$\phi^{-1}(\phi(U))=\bigcup_{k\in Ker(\phi)}kU$?

Question

Let $G$, $H$ be topological groups and let $\phi$ be a homomorphism between these two groups. Then is it true that if the image of an open $U$ by $\phi$ is $\phi(U)$ then we have $\phi^{-1}(\phi(U))=\bigcup_{k\in Ker(\phi)}kU$?

Answer 1

Yes, this is true, and you don't even need to assume $U$ is open. You just have the following equivalences:

$g\in \phi^{-1}(\phi(U))$ is true iff $\phi(g)\in\phi(U)$, which is true iff there exists $g'\in U$ such that $\phi(g)=\phi(g')$, which is true iff there is $g'\in U$ and $k\in \ker (\phi)$ such that $g=kg'$. In other words, $g\in\phi^{-1}(\phi(U))$ if and only if there is $k\in \ker(\phi)$ such that $g\in kU$. That is,$\phi^{-1}(\phi(U))=\bigcup_{k\in \ker(\phi)}kU$, as needed.

What you can say, if you want to say something about topology, is that this proves that if $U$ is indeed open, then $\phi^{-1}(\phi(U))$ is open as well, since it is a union of open sets.