# Is the Lie algebra morphism induced by injective/surjective Lie group morphism also injective/surjective?

I want to show that if $$f$$ is an injective/surjective Lie group homomorphism than $$f_*$$ is injective/surjective.

$$\require{AMScd}$$ $$\begin{CD}G @>f>> H\\ @AexpAA @AAexpA\\ Lie(G) @>>f_*> Lie(H) \end{CD}$$

Take open sets $$U,V,U', V'$$ such that $$V=exp(U), V'=exp(U'), f(V)=V', f_*(U)=U'$$ and $$exp|_U, exp|_V$$ are diffeomorphisms.

injectivity : Suppose $$f_*(x)=f_*(y)$$ (where $$f_*(x),f_*(y)\in V$$. Then take $$t$$ small enough such that $$tx,ty\in U$$. Then by the diagram, we have $$exp^{-1}(f(exp(tx)))=exp^{-1}(f(exp(ty)))$$ so $$tx=ty$$ i.e. $$x=y$$.

surjectivity : Consider $$y\in V\subset Lie(H)$$. Now chase back in the diagram a preimage of $$y$$ with respect to $$f_*$$. Then since every $$\hat{y}$$ can be written as $$ty$$ for some $$t\in \mathbb{R}$$ and some $$y \in V$$ we are done.

Is this correct

What about the converse? f_* injective/surjective $$\implies$$ $$f$$ injective/surjective?

• The homomorphism $C_2\to C_3$ is neither injective nor surjective, but the isomorphism on Lie algebras $0\to 0$ is an isomorphism. Thus for the converse you should probably restrict to connected Lie groups (you probably meant this anyway - but I thought it worth making explicit).
– tkf
Jul 19, 2020 at 2:06

To answer one of your questions: If $$f_*$$ is surjective and $$H$$ is connected then $$f$$ is surjective:

By commutativity of your diagram, $$f$$ must be surjective onto exp(Lie$$(H)$$), so the image of $$f$$ contains an open neighbourhood $$U\subseteq H$$ of $$e\in H$$. This will generate a subgroup $$H'\subset H$$, which is open: for any $$y\in H'$$, we have $$yU$$ a neighbourhood of $$y$$ contained in $$H'$$.

Then each coset of $$H'$$ is open (as multiplication by $$y\in H$$ is a homeomorphism), so the union of all non-trivial cosets is open, implying that $$H'$$ is closed.

We conclude that $$H'$$ is a non-empty open, closed, subset of a connected space $$H$$, so $$H=H'$$ and $$H'\subseteq$$im$$(f)$$.

As exemplified in the comments, the requirement that $$H$$ is connected is necessary. Otherwise, the inclusion of the connected component of $$e\in H$$ will not be surjective.

To answer another of your questions, even if $$G,H$$ are both simply connected, $$f_*$$ injective does not imply that $$f$$ is injective. Consider $$f\colon \mathbb{R}\to SU(2)$$ mapping: $$x\in \mathbb{R}\mapsto \left(\begin{array}{cc}e^{ix}&0\\0&e^{-ix}\end{array}\right)$$

• I don't understand the last point. Isn't $f_*$ in this case $x\in \mathbb{R}\mapsto \begin{pmatrix}ie^{ix} & 0\\ 0& -ie^{-ix} \end{pmatrix}$ which is also not injective? Jul 19, 2020 at 13:30
• Almost: $f_*$ is the linear map sending the tangent vector $\frac{\partial}{\partial x}$ at $0\in \mathbb{R}$ to $\left.\left(\begin{array}{cc}ie^{ix}&0\\0&-ie^{-ix}\end{array}\right)\right|_{x=0}=\left(\begin{array}{cc}i&0\\0&-i\end{array}\right)$. That is $$f_*\left(\lambda\frac{\partial}{\partial x}\right)=\lambda \left(\begin{array}{cc}i&0\\0&-i\end{array}\right),\qquad\forall\lambda\in \mathbb{R},$$ which is injective.
– tkf
Jul 19, 2020 at 13:57
• Okay, is there also a counter-example for surjectivity? Jul 19, 2020 at 16:09
• As the answer shows, if $H$ is connected then there are no counterexamples for surjectivity: $f_*$ surjective implies $f$ surjective. On the other hand let $f$ be the inclusion of Lie groups $\mathbb{R}\to \mathbb{R} \times C_2$. Then $f_*$ is an isomorphism, but $f$ is not surjective.
– tkf
Jul 19, 2020 at 17:25