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

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.

• I think you forgot to mention that $f\colon G_1\to G_2$ is a Lie homomorphism. Is there no assumption on dimensions of $G_1,G_2$? If they are equal, then clearly the theorem applies. Otherwise, the proposition will remain true, as having maximal rank is an open condition about the derivative, although there is some reasoning involved. – tomasz Oct 6 '14 at 7:30
• Actually you don't use the IFT but its conclusion: Rank Theorem. But I have some questions. First is $G_1$ compact? Second is $f$ a homomorphism. – gaoxinge Oct 6 '14 at 7:34
• @tomasz Sorry, yes $f$ is a morphism of Lie groups. There is no assumption on the dimensions of $G_i$. On second thought, don't Lie group homomorphisms always have constant rank, so if $df_e$ is surjective, that would imply $f$ is actually a smooth submersion, hence an open map? – dez Oct 6 '14 at 7:37
• @gaoxinge $G_1$ is not assumed to be compact, but $f$ is a homomorphism. This is actually corollary 2.10 of Kirillov's Lie Groups and Lie Algebras, and he mentions IFT, so I was confused. – dez Oct 6 '14 at 7:39

• 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.