What is the smallest class of spaces closed under finite homotopy colimits, finite homotopy limits, and splitting of homotopy-coherent idempotents?

Question

Finite CW complexes fail spectacularly to be closed under finite homotopy limits (e.g. $\Omega S^1 = \mathbb Z$). More subtly, they fail to be closed under homotopy retracts (by the Wall finiteness obstruction). In some sense, this is why one works with $\pi$-finite spaces when one needs a notion of "finite" space which is closed under finite limits.

But what happens if we bite the bullet and just close up under finite limits, colimits, and retracts? In the end, this is probably not the most natural thing to do (for instance, we might really want closure under extensions or something), but the question remains.

Questions:

1. Let $\mathcal F$ be the smallest class of spaces closed under finite homotopy colimits, finite homotopy limits, and splitting of homotopy-coherent idempotents. What is an explicit description of $\mathcal F$?

2. Fix a prime $p$, and let $\mathcal F_p$ be the smallest class of $p$-local spaces closed under finite homotopy colimits, finite homotopy limits, and splitting of homotopy-coherent idempotents. What is an explicit description of $\mathcal F_p$?

Notes:

• On the one hand, $\mathcal F$ is contained in the class of countable CW complexes.

• On the other hand, $\mathcal F$ contains all finite-dimensional countable CW complexes.

  To see this, note that by closure under finite limits and colimits, $\mathcal F$ contains the empty space and the one-point space, and thus by closure under retracts and finite colimits it contains all retracts of finite CW complexes. And $\mathcal F$ also contains $\mathbb Z = \Omega S^1$. Then by taking the homotopy pushout of the obvious maps $\mathbb Z \leftarrow S^0 \times \mathbb Z \to \mathbb Z$, we see that $\mathcal F$ contains $S^1 \times \mathbb Z$. Repeating, we see that $\mathcal F$ contains $S^n \times \mathbb Z$ for each $n$. By gluing, we obtain all finite-dimensional CW complexes with countably many cells.

• For example, plus constructions are obtained by attaching only 2-cells and 3-cells (the number of which is bounded in terms of the size of the fundamental group), so if $X \in \mathcal F$ and $P \subseteq \pi_1(X)$ is a perfect normal subgroup, then $X^+_P \in \mathcal F$, too. In particular, by taking $X = \vee^\omega S^1$ and suitable $P$, we get spaces in $\mathcal F$ with arbitrary countable fundamental group.

• There are also spaces in $\mathcal F$ such as $\Omega S^2$, with cells in arbitrarily large dimension. It's not clear to me exactly which infinite-dimensional countable CW complexes are in $\mathcal F$. For instance, what about $\mathbb C \mathbb P^\infty$ or $\mathbb R \mathbb P^\infty$?

• If we were working with just simply-connected spaces, the question would be much easier because the Eilenberg-Moore spectral sequence would always converge, and we could make some kind of argument about Serre classes. But I think the presence of non-simply-connected spaces complicates things.

Answer 1

(The discussion below is for pointed spaces.) I'll use $\mathcal{F}_*$ for the pointed version of your $\mathcal{F}$.

As Nicholas Kuhn says, this is related to the closed classes studied by E. Dror Farjoun and W. Chach\'olski. Including closure under limits in your collection actually makes it closer to the dual concept, which I have studied under the name of "resolving classes" and "resolving kernels".

I can't give a definitive answer to the question as posed, but I can show that the pointed version $\mathcal{F}_*$ contains no Eilenberg-MacLane spaces $K(G,n)$ for $n> 1$ or for $G$ finite abelian and $n = 1$

A resolving kernel is a collection of spaces of the form $$ \mathcal{R} = \{ Y \mid \mbox{$\mathrm{map}_*(X,Y) \sim *$ for all $X\in \mathcal{X}$}\} $$ for some collection of spaces $\mathcal{X}$.

Theorem: Let $\mathcal{A}$ be a collection of pointed spaces such that

• $\Sigma\mathcal{A}\subseteq \mathcal{R}$
• $\mathcal{A}\wedge \mathcal{A} \subseteq \mathcal{R}$

(up to weak equivalence). Write $\overline{\mathcal{A}}$ for the closure of $\mathcal{A}$ under "finite-type wedge" and extensions by cofibrations (and therefore under pushouts). If $\mathcal{R}$ is a resolving kernel and $\Sigma \mathcal{A} \subseteq \mathcal{R}$, then $\overline{\mathcal{A}}\subseteq \mathcal{R}$.

(A finite-type wedge is a wedge in which the connectivities of the summands increases to infinity.)

It is worth pointing (ha ha) out that the pointed case is quite different from the unpointed case, because getting even a single wedge into a resolving kernel is enough to bootstrap to the hypotheses of this theorem. Explicitly:

Theorem: If $X$ is a finite-type space with $\mathrm{map}_*(X, S^n\vee S^m) \sim *$ for any two $n, m > 1$, then $\mathrm{map}_*(X,K)\sim *$ for all simply-connected finite-dimensional CW complexes $K$. (If $\pi_1(X)$ is not a perfect group, then "simply-connected" can be dropped from the conclusion).

Now let's think about $\mathcal{F}_*$, the smallest class of pointed spaces containing all the spheres and closed under

• homotopy pushouts
• homotopy pullbacks
• homotopy retracts.

This is contained in $\mathcal{S}$,

Edit: Or maybe not! There are two classes and two closure properties: $\overline{\mathcal{A}}$ is closed under pushouts, while $\mathcal{R}$ is closed under pullbacks; and $\overline{\mathcal{A}}\subseteq \mathcal{R}$.

the smallest resolving kernel containing all the spheres. Let $\mathcal{M}_p$ denote the resolving kernel associated to $\mathcal{X} = \{ B\mathbb{Z}/p \}$ for your favorite prime $p$. Miller's theorem (the Sullivan conjecture) tells us that $$ \mathcal{F}_*\subseteq \mathcal{S} \subseteq \mathcal{M}_p. $$ But obviously $B\mathbb{Z}/p \not \in \mathcal{M}_p$, so we can conclude $$ B\mathbb{Z}/p \not\in \mathcal{F}_* $$ Similarly, $\mathcal{F}_*$ cannot contain Eilenberg-MacLane spaces $K(G,n)$ for any finite abelian group $G$ (such spaces also cannot see spheres by an easy extension of Miller's theorem). And using closure under forming homotopy fibers, we can see that $\mathcal{F}_*$ cannot contain $K(\mathbb{Z},n)$ for any $n$. Thus we have

Proposition: If $G$ is an abelian group, then $K(G,n) \not\in \mathcal{F}_*$ for $n > 1$.