# Question about Lie subgroups and homeomorphisms

I am reading about Lie groups and came across the definition of a Lie subgroup, which is simply an injective homomorphism of Lie groups H $$\hookrightarrow$$G. A comment is made that H may not be homeomorphic to its iamge. with the example f:$$\mathbb{Z} \rightarrow S^1$$, where n $$\mapsto e^{in}$$. The image is dense, which I understand.

Is the reason the example is not a homeomorphism because f$$^{-1}$$ is not continuous? For instance, there is no neighborhood of (1,0) whose preimage is a neighborhood of 0?

This example can be generalized to the torus if we think of the identification of the unit square and have a line through the origin with irrational slope. It seems this is an important example and I would like to be sure that I understand it. Thanks for any help and insight.

• $f$ is not surjective, for example $e^{i\pi}\in S^1$ is not in the image of $f$. – mouthetics Jan 2 at 16:09
• Sorry, I meant homeomorphic onto its image. I have edited. – Joel Pereira Jan 2 at 16:12

## 1 Answer

Yes, $$f^{-1}:\mathrm{Im}(f)\longrightarrow \mathbb{Z}$$ isn't continuous. For, $$\{ n\}$$ is open in $$\mathbb{Z}$$ but $$\{e^{in}\}$$ is not open in $$\mathrm{Im}(f)$$. If $$\{e^{in}\}$$ were open in $$\mathrm{Im}(f)$$ then $$\{e^{in}\}=\mathrm{Im}(f)\cap U$$ for some open subset $$U\subset S^1$$ (a posteriori $$U\neq \{e^{in}\}$$ as a singleton is not open in $$S^1$$). But this is impossible since $$\mathrm{Im}(f)$$ is dense in $$S^1$$.