p(\ul{x},{y},\ul{z})=q(\ul{x},{y},\ul{z},E(\ul{x}),E({y}),E(\ul{z}))$ for some $q\in \Z[\ul{x},{y},\ul{z},\ul{x}',\allowbreak {y}',\ul{z}']$
		(cf.\ e.g.\ Servi~\cite[Proposition~4.5.4]{servi} for details).
		Thus,  $$(K,E) \models \forall \ul{x} \forall y \ (g(\ul{x}) = y \leftrightarrow \exists \ul{z} \ p(\ul{x},{y},\ul{z}) = 0 ).$$
		In particular, we have
		\begin{align}\Kexp \models \exists\ul{z} \ p(\ul{a},{b},\ul{z}) = 0 \wedge \exists!y \exists \ul{z} \ p(\ul{a},{y},\ul{z}) = 0.\label{eq:khovanskiifct}\end{align}
		
		Let $f({y},\ul{z}) = p(\ul{a},{y},\ul{z}) \in A[{y},\ul{z}]^{E} = A[y,z_1,\ldots,z_\ell]^{E}$.
		We now apply \cite[Theorem~2.1]{macintyre} to the setting $n=r=\ell+1$, $k=\ecl^{E}_K(A)$ and $S=V=\{\underline{\alpha}\in K^n\mid f(\ul{\alpha})=0\}$: 
		There exist $f_0,\ldots,f_\ell \in k[{y},\ul{z}]^{E}$
	
		and $c,\ul{d} \in K$ such that $$f(c,\ul{d}) = f_0(c,\ul{d}) = \ldots = f_\ell(c,\ul{d}) = 0 \text{ and }\abs{J_{f_0,\ldots,f_\ell} (c,\ul{d})} \neq 0.$$
		This means that $(c,\ul{d})$ solves the Khovanskii system of $f_0,\ldots,f_\ell$, whence $c \in \ecl^{E}_K(k)$. By (\autoref{eq:khovanskiifct}) we already have $c = b$.
		In conclusion, $b\in \ecl^{E}_K(k)=\ecl^{E}_K(\ecl^{E}_K(A))=\ecl^{E}_K(A)=\ecl^{E}_K(B)$ (by Kirby~\cite[Lemma~3.3\ (3)]{kirby}).
	\end{proof}
	
	

	

	
	The aim of this note is to show that, under the assumption of Schanuel's Conjecture, the prime model of $\Texp$ is embeddable into any o-minimal $\EXP$-field (see \Autoref{cor:scprimeemb}). This result will be deduced from the more general \Autoref{thm:eclembeddable}. The proof of \Autoref{thm:eclembeddable} relies on K\H{o}nig's Infinity Lemma, which we introduce in the following.
	

		Let $V$ be a set and let $\mathcal{E} \subseteq \mathcal{P}(V)$ such that $|e| \in \{1,2\}$ for any $e \in E$.
		We call the pair $C = (V,\mathcal{E})$ a \textbf{graph} on $V$. The elements of $V$ are called the \textbf{vertices} of $C$ and the elements of $\mathcal{E}$ are called the \textbf{edges} of $C$. For any $x,y \in V$ we say that $x$ and $y$ are \textbf{neighbours} if $\{x,y\} \in \mathcal{E}$.
		Suppose that $V$ is infinite and let $C$ be a graph on $V$. Let $(v_n)_{n \in \N\cup\{0\}}$ be a sequence of pairwise distinct elements in $V$. We say that $v_0v_1\ldots$ is a \textbf{ray} in $C$ if for any $n \in \N_0$ we have $\{v_n,v_{n+1}\} \in \mathcal{E}$.
		The proof of the following can be found in Diestel~\cite[Lemma~8.1.2]{diestel}.
	
	\begin{konig}
		Let $(V_n)_{n\in \N\cup\{0\}}$ be an infinite family of disjoint non-empty finite sets and let $C$ be a graph on their union. Suppose that for any $n \in \N$, any vertex $x \in V_n$ has a neighbour $y \in V_{n-1}$. Then $C$ contains a ray $v_0v_1\ldots$ with $v_n \in V_n$ for any $n \in \N\cup\{0\}$.
	\end{konig}
	
	\begin{theorem}\thlabel{thm:eclembeddable}
		Let $(K_1,E_1)$ and $(K_2,E_2)$ be o-minimal $\EXP$-fields. Suppose that $(K_2,E_2) \models \Th_\exists(K_1,E_1)$. Then there exists an $\Lexp$-em\-bed\-ding of the ordered exponential field $(\ecl^{E_1}_{K_1}(\emptyset),E_1)$ into $(K_2,E_2)$.
	\end{theorem}
	
	\begin{proof}
		Let $A = \ecl^{E_1}_{K_1}\!(\emptyset)$. \Autoref{lem:expclosed} shows that $(F,E_1)$ is an ordered exponential field. 		
		Since there are only countably many Khovanskii systems with coefficients in $\Z^{E}$ and by \Autoref{lem:khovanskiifinite} each Khovanskii system only has finitely many solutions in $K_1$, the set $A$ is countable. Let $A = \{a_1,a_2,\ldots\}$ be an enumeration of $A$. Let $(c_k)_{k\in \N}$ be new constant symbols and expand $\Lexp$ to $\LL= \Lexp \cup \{c_k \mid k\in \N\}$. Now $(K_1,E_1)$ expands to an $\LL$-structure $\KK_1$ by interpreting $c_k^{\KK_1} = a_k$ for any $k \in \N$. Consider the atomic diagram $\Diag(A,E_1)$ of $(A,E_1)$ given by the set of all atomic $\LL$-formulas and negations of such satisfied in $\KK_1$. 
		By the Diagram Lemma (cf.\ Marker~\cite[Lemma~2.3.3~i)]{marker}), we need to interpret each $c_k$ in $(K_2,E_2)$ such that the $\LL$-expansion $$\KK_2 = \big(K_2,E_2,\big(c_k^{\KK_2}\big)_{k \in \N}\big)$$ satisfies
		$ \KK_2 \models \Diag(A,E_1)$, in order to obtain an $\Lexp$-embedding of $(A,E_1)$ into $(K_2,E_2)$.
		
		For convenience, we introduce the convention that any quantifier-free $\Lexp$-formula $\alpha$ is written as $\alpha(\ul{x})$, where $\ul{x} = (x_1,x_2,\ldots)$ and substituting for a variable that does not appear in $\alpha(\ul{x})$ does not change the formula. Moreover, an expression of the form $\alpha(\ul{c})$ stands for the $\LL$-sentence $\alpha(c_1,c_2,\ldots)$ and $\exists \ul{x} \ \alpha(\ul{x})$ stands for existential quantification over all free variables in $\alpha(\ul{x})$. 
		
		Since $\Diag(A,E_1)$ is countable, we can enumerate it by $$\Diag(A,E_1) = \{\varphi_1(\ul{c}),\varphi_2(\ul{c}),\ldots\},$$ where each $\varphi_i(\ul{x})$ is a quantifier-free $\Lexp$-formula. For each $i \in \N$, set $$\psi_i'(\ul{x}) =  \bigwedge_{\ell=1}^i \varphi_\ell(\ul{x})\text{ and }\psi_i = \psi_i'(\ul{c}) =\bigwedge_{\ell=1}^i \varphi_\ell(\ul{c}).$$
		Note that any model of $\psi_i$ satisfies the first $i$ sentences in the enumeration of $\Diag(A,E_1)$. Thus, for the $\LL$-theory $\Sigma = \{\psi_i \mid i \in \N\}$, we have $$\Sigma \vdash \{\varphi_1(\ul{c}),\varphi_2(\ul{c}),\ldots\}.$$ Hence, it suffices to interpret each $c_k$ in $K_2$ such that $\KK_2 \models \Sigma$.
	
		
		Let $k \in \N$ be arbitrary. Since $a_k \in \ecl^{E_1}_{K_1}\!(\emptyset)$, there is a Khovanskii system $S_k(y_1,\ldots,y_n)$ over $\Z^{E}$ such that $$(K_1,E_1) \models \exists \ul{y}\ S_k(a_k,y_2,\ldots,y_n).$$ 
		Since $(K_2,E_2) \models \Th_\exists(K_1,E_1)$, the set $$B_k = \setbr{b \in K_2 \mid (K_2,E_2) \models \exists \ul{y} \ S_k(b,y_2,\ldots,y_n)}$$ is non-empty. Moreover, $B_k$ is finite by \Autoref{lem:khovanskiifinite}.
		
		For any $n \in \N$, let 
		\begin{align*}X_n = &\Big\{\!(b_1,\ldots,b_n) \in K_2^n\ \Big| \\
				&\qquad (K_2,E_2) \models \exists \ul{x} \exists \ul{y}  \Big(\psi_n'(b_1,\ldots,b_n,x_{n+1},\ldots) \wedge \bigwedge_{\ell=1}^n S_\ell(b_\ell,\ul{y})\Big)\Big\}.\end{align*}
		Since $$(K_1,E_1) \models \exists \ul{x} \exists \ul{y}  \Big(\psi_n'(a_1,\ldots,a_n,x_{n+1},\ldots) \wedge \bigwedge_{\ell=1}^n S_\ell\brackets{a_\ell,\ul{y}}\Big)\!,$$
		we can argue as above to obtain that $X_n \neq \emptyset$. Moreover, since $X_n \subseteq B_1\times \ldots \times B_n$, we have that $X_n$ is finite.
		
		Let $i \in \N$, let $\pi_i$ denote the projection of a tuple onto the $i$-th coordinate and let $\widetilde{\pi}_i$ denote the projection of a tuple of length at least $i$ onto the first $i$ coordinates. Since for any $j \in \N$ we have that $\{\psi_{j+1}'\} \vdash \psi_j'$, we obtain
		\begin{align} X_i =  \widetilde{\pi}_i(X_i) \supseteq \widetilde{\pi}_i(X_{i+1}) \supseteq \ldots.\label{eq:descendingchain}\end{align}
		
		Let $$X = \setbr{\left.\!\brackets{b_k}_{k\in \N} \in K_2^{\N} \ \right| \ (b_1,\ldots,b_n) \in X_n \text{ for any }n \in \N}.$$
		We construct an infinite graph $C$ and then apply K\H{o}nig's Infinity Lemma to show that $X \neq \emptyset$. The set of vertices of $C$ is given by the disjoint union of all $X_i$ for $i \in \N$ plus some element $v_0$ not contained in any $X_i$. We construct the set of edges $\mathcal{E}$ of $C$ as follows. All vertices in $X_1$ are neighbours of $v_0$. Moreover, for any $i \in \N$, a vertex $w \in X_{i+1}$ is a neighbour of a vertex  $v \in X_i$ if $\widetilde{\pi}_i(w) = v$ or, in other words, if the tuples $v$ and $w$ coincide in the first $i$ components. By (\autoref{eq:descendingchain}), for any $n \in \N$ with $n\geq 2$, any vertex $w \in X_n$ has a neighbour $v \in X_{n-1}$. Moreover, any vertex in $X_1$ has $v_0$ as a neighbour. Thus, all conditions for K\H{o}nig's Infinity Lemma are satisfied, yielding that $C$ contains an infinite ray $v_0v_1\ldots$ with $v_i \in X_i$ for any $i \in \N$. 
		In particular, for any $i \in \N$ we have $$\widetilde{\pi}_i\brackets{v_{i+1}} = v_i.$$
		
		Finally, we can interpret $$c_i^{\KK_2} = \pi_i(v_i)$$ for any $i \in \N$. We can now show that $\KK_2 \models \Sigma$. Let $i \in \N$ and consider $\psi_i \in \Sigma$. Let $N \in \N$ with $N\geq i$ such that for any $j > N$ the constant $c_j$ does not appear in $\psi_i$. By construction of $v_0v_1\ldots$, we have that
		$$\KK_2 \models \exists \ul{x} \ \psi_N'(c_1,\ldots,c_N,x_{N+1},\ldots).$$
		But $\psi'_N(\ul{x})$ is of the form $\psi_i'(\ul{x}) \wedge \varphi_{i+1}(\ul{x}) \wedge \ldots \wedge \varphi_{N} (\ul{x})$.
		Thus, we have
		$$\KK_2 \models \psi_i'(c_1,\ldots,c_N) \wedge \exists \ul{x} \ \brackets{\varphi_{i+1}(\ul{x}) \wedge \ldots \wedge \varphi_{N} (\ul{x})}.$$
		Hence, $\KK_2 \models \psi_i$, as required.
	\end{proof}
	

	
	\section{Prime model of real exponentiation}
	
	\textbf{For the rest of this note, we denote by $(P,\exp)$ the prime model of $\Texp$.}
	Since $(P,\exp)\models \Texp$, any structure containing $(P,\exp)$ as a substructure already satisfies the existential theory $\Th_\exists(\Rexp)$.
	Thus, by \Autoref{prop:primekhovanskii} and \Autoref{thm:eclembeddable} we obtain the following.
		
	\begin{corollary}\thlabel{cor:primemodelemb} 
		There is an emebedding of $(P,\exp)$ into $(K,E)$ if and only if $(K,E) \models \Th_\exists(\Rexp)$.
	\end{corollary}
	
	\Autoref{cor:primemodelemb} shows that o-minimal $\EXP$-fields satisfying the existential theory of real exponentiation contain the prime model of real exponentiation as a substructure. Under the assumption of Schanuel's Conjecture, any o-minimal $\EXP$-field satisfies $\Th_{\exists}(\Rexp)$. This fact is basically due to Servi~\cite[page~104]{servi} (cf.\ also Krapp~\cite[Proposition~4.57]{krappdiss}).
	
	
	\begin{fact}\thlabel{fact:servi}
		Assume Schanuel's Conjecture. Then any o-minimal $\EXP$-field satisfies $\Th_\exists(\Rexp)$.
	\end{fact}

	\begin{remark}
		In light of the Transfer Conjecture, one would hope that Schanuel's Conjecture implies that any o-minimal $\EXP$-fields already satisfies the \emph{complete} theory $\Texp$ of real exponentiation. However, already for the weaker conclusion of \Autoref{fact:servi} it is sensible to assume Schanuel's Conjecture: As discussed in \cite[Section~4.5]{servi}, if the existential theory $\Th_\exists(\Rexp)$ is decidable, then so is the complete theory $\Texp$. Due to Fornasiero and Servi~\cite{fornasiero} as well as Hieronymi~\cite{hieronymi}, there exists a recursive axiomatisation of the class of all o-minimal $\EXP$-fields (cf.\ also \cite[Theorem~4.30]{krappdiss}). Hence, already the conclusion of \Autoref{fact:servi} would imply the decidability of $\Texp$.
	\end{remark}
	
	 As a result of \Autoref{cor:primemodelemb} and \Autoref{fact:servi}, we obtain the desired main result of this note.
	
	\begin{theorem}\thlabel{cor:scprimeemb}
		Assume Schanuel's Conjecture. Then $(P,\exp)$ embeds into any o-minimal $\EXP$-field.
	\end{theorem}
	
	
	
	\section{Residue exponential field}\label{sec:residue}
	
	In this final section, we discuss how our results above relate to another approach towards the Transfer Conjecture via exponential residue fields. Following the notation of Krapp~\cite{krappla}, we denote by $R_v$ the valuation ring of the natural valuation on $K$, by $I_v$ its valuation ideal and by $\ol{K}=R_v/I_v$ the residue field. In terms of the ordering on $K$, we have $R_v=\{a\in K\mid |a|<n\text{ for some }n\in \N\}$ (the convex hull of $\Z$ in $K$) and $I_v=\{a\in K\mid |a|<\tfrac 1n\text{ for all }n\in \N\}$. The ordering on $\ol{K}$ is given by $\ol{a}>0$ if and only if $a>0$ and $\ol{a}\neq 0$, where $a\in R_v$ and $\ol{a}=a+I_v$. Due to \cite[Proposition~3.2]{krappla}, the residue exponential $\ol{E}\colon \ol{K}\to \ol{K}^{>0}, \ol{a}\mapsto \ol{E(a)}$ is a well-defined exponential on $K$. Moreover, \cite[Theorem~3.3]{krappla} shows that the residue exponential field $(\ol{K},\ol{E})$ elementarily embeds into $\Rexp$. (Note that this $\Lexp$-embedding is unique, as any $\Lor$-embedding of an archimedean field into $\R$ is already unique.) In particular, $(\ol{K},\ol{E})$ is a model of $\Texp$. Hence, an affirmative answer to the following question is equivalent to the Transfer Conjecture (cf.\ \cite[Proposition~3.6]{krappla}).

	\begin{question}\thlabel{qu:2}
		Let $(K,E)$ be an o-minimal $\EXP$-field. Is $(\Kres,\ol{E})$ elementarily embeddable into $(K,E)$?
	\end{question}

	Just as for 
	\Autoref{qu:main}, the first approach towards \Autoref{qu:2} would be to show that the residue exponential field of $(K,E)$ is embeddable into $(K,E)$. Under the assumption of Schanuel's Conjecture, this is indeed the case for sufficiently saturated o-minimal $\EXP$-fields.
	
	\begin{proposition}\thlabel{prop:sat}
		Suppose that $(K,E)$ is $2^{\aleph_0}$-saturated.
		Then $(\Kres,\ol{E})=\Rexp$. Moreover, under the assumtion of Schanuel's Conjecture, $(\Kres,\ol{E})$ embeds into $(K,E)$.
	\end{proposition}

	\begin{proof}
		Since $K$ is $\omega$-saturated as an $\Lor$-structure, the residue field $\ol{K}$ is isomorphic to $\R$ via a unique isomorphism (see e.g.\ S.~Kuhlmann, F.-V.~Kuhlmann, Marshall and Zekavat~\cite[Theorem~6.2]{kuhlmann4}). By \cite[Theorem~3.3]{krappla}, we already obtain $(\Kres,\ol{E})=\Rexp$.
		
		Assuming Schanuel's Conjecture, \Autoref{fact:servi} implies that $(K,E)\models  \Th_\exists(\allowbreak \Rexp)$. Equivalently, $\Rexp$ satisfies the universal theory of $(K,E)$. 
		Thus, $\Rexp$ can be embedded into some $(K_1,E_1) \models \Th(K,E)$ (see e.g.\ Sarbadhikari and Srivastava~\cite[Proposition~2.4.2]{sarbadhikari}), where $K_1$ can be chosen of cardinality $|K_1| = \max\{|\R|,|\Lexp|\} = 2^{\aleph_0}$ by the Löwenheim--Skolem Theorem. Now $(K, E)$ is $\brackets{2^{\aleph_0}}^+\!$-universal, and thus $(K_1,E_1)$ elementarily embeds into $(K,E)$. Hence, $\Rexp$ embeds into $(K,E)$. 
	\end{proof}
	
	Let $(F,E)$ be a $2^{\aleph_0}$-saturated elementary extension of $(K,E)$. We obtain the following chain of substructures (where each elementary embedding is unique):
	$$(P,\exp)\preceq  (\Kres,\ol{E})\preceq \Rexp =(\ol{F},\ol{E})$$
	Assuming Schanuel's Conjecture, \Autoref{cor:scprimeemb} and \Autoref{prop:sat} respectively give us $\Lexp$-embeddings
	$$\varphi\colon (P,\exp)\hookrightarrow (K,E)\text{ and }\psi\colon \Rexp\hookrightarrow (F,E).$$
	In conclusion, we obtain the following picture:
	\begin{center}
	\tikzcdset{
		arrow style=tikz,
		diagrams={>={Straight Barb[scale=0.8]}}
	}

	\begin{tikzcd}
		  & (K,E) \arrow[r, hook, "\preceq"] & (F,E)\\
		(P,\exp) \arrow[ru, hook, "\varphi"] \arrow[r, hook, "\preceq"]& (\Kres,\ol{E}) \arrow[r, hook, "\preceq"]& (\ol{F},\ol{E})\arrow[u, hook, "\psi"]
	\end{tikzcd}
	\end{center}
	

\begin{acknowledgements}
	Several results of this work were part of my doctoral thesis \cite{krappdiss}. In this regard, I wish to thank my supervisor Salma Kuhlmann for her continuous help and support. I also thank Alessandro Berarducci for an insightful discussion about \cite[Corollaries~4.59 and 4.90]{krappdiss}, which encouraged me to compose this note.
\end{acknowledgements}


	
	\begin{footnotesize}
		