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I have again a new question about the automorphism of covering space and the universal cover of a topological space $B$. Actually, let $p : X \rightarrow B$ the universal cover of $B$. I take a normal subgroup $H$ of $Aut(p)$, and we consider : $p_H : X/H \rightarrow B$. I would like to prove :

$$Aut(p_H) \cong Aut(p)/H $$

I know that as $p_H$ is a Galois covering space, then : $Aut(p_H) \cong \pi_1(B,b)/(\pi_H)_*(\pi_1(X/H, x)$, where $x \in p_H^{-1}(b)$.

We know that $Aut(p) \cong \pi_1(B,b)$, let say by a isomorphism $f$, and then we would like to say that $f(H) = (p_H)_*(\pi_1(X/H, x))$, and that's where I'm stuck.

Someone could help me, please ?

Thank you !

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You don't have to use the isomorphism $Aut(p)\simeq\pi_1(B,b)$ here. When you see the result, you could think about creating a morphism $Aut(p)\to Aut(p_H)$ with kernel $H$. Lets try to do so. Let $f\in Aut(p)$, we want to find/create a function $f^\prime\in Aut(p_H)$. We can proceed as follows:

There is a unique $g\in Aut(p_H)$ such that the following diagram is commutative $$\require{AMScd} \begin{CD} X @>{f}>> X\\ @V{\pi}VV @VV{\pi}V\\ X/H @>g>> X/H \end{CD}$$ where $\pi:X\to X/H$ is the natural projection.

Let $f^\prime$ be this unique element. $f^\prime$ is just the factorization of $\pi\circ f$ by $\pi$ (but it is not obvious at first that such a factorization exists, this comes from the hypothesis that $H$ is normal in $Aut(p)$).

You can now check that the mapping $\phi:f\mapsto f^\prime$ is a morphism from $Aut(p)$ to $Aut(p_H)$, by using uniqueness in the preceding proposition. If we prove that $H=Ker(\phi)$ we are done.

First by definition of $H$, every element $f\in H$ verifies $\pi\circ f=\pi$ which can be rewritten as $\pi\circ f=id_{X/H}\circ\pi$ which means $\phi(f)=id_{X/H}$, so $H\subset Ker(\phi)$. Now if $f\in Ker(\phi)$, let $x\in X$. Because $\pi(x)=\pi(f(x))$ we have $x\sim_H f(x)$ so there exists $h\in H$ such that $h(x)=f(x)$. Hence $f$ and $h$ are two covering transformations which coincide at one point, so $f=h\in H$ which means $ker(\phi)\subset H$. We just proved $H=ker(\phi)$.

If you have difficulties to prove the proposition let me know and I can give you hints.

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