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Let $i:A \to X$ be an inclusion in the topological sense. I want to show the equivalence of the following two statements:

  1. There exists a retraction $r: X\times I \to A\times I \cup X\times \{0\}$

  2. $i:A \to X$ is a cofibration; therefore for every $f:X \to Y, h:A \times I \to Y$ with commuting $f \circ i = h (-, 0)$ there exists a homotopy $H: X \times I \to Y $ with $ H \circ (i \times id) =h$.

  3. "$\implies$" 1): By choosing $Y= A\times I \cup X\times \{0\} $, $h$ and $f$ as canonical inclusions from $X, A \times I$ to $A\times I \cup X\times \{0\} $, I get $H$ by the cofibration property: $X \times I \to Y$. My intention is to prove that $H$ is my retraction. Therefore it suffices to find an $s:Y \to X \times I$ such that $H \circ s = id_Y$. According to universal property it suffices to find $a: X \cong X \times {0} \to X \times I$ and $b:A \times I \to X \times I$ such that $H \circ a$ and $H \circ b$ were inclusions into $Y$. For $b$ I choose $i \times id$ so by definition of $h$ as inclusion I get $h = H \circ b$. My problem is now to find $a: X \to X \times I$ such that $H \circ a$ is an inclusion from $X$ to $Y$.

Now "$\implies$"2): There is given a retraction. Firstly I observe that the canonical inclusion maps $i_X: X \times \{0\} \cong X \to A\times I \cup X \times \{0\}$ and $i_{A \times I}: A \times I \to A\times I \cup X\times \{0\}$ can be extended to $r: X\times I\to A\times I \cup X\times \{0\}$ such that $r(i(a),t)=(a,t)$ and $r(x,0)=x$. This $r$ is a retraction of $s: A\times I \cup X\times \{0\} \to X\times I$ defined as $x \to (x,0)$ and $(a,t) \to (i(a),t)$. Therefore $f:X\times\{0\}\cong X \to Y$ and $h:A\times I\to Y$ are satisfying $h(a,0)=f(a,0) \cong f(a)$, which induces a unique map $\tilde H: A\times I \cup X\times \{0\} \to Y$ according to a universal property such that $\tilde Hi_X=f$ and $\tilde H i¬{ A \times I }=h$. Define $H=\tilde Hr:X\times I\to Y$, we have $H(x,0)=f$ and $H(a,t)=h(a,t)$.

My problem here is to use the assumed information because I can construct the $H$ for the given pair $f, h$ without using the $\textit{given}$ retraction. Where is the devil?

Quantum Chill
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user267839
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  • I think if you don't assume $A$ is closed, you need to do some work to prove that separate maps $X\to Y$ and $A\times I \to Y$, agreeing on $A$, induce a continuous map $A\times I \cup X\times {0}$. –  Oct 02 '17 at 19:22
  • Do you mean in $1)\rightarrow 2)$ where Tyrone glued them? By the way: Which glueing theorem it was applied? And how to argue if A would be not closed? – user267839 Oct 02 '17 at 22:02
  • Yes, exactly. I don't know any gluing theorem that applies (though there might be one). There is an argument in the appendix of Hatcher, Algebraic Topology. It comes from an old paper of Strom, that great point-set topologist. If you work in a nice category, like CGWH, then a cofibration is automatically closed, and the argument is easier. Though, you still have to prove that cofibrations are closed. –  Oct 02 '17 at 22:36
  • The argument in Hatcher (Strom's) is quite specific to the space $A\times I \cup X\times {0}$, so I don't see any way to generalize it to a gluing theorem involving a retraction $A\cup B \subset X \to A\cup B$. –  Oct 04 '17 at 17:02

1 Answers1

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For $2)\Rightarrow 1)$. Let $F:A\times I\hookrightarrow X\times 0\cup A\times I$ be the inclusion which defines a homotopy starting from the restriction of $i_0:X\cong X\times 0\hookrightarrow X\times 0\cup A\times I$. Since $A\hookrightarrow X$ is a cofibration there is an extension $r:X\times I\rightarrow X\times 0\cup A\times I$ of $F$ with the required properties.

For $1)\rightarrow 2)$. Assume given a map $f:X\rightarrow Y$ and a homotopy $F:A\times I\rightarrow Y$ starting from $f|_A$.We get an extension $F'=F\cup f:A\times I\cup X\times 0$ using the glueing theorem since these maps agree on $A\times 0$. Now define $H:X\times I\rightarrow Y$ by $H=F\circ r$. Then $H(x,0)=F'\circ r(x,0)=F'(x,0)=f(x)$ and $H(a,t)=F'\circ r(a,t)= F(a,t)$. So $H$ is a homotopy with the required properties for $A\hookrightarrow X$ to satisfy the $HEP$. That is, for it to be a cofibration.

Tyrone
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  • Hi, thank you for the answer. Ihave still a question about $1)\Rightarrow 2)$: Here we just know that there EXIST a retraction $r:X\times I \to X\times 0\cup A\times I$ (therefore there exist a $s: X\times 0\cup A\times I \to X\times I$ such that $s \circ r = id_{X\times 0\cup A\times I}$), but we don't know anything more (up to surj) about $r$, right? So how do you conclude the equations $F'\circ r(x,0)=F'(x,0)$ and $F'\circ r(a,t)= F(a,t)$? – user267839 Sep 30 '17 at 18:22
  • I think you mean $r\circ s=id_{X\times 0\cup A\times I}$. Without the statement specifying a nonstandard monomorphism $X\times 0\cup A\times I\rightarrow X\times I$ it will be assuming that $s$ is the standard subspace inclusion. The use of the word retraction (which is relative to a given embedding) over the word retract makes this clear. – Tyrone Sep 30 '17 at 19:19
  • ad $1)\rightarrow 2)$: which glueing theorem are you using? – user267839 Oct 02 '17 at 22:39
  • Possibly more commonly known as the pasting lemma. https://en.wikipedia.org/wiki/Pasting_lemma. – Tyrone Oct 03 '17 at 07:48
  • This assumes the two sets are both closed (or both open). –  Oct 03 '17 at 16:30
  • Firstly it should be noted that the theorem is only true for closed cofibrations, and is not true for non-closed cofibrations. All cofibrations in the category of weak Hausdorff complactly generated spaces are closed. Also if $X$ is Hausdorff then $A$ is necessarily closed in $X$. – Tyrone Oct 03 '17 at 19:40
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    Surprisingly, no. See prop A.18 in Hatcher: https://www.math.cornell.edu/~hatcher/AT/ATapp.pdf –  Oct 03 '17 at 22:00
  • Nice. I actually read the first of Strom's two papers on cofibrations (http://www.mscand.dk/article/view/10791/8812, http://www.mscand.dk/article/view/10877/8898) in which she makes the claim only for closed cofibrations. It's nice to see the full result, thanks for pointing it out! – Tyrone Oct 04 '17 at 07:44
  • I believe this is Lemma 3 in Strøm's Note on Cofibrations II –  Oct 04 '17 at 09:01