I'm a bit confused by Hatchers choice of words here. He says "The main classification theorem for covering spaces says that by associating the subgroup $p_{*}(\pi_{1}(\tilde{X},\tilde{x_{0}}))$ we obtain a bijection between all of the different connected covering spaces of $X$ and the conjugacy classes of subgroups of $\pi_{1}(X,x_{0})$.

But, if we keep track of the basepoint vertex $\tilde{x_{0}} \in \tilde{X}$, then this is a bijection between covering spaces $p:\tilde{X} \rightarrow X$ and actual subgroups of $\pi_{1}(X,x_{0})$.

My question concerns the highlighted portion. Here is what I think this means.

Fix a group $\pi_{1}(X,x_{0})$. Let $\tilde{x_{0}}\in p^{-1}(x)$. We associate $p_{*}(\pi_{1}(\tilde{X},\tilde{x_{0}}))$ to the covering space $p: \tilde{X} \rightarrow X$. If we fix $\tilde{x_{0}}$, then $p_{*}(\pi_{1}(\tilde{X},\tilde{x_{0}}))$ is the unique subgroup of $\pi_{1}(X,x_{0})$ associated with the covering space $p: \tilde{X} \rightarrow X$. This yields a bijection between subgroups $\pi_{1}(\tilde{X},\tilde{x_{0}}$ and covering spaces $p:\tilde{X} \rightarrow X$. If we do not fix $\tilde{x_{0}}$, then for different $\tilde{x_{0}}\in p^{-1}(x)$ we get different subgroups of $\pi_{1}(X,x_{0})$; in the latter case we will get a bijection between conjugacy classes of subgroups of $\pi_{1}(X,x_{0})$ and covering spaces $p: \tilde{X} \rightarrow X$.

Is this correct?


It depends on what you mean by "fix." What you want to do is include the basepoint as part of what a covering map is.

Basically, Hatcher means that there is a bijection between subgroups of $\pi_1(X,x_0)$ and based covering maps $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$.

So if two based covering maps $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$ and $q:(\tilde{Y},\tilde{y}_0)\to (X,x_0)$ satisfy $p_{\ast}(\pi_1(\tilde{X},\tilde{x}_0))=q_{\ast}(\pi_1(\tilde{X},\tilde{x}_0))$, then $p$ and $q$ are now equivalent in the stronger sense that there is a based homeomorphism $h:(\tilde{X},\tilde{x}_0)\to (\tilde{Y},\tilde{y}_0)$ such that $q\circ h=p$.

  • $\begingroup$ Okay that makes more sense, thank you. $\endgroup$ – TuoTuo Jul 22 '16 at 2:46

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