# On well-definedness of $f : Y \to X/H$ for Hatcher's Algebraic Topology, Exercise 1.3.24(a)

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.

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 $$p(x) = p\alpha(1) = pg\alpha(1) =p(gx)$$.
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.