Strong deformation retraction.

Question

Suppose $X$ is a path connected contractible space. So we know there is a homotopy $F:X\times I \rightarrow X$ with $F(x,0)=x$ and $F(x,1)=c$, for all $x\in X$ and some fixed $c\in X$. Since $X$ is path connected we could change the constant $c\in X$ as we wish, so $X$ deformation retracts to any $x\in X$.

My question is, if we include the hyposthesis of $X$ being a Hausdorff Normal space, and $x_o\in X$ is a non degenarate point (basically meaning that the inclusion $x_o\hookrightarrow X$ is a cofibration), then can we claim that $X$ strongly deformation retracts to $x_o$.

By "strongly" I mean that the homotopy $F:X\times I \rightarrow X$ would have the additional property that $F(x_o,t)=x_o$ for all $t\in I$, in other words it's an homotopy relative to $x_o$.

I know that, since $x_o$ is a non degenerate point, there is an open neighborhood $N$ of $x_o$ which strongly deformation retracts to $x_o$, but I see no reason for that also being the case for $X$ itself.

Since I don't know about the validity of this, ideas for proofs or counterexamples are both welcome.

Answer 1

Suppose that $i:A\subseteq X$ is a subspace inclusion which is both a cofibration and a homotopy equivalence. We place no other assumptions on $X$ or $A$. The assumptions grant a map $r':X\rightarrow A$ such that $r'i\simeq id_A$ and $ir'\simeq id_X$. Using the homotopy extension property we obtain a map $r:X\rightarrow A$ such that $r\simeq r'$ and $ri=id_A$. Necessarily $ir\simeq id_X$.

$A$ is a deformation retraction of $X$.

Now fix a homotopy $F:id_X\simeq ir$. Define $$G:X\times\{0,1\}\times I\cup A\times I\times I\rightarrow X$$ by putting $$G(x,0,t)=x,\qquad G(a,s,t)=F(a,(1-t)s),\qquad G(x,1,t)=F(r(x),1-t).$$ This is well-defined and continuous. We need the following.

Lemma: Let $X$ be a space and $A\subseteq X$ a cofibration. Then $X\times\{0,1\}\cup A\times I\subseteq X\times I$ is a cofibration. $\square$

The proof makes use of the fact that there is a self-homeomorphism of $I^2$ which maps $I\times\{0,1\}\cup\{0\}\times I$ onto $\{0\}\times I$. I'll leave details to your textbook.

Above we have defined $G$. Notice that its restriction to $(X\times\{0,1\}\cup A\times I)\times\{0\}$ is just the map $F$. Because of the lemma there exists a homotopy $\widetilde G:X\times I\times I\rightarrow X$ with $\widetilde G(x,s,0)=F(x,s)$ and $H|_{X\times\{0,1\}\times I\cup A\times I\times I}=G$. Put $$H_t(x)=\widetilde G(x,t,1).$$ This yields $H_0=id_X$, $H_1=ir$ and $H_t|_A=i$. Thus we have our conclusion.

A subspace inclusion $A\subset X$ which is both a cofibration and a homotopy equivalence is a strong deformation retract.

Of course if $X$ is contractible, then the inclusion $x_0\hookrightarrow X$ of any point is a homotopy equivalence. If this inclusion is a cofibration, then the outcome above applies. The Hausdorff and normality assumptions are not necessary.