Homeomorphism between neighborhoods in topological groups

Question

Let $G$ and $H$ be topological groups and $U$, $V$ neighborhoods of $e$ in $G$ and $H$ respectively. Note we are not assuming neighborhoods are open. Suppose $p:G\to H$ is a continuous group homomorphism such that $p|_U:U\to V$ is a homeomorphism.

a) Prove there exists open neighborhoods $U'$ and $V'$ of $e$ in $G$ and $H$ such that $p|_{U'}:U'\to V'$ is a homeomorphism.

b) Prove that $p$ is open

c) Prove that if $G$ is connected then the kernel of $p$ is in the centre of $G$. Hint, the kernel is a normal subgroup and is discrete.

My attempt for a) was to use the connected component $C_e$ of $e$ in G. Since this is open we can define $U'=\{cu|c\in C_e,u\in U\}$ and $V'$ similarly but with $C_e$ of $e\in H$ and $v\in V$. So these sets will be open but I don't think these work. If we take $cv\in V'$. Then by surjectivity of $p|_U$ we have a $u\in U$ such that $p(u)=v$. But we cannot guarantee this about $C_e$ so $p|_{U'}$ is not necessarily surjective. I think this may be the correct approach but i need to choose my elements of $G$ more carefully? Any help would be appreciated. I have no idea how one would do b) or c).

Answer 1

Let's see some failed attempts to answer item "a"... let $f = p|_U$. It would be nice to have a notation for restricting the domain and co-domain of a function... :-)

First try) Since $U$ is a neighbourhood of $e$, it contains an open neighbourhood $U'$ of $e$. Now take $V' = f(U')$. Since $f$ is a homeomorphism, $V'$ is open. Wait!!! $V'$ is open in $V$, not necessarily in $H$.

Second try) Since $V$ is a neighbourhood of $e$, it contains an open neighbourhood $V'$ of $e$. Now, take $U' = p^{-1}(V')$. Since $p$ is continuous, $U'$ is open. But wait! It might be that $U' \not \subset U$.

Your try) [...]

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For item "b"... in case of groups, I find it always best to talk about neighbourhoods instead of open sets. And then it would be nice to talk about a function being open at one point, the same way it can be continuous at one point. It is handy to know (see below) that a topological group homomorphism needs only to be open at one point to be open everywhere.

Denote by $\mathcal{G}_g$ the family of neighbourhoods of $g \in G$. And by $\mathcal{H}_h$ the neighbourhoods of $h \in H$.

Let us also use the notation \begin{equation*} \mathcal{A}|_B = \{A \cap B\,|\,A \in \mathcal{A}\}. \end{equation*} And notice that since $U$ and $V$ are neighbourhoods, \begin{align*} \mathcal{G}_e|_U &\subset \mathcal{G}_e \\ \mathcal{H}_e|_V &\subset \mathcal{H}_e. \end{align*}

Since $f$ is a homeomorphism, it takes $\mathcal{G}_e|_U$ to $\mathcal{H}_e|_V$ (not important here, but it is actually a bijection between those two families). Since a super set of a neighbourhood is also a neighbourhood, $p$ takes $\mathcal{G}_e$ to $\mathcal{H}_e$: \begin{equation*} p(\mathcal{G}_e) \subset \mathcal{H}_e. \end{equation*} This means that $p$ is open at $e$.

And since $p$ is a group homomorphism, \begin{equation*} p(\mathcal{G}_g) = p(g\mathcal{G}_e) = p(g)p(\mathcal{G}_e) \subset p(g) \mathcal{H}_e = \mathcal{H}_{p(g)}. \end{equation*} Therefore, $p$ is open everywhere, and therefore, open.

(this actually also proves item "a"!)

It all looked overly complicated. All that is being said, in a slightly complicated way is:

Just grab a neighbourhood of the identity $A \subset G$. Then, $A \cap U$ is taken to a neighbourhood of the identity of $H$. (fill the gaps)... therefore, $p$ is open at $e$.

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For item "c", let $K$ be the kernel. Notice that $K$ is discrete because if $p(g) = 1$, then no other element in $gU$, except for $g$ is in $K$: $$p(gU) = p(g)p(U) = p(U),$$ and $p|_U$ is injective.

To see that $K$ is in the center, for $k \in K$, consider the function \begin{align*} c: G &\rightarrow K \\ g &\mapsto gkg^{-1}. \end{align*} since $c$ is continuous, $G$ connected and $K$ discrete, $c$ is constant. Since $c(e) = k$, we have that for every $g \in G$, $gkg^{-1} = k$.