If $(W;\omega)$ is a well-pointed space that weakly contracts onto $\omega$, does it strongly contract onto $\omega$?

Question

We are given a well-pointed space $(W;\omega)$, by which I mean, for every pair of maps $f:W\to Y$, $h:I\to Y$, there exists at least one extension $G:W\times I\to Y$ which satisfies $G(w,0)=f(w),\,G(\omega,t)=h(t)$, for all $w$ and $t$. Equivalently, that means that there is a retract $r_0:W\times I\to(W\times\{0\})\cup(\{\omega\}\times I)$ where the RHS is given the subspace topology.

I have $H:W\times I\to W$ which has $H(w,0)=\omega$ and $H(w,1)=w$ for all $w\in W$, i.e. it is a weak contraction of $W$ onto $\omega$. I want, very much !, to be able to promote this to a strong contraction $\Gamma:W\times I\to W$, where $\Gamma(\omega,t)=\omega$ for all $t$ is satisfied. Apparently, this is doable, but I don't see how.

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Context: in the comments underneath this question of mine, a user suggest a resolution to my problem if we assumed a well-pointed space. I've been struggling for a long time to actually convert their suggestion into a full blown solution, but I've felt very close so I think I'm missing something quite obvious.

The problem was, I had a weak ambient contraction $\Gamma:B\times I\to A$ (with $B$ a subspace of $A$) onto some point $\rho$, and I knew $(B;\rho)$ to be well-pointed ($(A;B)$ also has the homotopy extension property) and it was crucial that I find a way to 'promote' $\Gamma$ to a strong contraction, i.e. to ensure that $\Gamma(\rho,t)=\rho$ for all times $t$.

I was getting lost in specifics, so I felt it appropriate to abstract the problem. I have solved a similar problem already:

If $f:(S^n;(1,0,\cdots,0))\to(A;a)$ is a based map and $f\simeq c_a$ via some unbased homotopy, we can always promote this to a based homotopy.

The solution involved observing that the nullhomotopy $H:f\simeq c_a$ factors through $D^{n+1}=S^n\times I/(S^n\times\{0\})$ as some $\tilde{H}$ and that we could insert a map $r:D^{n+1}\to D^{n+1}$ which 'zips' the disk in the following way; it is the identity on $S^n$ and on the segment $(t,0,\cdots,0),0\le t\le1$, it is a constant map to $(1,0,\cdots,0)$. Then the composite $\tilde{H}\circ r:D^{n+1}\to A$ corresponds to a based nullhomotopy $f\simeq c_a$ when you precompose it with the quotient map $S^n\twoheadrightarrow D^{n+1}$.

This has been weighing on my mind, as I think that some variant of this solves the problem at hand. For the last hour I have been fiddling with cones and reduced cones, trying to 'draw the right diagram'. That's because the map $r$, which is a key player in solving the above problem, is essentially a quotient map for the reduced cone. The cone $CY$ of a space is $Y\times I/(Y\times\{0\})$ and the reduced cone $\Gamma Y$ of a based space $(Y;y_0)$ is the quotient space $Y\times I/(Y\times\{0\}\cup\{y_0\}\times I)=CY/(\{y_0\}\times I)$. We can easily check that $CS^n\cong D^{n+1}$ and $\Gamma S^n$ is found homeomorphic to $D^{n+1}$ also, via the above $r$.

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If I could find a continuous injection $\Gamma W\hookrightarrow CW$ that preserved the top $W\times\{1\}$ and the trivial point $\ast$, then I think that I would be done. For the composite: $$W\times I\twoheadrightarrow\Gamma W\hookrightarrow CW\overset{\overline{H}}{\longrightarrow}W$$Would be a strong contraction of $W$ onto $\omega$, where $\overline{H}$ is the map induced by $H:W\times I\to W$. This is related to the solved case of $W=S^n$ as the intermediary $r$ was essentially just $CS^n\twoheadrightarrow\Gamma S^n\hookrightarrow CS^n$. I am pretty sure this is equivalent to finding a retraction: $$W\times I\to(W\times\partial I)\cup(\{\omega\}\times I)$$

At least, finding such a retraction would also solve my problem as it would imply the maps $W\times\{0\}\to\{\text{pt}\}\overset{\omega}{\hookrightarrow}W, W\times\{1\}\cong W,\{\omega\}\times I\overset{\pi_L}{\longrightarrow}\{\omega\}\hookrightarrow W$ would have an extension $\Gamma:W\times I\to W$, which would be precisely a strong contraction onto $\omega$. I also know that there exist retractions $R:I\times I\to(I\times\partial I)\cup(\{0\}\times I)$. That's tantalising: it seems almost as if I could replace the first component with $W$, in such a way that $\{0\}\mapsto\{\omega\}$... so the composite $\langle H\circ(1\times\pi_LR),\pi_RR\rangle:W\times I\times I\to W\times I$ looks promising. This maps $W\times\{0\}\times I$ to $\{\omega\}\times I$, identically on the second coordinate. It maps $W\times\{1\}\times I$ to $W\times I$ identically on both coordinates, and it maps $(w,s,t)\in W\times I\times\partial I$ to $(H(w,s),t)$. This doesn't quite do what I want, since I need a way to guarantee $\omega$ is stuck at all times. I have $r_0:W\times I\to(W\times\{1\})\cup(\{\omega\}\times I)$ at my disposal too, but I can't make everything fit together.

I would highly appreciate any help with this.

Answer 1

$\newcommand{\K}{\mathcal{K}}$This is essentially proposition $0.19$ in Hatcher's "Algebraic Topology", I now realise. It involves homotopies of homotopies, and try as I might to do this myself, I failed because the proof is actually multistage, constructing a few intermediate maps.

The proposition:

Suppose $A,X,Y$ are all spaces and that there are subspace inclusions $i\hookrightarrow X,j:A\hookrightarrow Y$ which allow us to view $A\subset X$ and $A\subset Y$. Suppose $(X;A)$ and $(Y;A)$ both have the homotopy extension property, and suppose there is a homotopy equivalence $f:X\to Y$ with $f\circ i=j$. (Hatcher writes, $f|_A=1$, but I find this misleading).

Then, there exists $g:Y\to X$ such that $g\circ f\simeq 1_X,\,\text{rel }A$ and $f\circ g\simeq1_Y,\,\text{rel }A$.

This applies to my situation with $A=Y=\{\omega\}$ and $X=W$. The map $f$ is the only one possible, and is a homotopy equivalence because of the weak contraction $H$. To say $f\circ g\simeq 1_{\{\omega\}},\,\text{rel }\{\omega\}$ is highly trivial, but the implication: $g\circ f\simeq1_W,\,\text{rel }\{\omega\}$ means that there is a homotopy $\Lambda:W\times I\to W$ with $\Lambda(w,0)=w$, $\Lambda(w,1)=g(f(w))=g(\omega)=\omega$ and $\Lambda(\omega,t)=\omega$ for all $t$ and $w$. That is, $\Lambda$ is a strong contraction of $W$ onto $\{\omega\}$.

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The proof simplifies considerably when $A$ and $Y$ are both points. Redefine the contraction $H$ to be $\omega$ at time zero and to be the identity at time one.

The map $g$ is just the inclusion $\{\omega\}\hookrightarrow W$, and the map $f$ just the constant $w\mapsto\omega$. Define the loop $\gamma:I\to W$ by $t\mapsto H(\omega,t)$. We define: $$k:W\times I\to W,\,(w,t)\mapsto\begin{cases}\gamma(1-2t)&0\le t\le\frac{1}{2}\\H(w,2t-1)&\frac{1}{2}\le t\le1\end{cases}$$

Let $k':I\to W$ be the loop $t\mapsto k(\omega,t)$. Note that $k'(t)=k'(1-t)$ for all $t$. $k'$ is essentially the loop concatenation $\gamma\cdot\overline{\gamma}$, which is zero in $\pi_1(W;\omega)$. Indeed, the proof involves identifying this loop with the null loop. We define a map: $$\kappa:I\times I\to W,\,(t,u)\mapsto\begin{cases}k'(t)&|1-2t|\ge u\\k'\left(\frac{1-u}{2}\right)=k'\left(\frac{1+u}{2}\right)&|1-2t|\le u\end{cases}$$

The homotopy extension property applied to $(W\times I;\{\omega\}\times I)$ finds: $$\K:W\times I\times I\to W$$Which maps $(w,t,0)\mapsto k(w,t)$ and $(\omega,t,u)\mapsto\kappa(t,u)$.

We get a strong contraction onto $\{\omega\}$ via the map: $$W\times I\to W,(w,t)\mapsto\begin{cases}\mathcal{K}(w,0,3t)&0\le t\le\frac{1}{3}\\\mathcal{K}(w,3t-1,1)&\frac{1}{3}\le t\le\frac{2}{3}\\\mathcal{K}(w,1,3(1-t))&\frac{2}{3}\le t\le1\end{cases}$$

There is a clever creation of a contour on which $\omega$ is stable, on the rim of the square $I\times I$. There is also nothing special about having a codomain $W$; if $H$ is an ambient contraction of $W$ in some $W'\supset W$, the proof would still work (importantly, for my linked question).

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A nice generalisation of this, taken by copying the proof, is that if $(W;\omega)$ is well-pointed then any map $f:W\to X$, for any space $X$, finds itself nullhomotopic if and only if it is based nullhomotopic as a map $(W;\omega)\to(X;f(\omega))$. I didn't check this in detail but I'm pretty confident of it...