[proofplan] Fix $X\in\mathfrak g$ and compare two curves in $H$: the image under $\varphi$ of the exponential one-parameter subgroup generated by $X$, and the exponential one-parameter subgroup generated by $d\varphi(X)$. The homomorphism property of $\varphi$ makes the first curve a one-parameter subgroup of $H$. The chain rule identifies its initial tangent vector with $d\varphi(X)$, so the defining uniqueness property of the Lie group exponential gives equality of the two curves for all real parameters. Setting $t=1$ then gives the stated special case. [/proofplan]

[step:Define the image curve and verify that it is a one-parameter subgroup of $H$] Fix $X\in\mathfrak g$. Define the smooth curve \begin{align*} \gamma:\mathbb R\to H,\qquad t\mapsto \varphi(\exp_G(tX)). \end{align*} The curve is smooth because the map $t\mapsto \exp_G(tX)$ is smooth and $\varphi$ is smooth. For $s,t\in\mathbb R$, the exponential curve \begin{align*} \alpha_X:\mathbb R\to G,\qquad u\mapsto \exp_G(uX) \end{align*} is the one-parameter subgroup of $G$ with initial tangent vector $X$. Hence \begin{align*} \exp_G((s+t)X)=\exp_G(sX)\exp_G(tX). \end{align*} Since $\varphi$ is a [group homomorphism](/page/Group%20Homomorphism), \begin{align*} \gamma(s+t)=\varphi(\exp_G((s+t)X)). \end{align*} Using the one-parameter subgroup identity in $G$ and then the homomorphism property of $\varphi$ gives \begin{align*} \gamma(s+t)=\varphi(\exp_G(sX)\exp_G(tX)). \end{align*} Therefore \begin{align*} \gamma(s+t)=\varphi(\exp_G(sX))\varphi(\exp_G(tX))=\gamma(s)\gamma(t). \end{align*} Also, \begin{align*} \gamma(0)=\varphi(\exp_G(0))=\varphi(e_G)=e_H. \end{align*} Thus $\gamma$ is a smooth one-parameter subgroup of $H$. [/step]

[step:Compute the initial tangent vector of the image curve]We compute the derivative of $\gamma$ at $0$. Let \begin{align*} \alpha_X:\mathbb R\to G,\qquad t\mapsto \exp_G(tX). \end{align*} Then $\gamma=\varphi\circ\alpha_X$. Identifying $T_0\mathbb R$ with $\mathbb R$ by the standard coordinate vector $1\in T_0\mathbb R$, the defining property of the exponential gives \begin{align*} d(\alpha_X)_0(1)=X. \end{align*} By the chain rule for smooth maps, \begin{align*} d\gamma_0(1)=d\varphi_{e_G}(d(\alpha_X)_0(1)). \end{align*} Substituting $d(\alpha_X)_0(1)=X$ and $d\varphi=d\varphi_{e_G}$ yields \begin{align*} d\gamma_0(1)=d\varphi(X). \end{align*} Thus the initial tangent vector of the one-parameter subgroup $\gamma$ is $d\varphi(X)$.[/step]

[guided]We now identify which element of $\mathfrak h=T_{e_H}H$ generates the one-parameter subgroup $\gamma$. The curve already has the form \begin{align*} \gamma=\varphi\circ\alpha_X, \end{align*} where \begin{align*} \alpha_X:\mathbb R\to G,\qquad t\mapsto \exp_G(tX). \end{align*} The exponential map on $G$ is characterized so that $\alpha_X$ is the one-parameter subgroup whose initial velocity is $X$. With the standard identification of $T_0\mathbb R$ with $\mathbb R$ through the coordinate vector $1\in T_0\mathbb R$, this means \begin{align*} d(\alpha_X)_0(1)=X. \end{align*} Because $\varphi:G\to H$ is smooth, the ordinary chain rule applies to the composition $\gamma=\varphi\circ\alpha_X$. At $0\in\mathbb R$, the point $\alpha_X(0)$ is $e_G$, so the chain rule gives \begin{align*} d\gamma_0(1)=d\varphi_{\alpha_X(0)}(d(\alpha_X)_0(1)). \end{align*} Since $\alpha_X(0)=e_G$, this becomes \begin{align*} d\gamma_0(1)=d\varphi_{e_G}(d(\alpha_X)_0(1)). \end{align*} Now substitute the defining initial-velocity identity $d(\alpha_X)_0(1)=X$: \begin{align*} d\gamma_0(1)=d\varphi_{e_G}(X). \end{align*} By the notation in the statement, $d\varphi=d\varphi_{e_G}$, so \begin{align*} d\gamma_0(1)=d\varphi(X). \end{align*} Thus the image curve $\gamma$ is not merely some one-parameter subgroup of $H$; it is precisely a one-parameter subgroup with initial tangent vector $d\varphi(X)$.[/guided]

[step:Use uniqueness of one-parameter subgroups with prescribed initial tangent vector] Define the exponential one-parameter subgroup in $H$ generated by $d\varphi(X)$ by \begin{align*} \beta:\mathbb R\to H,\qquad t\mapsto \exp_H(t\,d\varphi(X)). \end{align*} By the defining property of the Lie group exponential, $\beta$ is the unique smooth one-parameter subgroup of $H$ whose initial tangent vector is $d\varphi(X)$. The previous steps show that $\gamma$ is also a smooth one-parameter subgroup of $H$ and that its initial tangent vector is $d\varphi(X)$. Therefore uniqueness gives \begin{align*} \gamma(t)=\beta(t) \end{align*} for every $t\in\mathbb R$. Unwinding the definitions of $\gamma$ and $\beta$ gives \begin{align*} \varphi(\exp_G(tX))=\exp_H(t\,d\varphi(X)) \end{align*} for every $t\in\mathbb R$. [/step]

[step:Set $t=1$ to obtain the particular case] Taking $t=1$ in the identity already proved gives \begin{align*} \varphi(\exp_G(1\cdot X))=\exp_H(1\cdot d\varphi(X)). \end{align*} Since scalar multiplication by $1$ is the identity on the vector spaces $\mathfrak g$ and $\mathfrak h$, this is exactly \begin{align*} \varphi(\exp_G X)=\exp_H(d\varphi(X)). \end{align*} This completes the proof. [/step]