2
$\begingroup$

I am trying to prove that $X/H \rightarrow X/G$ is a normal covering if and only if $H$ is a normal subgroup of $G$.

In my particular case, $X$ is the universal covering of $X/G$ and $G$ is the deck transformation group of $X$ over $X/G$. I do not want to deal with the identification between deck transformation groups and the fundamental group where the proof is given in a lot of sources.

I found this answer that deals with a very similar problem to mine in item 3.

Group action and covering spaces

However, I can not see why for a deck transformation $\phi \in G= Deck(X \rightarrow X/G)$ there is an induced deck transformation $\tilde{\phi} \in Deck(X/H \rightarrow X/G)$. My main concern is why the map $\phi$ preserves $H$-equivalence classes and gives a well defined map $X/H \rightarrow X/G$.

$\endgroup$
  • $\begingroup$ This may be a shortcut, but see Hatcher's Algebraic Topology, Prop. 1.39 $\endgroup$ – Igor Sikora May 31 '18 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.