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Question: Suppose that $\varphi:G\to H$ is a Lie group homomorphism such that $G$ is simply connected and $\varphi_*:{\frak g}\to{\frak h}$ is injective. Is $\varphi$ injective?

Since Lie group homomorphisms have constant rank and $\varphi_*$ is injective it is automatic that $(d\varphi)_g$ is injective at all point $g\in G$. I am not sure if that helps. We must somehow use that it is simply connected since for example a non-trivial covering $\tilde{G}\to G$ would produce an injective Lie algebra homomorphism without itself being injective.

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  • $\begingroup$ You seem to have the answer already, as a nontrivial covering is exactly such a map, so you only need prove that one exists, i.e., give a counterexample. $\endgroup$ – Travis Willse Feb 27 '16 at 8:57
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No. For example, $$\varphi:{\Bbb R}\to S^1,\quad t\mapsto e^{it},$$ induces a bijective Lie algebra homomorphism.

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