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 $\phi^{-1}(\phi(U))=\bigcup_{k\in Ker(\phi)}kU$?
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If $g \in \phi^{-1}(\phi(U))$, then $\exists u \in U:\phi(g) = \phi(u) $, so that $\phi(gu^{-1}u)=\phi(gu^{-1})\phi(u)=\phi(u), \phi(gu^{-1})=e$. So $g=hu$ where $h=gu^{-1}\in\ker(\phi)$. It seems unrelated to topology. – Colliot Jul 08 '20 at 18:15
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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.
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