TITLE: Question on Lie group homomorphism and Exponential Map Commutation
QUESTION [5 upvotes]: I have a question regarding the theorem which states that, if $\phi: G\rightarrow H$ is a Lie group homomorphism, then for a $X\in\mathfrak{g}$, $\phi\exp{X}=\exp{D\phi X}$, where $D\phi$ is the differential of $D\phi$.
The question is whether the above relation ($\phi\exp{X}=\exp{D\phi X}$) holds only for Lie group homomorphism $\phi$, since the proof I have come up with does not necessarily require $\phi$ to be a Lie algebra homomorphism.
My proof is as follows.

I first identified $\mathfrak{g}$ with $ T_e{G}$, and tangent vector as an equivalence class of a curve divided by the tangency at $e\in G$.

Thus I identified $\exp{tX}$ as a tangent vector of $e\in G$ then used the following equation. $\phi\exp{tX}=[\phi\circ X]=[D\phi X]=\exp(tD\phi X)$.

Then, I plugged 1 in t.


If the above proof is correct, it seems $\phi$ need not be a group homomorphism, but I am not sure about my proof.
Could somebody please make it clear?
Thank you in advance.

REPLY [3 votes]: The statement is false if you don't assume that $\phi$ is a group homomorphism. If $\phi$ is just any differentiable map from $G$ into $H$, you have no reason to suppose that $\phi(e_G)=e_H$ and therefore $\phi\circ X$ doesn't have to be a curve passing through $e_H$.
As an example, consider the map$$\begin{array}{rccc}\phi\colon&GL_2(\Bbb R)&\longrightarrow&(\Bbb R\setminus\{0\},\times)\\&M&\mapsto&\operatorname{tr}(M)-1.\end{array}$$Note that, in this case, we do have $\phi(\operatorname{Id}_2)=1$. But$$\phi\left(\exp\left(\begin{bmatrix}ta&0\\0&tb\end{bmatrix}\right)\right)=\phi\left(\begin{bmatrix}e^{ta}&0\\0&e^{tb}\end{bmatrix}\right)=e^{ta}+e^{tb}-1,$$and, since $D\phi_{\operatorname{Id}_2}=\operatorname{tr}$,$$\exp\left(tD\phi_{\operatorname{Id}_2}\left(\begin{bmatrix}a&0\\0&b\end{bmatrix}\right)\right)=e^{t(a+b)},$$which is different from $e^{ta}+e^{tb}-1$.