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.

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

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 $S^1$.


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