# Can filtered colimits be computed in the homotopy category?

For $$\mathcal{S}$$ the $$(\infty,1)$$-category of spaces its homotopy category $$h\mathcal{S}$$ does not have pushouts or pullbacks. Even if it does, they won't always agree with the (homotopy) pushouts or pullbacks in $$\mathcal{S}$$.

Generally, filtered colimits are much better behaved than general ones. For example filtered colimits in the (Quillen) model category $$\mathit{sSet}$$ compute homotopy colimits in $$\mathcal{S}$$. In light of this I was wondering the following:

Does $$h\mathcal{S}$$ have filtered colimits and, if so, does the functor $$\mathcal{S} \to h\mathcal{S}$$ preserve them?

More generally one could ask the same for any presentable $$(\infty,1)$$-category $$\mathcal{C}$$; I particularly care about the case of the derived category $$\mathcal{D}(R)$$ of a ring $$R$$.

No, for a diagram $$X: I \to \mathcal{S} \to h\mathcal{S}$$ the colimit in $$h\mathcal{S}$$ would satisfy $$[\mathrm{colim} X(i),Y] \cong \lim [X(i),Y]$$ where brackets denote morphisms in $$h\mathcal{S}$$. Informally, the set of maps in $$h\mathcal{S}$$ from the (homotopy) colimit in $$\mathcal{S}$$ should involve also higher derived limits.

For a concrete example, you could take $$I = (\mathbb{N},<)$$, $$X(i) \simeq S^1$$ for all $$i$$, $$X(i) \to X(i+1)$$ a map of degree 2. Then $$\mathrm{colim}(X(i)) \simeq K(\mathbb{Z}[1/2],1)$$ is not the colimit in $$h\mathcal{S}$$ since it has distinct morphisms to $$Y = K(\mathbb{Z},2)$$ which become equal upon restricting to each $$X(i)$$.

Similar examples exist in $$D(\mathbb{Z})$$.

• Thanks! I suppose the colimit of the diagram you're suggesting is $K(\mathbb{Z}[\frac{1}{2}],1)$ and you're using that $H^2(\mathbb{Z}[\frac{1}{2}]; \mathbb{Z})$ is non-trivial? – J. Steinebrunner Feb 11 at 3:37

As has already been said, the homotopy category does not admit filtered colimits in general, but it’s much worse than that. Even colimits in an $$\infty$$-category which don’t give rise to colimits in the homotopy category sometimes do give rise to weak colimits. (A weak colimit cocone gives the existence, but not the uniqueness, of the factorizations a colimit cocone gives.) This is the case, for instance, with sequential colimits, as well as those along any free category, at least in spaces.

So one might ask whether at least every filtered colimit in $$\mathcal S$$ gives rise to a weak filtered colimit in $$h\mathcal S$$. Alas, this is still not true. In our paper kindly referenced by Tim, Christensen and I give an $$\aleph_1$$-indexed sequence of spaces whose homotopy colimit is not a weak colimit in the homotopy category, namely the sequence mapping a countable ordinal $$\alpha$$ to the wedge of $$\alpha$$ circles. The homotopy colimit is a wedge of $$\aleph_1$$ circles, and the problem is that a map out of that just requires too much coherence to be constructed out of a cocone over countable wedges in $$h\mathcal S$$. So there is not much hope for filtered colimits in $$h\mathcal S$$. I expect the same counterexample would work, though I have no idea how to make the argument, in higher homotopy categories $$h_n\mathcal S$$.

Regarding “minimal” or “distinguished” weak colimits, the general idea is that you want some weak colimits which are distinguished up to at least non-unique isomorphism, as occurs for cones in triangulated categories. Since “homotopy colimits” of sequences in triangulated categories with countable coproduct a are constructed out of those coproduct together with cones, they are also distinguished in this sense.

It is possible to get at the idea of minimal weak colimit of at least a filtered diagram in a category which may not be triangulated, but which has some set of objects detecting isomorphisms, by asking that $$Hom(S,\mathrm{wcolim} D_i)\cong \mathrm{colim} Hom(S, D_i)$$ for every $$S$$ in your isomorphism-detecting set. Such weak colimits are then indeed determined up to isomorphism, and they’re also nice because they see the objects $$S$$ as compact.

However, this is not to say that such distinguished weak colimits are common! Our diagram from above actually admits no weak colimit which views even $$S^1$$ as compact in this way. (Though note that some weak colimit always exists-homotopy pushouts give weak pushouts, coproducts exist, and then the usual construction applies.)

If your category actually has a set of compact generators in a model, as for $$D(R)$$, then a distinguished weak colimit must come from a homotopy colimit. Franke gives an argument, cited in our paper, that on these grounds distinguished weak colimits of uncountable chains should essentially never exist in $$D(R)$$. The problem is that there’s a spectral sequence converging to homs out of a homotopy colimit indexed by $$J$$ whose $$E_2$$ page involves the derived functors $$R^n\mathrm{lim}^J$$ for all $$n$$. These derived functors were shown by Osofsky to be non vanishing up through $$n$$ when $$J=\aleph_n$$, and the homotopy colimit is a weak colimit only if the spectral sequence collapses, so this probably shouldn’t happen. However Franke doesn’t give an argument that there couldn’t in principle be enough unlikely differentials to produce the collapse.

Christensen and I tried for a while to work with the analogous spectral sequence for spaces, but it seemed to require a proficiency with calculating higher derived limits unsupported by the literature-Osofsky gives a special example, and for all I can tell nobody else has ever calculated a derived limit over $$\aleph_n$$. So our approach turns out to be entirely different and doesn’t immediately apply to the stable case. Thus I think it’s unknown, though highly doubtful, whether $$D(R)$$ admits minimal filtered colimits in general.

Ironically, I was wondering something similar earlier this week (the irony is that I was sitting next to the OP while doing my wondering). Here's another reason why this can't be the case. In general, if $$C\subseteq D$$ is a full subcategory and every object in $$D$$ is a colimit of a diagram in $$C$$, then it follows that $$C$$ is a strong generator for $$D$$ -- i.e. the hom functors $$\{Hom(c,-): D \to Set\}_{c \in C}$$ are jointly faithful and conservative. If $$C$$ is essentially small, then one can take a product (or coproduct) of these functors to obtain a faithful, conservative functor $$D \to Set$$.

At the $$\infty$$-categorical level, every space $$X$$ is the filtered colimit $$X = \varinjlim_i X_i$$ of finite complexes $$X_i$$. If this colimit were also a colimit in the homotopy category, then $$X$$ would be a colimit of finite complexes in the homotopy category. Since the homotopy category of finite complexes is essentially small, we would get a faithful, conservative functor $$hS \to Set$$ by the above.

But as Freyd famously showed, there does not exist a faithful functor $$hS \to Set$$! I learned recently that Carlson and Christensen have shown there also does not exist a set of spaces such that the corresponding functor $$hS \to Set$$ is conservative!

We can conclude something stronger from this: there does not exist a regular cardinal $$\lambda$$ such that $$\lambda$$-filtered colimits in $$S$$ can be computed in the homotopy category. More generally, let $$A$$ be a $$\lambda$$-accessible $$\infty$$-category, and suppose that $$A \to hA$$ preserves $$\lambda$$-filtered colimits. Then, since $$A \to hA$$ is obtained by applying the filtered-colimit-preserving functor $$\pi_0: Spaces \to Sets$$ to the homsets, one sees that any $$\lambda$$-presentable object in $$A$$ is also $$\lambda$$-presentable in $$hA$$. Moreover, every object is a $$\lambda$$-filtered colimit of such objects. It follows that the functor $$hA \to Ind_\lambda(hA_\lambda)$$ is fully faithful (here $$A_\lambda \subseteq A$$ is the full subcategory of $$\lambda$$-presentable objects); by Rosicky's theorem (see below), this functor is also essentially surjective. So $$hA = Ind_\lambda(hA_\lambda)$$ is also $$\lambda$$-accessible! (As an alternative to Rosicky's theorem, if we assume that $$hA$$ has $$\lambda$$-filtered colimits, then it since every object is a $$\lambda$$-filtered colimit of $$\lambda$$-presentable objects, it follows directly that $$hA$$ is accessible) That is, if you have an accessible $$\infty$$-category and you can compute sufficiently-filtered colimits in the homotopy category, then the homotopy category is also accessible. Beyond the existence of classic counterexamples like Spaces, my sense is that this situation is comparatively rare in practice (except when $$A$$ was a 1-category to start with). In particular, I think it's rare for the homotopy derived category of a ring to be accessible. I don't know about the existence of filtered colimits in these homotopy categories though, but I suspect they typically don't exist.

A notable exception would be the derived category of field $$D(k)$$. In this case, the homotopy category is the category of graded $$k$$-vector spaces, which is accessible (indeed, it's locally presentable -- though not all colimits are preserved by the forgetful functor) and the forgetful functor can be identified with homology, which does preserve filtered colimits.

I haven't thought about this much -- it's possible that there are a lot of rings $$R$$ for which the homotopy category of $$D(R)$$ is accessible...

I think there's a lot more to be said about understanding (highly) filtered colimits by looking at the homotopy category.

• One commonly-used trick is that (co)products can be computed in the homotopy category, so if you have a triangulated category with countable coproducts, then you can compute a sequential colimit $$\varinjlim_n X_n$$ as the cofiber (using the triangulated structure) of the map $$\oplus_n X_n \to \oplus_n X_n$$ given by the difference of the identity and the coproduct of the maps $$X_n \to X_{n+1}$$. This allows you to compute countable filtered colimits using only the homotopy category + triangulated structure, by passing to a cofinal chain. But I don't think this generalizes to larger filtered colimits.

• There are Brown representability-type results which look at the functor $$ho(Ind_\lambda(C)) \to Ind_\lambda(ho(C))$$. Rosicky showed that this functor is always essentially surjective (perhaps after passing to a higher $$\lambda$$?). In some cases, Adams apparently showed that this functor is also full, and I think there's more work on this in the triangulated categories literature. But maybe this is not so close to the original question.

• Another tangentially related point -- not every idempotent in the homotopy category splits! (This is also a theorem of Freyd). The splitting of an idempotent is an example of a colimit of a diagram which is $$\lambda$$-filtered for every $$\lambda$$. This is only tangentially related, because if you know that your idempotent arises from a genuine homotopy-coherent idempotent (i.e. an idempotent in the $$\infty$$-category of spaces) then it does split; in particular the splitting in the homotopy category exists and agrees with the $$\infty$$-categorical splitting.

• I think in the triangulated categories literature, people deal with this issue by various "weak colimit" and "minimal weak colimit" constructions, but I don't know exactly how these work. Perhaps an expert can comment.

• Maybe it's implicit in things I've said above, but with these sorts of questions there can be a big difference between $$\infty$$-categories like $$Spaces$$ and $$\infty$$-categories like $$D(R)$$. For example, there does exist a conservative functor $$D(R) \to Set$$ (even already there is a conservative functor $$h(pointed spaces) \to Set$$ by Whitehead's theorem) -- although I think still not a faithful functor.

• Fun fact, in case you intended the implication in your last point that $h\mathcal S$ admits no conservative functor whatsoever to Set: actually every locally small category admits such a functor! I didn’t believe that when I heard it, but it’s yet another coup de Freyd. Of course there is still a big gap between $h\mathcal S$ and $D(R)$ in admitting such a functor which is a coproduct of representables. – Kevin Arlin Feb 10 at 7:32
• @KevinCarlson Whoa -- that's surprising! When you say "every locally small $C$ admits a functor $C \to Set$", do we have to assume that $C$ has no more isomorphism classes of objects than the size of the universe? – Tim Campion Feb 10 at 7:54
• Nope! As, er, tcamps recalls in the comments here: math.stackexchange.com/a/2128004/31228... the functor is roughly “the set of split subobjects with a disjoint basepoint.” In particular any object lacking nontrivial split subobjects will be mapped to a two-point set, so the functor is not injective on isomorphism classes even when applied to Set itself. If there are a large number of such objects, then the functor will have a large fiber on isomorphism classes while still being conservative. – Kevin Arlin Feb 10 at 14:56
• Haha! I can't believe I forgot such a striking fact! – Tim Campion Feb 10 at 15:59