[proofplan] We use exponential coordinates near the identity in both Lie groups. In those coordinates, the local multiplication law is described by the Baker--Campbell--Hausdorff series. Since $\Phi$ preserves Lie brackets, it carries every iterated bracket in the BCH series for $\mathfrak g$ to the corresponding iterated bracket in $\mathfrak h$. Therefore the coordinate map $X\mapsto \Phi X$ intertwines the local group laws, and conjugating back by the two exponential maps gives the required local diffeomorphism. [/proofplan]

[step:Choose exponential coordinates compatible with $\Phi$]Let \begin{align*} \exp_G:\mathfrak g\to G \end{align*} and \begin{align*} \exp_H:\mathfrak h\to H \end{align*} denote the Lie group exponential maps. Since $G$ and $H$ are finite-dimensional real Lie groups, the standard local diffeomorphism theorem for the Lie group exponential map applies to $\exp_G$ and $\exp_H$ at $0$. Hence there exist open neighbourhoods $A_0\subset\mathfrak g$ of $0$ and $B_0\subset\mathfrak h$ of $0$ such that \begin{align*} \exp_G|_{A_0}:A_0\to \exp_G(A_0) \end{align*} and \begin{align*} \exp_H|_{B_0}:B_0\to \exp_H(B_0) \end{align*} are diffeomorphisms onto open neighbourhoods of $e_G$ and $e_H$, respectively. Since $\Phi:\mathfrak g\to\mathfrak h$ is a linear isomorphism and $B_0$ is open with $0\in B_0$, the set \begin{align*} A_1:=A_0\cap \Phi^{-1}(B_0) \end{align*} is an open neighbourhood of $0$ in $\mathfrak g$. Shrinking $A_1$ if necessary, choose an open neighbourhood $A\subset A_1$ of $0$ such that the BCH product in $\mathfrak g$ is defined on the required local domain inside $A$, and such that the corresponding BCH product in $\mathfrak h$ is defined on $\Phi(A)\times\Phi(A)$. This shrinking is possible by the local BCH theorem for finite-dimensional Lie algebras, applied to $\mathfrak g$ and $\mathfrak h$; see [citetheorem:8798]. Define \begin{align*} U:=\exp_G(A) \end{align*} and \begin{align*} V:=\exp_H(\Phi(A)). \end{align*} Then $U$ is an open neighbourhood of $e_G$ in $G$, and $V$ is an open neighbourhood of $e_H$ in $H$.[/step]

[guided]The first task is to put both groups into coordinates where their local products can be compared. The natural coordinates near the identity are exponential coordinates. Let \begin{align*} \exp_G:\mathfrak g\to G \end{align*} and \begin{align*} \exp_H:\mathfrak h\to H \end{align*} be the exponential maps of the Lie groups $G$ and $H$. We use the standard local theorem that the exponential map is a diffeomorphism from a neighbourhood of $0$ in the [Lie algebra](/page/Lie%20Algebra) onto a neighbourhood of the identity in the Lie group. Thus there are open neighbourhoods $A_0\subset\mathfrak g$ and $B_0\subset\mathfrak h$ of $0$ such that \begin{align*} \exp_G|_{A_0}:A_0\to \exp_G(A_0) \end{align*} and \begin{align*} \exp_H|_{B_0}:B_0\to \exp_H(B_0) \end{align*} are diffeomorphisms onto open identity neighbourhoods. This is the coordinate chart input for the proof. We also need the [linear map](/page/Linear%20Map) $\Phi$ to carry the chosen domain in $\mathfrak g$ into the chosen domain in $\mathfrak h$. Since $\Phi$ is continuous and invertible, the set \begin{align*} A_1:=A_0\cap \Phi^{-1}(B_0) \end{align*} is an open neighbourhood of $0$ in $\mathfrak g$. If $X\in A_1$, then $X\in A_0$ and $\Phi X\in B_0$, so both $\exp_G X$ and $\exp_H(\Phi X)$ are represented in the chosen exponential coordinate charts. Finally, we shrink $A_1$ to an open neighbourhood $A$ of $0$ on which the BCH multiplication law is valid. The relevant local BCH result is [citetheorem:8798]: for each finite-dimensional Lie algebra, after shrinking to a sufficiently small neighbourhood of $0$, the expression $\operatorname{BCH}(X,Y)$ is defined whenever $X$, $Y$, and their product are kept in that neighbourhood, and it gives the local Lie group law. We apply this theorem to the finite-dimensional Lie algebras $\mathfrak g$ and $\mathfrak h$. Because $\Phi$ is a linear homeomorphism, we choose $A\subset A_1$ small enough that whenever $X,Y\in A$ and $\operatorname{BCH}_{\mathfrak g}(X,Y)\in A$, the pair $(\Phi X,\Phi Y)$ lies in the BCH domain for $\mathfrak h$ and $\operatorname{BCH}_{\mathfrak h}(\Phi X,\Phi Y)\in \Phi(A)$. Now set \begin{align*} U:=\exp_G(A) \end{align*} and \begin{align*} V:=\exp_H(\Phi(A)). \end{align*} Because the restricted exponential maps are diffeomorphisms on these neighbourhoods, $U$ and $V$ are open neighbourhoods of $e_G$ and $e_H$.[/guided]

[step:Define the candidate local diffeomorphism]Define \begin{align*} F:U\to V,\quad F(\exp_G X):=\exp_H(\Phi X)\text{ for }X\in A. \end{align*} This is well-defined because $\exp_G|_A$ is injective. Equivalently, \begin{align*} F=\exp_H\circ \Phi\circ (\exp_G|_A)^{-1}. \end{align*} Since $\exp_G|_A$, $\Phi$, and $\exp_H|_{\Phi(A)}$ are diffeomorphisms onto their images, $F$ is a diffeomorphism. Its inverse is \begin{align*} F^{-1}=\exp_G\circ \Phi^{-1}\circ (\exp_H|_{\Phi(A)})^{-1}. \end{align*}[/step]

[guided]The intended map is forced by the requirement that its differential at the identity be $\Phi$: in exponential coordinates it must send the coordinate vector $X\in A\subset\mathfrak g$ to the coordinate vector $\Phi X\in \Phi(A)\subset\mathfrak h$. Thus define \begin{align*} F:U\to V,\quad F(\exp_G X):=\exp_H(\Phi X)\text{ for }X\in A. \end{align*} This formula is well-defined because $\exp_G|_A:A\to U$ is injective, so each element of $U$ has a unique exponential coordinate $X\in A$. In composition form, \begin{align*} F=\exp_H\circ \Phi\circ (\exp_G|_A)^{-1}. \end{align*} Each factor in this composition is a diffeomorphism onto its image: $\exp_G|_A$ and $\exp_H|_{\Phi(A)}$ by the chosen exponential coordinates, and $\Phi$ because it is a linear isomorphism between finite-dimensional real vector spaces. Therefore $F$ is a diffeomorphism from $U$ onto $V$. Its inverse is explicitly \begin{align*} F^{-1}=\exp_G\circ \Phi^{-1}\circ (\exp_H|_{\Phi(A)})^{-1}. \end{align*}[/guided]

[step:Show that $\Phi$ preserves the BCH product]Let $\operatorname{BCH}_{\mathfrak g}$ denote the BCH series formed using the bracket on $\mathfrak g$, and let $\operatorname{BCH}_{\mathfrak h}$ denote the BCH series formed using the bracket on $\mathfrak h$. Since $\Phi$ is a Lie algebra homomorphism, for every $X,Y\in\mathfrak g$, \begin{align*} \Phi([X,Y])=[\Phi X,\Phi Y]. \end{align*} By induction on the bracket depth, $\Phi$ carries every iterated bracket in $X$ and $Y$ to the corresponding iterated bracket in $\Phi X$ and $\Phi Y$. Since the BCH series is a universal Lie series in iterated brackets, on the chosen BCH convergence neighbourhood from [citetheorem:8798] we have \begin{align*} \Phi(\operatorname{BCH}_{\mathfrak g}(X,Y))=\operatorname{BCH}_{\mathfrak h}(\Phi X,\Phi Y). \end{align*}[/step]

[guided]We now verify that the coordinate change $X\mapsto \Phi X$ respects the BCH products. Let $\operatorname{BCH}_{\mathfrak g}$ be the BCH series computed with the Lie bracket of $\mathfrak g$, and let $\operatorname{BCH}_{\mathfrak h}$ be the BCH series computed with the Lie bracket of $\mathfrak h$. The theorem [citetheorem:8798] gives the local BCH domains after the shrinking performed above, so the following expressions are defined for the pairs used in the multiplication argument. Because $\Phi:\mathfrak g\to\mathfrak h$ is a Lie algebra homomorphism, for every $X,Y\in\mathfrak g$, \begin{align*} \Phi([X,Y])=[\Phi X,\Phi Y]. \end{align*} Applying this identity repeatedly shows, by induction on bracket depth, that every iterated bracket in $X$ and $Y$ is carried by $\Phi$ to the same iterated bracket in $\Phi X$ and $\Phi Y$. For example, the bracket $[X,[X,Y]]$ is sent to \begin{align*} \Phi([X,[X,Y]])=[\Phi X,\Phi([X,Y])]=[\Phi X,[\Phi X,\Phi Y]]. \end{align*} The BCH series is universal: its terms are fixed rational linear combinations of these iterated brackets. Therefore applying $\Phi$ term by term to the BCH series in $\mathfrak g$ gives exactly the BCH series in $\mathfrak h$ evaluated at $\Phi X$ and $\Phi Y$. Hence, on the chosen convergence neighbourhood, \begin{align*} \Phi(\operatorname{BCH}_{\mathfrak g}(X,Y))=\operatorname{BCH}_{\mathfrak h}(\Phi X,\Phi Y). \end{align*}[/guided]

[step:Translate the BCH identity back to group multiplication] Let $g,h\in U$ and suppose $gh\in U$. Since $\exp_G|_A:A\to U$ is a diffeomorphism, there are unique $X,Y,Z\in A$ such that \begin{align*} g=\exp_G X,\quad h=\exp_G Y,\quad gh=\exp_G Z. \end{align*} By the BCH description of the local multiplication law in exponential coordinates from [citetheorem:8798], and by the choice of $A$, the condition $gh\in U$ implies \begin{align*} Z=\operatorname{BCH}_{\mathfrak g}(X,Y). \end{align*} Therefore \begin{align*} F(gh)=F(\exp_G Z)=\exp_H(\Phi Z). \end{align*} Using the BCH preservation identity from the previous step gives \begin{align*} \Phi Z=\operatorname{BCH}_{\mathfrak h}(\Phi X,\Phi Y). \end{align*} Applying the BCH multiplication formula in $H$ gives \begin{align*} \exp_H(\Phi Z)=\exp_H(\Phi X)\exp_H(\Phi Y). \end{align*} By the definition of $F$, \begin{align*} \exp_H(\Phi X)\exp_H(\Phi Y)=F(g)F(h). \end{align*} Combining these equalities yields \begin{align*} F(gh)=F(g)F(h). \end{align*} [/step]

[step:Conclude that $F$ is a local Lie group isomorphism] We have constructed open neighbourhoods $U\subset G$ of $e_G$ and $V\subset H$ of $e_H$, and a diffeomorphism $F:U\to V$. The preceding step proves that for all $g,h\in U$ satisfying $gh\in U$, \begin{align*} F(gh)=F(g)F(h). \end{align*} Thus $F$ is the required local product-preserving diffeomorphism. [/step]