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I'm trying to prove the following statement:

Suppose $G$ and $\tilde G$ is connected and locally path connected topological groups, and $p:\tilde G\rightarrow G$ is a covering map and a homomorphism between topological groups. Prove that: the kernel of $p$ is isomorphic to the covering transformation group $Aut(\tilde G,p)$.

Here are my attempts.Since the fundamental group of a topological group is an abelian group, we have $p_*(\pi_1(\tilde G))$ is a normal subgroup of $\pi_1(G)$. So we have the group isomorphism: $$ Aut(\tilde G,p)\cong \pi_1(G)/p_*(\pi_1(\tilde G)) $$ But how to prove the right-hand-side isomorphic to $Ker\ p$?

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    $\begingroup$ Do you mean $p\colon\tilde G\to G$? $\endgroup$
    – Kenta S
    Jun 8, 2022 at 7:07
  • $\begingroup$ yes, I edit it. $\endgroup$
    – Robin
    Jun 8, 2022 at 10:44
  • $\begingroup$ It is probably more useful to evaluate a covering transformation at the neutral element of $\tilde{G}$ to obtain a group homomorphism $\mathrm{Aut}(\tilde{G},p)\to \mathrm{ker}\, p$ and to show that this is an isomorphism. $\endgroup$ Jun 8, 2022 at 10:49

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