[proofplan] We first use one-parameter subgroups to construct a real-[linear map](/page/Linear%20Map) $A:\mathfrak g\to\mathfrak h$ satisfying $\varphi(\exp_G(tX))=\exp_H(tA(X))$. Homogeneity follows by reparametrising one-parameter subgroups, and additivity follows from the Lie-Trotter product formula together with continuity of $\varphi$. We then choose product-of-exponentials coordinates near the identity in $G$; in those coordinates $\varphi$ is explicitly a product of matrix exponentials in $H$, hence smooth. Finally, the homomorphism property and smoothness of left translations propagate smoothness from a neighbourhood of the identity to all of $G$. [/proofplan]

[step:Associate a Lie algebra element to each one-parameter subgroup]Let $\exp_G:\mathfrak g\to G$ and $\exp_H:\mathfrak h\to H$ denote the matrix exponential maps restricted to the Lie algebras of $G$ and $H$. Let $e_G\in G$ and $e_H\in H$ denote the identity elements. For $X\in\mathfrak g$, define the map \begin{align*} \gamma_X:\mathbb R&\to H \end{align*} by \begin{align*} \gamma_X(t)&=\varphi(\exp_G(tX)). \end{align*} Since $X\in\mathfrak g$, the curve $t\mapsto \exp_G(tX)$ is a smooth one-parameter subgroup of $G$ by the definition of the matrix [Lie algebra](/page/Lie%20Algebra) and the matrix exponential. Since $\varphi$ is a continuous [group homomorphism](/page/Group%20Homomorphism), $\gamma_X$ is a continuous group homomorphism from $(\mathbb R,+)$ to $H$. By [citetheorem:8784], every continuous group homomorphism $\gamma:\mathbb R\to H$ has the form $\gamma(t)=\exp_H(tY)$ for a unique $Y\in\mathfrak h$. Applying this theorem to $\gamma_X$, there is a unique element $A(X)\in\mathfrak h$ such that \begin{align*} \varphi(\exp_G(tX))=\exp_H(tA(X)) \end{align*} for every $t\in\mathbb R$. This defines a map \begin{align*} A:\mathfrak g&\to\mathfrak h. \end{align*}[/step]

[guided]Fix $X\in\mathfrak g$. The reason to look at the curve $t\mapsto \varphi(\exp_G(tX))$ is that $\exp_G(tX)$ is the one-parameter subgroup generated by $X$, so it is the part of $G$ whose behaviour should determine the infinitesimal map of $\varphi$. Define \begin{align*} \gamma_X:\mathbb R&\to H \end{align*} by \begin{align*} \gamma_X(t)&=\varphi(\exp_G(tX)). \end{align*} The map $t\mapsto \exp_G(tX)$ is a smooth group homomorphism from $(\mathbb R,+)$ to $G$, because $X$ lies in the Lie algebra $\mathfrak g$ of the matrix Lie group $G$. The map $\varphi$ is continuous by hypothesis, so $\gamma_X$ is continuous. It is also a group homomorphism, since for $s,t\in\mathbb R$, \begin{align*} \gamma_X(s+t)=\varphi(\exp_G((s+t)X)). \end{align*} Using the one-parameter subgroup law in $G$ gives \begin{align*} \exp_G((s+t)X)=\exp_G(sX)\exp_G(tX). \end{align*} Since $\varphi$ is a group homomorphism, \begin{align*} \gamma_X(s+t)=\varphi(\exp_G(sX))\varphi(\exp_G(tX))=\gamma_X(s)\gamma_X(t). \end{align*} Now apply the standard classification theorem for continuous one-parameter subgroups of matrix Lie groups. Its hypotheses are satisfied because $H$ is a matrix Lie group and $\gamma_X:\mathbb R\to H$ is a continuous group homomorphism. Therefore there is a unique element $A(X)\in\mathfrak h$ such that \begin{align*} \gamma_X(t)=\exp_H(tA(X)) \end{align*} for every $t\in\mathbb R$. Equivalently, \begin{align*} \varphi(\exp_G(tX))=\exp_H(tA(X)). \end{align*} This uniqueness allows us to define a map \begin{align*} A:\mathfrak g&\to\mathfrak h \end{align*} by assigning to each $X\in\mathfrak g$ the unique generator of the one-parameter subgroup $\varphi(\exp_G(tX))$.[/guided]

[step:Prove that the induced map on Lie algebras is real-linear]First let $c\in\mathbb R$ and $X\in\mathfrak g$. For every $t\in\mathbb R$, \begin{align*} \varphi(\exp_G(t(cX)))=\varphi(\exp_G((tc)X))=\exp_H(tcA(X))=\exp_H(t(cA(X))). \end{align*} By the uniqueness in the classification of one-parameter subgroups, \begin{align*} A(cX)=cA(X). \end{align*} Now let $X,Y\in\mathfrak g$. We use the Lie-Trotter product formula in matrix Lie groups: \begin{align*} \exp_G(t(X+Y))=\lim_{k\to\infty}\left(\exp_G\left(\frac{tX}{k}\right)\exp_G\left(\frac{tY}{k}\right)\right)^k \end{align*} in $G$, for every $t\in\mathbb R$. Applying the continuous map $\varphi:G\to H$ and using that $\varphi$ is a group homomorphism gives \begin{align*} \varphi(\exp_G(t(X+Y)))=\lim_{k\to\infty}\left(\varphi\left(\exp_G\left(\frac{tX}{k}\right)\right)\varphi\left(\exp_G\left(\frac{tY}{k}\right)\right)\right)^k. \end{align*} By the defining property of $A$, \begin{align*} \varphi(\exp_G(t(X+Y)))=\exp_H(tA(X+Y)) \end{align*} and \begin{align*} \varphi\left(\exp_G\left(\frac{tX}{k}\right)\right)\varphi\left(\exp_G\left(\frac{tY}{k}\right)\right)=\exp_H\left(\frac{tA(X)}{k}\right)\exp_H\left(\frac{tA(Y)}{k}\right). \end{align*} Applying the Lie-Trotter product formula in $H$ to $A(X),A(Y)\in\mathfrak h$ yields \begin{align*} \lim_{k\to\infty}\left(\exp_H\left(\frac{tA(X)}{k}\right)\exp_H\left(\frac{tA(Y)}{k}\right)\right)^k=\exp_H(t(A(X)+A(Y))). \end{align*} Therefore \begin{align*} \exp_H(tA(X+Y))=\exp_H(t(A(X)+A(Y))) \end{align*} for every $t\in\mathbb R$. By uniqueness of the generator of a continuous one-parameter subgroup, \begin{align*} A(X+Y)=A(X)+A(Y). \end{align*} Thus $A:\mathfrak g\to\mathfrak h$ is real-linear.[/step]

[guided]We prove real-linearity in two separate parts: homogeneity and additivity. First fix $c\in\mathbb R$ and $X\in\mathfrak g$. For every $t\in\mathbb R$, the scalar parameter in the one-parameter subgroup can be reparametrised as \begin{align*} \varphi(\exp_G(t(cX)))=\varphi(\exp_G((tc)X)). \end{align*} By the defining property of $A(X)$, this equals \begin{align*} \exp_H(tcA(X))=\exp_H(t(cA(X))). \end{align*} Thus the two one-parameter subgroups generated by $A(cX)$ and by $cA(X)$ agree for every $t\in\mathbb R$. The uniqueness part of the classification theorem for continuous one-parameter subgroups gives \begin{align*} A(cX)=cA(X). \end{align*} Now fix $X,Y\in\mathfrak g$. The point of the Lie-Trotter product formula is that it expresses the exponential of a sum using only exponentials of the two summands, which are precisely the curves whose images under $\varphi$ we already understand. For every $t\in\mathbb R$, the Lie-Trotter product formula in the matrix Lie group $G$ gives \begin{align*} \exp_G(t(X+Y))=\lim_{k\to\infty}\left(\exp_G\left(\frac{tX}{k}\right)\exp_G\left(\frac{tY}{k}\right)\right)^k. \end{align*} The map $\varphi:G\to H$ is continuous by hypothesis, so it preserves this limit after applying it. Since $\varphi$ is a group homomorphism, it also preserves each finite product. Therefore \begin{align*} \varphi(\exp_G(t(X+Y)))=\lim_{k\to\infty}\left(\varphi\left(\exp_G\left(\frac{tX}{k}\right)\right)\varphi\left(\exp_G\left(\frac{tY}{k}\right)\right)\right)^k. \end{align*} By the definition of $A$, the left-hand side is \begin{align*} \varphi(\exp_G(t(X+Y)))=\exp_H(tA(X+Y)), \end{align*} and each factor in the right-hand product satisfies \begin{align*} \varphi\left(\exp_G\left(\frac{tX}{k}\right)\right)\varphi\left(\exp_G\left(\frac{tY}{k}\right)\right)=\exp_H\left(\frac{tA(X)}{k}\right)\exp_H\left(\frac{tA(Y)}{k}\right). \end{align*} Because $A(X),A(Y)\in\mathfrak h$, we may apply the Lie-Trotter product formula in $H$ to obtain \begin{align*} \lim_{k\to\infty}\left(\exp_H\left(\frac{tA(X)}{k}\right)\exp_H\left(\frac{tA(Y)}{k}\right)\right)^k=\exp_H(t(A(X)+A(Y))). \end{align*} Combining the preceding identities gives \begin{align*} \exp_H(tA(X+Y))=\exp_H(t(A(X)+A(Y))) \end{align*} for every $t\in\mathbb R$. The two sides are continuous one-parameter subgroups of $H$ with generators $A(X+Y)$ and $A(X)+A(Y)$, respectively. By uniqueness of the generator, these generators are equal: \begin{align*} A(X+Y)=A(X)+A(Y). \end{align*} Together with homogeneity, this proves that $A:\mathfrak g\to\mathfrak h$ is real-linear.[/guided]

[step:Write the homomorphism in product exponential coordinates near the identity]Let $r=\dim_{\mathbb R}\mathfrak g$, and choose an ordered real basis $X_1,\dots,X_r$ of $\mathfrak g$. Define \begin{align*} \Theta:\mathbb R^r&\to G \end{align*} near $0\in\mathbb R^r$ by \begin{align*} \Theta(t_1,\dots,t_r)&=\exp_G(t_1X_1)\cdots\exp_G(t_rX_r). \end{align*} The differential $d\Theta_0:\mathbb R^r\to\mathfrak g$ sends the standard basis vector of $\mathbb R^r$ in the $i$-th position to $X_i$. Hence $d\Theta_0$ is a linear isomorphism. By the [inverse function theorem](/theorems/51), after restricting to an open neighbourhood $U\subset\mathbb R^r$ of $0$, the map $\Theta:U\to V$ is a diffeomorphism onto an open neighbourhood $V\subset G$ of $e_G$. For $t=(t_1,\dots,t_r)\in U$, the homomorphism property gives \begin{align*} \varphi(\Theta(t))=\varphi(\exp_G(t_1X_1))\cdots\varphi(\exp_G(t_rX_r)). \end{align*} Using the defining property of $A$ on each basis element, \begin{align*} \varphi(\Theta(t))=\exp_H(t_1A(X_1))\cdots\exp_H(t_rA(X_r)). \end{align*} The right-hand side is a smooth map $U\to H$, because matrix exponentiation is smooth and multiplication in the matrix Lie group $H$ is smooth. Hence $\varphi|_V$ is smooth.[/step]

[guided]The local smoothness of $\varphi$ is proved by choosing coordinates in which every group element near the identity is written as a product of exponentials. Let \begin{align*} r=\dim_{\mathbb R}\mathfrak g, \end{align*} and choose an ordered real basis $X_1,\dots,X_r$ of $\mathfrak g$. Define \begin{align*} \Theta:\mathbb R^r&\to G \end{align*} near $0\in\mathbb R^r$ by \begin{align*} \Theta(t_1,\dots,t_r)&=\exp_G(t_1X_1)\cdots\exp_G(t_rX_r). \end{align*} We compute the differential of $\Theta$ at $0$. If $e_i\in\mathbb R^r$ denotes the $i$-th standard basis vector, then the curve $s\mapsto \Theta(se_i)$ is exactly \begin{align*} s\mapsto \exp_G(sX_i). \end{align*} Its velocity at $s=0$ is $X_i\in T_{e_G}G=\mathfrak g$. Thus \begin{align*} d\Theta_0(e_i)=X_i \end{align*} for every $i\in\{1,\dots,r\}$. Since $X_1,\dots,X_r$ is a basis, $d\Theta_0$ is a linear isomorphism. The inverse function theorem now applies to the smooth map $\Theta$. Therefore there is an open neighbourhood $U\subset\mathbb R^r$ of $0$ such that $\Theta$ restricts to a diffeomorphism from $U$ onto an open neighbourhood $V\subset G$ of $e_G$. In these coordinates, $\varphi$ has an explicit formula. For $t=(t_1,\dots,t_r)\in U$, the definition of $\Theta$ and the homomorphism property of $\varphi$ give \begin{align*} \varphi(\Theta(t))=\varphi(\exp_G(t_1X_1))\cdots\varphi(\exp_G(t_rX_r)). \end{align*} For each basis vector $X_i$, the defining property of $A$ says \begin{align*} \varphi(\exp_G(t_iX_i))=\exp_H(t_iA(X_i)). \end{align*} Substituting this into the preceding product yields \begin{align*} \varphi(\Theta(t))=\exp_H(t_1A(X_1))\cdots\exp_H(t_rA(X_r)). \end{align*} This formula proves smoothness near the identity. Indeed, for each fixed $A(X_i)\in\mathfrak h$, the map $t_i\mapsto \exp_H(t_iA(X_i))$ is smooth, and multiplication in the matrix Lie group $H$ is smooth. Therefore the composite map \begin{align*} t\mapsto \exp_H(t_1A(X_1))\cdots\exp_H(t_rA(X_r)) \end{align*} is smooth from $U$ to $H$. Since $\Theta:U\to V$ is a diffeomorphism, this is exactly the coordinate representation of $\varphi|_V$, so $\varphi$ is smooth on $V$.[/guided]

[step:Translate local smoothness to every point of the group] Let $g\in G$. Since $V\subset G$ is an open neighbourhood of $e_G$, the set $gV=\{gv:v\in V\}$ is an open neighbourhood of $g$, because left multiplication by $g$ is a diffeomorphism of the matrix Lie group $G$. For $x\in gV$, define $v=g^{-1}x\in V$. Then \begin{align*} \varphi(x)=\varphi(gv)=\varphi(g)\varphi(v). \end{align*} Thus on $gV$, \begin{align*} \varphi=L_{\varphi(g)}^H\circ \varphi|_V\circ L_{g^{-1}}^G, \end{align*} where $L_{g^{-1}}^G:G\to G$ is left multiplication by $g^{-1}$ and $L_{\varphi(g)}^H:H\to H$ is left multiplication by $\varphi(g)$. Both left translations are smooth, and $\varphi|_V$ is smooth by the previous step. Therefore $\varphi$ is smooth on the neighbourhood $gV$ of $g$. Since $g\in G$ was arbitrary, $\varphi$ is smooth on all of $G$. [/step]