[proofplan] We view $\mathfrak h$ and $\mathfrak m$ as finite-dimensional real vector spaces and use the given matrix Lie group structure on $G$. The map $(X,Y)\mapsto \exp(X)\exp(Y)$ is smooth because the matrix exponential is smooth on the [Lie algebra](/page/Lie%20Algebra) and multiplication is smooth on $G$. Its differential at $(0,0)$ is the direct-sum addition map $(X,Y)\mapsto X+Y$, which is an isomorphism from $\mathfrak h\oplus\mathfrak m$ onto $\mathfrak g=T_I G$. The [inverse function theorem](/theorems/51) then gives the required adapted exponential coordinates. [/proofplan]

[step:Define the local product map and verify smoothness] Let \begin{align*} \Psi:\mathfrak h\times\mathfrak m\to G,\qquad \Psi(X,Y)=\exp(X)\exp(Y). \end{align*} Here $\mathfrak h\times\mathfrak m$ is regarded as a finite-dimensional real [vector space](/page/Vector%20Space) with its product smooth structure. Since $\mathfrak h,\mathfrak m\subset\mathfrak g$, the matrix exponential satisfies $\exp(X)\in G$ and $\exp(Y)\in G$ for all $X\in\mathfrak h$ and $Y\in\mathfrak m$. Therefore $\Psi$ is well-defined. The matrix exponential restricted to $\mathfrak g$ is smooth, and multiplication $G\times G\to G$ is smooth for a matrix Lie group. Thus $\Psi$ is a smooth map from the manifold $\mathfrak h\times\mathfrak m$ to the manifold $G$. [/step]

[step:Compute the differential at the origin as the direct-sum addition map]We compute the differential \begin{align*} d\Psi_{(0,0)}:\mathfrak h\times\mathfrak m\to T_I G. \end{align*} Let $A\in\mathfrak h$ and $B\in\mathfrak m$. Define the smooth curve \begin{align*} c:\mathbb R\to G,\qquad c(t)=\Psi(tA,tB)=\exp(tA)\exp(tB). \end{align*} Using the first-order expansion of the matrix exponential in the ambient vector space $M(n,\mathbb C)$, \begin{align*} \exp(tA)=I+tA+o(t) \end{align*} and \begin{align*} \exp(tB)=I+tB+o(t), \end{align*} we obtain \begin{align*} c(t)=(I+tA+o(t))(I+tB+o(t))=I+t(A+B)+o(t). \end{align*} Hence the ambient derivative of $c$ at $0$ is $A+B$. Under the standard identification $T_I G=\mathfrak g$ for a matrix Lie group, this gives \begin{align*} d\Psi_{(0,0)}(A,B)=A+B. \end{align*}[/step]

[guided]The point of this step is to check exactly what the candidate coordinate map does infinitesimally. We must compute the derivative as a tangent vector to $G$ at $I$, and because $G$ is a matrix Lie group, tangent vectors at $I$ are represented by matrices in $\mathfrak g$. Fix $A\in\mathfrak h$ and $B\in\mathfrak m$. To evaluate the differential in the direction $(A,B)$, we use the curve \begin{align*} c:\mathbb R\to G,\qquad c(t)=\Psi(tA,tB)=\exp(tA)\exp(tB). \end{align*} This curve satisfies $c(0)=I$, so its velocity at $0$ is an element of $T_I G$. The matrix exponential has the first-order expansion \begin{align*} \exp(tA)=I+tA+o(t) \end{align*} in the ambient vector space $M(n,\mathbb C)$ as $t\to 0$, and similarly \begin{align*} \exp(tB)=I+tB+o(t). \end{align*} Multiplying these two expansions in the matrix algebra gives \begin{align*} c(t)=(I+tA+o(t))(I+tB+o(t)). \end{align*} The product of the two first-order terms is $t^2AB$, which is absorbed into $o(t)$ after division by $t$ and passage to the limit. Therefore \begin{align*} c(t)=I+t(A+B)+o(t). \end{align*} Thus the velocity of $c$ at $0$ is the matrix $A+B$. Since the tangent space at the identity of the matrix Lie group is identified with its Lie algebra, $T_I G=\mathfrak g$, we have \begin{align*} d\Psi_{(0,0)}(A,B)=A+B. \end{align*} This computation uses only the vector-space splitting and does not require $\mathfrak h$ or $\mathfrak m$ to be Lie subalgebras.[/guided]

[step:Use the direct-sum decomposition to identify the differential as an isomorphism] The map \begin{align*} L:\mathfrak h\times\mathfrak m\to\mathfrak g,\qquad L(A,B)=A+B \end{align*} is linear. Since $\mathfrak g=\mathfrak h\oplus\mathfrak m$, every $Z\in\mathfrak g$ has a unique representation $Z=A+B$ with $A\in\mathfrak h$ and $B\in\mathfrak m$. Hence $L$ is bijective. Therefore \begin{align*} d\Psi_{(0,0)}:\mathfrak h\times\mathfrak m\to T_I G \end{align*} is a linear isomorphism. [/step]

[step:Apply the inverse function theorem to obtain adapted exponential coordinates] The inverse function theorem for smooth maps between finite-dimensional smooth manifolds applies to $\Psi$ at $(0,0)$ because $\Psi$ is smooth and $d\Psi_{(0,0)}$ is a linear isomorphism. Therefore there exists an open neighbourhood $W\subset \mathfrak h\times\mathfrak m$ of $(0,0)$ such that $\Psi|_W$ is a diffeomorphism from $W$ onto an open neighbourhood $U_G\subset G$ of $I$. Choose open neighbourhoods $U_{\mathfrak h}\subset\mathfrak h$ of $0$ and $U_{\mathfrak m}\subset\mathfrak m$ of $0$ such that \begin{align*} U_{\mathfrak h}\times U_{\mathfrak m}\subset W. \end{align*} Replacing $U_G$ by the open subset $\Psi(U_{\mathfrak h}\times U_{\mathfrak m})$ of $G$, the restricted map \begin{align*} \Phi:U_{\mathfrak h}\times U_{\mathfrak m}\to U_G,\qquad \Phi(X,Y)=\exp(X)\exp(Y) \end{align*} is a diffeomorphism. This is exactly the asserted adapted exponential coordinate chart. [/step]