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I would like to show that if $i:A\to X$ is a cofibration and $f:A\to B$ is a homotopy equivalence, then the induced map $k:X\to X\cup_AB$ is again a homotopy equivalence. $\require{AMScd}$ $$ \begin{CD} A @>i>> X \\ @VfVV @VVkV \\ B @>>> X\cup_AB \end{CD} $$ What I have done: Choose a homotopy inverse of $f$, say $g$, and choose a homotopy $H:\mathrm{id}_{A}\simeq g\circ f$. Apply the HEP to $i\circ H:A\times I\to X$ and $\mathrm{id}_{X}:X\to X$ to obtain a homotopy $\tilde H:\mathrm{id}_{X}\simeq \tilde H_1$. Then $\tilde H_1:X\to X$ and $i\circ g:B\to X$ are compatible with the pushout, so they induce a map $\ell:X\cup_A B\to X$. I would like show to that $\ell$ is the inverse to $k$. Certainly $\ell\circ k\simeq\mathrm{id}_{X}$ via $H$, but I can't show the other direction $k\circ\ell\simeq\mathrm{id}_{X\cup_AB}$.

Note: this question was asked here, but the answer was just a list of references. I would really like an explicit proof. Any help would be appreciated. Thanks in advance!

Sardines
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  • It looks like this has been asked before: https://math.stackexchange.com/questions/1002800/homotopy-equivalences-in-pushout-square-with-cofibration?rq=1 it’s not really my area, but the answer was essentially a list of references, so it may be fairly tricky – jackson Sep 20 '23 at 04:15
  • @jackson Right... I would really like an explicit proof. – Sardines Sep 20 '23 at 23:01
  • if your spaces are nice enough (e.g. CW complexes) then this follows (abstractly) from the model structure of spaces (it's left proper) – Daniel Teixeira Sep 20 '23 at 23:11
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    In order to get this right, I need to verify your category of spaces and your definition of cofibration. –  Sep 21 '23 at 11:24
  • @JustinYoung A priori I don't have assumptions on my spaces. By "cofibration" I mean an inclusion that satisfies the homotopy extension property. Nevertheless, if you need more assumptions (say, assuming spaces are CGWH) to make this statement work, please feel free to do so. – Sardines Sep 21 '23 at 15:10

1 Answers1

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Here's another more direct approach. We'll use the following proposition:

Let $f: A \to B$ be a map. TFAE:

  1. $f$ is a homotopy equivalence.
  2. The induced map $f^*: Map(B,Z) \to Map(A,Z)$ is a homotopy equivalence for all $Z$, where the map is given by $h \mapsto h \circ f$.
  3. The induced map $f^*: [B,Z] \to [A,Z]$ is a bijection for all $Z$.

We use property 3. Let $W = X \cup_A B$. We want to show that $g^*:[W,Z] \to [X,Z]$ is a bijection for all $Z$. I'll just sketch the idea of surjectivity (and injectivity is very similar).

Let $z: X \to Z$ be a map. Consider $z \circ i: A \to Z$. Since $f$ is a homotopy equivalence, there exists a map $z': B \to Z$ such that $z' \circ f \sim z \circ i$. Since $i$ is a cofibration, there exists a solution $\tilde{h}: X \to Z^I$ to the homotopy extension problem

\begin{CD} A @>{h}>> Z^I\\ @VVV @VV{ev_0}V \\ X @>>> Z \end{CD}

Use this to define the map $w: W \to V$ such that $w \circ g \sim v$. Namely, consider a homotopy $A \times I \to Z$ and create a map $z'' \sim z$ such that $z' \circ f \sim z'' \circ i$. Define $w$ in terms of $z'$ and $z''$ and this gives the appropriate map [details left to you].

Injectivity is a similar idea.

Alvin Jin
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  • Let me add the argument for $3\to 1$ since that's what you need for this argument. First, there is a map $g: B\to A$ so that $g\circ f \simeq 1_A$. Now, observe that $f\circ g\circ f \simeq f$, and therefore $f^(f\circ g) = f^(1_B)$, from which it follows that $f\circ g \simeq 1_B$ by injectivity. –  Sep 29 '23 at 12:24
  • Final point: this argument generalizes (given in Hirschhorn, for example) to show that a pushout square of cofibrant objects in any model category satisfies the property given above. In the Strom model category (technically there are two, one with all spaces and closed cofibrations, and one with CGWH and standard cofibrations) all objects are cofibrant, and so that argument implies this result. –  Sep 29 '23 at 12:28