I'd like to prove the following statement:
Suppose that $G_{1}$ and $G_{2}$ are Lie groups with corresponding Lie algebras $\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$. Suppose that $f : G_{1} \to G_{2}$ is a smooth homomorphism. If $df_{I} : \mathfrak{g}_{1} \to \mathfrak{g}_{2}$ is bijective, then $df_{g} : \mathcal{T}_{g}\left( G_{1} \right) \to \mathcal{T}_{f(g)}\left( G_{2} \right)$ is bijective for all $g \in G_{1}$
This seems to be a commonly-used result in the textbooks I'm reading, but I can't find a proof of it. How do you prove this?
If I take any point $g \in G_{1}$, and a tangent vector $x \in \mathcal{T}_{g}(G_{1})$, then I know that: $$ df_{g}(x) = (f \circ \gamma)^{\prime}(0) $$ Where $\gamma : \mathbb{R} \to G_{1}$ is a curve such that $\gamma(0)=I$ and $\gamma^{\prime}(0) = x$. My thinking is a somehow need to "translate" $\mathfrak{g}_{1}$ to $\mathcal{T}_{g}(G_{1})$....any help is much appreciated.
If you look at the equation $f(gh)=f(g)f(h)$, then this can be written in terms of left translations as $f\circ\lambda_g=\lambda_{f(g)}\circ f$ and thus as $f=\lambda_{f(g)}\circ f\circ\lambda_{g^{-1}}$. Differentiating this equation in the point $g$, you get $df(g)$ written as a composition of $df(e)$ with two maps which are linar isomorphisms since left translations are diffeomorphisms.