Isomorphic Covering Spaces without regard to basepoints and Conjugacy Classes

Question

I have been studying Algebraic Topology from Allen Hatcher (pg.67) and I am confused with a statement in the text; first I shall give some background.

We know that:

If $X$ is path-connected, locally path-connected space, then two path connected covering spaces $p_1:\widetilde{X}_1\rightarrow X,p_2:\widetilde{X}_2\rightarrow X,$ are isomorphic via the isomorphism $f:\widetilde{X}_1\rightarrow\widetilde{X}_2$ such that $\widetilde{x}_1\in p^{-1}(x_0)\mapsto\widetilde{x}_2\in p^{-1}(x_0)$ if and only if $p_{1_*}(\pi_1(\widetilde{X}_1,\widetilde{x}_1))=p_{2_*}(\pi_1(\widetilde{X}_2,\widetilde{x}_2)).$

Now the next theorem says:

Theorem: Let $X$ be path-connected, locally path-connected, and semilocally simply-connected. Then there is a bijection between the set of basepoint-preserving isomorphism classes of path-connected covering spaces $p:\left(\widetilde{X}, \tilde{x}_{0}\right) \rightarrow\left(X, x_{0}\right)$ and the set of subgroups of $\pi_{1}\left(X, x_{0}\right),$ obtained by associating the subgroup $p_{*}\left(\pi_{1}\left(\tilde{X}, \tilde{x}_{0}\right)\right)$ to the covering space $\left(\widetilde{X}, \tilde{x}_{0}\right) .$ If basepoints are ignored, this correspondence gives a bijection between isomorphism classes of path-connected covering spaces $p: \widetilde{X} \rightarrow X$ and conjugacy classes of subgroups of $\pi_{1}\left(X, x_{0}\right).$

The first statement is no issue, it follows quite easily; the proof of the second starts as -

We show that for a covering space $p:\left(\widetilde{X}, \tilde{x}_{0}\right) \rightarrow\left(X, x_{0}\right),$ changing the basepoint $\tilde{x}_{0}$ within $p^{-1}\left(x_{0}\right)$ corresponds exactly to changing $p_{*}\left(\pi_{1}\left(\widetilde{X}, \tilde{x}_{0}\right)\right)$ to a conjugate subgroup of $\pi_{1}\left(X, x_{0}\right) .$

The proof of this makes sense to me, but I just don't see how this is answering the proof required and asked in the Theorem. I know I have to use the first result, but I cannot wrap my head around what's happening with and without the fixing of basepoints. Any help will be very appreciated.

Answer 1

You know that two path connected covering spaces $p:\left(\widetilde{X}, \tilde{x}_{0}\right) \rightarrow\left(X, x_{0}\right)$ and $p':\left(\widetilde{X'}, \tilde{x'}_{0}\right) \rightarrow\left(X, x_{0}\right)$ are basepoint-preserving isomorphic if and only if $p_*(\pi_1(\widetilde{X}, \tilde{x}_{0})) = p'_*(\pi_1(\widetilde{X'}, \tilde{x'}_{0}))$. You also know that $p_*(\pi_1(\widetilde{X},\tilde{x}_{0}))$ is conjugate to a subgroup $H \subset \pi_1(X,x_0)$ if and only if $H = p_*(\pi_1(\widetilde{X},\tilde{x}_{1}))$ for some $\tilde{x}_{1} \in p^{-1}(x_0)$.

Dropping basepoints, we consider the covering spaces $p: \widetilde{X}\rightarrow X$ and $p':\widetilde{X'} \rightarrow X$ and ask when they are isomorphic.

1. Let $p, p'$ be isomorphic. Let $h : \widetilde{X} \to \widetilde{X'}$ be an isomorphism and let $\tilde{x}_{1} = h^{-1}(\tilde{x}'_{0})$. Then $p:\left(\widetilde{X}, \tilde{x}_{1}\right) \rightarrow\left(X, x_{0}\right)$ and $p':\left(\widetilde{X'}, \tilde{x}'_{0}\right) \rightarrow\left(X, x_{0}\right)$ are basepoint-preserving isomorphic via $h$ so that $p_*(\pi_1(\widetilde{X}, \tilde{x}_{1})) = p'_*(\pi_1(\widetilde{X'}, \tilde{x'}_{0}))$. But $p_*(\pi_1(\widetilde{X}, \tilde{x}_{1}))$ and $p_*(\pi_1(\widetilde{X}, \tilde{x}_{0}))$ are conjugate, thus also $p_*(\pi_1(\widetilde{X}, \tilde{x}_{0}))$ and $p'_*(\pi_1(\widetilde{X'}, \tilde{x'}_{0}))$ are conjugate.

2. Let $p_*(\pi_1(\widetilde{X}, \tilde{x}_{0}))$ and $p'_*(\pi_1(\widetilde{X'},\tilde{x'}_{0}))$ be conjugate. Then $p'_*(\pi_1(\widetilde{X'}, \tilde{x'}_{0})) = p_*(\pi_1(\widetilde{X}, \tilde{x}_{1}))$ for some $\tilde{x}_{1} \in p^{-1}(x_0)$. Hence $p : (\widetilde{X}, \tilde{x}_{1})) \to (X,x_0)$ and $p' : (\widetilde{X'}, \tilde{x'}_{0})) \to (X,x_0)$ are are basepoint-preserving isomorphic. This implies that $p: \widetilde{X}\rightarrow X$ and $p':\widetilde{X'} \rightarrow X$ are isomorphic.