# the kernel of the covering map is isomorphic to the covering transformation group of topological group

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 $$Ker\ p$$?

• Do you mean $p\colon\tilde G\to G$? Jun 8, 2022 at 7:07
• yes, I edit it. Jun 8, 2022 at 10:44
• 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. Jun 8, 2022 at 10:49