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

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.

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 $$r$$.

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.

$$\newcommand{\K}{\mathcal{K}}$$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 $$\{\omega\}$$.

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).

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...