If $df_e: T_eG_1\to T_eG_2$ is surjective, and $G_2$ connected Lie group, then $f$ is surjective.

Question

Suppose $f$ is a morphism of Lie groups, and $df_e\colon T_eG_1\to T_eG_2$ is a surjective map of the tangent spaces of two Lie groups, where $G_2$ is connected.

I read that by the Inverse Function Theorem, $df_e$ surjective implies $f$ is surjective onto a neighborhood $U$ of $e$ in $G_2$. Then since $G_2$ is connected, it is well known that $U$ generates $G_2$. Since $U$ is the image of $f$, hence a subgroup, $U=G_2$, and $f$ is surjective.

My question is, how does the Inverse function theorem come into play? I know that the IFT for manifolds says that if $df_p\colon T_pM\to T_{f(p)}N$ is invertible, then there exist connected nhbds $U\ni p$ and $V\ni f(p)$ such that $f|U\colon U\to V$ is a diffeomorphism. I don't see how we can apply it if $df_e$ is just known to be surjective.

Answer 1

• First $f(G_1)$ is open in $G_2$.

Lemma Every submersion is an open map.
Proof There are clues in Lee's book: Introduction to Smooth Manifolds, P169. The outline of proof is below. $$\text{Inverse Function Theorem}\Rightarrow\text{Rank Theorem}\Rightarrow\text{Lemma}$$

If $df_e:T_eG_1\rightarrow T_eG_2$is surjective map(or equivalently $\text{rank}_ef=\dim G_2$, or $f$ is a submersion at $e$), then by using $L_g$ of $G_1$ and $L_{f(g)}$ of $G_2$, we can get the conclusion $df_g:T_gG_1\rightarrow T_{f(g)}G_2$is surjective map(or equivalently $\text{rank}_gf=\dim G_2$, or $f$ is a submersion).

Now from the Lemma, $f(G_1)$ is open in $G_2$.

• $f(G_1)$ is closed in $G_2$.

Lemma Let $G$ be a topological group and $H$ is its open subgroup. Then we have $H$ is closed in $G$.
Proof If $H$ is open, then $gH$ is also open. Because the complement of $H$ is the union $gH$, $H$ is closed in $G$.

Since $f(G_1)$ is open in $G_2$, we get the conclusion $f(G_1)$ is closed in $G_2$.

• $f(G_1)=G_2$

Hint $G_2$ is connected.