The trivial subgroup is closed (in the sense of topology)

Question

I try to understand the proof of the below statement:

Statement: Let $\phi: G \rightarrow H$ be a homomorphism of Lie groups. Then the kernel of $\phi$ is a closed subgroup of $G$.

Proof: Put $K= $ Ker $\phi$. Then $K$ is a subgroup of $G$. Now $\phi$ is continuous and $\lbrace e_{H} \rbrace$ is a closed subset of $H$. Hence, $K = \phi^{-1}(\lbrace e_{H}\rbrace)$ is a closed subset of $G$.

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My confusion: I can't see why $\lbrace e_{H} \rbrace$ is a closed subset of $H$. How is this "straight forward"?

Answer 1

Lie groups are differentiable manifolds, particularly they are Hausdorff spaces. In a Hausdorff space, every singleton is closed. Hence every $\{h\}\subseteq H$ is closed, e.g. $\{e_H\}$ is closed.

Answer 2

For a more specific proof, we prove that the complement of $e_H$ is open by showing every point in the complement has an open neighborhood around it not containing $e_H$. Suppose $x\in H$ is a point not equal to $e_H$. Let $U$ be an open neighborhood of $x$ homeomorphic to $\mathbb{R}^n$. If $e_H$ is not in $U$, then we are done. Otherwise we may put a metric $d$ on $U$ that comes from the Euclidean metric via the homeomorphism with $\mathbb{R}^n$; suppose $d(x,e_H)=\epsilon$. Then the open ball of radius $\epsilon/2$ around $x$ is open in $H$ and does not contain $e_H$, so we are done.