# Cohomology theories as colimits

I am looking for examples of cohomology theories that can be written as (filtered, or another nice class of) colimits of "simpler" functors, i.e. which $\{h^n : {\bf Top}^2 \to {\bf Ab}\}_n$ are such that $$h^n(X) \cong \text{colim}_j\; h_j^n(X)$$ for a suitable diagram ${\cal J}\to [{\bf Top}^2,{\bf Ab}]$.

Of course this is a really vague question:

1. A "cohomology theory" (and the category thereof) is what (for example) Rudyak I.3.8 defines as such.
2. I'm not asking that the $h^n_j$ are cohomology theories themselves, but you can assume this additional requirement.
3. You're quite free to interpret the word "simple" in the way to like more. I'm in fact explicitly asking for which meaning of "simple" this question has a good answer.

The question remains a bit vague: whatever $h^n(X)$ is, you can take a presentation for this abelian group and say that it is a colimit. Nevertheless I think that asking for a colimit of functors is a bit more restrictive and avoids trivial cases.

My feeling is that the answer is always "quasi-affirmative": a cohomology theory, i.e. a spectrum, belongs to a presentable quasicategory. But spectra and cohomology theories aren't really the same thing.

• Well any spectrum $X$ is obtained as a filtered homotopy colimit like $\text{hocolim}\Sigma^{-n}\Sigma^{\infty}X_n$ for some spaces $X_n$. If you restrict what you mean by 'cohomology theory' to take as input only finite complexes, then this would present an arbitrary cohomology theory in terms of one built from cohomology theories associated to suspension spectra. It's up to you to decide whether that's any easier... – Dylan Wilson Aug 24 '17 at 19:22
• Do suspension spectra are finitely presentable objects in the category of spectra? If yes, is this inclusion proper? – fosco Aug 25 '17 at 9:28
• No, not all suspension spectra are finitely presentable- that's not what I was using. Suspension spectra of finite complexes are, though. – Dylan Wilson Aug 25 '17 at 9:30
• Thanks, that's what I had in mind: $\Sigma^\infty$ must send fp objects into fp objects since $\Omega^\infty$ preserves filtered (in fact sifted) colimits. I doubt this exhausts all the fp objects: is there a characterization of some sort for the subcategory ${\rm Sp}_{<\omega}$ of fp spectra? – fosco Aug 25 '17 at 9:51
• Those (and there desuspensions, of course) are actually all of them. – Dylan Wilson Aug 25 '17 at 10:49

Suppose you just look at colimits indexed by $\mathbb{N}$, and fix the cohomological degree $n$. I claim that there must exist $j$ such that the map $h_j^n(X)\to h^n(X)$ is surjective for all $X$. Indeed, if not, we can choose spaces $X_j$ and classes $a_j\in h^n(X_j)$ that do not come from $h^n_j(X_j)$. Put $X=\coprod_jX_j$. As $h$ is assumed to be a cohomology theory, there is a unique element $a\in h^n(X)$ with $a|_{X_j}=a_j$ for all $j$. If this comes from some $a'\in h^n_j(X)$ then we see that $a'|_{X_j}$ maps to $a_j$, contrary to assumption. (We do not need to assume that $h_j^n$ converts coproducts to products for this, we are just using functoriality.)
I suspect that under mild assumptions one can prove something stronger: there is a cofinal sequence $j_1<j_2<\dotsb$ and functors $u_k$ such that $h_{j_k}^n=h^n\oplus u_k$ with the map $h_{j_k}\to h_{j_{k+1}}$ being zero on $u_k$. This means that there are no really interesting examples. All this depends on the colimit formula being valid for infinite complexes, of course.
I think by restriction to finite complexes of the source of a given map $f:X\to BO$, one obtains a family of spectra which filter Thom spectrum of $f$. As famous examples, we may consider Ravenel spectra $X(n)$ as well as various Madsen-Tillmann spectra such as $MTO(n)$. Of course, I don't claim this gives all possible examles. This is perhaps dual to what you ask for.