I refer to Exercise 1.3.24(a) of Hatcher's Algebraic Topology:

Given a covering space action of a group $G$ on a path-connected, locally path-connected space $X$, each subgroup $H \subseteq G$ determines a composition of covering spaces $X \to X/H \to X/G$. Show that every path-connected covering space between $X$ and $X/G$ is isomorphic to $X/H$ for some subgroup $H \subseteq G$.

The common tactic is to define $H \subseteq G$ as the subgroup of deck transformations of the covering map $p : X \to Y$, which induces a second covering map $p' : X \to X/H$, then show that $p$ and $p'$ are isomorphic. Then, define maps $f : Y \to X/H$ and $g : X/H \to Y$ as follows: For $x \in p^{-1}(y)$, define $f(y) = Hx$, and $g(Hx) = p(x)$. We conclude by showing that they are well-defined, continuous, and induces the required isomorphism.

My main issue lies in the well-definedness part. The well-definedness of $g$ is clear, but I struggle to show the same for $f$. Given $x_1,x_2 \in X$ such that $p(x_1) = p(x_2) = y$, we want to show that $Hx_1 = Hx_2$, or equivalently $x_2 = hx_1$ for some $h \in H$. Since $p(x_1) = p(x_2)$, the normality of the covering space $X \to X/G$ ensures the existence of $g \in G$ such that $gx_1 = x_2$.

Now if $g \in H$, then we are done. However, since $p : X \to Y$ may not be normal, I'm not sure how to go about showing this.

Any help is appreciated.


2 Answers 2


Showing that $g\in H$ amounts to showing that for all $x\in X, p(gx) = p(x)$. So fix $x\in X$. Since $X$ is path connected, there is a path $\alpha\colon [0,1]\to X$ from $x_1$ to $x$. Let $q$ denote the covering map $X\to X/G$. Then $q\alpha = qg\alpha$ and both $p\alpha$ and $pg\alpha$ are lifts, both starting at $y$. Since $Y\to X/G$ is a covering, uniqueness of lifts implies that $p(x) = p\alpha(1) = pg\alpha(1) =p(gx)$.


I am also working on this exercise and I met with exactly the same problem. But it turns out to be very simple.

Notations: $p:X\rightarrow Y$, $q:X\rightarrow X/G$, $r:Y\rightarrow X/G$

We want to show that the covering map $p$ is normal. Suppose it's not normal, then $\exists x_1, x_2\in p^{-1}(y)$ s.t. $p_*\pi_1(X,x_1)\neq p_*\pi_1(X,x_2)$. Note that $r_*$ is injective since it's a covering map. So $r_*p_*\pi_1(X,x_1)\neq r_*p_*\pi_1(X,x_2)$. The diagram commutes, so $q_*\pi_1(X,x_1)\neq q_*\pi_1(X,x_2)$, which contradicts with the fact that $q_*$ is a normal covering map.


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