Let $Y$ be a CW-complex and $f: Y^2 \to Y$ be the inclusion of its two skeleton. We define $f^*Z := \{(y, z) \in Y^2 \times Z \ | \ y = f(y) = p(z)\}$ and $f^*p : f^*Z \to Y^2$ the restriction of the projection $Y^2 \times Z \to Y^2$.


Show that for every covering map $q: W \to Y^2$, there exists a covering $p: Z \to Y$ and a homeomorphism $g: W \to f^*Z$ such that $q = (f^*p)\circ g$.

Possible approach

A CW complex has a universal cover and $\pi_1(Y^2) = \pi_1(Y)$, so there is a bijection between coverings of $Y^2$ and $Y$ by the classification theorem of coverings. Take $p$ to be the covering corresponding to $q$. Also this is part b of a larger question, part a was to prove covering maps are stable under pullback.

  • $\begingroup$ What is $Z$, and what is $p$? $\endgroup$ – Santana Afton Apr 5 at 11:43
  • $\begingroup$ $Z$ is a topological space and $p$ is a covering map. They are not given, we must prove that they exist such that they satisfy the requirements. $\endgroup$ – Pel de Pinda Apr 5 at 13:41

In your question you describe the general construction of a pullback of a map $p : Z \to Y$ along a map $f : X \to Y$. Pullback diagrams are characterized by a universal property, so pullbacks are unique up to homeomorphism. Now it is easy to see that if $f$ is the inclusion of a subspace, then the restriction $p_ X = p \mid_{p^{-1}(X)} : p^{-1}(X) \to X$ and the inclusion $f_X : p^{-1}(X) \to Z$ build a pullback diagram with $p,f$. Hence instead of $f^*p$ we may take $p_X$ which is somewhat easier to visualize.

Your question therefore in fact means to find an extension of $q : W \to Y^2$ to a covering $p : W' \to Y$ (i.e. $W \subset W'$ and $p \mid_W = q$).

$Y^3$ is obtained from $Y^2$ by attaching $3$-cells, similarly we get $Y^4, Y^5, \dots$. Now each attaching map $\phi : S^2 \to Y^2$ lifts to $\tilde{\phi} : S^2 \to W$ because $\pi_1(S^2) = 0$. More precisely, for each $w \in q^{-1}(\phi(*))$ we get a unique lift $\tilde{\phi}_w : S^2 \to W$ such that $\tilde{\phi}_w(*) = w$. Use the collection of all these $\tilde{\phi}_w$ to attach $3$-cells to $W$. This yields a CW-complex $W^3 \supset W$ and an extension $q^3 : W^3 \to Y^3$ of $q$ (the open $3$-cell of $W^3$ belonging to $\tilde{\phi}_w$ is mapped homeomorphically onto the the open $3$-cell of $Y^3$ beloging to $\phi$). Preceeding skeletonwise we get the desired extension $p : W' \to Y$.


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