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A pair $(X, A)$ is a good pair (as defined in Hatcher) if $A$ is a deformation retract of some neighborhood $N$ in $X$. Suppose $\pi : Y \to X$ be a fibration, and let $B = p^{-1}(A)$. Hatcher claims in his proof of Theorem 5.3 (see page 530 here) that $B$ is a deformation retraction of $\pi^{-1}(N)$ in the weak sense, meaning that the deformation retraction need not fix the subspace $B$. His argument is that, given a deformation retraction of $N$ onto $A$, by the homotopy lifting property this lifts to a weak deformation retraction of $\pi^{-1}(N)$ onto $B$. He uses this to conclude that the inclusion map $B \hookrightarrow \pi^{-1}(N)$ is a homotopy equivalence. However, I can't see why the lifted homotopy necessarily ends at a retraction $\pi^{-1}(N) \to B$; namely, it may not fix the subspace $B$. Would someone be able to clarify his argument?

There is a similar question here, which is more general. However, I have the same issue with the answer given; I don't see why the lift $H$ would fix $E_1$.

Frank
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1 Answers1

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There is indeed no immediate reason for a lift to necessarily end at a retraction $\pi^{-1}(N)\rightarrow B$. However, when Hatcher says "where the latter deformation retraction is in the weak sense that points in the subspace need not be fixed during the deformation", this (perhaps counter-intuitively) includes the end value. In other words, all he claims is that there is a homotopy $H\colon\pi^{-1}(N)\times I\rightarrow\pi^{-1}(N)$ such that $H_0=\mathrm{id}_{\pi^{-1}(N)}$, $H_t(B)\subseteq B$ for all $t\in I$ and $H_1(\pi^{-1}(N))\subseteq B$. The lift clearly does satisfy this and this in fact suffices to conclude that $B\hookrightarrow\pi^{-1}(N)$ is a homotopy equivalence, which I leave as an exercise. Two alternatives:

  1. You could alternatively argue that if the bottom map in such a fibration pullback square is a weak homotopy equivalence, then so is the top map (this follows e.g. by the naturality of the long exact fibration sequence). The argument in Hatcher only needs a weak homotopy equivalence (plus, a weak homotopy equivalence between CW-complexes is a homotopy equivalence anyway).

  2. If you additionally assume that $A\hookrightarrow N$ is a closed cofibration (which is the case in Hatcher's application), you can in fact argue that $B\hookrightarrow\pi^{-1}(N)$ is a strong deformation retraction. Indeed, in that case $B\hookrightarrow\pi^{-1}(N)$ is a closed cofibration too (this is the subject of the linked question and while the argument there is indeed incomplete, it can easily be completed, see Theorem 12 in Strøm's Note on Cofibrations II). Then, applying the "covering homotopy extension property" (which is proven originally in Strøm's Note on Cofibrations I and also covered in many algebraic topology texts), you get a homotopy $\pi^{-1}(N)\times I\rightarrow\pi^{-1}(N)$ covering the homotopy $\pi^{-1}(N)\times I\rightarrow N\times I\rightarrow N$ and extending the constant homotopy $B\times I\rightarrow B$, i.e. a strong deformation retraction of $\pi^{-1}(N)$ onto $B$.

Thorgott
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  • Thanks for this answer! At the end did you mean to write $N \times I \to N$ instead of $\pi^{-1}(N) \times I \to N \times I \to N$? – Frank Nov 29 '23 at 02:39
  • Not quite, you have a deformation retraction $N\times I\rightarrow N$ of $N$ onto $A$ and you pre-compose this with the map $\pi^{-1}(N)\times I\rightarrow N\times I$ that is the projection in the first factor and the identity in the second factor. This is the thing you lift through the fibration $\pi^{-1}(N)\rightarrow N$ with initial value the identity of $\pi^{-1}(N)$ in order to get the homotopy $H\colon \pi^{-1}(N)\times I\rightarrow\pi^{-1}(N)$. – Thorgott Nov 29 '23 at 12:56
  • Ah I see, that makes sense. – Frank Nov 29 '23 at 21:19