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

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.

• Using $\Sigma \Omega \Sigma S^1$, you can get $X = \bigvee_{n > 1} S^n$. Using your trick, you can get the pullback of $X \vee S^1 \to S^1 \leftarrow \ast$, which is $\bigvee X$. Wedging with the spaces you constructed, you can get something with countably many spheres in each dimension. And with a swindle argument you can map that to itself and cone off as many spheres as you want in each degree. So I suspect it is "countable complexes"? – Tyler Lawson Oct 2 at 23:29
• Ah, the Snaith splitting, of course, thanks! So there are no obstructions on the number of cells... this yields all spaces which can be obtained from countable wedges of spheres via finitely many steps of gluing. Still a long way to go to get all countable complexes, but it feels much closer now! I'm still not sure how to construct even $\mathbb R \mathbb P^\infty$ or $\mathbb C \mathbb P^\infty$, though. – Tim Campion Oct 2 at 23:40
• This is true, there's still a lot to go. If you stuck to the 1-connected case, homotopy pullbacks don't wreck the property "the Poincare series of $H_* X$ is a rational function". There's quite a bit of daylight between those two extremes... – Tyler Lawson Oct 2 at 23:48
• The papers of Wojciech Chacholski are relevant to these sorts of questions; see e.g. Chachólski, Wojciech, Desuspending and delooping cellular inequalities, Invent. Math. 129 (1997), no. 1, 37–62. He and folks like Dror look at questions like yours quite systematically. – Nicholas Kuhn Oct 3 at 1:17
• @Tyrone I don't really care what the point-set level manifestation of $\mathcal F$ is -- I'm happy for it to contain all weakly contractible spaces. If you're unhappy with this formulation, then assume that every categorical term I've used is prefixed by "$\infty$". Since we are doing homotopy theory, "finite colimit" means "colimit indexed by a category whose nerve has finitely many nondegenerate simplices", so $\mathbb Z / 2$ is not considered finite. Equivalently, a class of objects is closed under finite colimits iff it contains the initial object and is closed under pushouts. – Tim Campion Oct 3 at 16:00

(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 $$n > 1$$.

• After spending some time thinking about this, I felt tantalizingly close to showing $\mathcal M_p$ itself to be closed under suspension and smash, but I'm really not sure it is. The best I can say is I'm pretty sure that if $BC_p \to A \wedge B$ is a pointed map with $A,B \in \mathcal M_p$, then it factors through the Bockstein $BC_p \to \mathbb C \mathbb P^\infty$. The argument I have in mind invokes Ganea's Theorem, which is kind of fun. Anyway, the first stated Theorem is quite suggestive... – Tim Campion Oct 11 at 13:31
• Just to be clear for the casual reader -- as clarified in the edit, there is a rather large gap in Jeff Strom's argument, so the question is still wide open. – Tim Campion Oct 11 at 14:00