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 $H_n$.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Dec 22, 2022 at 16:47
  • $\begingroup$ Hi FShrike. Did you resolve the problem in the end? $\endgroup$
    – Tyrone
    Dec 28, 2022 at 9:00
  • $\begingroup$ @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 $\endgroup$
    – FShrike
    Dec 28, 2022 at 9:08


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