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Question: Suppose that $\varphi:G\to H$ is a surjective Lie group homomorphism whose differential $\varphi_*:{\frak g}\to{\frak h}$ is a Lie algebra isomorphism. Is $\varphi$ necessarily a smooth covering map?

Since Lie group homomorphism have constant rank, it follows that $\varphi$ is a local diffeomorphism. Hence, it is a smooth covering map if and only if it is a topological covering map. But is it?

I tried to get a counterexample but I couldn't figure out any.

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  • $\begingroup$ There isn't any nonzero map from $S^1$ to $\mathbb{R}$ that's a Lie group homomorphism. In fact it suffices to require that $\varphi$ is a Lie algebra isomorphism and that $H$ is connected; $\varphi$ is then automatically surjective. $\endgroup$ – Qiaochu Yuan Mar 1 '16 at 16:17
  • $\begingroup$ @QiaochuYuan Right, thanks. I deleted this stupid remark. $\endgroup$ – SHP Mar 1 '16 at 16:26
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Yes, it is a covering map.

First, let $K\subseteq G$ denote the kernel of this map, which is, of course, a normal subgroup. Then, the Lie algebra $\mathfrak{k}\subseteq \mathfrak{g}$ is an ideal in $\mathfrak{g}$. Then the map $\phi_\ast$ indcues an isomorphism $\overline{\phi}_\ast:\mathfrak{g}/\mathfrak{k}\rightarrow \mathfrak{h}$. Since the map $\phi_\ast:\mathfrak{g}\rightarrow \mathfrak{h}$ factors through $\overline{\phi}_\ast$, and $\phi_\ast$ is an isomorphism, it follows that $\mathfrak{k}$ is trivial. In other words, $K$ is discrete.

Now, it's a general fact (see, for example, Brocker and tom Dieck's Representations of Compact Lie Groups, Theorem 4.3, page 33) that is $G'$ is closed subgroup $G$, then the natural map $G\rightarrow G/G'$ is a principal bundle with fiber $G'$. Taking $K = G'$, we get a principal bundle $K\rightarrow G\rightarrow G/K = H$. Since the fiber is $K$ is discrete, this is a covering.

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