# Retraction induces Cofibration

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

1) There existits 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 exist a homotopy $H: X \times I \to Y$ with $H \circ (i \times id) =h$.

2) "$\implies$" 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 suffies to find a $s:Y \to X \times I$ such that $H \circ s = id_Y$. According to univ. property it suffies 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 to $Y$. 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 a inclusion from $X$ to $Y$.

Now 1) "$\implies$"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 extend 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 to $s: A\times I \cup X\times \{0\} \to X\times I$ 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 $h(a,0)=f(a,0) \cong f(a)$, what induces a unique map $\tilde H: A\times I \cup X\times \{0\} \to Y$ according univ 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 asumption information because I can construct the $H$ for the given pair $f, h$ without using the GIVEN retraction. Where is the devil?

• 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 $A\times I \cup X\times \{0\}$. – Justin Young Oct 2 '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? – KarlPeter Oct 2 '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. – Justin Young Oct 2 '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 $A\cup B \subset X \to A\cup B$. – Justin Young Oct 4 '17 at 17:02

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.
• 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 $F'\circ r(a,t)= F(a,t)$? – KarlPeter 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? – KarlPeter Oct 2 '17 at 22:39