# On the contractibility of $S^\infty$ and more generally on the contractibility of colimit spaces of type $A_0\hookrightarrow A_1\hookrightarrow\cdots$

There is an old MO thread as well as several MSE threads concerning the contractibility of $$S^\infty$$. Interpretations of the notation vary, so I here define $$S^\infty$$ as follows:

For every $$n\in\Bbb N_0$$ consider the standard $$n$$-sphere $$S^n\subset\Bbb R^{n+1}$$ and let $$\iota_n:S^n\hookrightarrow S^{n+1}$$ denote the embedding: $$S^n\ni\vartheta\mapsto(\vartheta,0)\in S^{n+1}\subseteq\Bbb R^{n+1}\times\Bbb R$$Consider the diagram: $$S^0\overset{\iota_0}{\hookrightarrow}S^1\overset{\iota_1}{\hookrightarrow}S^2\overset{\iota_2}{\hookrightarrow}\cdots\overset{\iota_{n-1}}{\hookrightarrow}S^n\overset{\iota_n}{\hookrightarrow}S^{n+1}\overset{\iota_{n+1}}{\hookrightarrow}\cdots$$In the category $$\mathsf{Top}$$. Let $$S^\infty$$ be its colimit.

So I don't attribute any particular norm on $$S^\infty$$, it is really just an abstract CW complex with cellular decomposition $$(e_0\cup e_0)\cup(e_1\cup e_1)\cup(e_2\cup e_2)\cup\cdots$$ and $$k$$-skeleton $$S^k$$ for every $$k\ge-1$$.

This answer makes the following claim with the following (disputed) proof (I have changed it slightly to suit my preferences):

Suppose $$A_0\hookrightarrow A_1\hookrightarrow A_2\hookrightarrow\cdots$$ is a chain of subspace inclusions where every pair $$(A_{n+1};A_n)_{n\in\Bbb N_0}$$ has the homotopy extension property. Let $$A$$ be the colimit of the associated diagram. Suppose there is an ambient contraction $$\Gamma_n:A_n\times I\to A_{n+1}$$ for every $$n$$, and let $$p_n$$ denote the point in $$A_{n+1}$$ to which $$\Gamma_n(-,1)$$ identically maps $$A_n$$.

Claim: the space $$A$$ is contractible. The proof:

By basic colimit properties (and cocontinuity of $$(-)\times I$$, since $$I$$ is strongly locally compact Hausdorff (!)), $$(A;A_n)$$ inherits the homotopy extension property. Using this, we get a map $$\alpha_n:A\times I\to A$$ by defining the component $$A\to A$$ at time zero to be the identity, and letting the map $$A_n\times I\to A$$ be the composite $$A_n\times I\overset{\Gamma_n}{\longrightarrow}A_n\hookrightarrow A$$.

We define a map $$f:A\times I\to A$$ by asserting that the restriction of $$f$$ to every $$A\times[1-2^{-n},1-2^{-(n+1)}]$$, $$n\ge0$$, is the map with every $$t$$-time slice equal to the composite: $$\alpha_{n+1}(-,2-2^{n+1}(1-t))\circ\alpha_n(-,1)\circ\alpha_{n-1}(-,1)\circ\cdots\circ\alpha_0(-,1):A\to A$$

These partial definitions agree on every common intersection $$\{1-2^{-k}\}_{k\in\Bbb N_0}$$, so define a continuous function $$A\times[0,1)$$ by a standard gluing lemma.

Here is the rub: I was not the only one to notice the author did not handle the behaviour of $$f$$ at time $$1$$. This is crucial, since we need $$f$$ to collapse $$A$$ to a single, well-defined point, and to do so continuously. As it stands, if one fixes an $$a\in A$$ and tracks the point $$f(a,t)$$ as $$t$$ varies through $$1-2^{-k},k\ge0$$, we see that eventually we get (a tail of) the sequence $$(p_n)_{n\in\Bbb N_0}$$.

I tried to cover this up in my own notes with the additional hypothesis (which applies to the contraction of $$S^\infty$$):

We further demand that there is some accumulation point $$\rho$$ of the sequence $$(p_n)_{n\in\Bbb N_0}$$.

So that we might then set $$f(a,1)=\rho$$ for all $$a\in A$$. I proved to myself that this makes $$f$$ continuous. In the case of $$S^\infty$$ we can pick $$\rho$$ to be [the equivalence class of] $$1\in S^0$$ so that $$p_n$$ is a constant sequence with value $$\rho$$, so I'm not so worried about that, but I am worried about rescuing the theorem, since it seems useful/interesting (if true...).

My proof:

Define $$f$$ as above. $$f$$ is continuous at every point in $$A\times[0,1)$$ so it suffices to demonstrate continuity at the points of $$A\times\{1\}$$. Say $$\mathcal{O}\ni\rho$$ is open in $$A$$. By definition, there is an increasing sequence $$(n_k)_{k\in\Bbb N_0}\subseteq\Bbb N$$ with $$\rho$$ a topological limit of the subsequence $$(p_{n_k})_{k\in\Bbb N_0}$$. For some $$K\in\Bbb N_0$$, we get $$\rho_{n_k}\in\mathcal{O}$$ for all $$k\ge K$$.

For every $$k\ge K$$ define $$W_k:=\alpha_{n_k}^{-1}(\mathcal{O})\cap A\times(1-2^{-n_k},1]$$, which we know are open neighbourhoods of $$A_{n_k}\times\{1\}$$ in $$A\times I$$ since $$\rho_{n_k}\in\mathcal{O}$$. Taking unions in $$k$$ of the $$W_k$$, we get an open neighbourhood of $$A\times\{1\}$$ which shows $$f$$ to be continuous.

But even as I write this question, I have begun to realise my proof is wrong since it is not given that the $$\alpha_{n_k+1}$$, when acting over $$\alpha_{n_k}(W_k)$$, will keep all images contained in $$\mathcal{O}$$.

It is easy to 'make it true' by demanding that the sequence $$p_n$$ is constant and that each contraction is a strong contraction (this still saves the case of $$S^\infty$$).

So I suppose now my question is - what are some weaker hypotheses with which we can salvage this theorem (what are the weakest?)? Or is the theorem true as originally stated, just with an incomplete proof?

EDIT: I thank the user @CheerfulParsnip for inspiring me. With a slightly tweaked set of hypotheses, we can find $$A$$ contractible at the cost of not being able to track down an explicit contraction.

Suppose every $$A_i$$ is a CW complex and that each $$A_i\hookrightarrow A_{i+1}$$ is a subcomplex inclusion. Then, $$(A_{k+1};A_k)$$ always has the homotopy extension property and $$A$$ inherits a natural CW structure. The idea is to use the fact that these are all CW complexes to employ Whitehead's theorem. We don't need to suppose anything about the sequence $$(p_n)_{n\in\Bbb N_0}$$ which is good news!

Let $$n\in\Bbb N$$ be arbitrary and $$a\in A$$ an arbitrary base point. Let $$f:(S^n;(1,0,\cdots,0))\to(A;a)$$ be any based map. For compactness reasons, there is an $$m\in\Bbb N_0$$ such that we can factor $$f$$ as $$f:S^n\to A_m\hookrightarrow A$$. Then consider: \begin{align}H:S^n\times I&\to A\\(\theta,t)&\mapsto\begin{cases}\Gamma_m(f(\theta),2t)&0\le t\le\frac{1}{2}\\\Gamma_m(a,2(1-t))&\frac{1}{2}\le t\le1\end{cases}\end{align}

$$H$$ provides $$f\simeq c_a$$, however this homotopy is not necessarily based. Hopefully that's enough to show $$\pi_n(A;a)$$ is trivial.

It is, as it turns out:

Let an overhead bar denote time reversal. Let $$q:S^n\times I\twoheadrightarrow D^{n+1}$$ be the canonical quotient map for $$S^n\times I/(S^n\times\{1\})$$, with $$q(\theta,t)=(1-t)\cdot\theta$$. Let $$r:D^{n+1}\to D^{n+1}$$ map $$x=(x_0,x_1,\cdots,x_n)\mapsto(x_0+1-\|x\|,x_1,\cdots,x_n)$$.

We know $$H$$ factors through $$q$$ by some $$\tilde{H}$$ and we can then define $$G:S^n\times I\to A$$ as the composite: $$S^n\times I\overset{q}{\twoheadrightarrow}D^{n+1}\overset{r}{\longrightarrow}D^{n+1}\overset{\tilde{H}}{\longrightarrow}A$$

We have that $$G:f\simeq c_{f(a)}$$ is now a based homotopy, since $$r$$ is constant on the fibre $$q^{-1}(1,0,\cdots,0)$$ - we might even abstractly take $$r$$ from the reduced cone quotient - hence $$[f]=0$$ in $$\pi_n(A;a)$$. Yet $$f$$ was arbitrary, so $$\pi_n(A;a)$$ is trivial (for every $$a\in A$$).

It remains to check $$\pi_0(A)$$ is trivial; but if $$a,b\in A$$ are any two points, there is some $$A_m$$ containing them both and $$a\simeq p_m\simeq b$$ in $$A_{m+1}$$ hence also in $$A$$ in path connections, so $$\pi_0(A)$$ is indeed trivial. By Whitehead's theorem, $$A$$ must be contractible.

So that's one possible resolution (assuming the lack of a based homotopy can be circumvented).

I also thank the user @Tyrone for suggesting another set of reasonable hypotheses. We suppose exactly the hypotheses of the original claim, and further suppose that there is some $$\rho\in A_0$$ such that $$(A_0;\rho)$$ is a well-pointed space (all hypotheses hold, for instance, if $$A$$ is any CW complex with skeletal filtration $$A_0,A_1,\cdots$$ and each $$A_i$$ is weakly contractible as a subcomplex in $$A_{i+1}$$).

We may, similarly to the above, construct a new map: \begin{align}H_n:A_n\times I&\to A_{n+1}\\(a,t)&\mapsto\begin{cases}\Gamma_n(a,2t)&0\le t\le\frac{1}{2}\\\Gamma_n(\rho,2(1-t))&\frac{1}{2}\le t\le1\end{cases}\end{align}

For every $$n\in\Bbb N_0$$, and $$H_n$$ is an ambient weak contraction of $$A_n$$ onto the point $$\rho$$. By assumption, and chaining homotopy extension properties, we find that $$(A_n;\rho)$$ is well-pointed for all $$n$$.

Using this (modulo making a minor change to the proof), we can promote $$H_n$$ to a strong ambient contraction $$A_n$$ onto $$\rho$$.

Now we have ourselves a family of strong ambient contractions onto the same point $$\rho\in A_0\subseteq A_n$$, and it is easy to take the $$f$$ from the original half-proof, set $$f(a,1)=\rho$$ for all $$a\in A$$, and check that it is continuous after first replacing $$\Gamma_n$$ with $$H_n$$.

• Comments are not for extended discussion; this conversation has been moved to chat. Dec 22, 2022 at 16:47
• Hi FShrike. Did you resolve the problem in the end? Dec 28, 2022 at 9:00
• @Tyrone Yes, thank you. It was pointed out to me that the issue of: "a contractible well-pointed space is strongly contractible" is essentially a corollary of a certain lemma involving cofibrations that I'd already learned from Hatcher. Annoying that I forgot about this. See here for what I mean. Thanks again for your time and ideas Dec 28, 2022 at 9:08