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 $X/H \rightarrow X/G$.

  • $\begingroup$ This may be a shortcut, but see Hatcher's Algebraic Topology, Prop. 1.39 $\endgroup$ May 31, 2018 at 21:57


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