Homomorphism of matrix groups is determined by its differential

Question

May I use some help on how to prove the Ex 7.15 in page 111 from "Matrix groups for undergraduate, 1st" by Tapp ?

Let $G_1$ and $G_2$ be matrix groups with Lie algebras $g_1$ and $g_2$ respectively. Let $f:G_1\rightarrow G_2$ be a $C^1$ homomorphism. Denote the differential of $f$ at unit element $I$ by $df_I:g_1\rightarrow g_2$. Prove that for all $v\in g_1$, $$f(e^v)=e^{df_I(v)}.$$

I attempted to expand the exponential on both sides using the formula

$e^v = I+v+1/2v^2+1/6v^3 +...$

and got

$f(v)=df_I(v)$

which I have no clue to prove.

Any tips ?

Answer 1

If $v\in g_1$ then $t\mapsto e^{tv}$ is a Lie group homomorphism from $\Bbb R$ (under addition) to $G_1$, and all Lie group homomorphisms from $\Bbb R$ to $G_1$ are of this form. Take the composition with this with $f$ and we get the a Lie group homomorphism from $\Bbb R$ to $G_2$. It must equal $t\mapsto e^{tw}$ for some $w\in G_2$. But considering the derivative ay zero gives us $w=df(v)$. Thus $$f(e^{tv})=e^{t df(v)}.$$ Now take $t=1$.