[proofplan] We first construct a local homomorphism near the identity by choosing exponential coordinates in $G$ and $H$ and using the Baker--Campbell--Hausdorff local group law. The bracket-preserving property of $A$ makes the coordinate formula $X\mapsto A(X)$ compatible with the BCH product, so it defines a local [group homomorphism](/page/Group%20Homomorphism). A standard [monodromy theorem](/theorems/3371) for local Lie group homomorphisms then extends this local homomorphism uniquely along paths starting at $e_G$; simple connectedness makes the extension independent of the path. The resulting global map is smooth, multiplicative, has differential $A$, and uniqueness follows because homomorphisms from a connected Lie group are determined by their differentials. [/proofplan]

[step:Build the local homomorphism in exponential coordinates]Let \begin{align*} \exp_G:\mathfrak g\to G \end{align*} denote the exponential map of the Lie group $G$, and let \begin{align*} \exp_H:\mathfrak h\to H \end{align*} denote the exponential map of the Lie group $H$. Choose open neighbourhoods $U_{\mathfrak g}\subset\mathfrak g$ of $0$, $V_G\subset G$ of $e_G$, $U_{\mathfrak h}\subset\mathfrak h$ of $0$, and $V_H\subset H$ of $e_H$ such that \begin{align*} \exp_G|_{U_{\mathfrak g}}:U_{\mathfrak g}\to V_G \end{align*} and \begin{align*} \exp_H|_{U_{\mathfrak h}}:U_{\mathfrak h}\to V_H \end{align*} are diffeomorphisms. Shrink $U_{\mathfrak g}$, if necessary, so that $A(U_{\mathfrak g})\subset U_{\mathfrak h}$. Define the map \begin{align*} f:V_G\to V_H \end{align*} by \begin{align*} f(\exp_G X)=\exp_H(A X) \end{align*} for $X\in U_{\mathfrak g}$. This is well-defined because $\exp_G|_{U_{\mathfrak g}}$ is injective, and it is smooth because it is the composition \begin{align*} V_G\xrightarrow{(\exp_G|_{U_{\mathfrak g}})^{-1}}U_{\mathfrak g}\xrightarrow{A}U_{\mathfrak h}\xrightarrow{\exp_H|_{U_{\mathfrak h}}}V_H. \end{align*}[/step]

[guided]We need a candidate group homomorphism before we can extend anything globally. Since a Lie group is locally modelled on its [Lie algebra](/page/Lie%20Algebra) near the identity, the natural candidate is obtained by moving from $G$ to $\mathfrak g$, applying $A$, and moving back into $H$. Let \begin{align*} \exp_G:\mathfrak g\to G \end{align*} and \begin{align*} \exp_H:\mathfrak h\to H \end{align*} denote the exponential maps of $G$ and $H$, respectively. Choose open neighbourhoods $U_{\mathfrak g}\subset\mathfrak g$ of $0$ and $V_G\subset G$ of $e_G$ such that \begin{align*} \exp_G|_{U_{\mathfrak g}}:U_{\mathfrak g}\to V_G \end{align*} is a diffeomorphism. Choose similarly open neighbourhoods $U_{\mathfrak h}\subset\mathfrak h$ of $0$ and $V_H\subset H$ of $e_H$ such that \begin{align*} \exp_H|_{U_{\mathfrak h}}:U_{\mathfrak h}\to V_H \end{align*} is a diffeomorphism. Since $A:\mathfrak g\to\mathfrak h$ is linear and continuous between finite-dimensional vector spaces, after shrinking $U_{\mathfrak g}$ we may assume $A(U_{\mathfrak g})\subset U_{\mathfrak h}$. Now define \begin{align*} f:V_G\to V_H \end{align*} by the rule \begin{align*} f(\exp_G X)=\exp_H(A X) \end{align*} for $X\in U_{\mathfrak g}$. This definition does not depend on a choice of $X$, because $\exp_G|_{U_{\mathfrak g}}$ is one-to-one. The formula also shows smoothness: in exponential coordinates, $f$ is exactly the [linear map](/page/Linear%20Map) $A$, and therefore \begin{align*} f=\exp_H|_{U_{\mathfrak h}}\circ A\circ(\exp_G|_{U_{\mathfrak g}})^{-1}. \end{align*} Thus $f$ is a smooth map from a neighbourhood of $e_G$ to a neighbourhood of $e_H$.[/guided]

[step:Use the BCH formula to prove the local homomorphism law] Let \begin{align*} \operatorname{BCH}_{\mathfrak g}:\Omega_{\mathfrak g}\to\mathfrak g \end{align*} denote the local Baker--Campbell--Hausdorff map of $\mathfrak g$, defined on an open neighbourhood $\Omega_{\mathfrak g}\subset\mathfrak g\times\mathfrak g$ of $(0,0)$. Let \begin{align*} \operatorname{BCH}_{\mathfrak h}:\Omega_{\mathfrak h}\to\mathfrak h \end{align*} denote the corresponding local Baker--Campbell--Hausdorff map of $\mathfrak h$, defined on an open neighbourhood $\Omega_{\mathfrak h}\subset\mathfrak h\times\mathfrak h$ of $(0,0)$. Shrink $U_{\mathfrak g}$ once more so that whenever $X,Y\in U_{\mathfrak g}$ and $\exp_G X\exp_G Y\in V_G$, the pair $(X,Y)$ lies in $\Omega_{\mathfrak g}$, the element $\operatorname{BCH}_{\mathfrak g}(X,Y)$ lies in $U_{\mathfrak g}$, the pair $(A X,A Y)$ lies in $\Omega_{\mathfrak h}$, and \begin{align*} \exp_G X\exp_G Y=\exp_G(\operatorname{BCH}_{\mathfrak g}(X,Y)). \end{align*} The BCH local group law is available in sufficiently small exponential coordinates by [citetheorem:8798]. Because $A$ is a Lie algebra homomorphism of real Lie algebras, it carries every iterated bracket in $\mathfrak g$ to the corresponding iterated bracket in $\mathfrak h$. The BCH series is a universal convergent Lie series on these local domains, so applying $A$ term by term gives \begin{align*} A(\operatorname{BCH}_{\mathfrak g}(X,Y))=\operatorname{BCH}_{\mathfrak h}(A X,A Y). \end{align*} Therefore, for $X,Y\in U_{\mathfrak g}$ with $\exp_G X\exp_G Y\in V_G$, \begin{align*} f(\exp_G X\exp_G Y)=\exp_H(A(\operatorname{BCH}_{\mathfrak g}(X,Y))). \end{align*} Using the BCH compatibility above, this becomes \begin{align*} f(\exp_G X\exp_G Y)=\exp_H(\operatorname{BCH}_{\mathfrak h}(A X,A Y)). \end{align*} By the BCH product formula in $H$, \begin{align*} f(\exp_G X\exp_G Y)=\exp_H(A X)\exp_H(A Y)=f(\exp_G X)f(\exp_G Y). \end{align*} Thus $f$ is a local Lie group homomorphism near $e_G$. [/step]

[step:Continue the local homomorphism along paths from the identity] We use the following standard monodromy theorem for local Lie group homomorphisms: if $G$ is connected and simply connected and $f:V_G\to H$ is a smooth local homomorphism on a neighbourhood $V_G$ of $e_G$, then there exists a unique smooth Lie group homomorphism $\varphi:G\to H$ whose restriction to some neighbourhood of $e_G$ contained in $V_G$ equals $f$. This is the local-to-global continuation theorem for Lie group homomorphisms; it is the only external continuation result used here. For completeness, we recall the construction. Let $\gamma:[0,1]\to G$ be a continuous path with $\gamma(0)=e_G$. Choose a subdivision \begin{align*} 0=t_0<t_1<\cdots<t_m=1 \end{align*} such that each increment $\gamma(t_{j-1})^{-1}\gamma(t_j)$ lies in $V_G$. Define the continuation value along this subdivided path by \begin{align*} F(\gamma;t_0,\dots,t_m)=f(\gamma(t_0)^{-1}\gamma(t_1))f(\gamma(t_1)^{-1}\gamma(t_2))\cdots f(\gamma(t_{m-1})^{-1}\gamma(t_m)). \end{align*} The local homomorphism law for $f$ implies that refining the subdivision does not change this product. The same local law also implies invariance under a homotopy of paths with fixed endpoints, by applying the construction on a sufficiently fine rectangular subdivision of the homotopy square. Since $G$ is simply connected, every two paths from $e_G$ to the same endpoint are homotopic with fixed endpoints. Thus the value \begin{align*} \varphi(g):=F(\gamma;t_0,\dots,t_m) \end{align*} is independent of the chosen path $\gamma$ from $e_G$ to $g$ and of the subdivision. The standard continuation theorem then gives a smooth map \begin{align*} \varphi:G\to H \end{align*} which agrees with $f$ near $e_G$. [/step]

[step:Verify that the continued map is a Lie group homomorphism] Let $g_1,g_2\in G$. Choose continuous paths \begin{align*} \gamma_1:[0,1]\to G \end{align*} and \begin{align*} \gamma_2:[0,1]\to G \end{align*} with $\gamma_1(0)=e_G$, $\gamma_1(1)=g_1$, $\gamma_2(0)=e_G$, and $\gamma_2(1)=g_2$. Define a path \begin{align*} \gamma_{12}:[0,1]\to G \end{align*} by first following $\gamma_1$ from $e_G$ to $g_1$ and then following $t\mapsto g_1\gamma_2(t)$ from $g_1$ to $g_1g_2$. The continuation product along the first part gives $\varphi(g_1)$, and the continuation product along the translated second part gives $\varphi(g_1)\varphi(g_2)$ because the local homomorphism law identifies the translated increments with the same local values multiplied after the already accumulated value. Therefore \begin{align*} \varphi(g_1g_2)=\varphi(g_1)\varphi(g_2). \end{align*} Also $\varphi(e_G)=e_H$, since the constant path at $e_G$ has empty continuation product. Hence $\varphi:G\to H$ is a Lie group homomorphism. Smoothness is part of the local-to-global continuation theorem and follows locally from the fact that every point of $G$ is reached by left translating the identity neighbourhood on which $\varphi$ is represented by the smooth local formula. [/step]

[step:Compute the differential at the identity] For $X\in\mathfrak g$, choose $\varepsilon>0$ such that $tX\in U_{\mathfrak g}$ for all $t\in(-\varepsilon,\varepsilon)$. On this interval, \begin{align*} \varphi(\exp_G(tX))=f(\exp_G(tX))=\exp_H(tA X). \end{align*} The curve \begin{align*} c_X:(-\varepsilon,\varepsilon)\to G \end{align*} defined by \begin{align*} c_X(t)=\exp_G(tX) \end{align*} has tangent vector $c_X'(0)=X$. Applying the definition of the differential to the composed curve $\varphi\circ c_X$ gives \begin{align*} d\varphi_{e_G}(X)=\frac{d}{dt}\bigg|_{t=0}\exp_H(tA X)=A X. \end{align*} Since this holds for every $X\in\mathfrak g$, we have \begin{align*} d\varphi_{e_G}=A. \end{align*} [/step]

[step:Use connectedness to prove uniqueness] Let \begin{align*} \psi:G\to H \end{align*} be a smooth Lie group homomorphism such that \begin{align*} d\psi_{e_G}=A. \end{align*} Then \begin{align*} d\psi_{e_G}=d\varphi_{e_G}. \end{align*} Since $G$ is connected, homomorphisms from a connected Lie group are determined by their differentials by [citetheorem:8808]. Therefore $\psi=\varphi$. Thus there exists a unique smooth Lie group homomorphism $\varphi:G\to H$ satisfying $d\varphi_{e_G}=A$. [/step]