In Hatcher's Algebraic Topology, exercise 0.26 asks to show that if $(X,A)$ is a pair with the homotopy extension property, then the retraction $r:X\times I\rightarrow Y=X\times\{0\}\cup A\times I$ can be upgraded to a deformation retraction. This part I get.
But then the exercise ends by asking me to show the proof of proposition 0.18 now works for $(X,A)$, where the proof is given for CW complexes in the text. To recall what this proof looks like, we have two homotopic maps $f,g:A\rightarrow B$. A homotopy for these, $F:A\times I\rightarrow B$, is used to build the adjunction space $(X\times I)\cup_F B$. The proof then uses the deformation retraction of $X\times I$ to $Y$ to get a deformation retraction of $(X\times I)\cup_F B$ to $X\cup_f B$.
But I can't seem to make this work. Clearly there is a deformation retraction of $(X\times I)\cup_F B$ to a subspace $Z$. I don't see why this is homeomorphic to $X\cup_f B$ though. There is an obvious diagram
$$\require{AMScd}
\begin{CD}
X\sqcup B @>{i}>> (X\times I)\sqcup B\\
@VVV @VVqV \\
X\cup_f B @>>> Z\subset (X\times I)\cup_F B
\end{CD}$$
To conclude the bottom arrow is a homeomorphism however would require $qi$ to be a quotient map. For that to be true, then I would need to show for $C\subset Z$, if $(qi)^{-1}(C)$ is closed, then $C$ is closed in $Z$. Unraveling this, we have $q^{-1}(C)=S\sqcup T\subset (X\times I)\sqcup B$. We need to find a closed subset $K\subset X\times I$ such that $K\cap Y=S$. We can write $S_0=S\cap (X\times\{0\})$, so that $S=S_0\cup F^{-1}(T)$, where $S_0$ is closed in $X\times I$ (by hypothesis) and $F^{-1}(T)$ is closed in $A\times I$.
A natural choice for $K$ is $r^{-1}(S)$. But that only works if $S$ is closed in $Y$, and I don't see why that needs to be true. In fact, I think this is the crux of the problem: why would $S_0\cup F^{-1}(T)$ be closed in $Y$?
Hatcher acknowledges the topological subtlety of 0.18's proof in the errata (bottom of first page). In fact, this line of reasoning is similar to showing the restriction $q:Y\sqcup B\rightarrow Z$ is a quotient map. This is true, but is easier because we get $S$ is closed in $Y$ by hypothesis.
So, does the proof of proposition 0.18 still work if we don't assume $A$ is closed?
Best Answer
Here's a solution to exercise 0.26. Below $(X,A)$ is a pair with the homotopy extension property, and I'll write $Y=X\times\{0\}\cup A\times I$. The HEP implies there's a retraction $r:X\times I\rightarrow Y$.
$(X\times I, Y)$ has the HEP as well
Let $L$ be the "lower left" half of the boundary of $I\times I$, and $U$ the "upper right" half of the boundary. There is a deformation retraction of $I\times I$ to $L$, given by projecting from the point $(2,2)$ in the plane. For a point $(a,b)\in I\times I$, the line from $(2,2)$ to $(a,b)$ will hit a single point of $U$ (call it $\mu(a,b)$) and a single point of $L$ (call it $\ell(a,b))$. Then we can reparameterize the segment from $U$ to $L$: \begin{equation*} P_{a,b}(t) = (1-t)\mu(a,b) + t\ell(a,b) \end{equation*} Let $q(a,b)$ be the time when this segment hits $(a,b)$: $P_{a,b}(q(a,b)) = (a,b)$.
We need a retraction from $X\times I^2$ to $X\times I\times\{0\}\cup Y\times I=X\times L\cup Y\times I^2$. Writing $(y,t)=r(x,q(a,b))$, this is given by \begin{equation*} (x,a,b)\mapsto (y,P_{a,b}(t)) \end{equation*}
$Y$ is a weak deformation retract of $X\times I$
We can define a weak deformation retraction (exercise 0.4) as \begin{equation*} H_t(x,s) = (r(x,t), s(1-t)) \end{equation*}
$X\times I$ deformation retracts to $Y$
This now follows from exercise 0.4 and corollary 0.20.
Proposition 0.18 holds for $(X,A)$
Just as in the proof of proposition 0.18, the deformation retraction $D_t:X\times I\rightarrow Y$ induces a deformation retraction on $(X\times I)\cup_F X_0$ to a subspace $Z$, which fits into the diagram below:
$$\require{AMScd} \begin{CD} X\sqcup X_0 @>{i}>> Y\sqcup X_0\\ @VVV @VV{q}V \\ X\cup_f X_0 @. Z \end{CD}$$
Here $i(x)=(x,0)$. To follow the rest of the proof of 0.18, we want to conlude that $qi$ is a quotient map, since then there will be an induced homeomorphism between $X\cup_f X_0$ and $Z$.
Suppose $C\subset Z$ is such that $(qi)^{-1}(C)$ is closed in $X\sqcup X_0$; we want to show $C$ is closed in $Z$. Let $q^{-1}(C) = S\sqcup T\subset Y\sqcup X_0$. $T$ is closed in $X_0$ and $S\cap (X\times\{0\})$ is closed in $X\times\{0\}$ by hypothesis. $S\cap (A\times I)=F^{-1}(T)$ is closed in $A\times I$ by continuity. The proof of proposition A.18 in the appendix then shows $S$ is closed in $Y$.
Thus $M=D_1^{-1}(S)\subset X\times I$ is closed, and $M\cap Y=S$. This means $M\sqcup T$ is saturated, so that $q(M\sqcup T)$ is closed. But then \begin{align*} q(M\sqcup T)\cap Z &= q((M\sqcup T)\cap q^{-1}(Z))\\ &= q((M\sqcup T)\cap (Y\sqcup B))\\ &= q(S\sqcup T)\\ &= C \end{align*} And so $C$ is closed in $Z$ as desired.