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This is an exercise I've been working on from Hatcher (1.3.32):

Consider covering spaces $p: \tilde{X} \to X$ with $\tilde{X}$ and $X$ connected CW complexes, the cells of $\tilde{X}$ projecting homeomorphically onto $X$ cells. Restricting $p$ to the 1-skeleton then gives a covering space $\tilde{X}_1^1 \to X^1$. Show the following:

(a) Two such covering spaces $\tilde{X}_1 \to X$ and $\tilde{X}_2 \to X$ are isomorphic iff the restrictions $\tilde{X}_1^1 \to X$ and $\tilde{X}_2^1 \to X$ are isomorphic.

(b) $\tilde{X} \to X$ is normal iff $\tilde{X}^1 \to X^1$ is normal.

(c) The groups of deck transformations of $\tilde{X} \to X$ and $\tilde{X}^1_1 \to X^1$ are isomorphic via the restriction map.

For (a) and (b), one direction of the iff is trivial.

For (b), I think I can show the converse: Assume $\tilde{X}^1 \to X^1$ is normal. Then given $\tilde{x_1}, \tilde{x}_2 \in \tilde{X}$ such that $p(\tilde{x}_1) = p(\tilde{x}_2) := x \in X$, find a path $\ell$ in $X$ from $x$ to a point $x_0$ in the 1-skeleton. Then there exist unique lifts $\tilde{\ell}_1$ and $\tilde{\ell}_2$ beginning at $\tilde{x}_1$ and $\tilde{x}_2$ respectively, and ending at some points $\hat{x}_1$ and $\hat{x}_2$ in the 1-skeleton of $\tilde{X}$. By assumption, there's a deck transformation of the 1-skeleton taking $\hat{x}_1$ to $\hat{x}_2$. By (a), this would correspond to a deck transformation $h$ of $\tilde{X}$. Uniqueness of lifts and the fact that $ph = p$ would force this deck transformation to take $\tilde{x_1}$ to $\tilde{x_2}$.

For (c), clearly the restriction map from deck transformations of $\tilde{X}$ to deck transformations of $\tilde{X}^1$ is injective since a deck transformation (of a connected space) is determined by its action on a point. Preservation of compositions also clear. Surjectivity would follow from (a).

So what's left is proving the nontrivial implication in (a): given a covering isomorphism $\phi: \tilde{X}_1^1 \to \tilde{X}_2^1$, how can I produce an isomorphism $\psi: \tilde{X}_1 \to \tilde{X}_2$?

I want $\psi |_{\tilde{X}_1^1} = \phi$. The space $\tilde{X}_1$ (and $\tilde{X}_2$) is built up by attaching disks; at the first step, disks $D_\alpha$ are added via attaching maps $\theta_\alpha: \partial D_\alpha \to \tilde{X}_1^1$.

This is where I get stuck. I think I want to argue that $\tilde{X}_2$ is built up from its 1-skeleton by attaching disks via maps $\psi \circ \theta_\alpha$, but I'm not sure...

Alternatively, I thought about trying an argument based on the fact the the fundamental group of a CW complex is the group of its one skeleton modded out by relations from its 2-skeleton, and using the fact that coverings are isomorphic iff the projections of their fundamental groups are the same, but I couldn't see my way through that argument either.

Any advice on how to proceed would be appreciated. Thanks.

ec92
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2 Answers2

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Just some thoughts on (a). I'm not certain if this is all true, but it was too long to be a comment:

  • Using the classification theorem you mentioned + the fact that the fundamental group is determined by the 2-skeleton, it is enough to show that the fundamental groups of $\tilde{X}^2_1$ and $\tilde{X}^2_2$ (actually, their images in $\pi_1(X^2)$) are either isomorphic, or conjugate (with respect to $\pi_1(X^2)\cong \pi_1(X))$. The difference being whether you want the isomorphisms to preserve basepoint.
  • Since all one dimensional CW complexes are graphs, we can think of the isomorphism $\tilde{X}^1_1\to\tilde{X}^1_2$ as a graph isomorphism. Also note that the fundamental groups of graphs are free.
  • Choose a presentation $\pi_1(X)=\langle g_\alpha | r_\beta \rangle$
  • Attach a 2-cell to $X^1$ along a word corresponding to one of the relations $r$ of $\pi_1(X)$ to from a new space $X^1\cup_re^2$
  • For a covering space of a graph (which is also a graph), the vertices and edges are the lifts of vertices and edges in the base graph. Thus the introduction of the 2-cell in $X^1$ introduces 2-cells in both covering spaces $\tilde{X}^1_1$ and $\tilde{X}^1_2$, and the resulting complexes (call them $Y_1,Y_2$ for convenience) are still covering maps of $X^1\cup_re^2$ (intuitive, but perhaps you should check).
  • [Guess] Since $\tilde{X}^1_1$ and $\tilde{X}^1_2$ are isomorphic covering spaces, I believe this forces the 2-cells to be attached along the isomorphic edges (draw a commuting diagram). This should show both $Y_1\cong Y_2$ and $\pi_1(Y_1)\cong \pi_1(Y_2)$ (or conjugate, if basepoint is not preserved).
  • If my previous assertion is true, you just have to attach the 2-cells corresponding to the rest of the relations, which will get you an isomorphism between the $\tilde{X}^2_1$ and $\tilde{X}^2_2$, or more rigorously, show that the fundamental groups of $\tilde{X}^2_1$ and $\tilde{X}^2_2$ are isomorphic/conjugate.

I realise parts of what I said was not concise, redundant, or even wrong. However, I hope that you get some ideas out of it.

fixedp
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There is a simple approach to the problem (part (a) non trivial direction).

Suppose there is a graph isomorphism $\phi: {X_1}^1 \to {X_2}^2$ between the 1-skeletons. We need to extend this isomorphism to the whole CW complex covering. For that we start with a 2-skeleton, say ${X_i}^2; \, i=1,2.$

Let $x$ be a point in the 2-skeleton ${X_1}^2 $ and $y$ be a point in 1-skeleton ${X_1}^1$, then there is path $\gamma$ from $y$ to $x$. So, $p(\gamma)$ is path in X with starting point in $X^1$.

Under the other covering map $q$, lift this path $p(\gamma)$, say $\gamma_q$, to starting point $\phi(y) \in {X_2}^1$. Now this lift is unique by unique lifting criterion. So, define extension $\phi : {X_1}^2 \to {X_2}^2$ by $\phi(x)= \gamma_q(1)$.

A similar approach can be used to extend map $\phi : X_1 \to X_2$.

Prerak Deep
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