# Isomorphic Covering Spaces without regard to basepoints and Conjugacy Classes

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.

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.