## August 24, 2012

### Brown representability

#### Posted by Mike Shulman

Time for more debunking of pervasive mathematical myths. Quick, state the Brown representability theorem!

Here’s (roughly) the original version of the theorem that Brown proved:

Theorem. Let $ℋ$ denote the homotopy category of pointed connected CW complexes, and let $F:{ℋ}^{\mathrm{op}}\to {\mathrm{Set}}_{*}$ be a functor to pointed sets which takes coproducts to products, and homotopy pushouts to weak pullbacks (i.e. spans with the existence, but not the uniqueness, property of a pullback). Then $F$ is a representable functor.

(The restriction to CW complexes is, of course, equivalent to saying we invert weak homotopy equivalences among all spaces.) As a graduate student, it was a revelation to me to realize that Brown representability is closely related, at least in spirit, to the Adjoint Functor Theorem. In fact, although this isn’t always made explicit, an essential step in the usual proof of the (special, dual) AFT is the following:

Lemma. Let $𝒞$ be cocomplete, locally small, and well-copowered, with a small generating set, and let $F:{𝒞}^{\mathrm{op}}\to \mathrm{Set}$ be a functor that takes colimits in $𝒞$ to limits. Then $F$ is representable.

Clearly there is a family resemblance. But how does Brown’s theorem get away with such weaker assumptions on limits? The category $ℋ$ is certainly not cocomplete — it only has weak colimits — but we still get a strong representability theorem, not merely a “weak” one of some sort. The answer is that Brown uses a stronger generation property, as can be seen from the following abstract version of Brown’s original theorem (also due to Brown):

Theorem. Let $𝒦$ be a category with coproducts, weak pushouts, and weak sequential colimits, which admits a strongly generating set $𝒢$, closed under finite coproducts and weak pushouts, and such that for $X\in 𝒢$ the hom-functor $𝒦\left(X,-\right)$ takes the weak sequential colimits to actual colimits. Then if $F:{𝒦}^{\mathrm{op}}\to \mathrm{Set}$ takes coproducts to products and weak pushouts to weak pullbacks, it is representable.

Recall that $𝒢$ is a generating set if the hom-functors $𝒦\left(X,-\right)$ are jointly faithful, i.e. detect equality of parallel arrows, and a strongly generating set if these functors are moreover jointly conservative, i.e. detect invertibility of arrows. In the homotopy category of pointed connected spaces, the pointed spheres $\left\{{S}^{n}\mid n\ge 1\right\}$ are of course strongly generating, since by definition, mapping out of them detects homotopy groups.

Now in surprisingly many places, you may find Brown’s original representability theorem quoted without the adjective “connected”. You may even be tempted to conjecture that it should also still be true without the adjective “pointed”. However, there is no obvious strongly generating set in either of these cases! If we add ${S}^{0}$, then we can detect ${\pi }_{0}$-isomorphisms of pointed spaces, but the other pointed spheres only detect higher homotopy groups of the basepoint component. On the other hand, mapping out of something like $\left({S}^{n}{\right)}_{+}$ doesn’t detect homotopy groups of other components either, since in that case the loops are “free” rather than based — and the unpointed spheres ${S}^{n}$ have the same problem in the unpointed homotopy category.

(This should, of course, be contrasted with the fact that the $\left(\infty ,1\right)$-category of unpointed spaces does have a strong generator in the $\left(\infty ,1\right)$-categorical sense: namely, the point. Strong-generation is not preserved on passage to homotopy categories!)

In fact, as pointed out recently on MO by Karol Szumiło (thanks Karol!), Brown representability for non-connected and unpointed spaces is false! This was proven back in 1993 by Peter Freyd and Alex Heller. Their construction begins with the group $G$ that is freely generated by an endo-homomorphism $\varphi :G\to G$ and an element $b\in G$ such that $\varphi \left(\varphi \left(x\right)\right)=b\cdot \varphi \left(x\right)\cdot {b}^{-1}$ for all $x\in G$.

(Aside: This is a slightly unusual sort of “free generation”. Exercise for homotopy type theorists in the audience: define this group $G$ as a higher inductive type.)

Now $\varphi$ induces an endomorphism $B\varphi :BG\to BG$, and since free homotopies of continuous maps between classifying spaces of groups correspond to conjugations, $\varphi$ is idempotent in the homotopy category of unpointed spaces. Freyd and Heller proved that this idempotent does not split. (I have not attempted to understand their proof of this, but I find it believable.) Now, however, $\varphi$ induces an idempotent of the representable functor $\left[-,BG\right]$, which does split (since idempotents split in $\mathrm{Set}$); let $F$ be the splitting. Since $F$ is a retract of a representable, which takes (weak) colimits to (weak) limits, it does the same; but it is not itself representable because $\varphi$ does not split.

Finally, $B{\varphi }_{+}:B{G}_{+}\to B{G}_{+}$ gives a similar counterexample in pointed, non-connected spaces.

Now the original, and probably still most common, use of the Brown representability theorem is to produce spectra which represent cohomology theories. In that case, the suspension axiom ${H}^{n}\left(X\right)\cong {H}^{n+1}\left(\Sigma X\right)$ (for reduced cohomology) means that the whole thing is determined by its behavior on pointed connected spaces (since $\Sigma X$ is connected), so this is not a problem. There are also different tricks that appear to work for any single functor as long as it lands in (abelian) groups rather than sets. Note also that the collection of spheres $\left\{{S}^{n}\mid n\in ℤ\right\}$ is strongly generating in the category of all spectra. But outside of those domains, Brown representability is subtler than I used to think.

(I was first made aware of this issue several years ago, when Peter May and Johann Sigurdsson ran into it while trying to use the abstract version cited above to produce right adjoints to derived pullback and smash product functors in parametrized homotopy theory. However, I didn’t realize until now that the nonconnected and unpointed versions were actually false even in the classical case of ordinary spaces.)

Posted at August 24, 2012 5:55 PM UTC

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### Re: Brown representability

Their construction begins with the group $G$ that is freely generated by an endo-homomorphism $\varphi :G\to G$ and an element $b\in G$ such that $\varphi \left(\varphi \left(x\right)\right)=b\cdot \varphi \left(x\right)\cdot {b}^{-1}$ for all $x\in G$.

…and that group is, if I’m not mistaken, Thompson’s group $F$.

(Freyd seems to prefer not to call it Thompson’s group, because he believes he discovered it first, and maybe he did. But everyone else calls it Thompson’s group.)

Posted by: Tom Leinster on August 25, 2012 1:41 AM | Permalink | Reply to this

### Re: Brown representability

Hah, I was going to point out the appearance of F, but Tom has (fittingly) beaten me to it.

Posted by: Yemon Choi on August 25, 2012 9:21 AM | Permalink | Reply to this

### Re: Brown representability

Does your abstract characterisation of F have anything to do with its appearance in this post?

Posted by: David Corfield on August 25, 2012 10:38 PM | Permalink | Reply to this

### Re: Brown representability

Ah, excellent! I wasn’t sure whether this post was worth writing, but you’ve proven that it absolutely was, because otherwise I might never have realized that.

Posted by: Mike Shulman on August 26, 2012 1:57 AM | Permalink | Reply to this

### Re: Brown representability

Clearly there is a family resemblance.

Are there other members of the family?

Posted by: David Corfield on August 25, 2012 10:23 PM | Permalink | Reply to this

### Re: Brown representability

Well, there are variants of the classical AFT. E.g. it’s enough for the category to be total, rather than cocomplete with a generating set. And there are versions for indexed categories and enriched categories, and more generally for objects in a Yoneda structure.

Posted by: Mike Shulman on August 26, 2012 4:37 AM | Permalink | Reply to this

### Re: Brown representability

That’s a very nice summary of the whole story. However, it occurred to me that the notion of strongly generating set is not the right one here. Namely, the functors represented by spheres are not jointly faithful in the homotopy category of based connected spaces. Think about any nontrivial cohomology operation of nonzero degree, it is represented by a map $K\left(A,m\right)\to K\left(B,n\right)$ with $m. It induces zero on homotopy groups just like the trivial map but is nontrivial. On the other hand the functors represented by spheres are jointly conservative and it seems to be all that is relevant to the Representability Theorem.

I also want to point out that an earlier paper of Heller that I mentioned in the comment to your MO question also gives a counterexample in the non-based case which is even more striking since it is a half-exact functor which is not even a retract of a representable functor. I admit that I don’t really understand the construction, but it seems that the problem is that the fundamental group of the representing space would have to be proper class.

Posted by: Karol Szumiło on August 25, 2012 10:41 PM | Permalink | Reply to this

### Re: Brown representability

the notion of strongly generating set is not the right one here

An excellent point.

It may be worth mentioning at this point, in case some reader is unaware, that there is no unanimity in the literature on the meaning of “generating set”. I like the definitions I gave in the post above, because they are systematic: $𝒢$ is generating if for all $X$, the family of all morphisms $G\to X$ for $G\in 𝒢$ is jointly epimorphic, and strongly generating if that family is jointly strong-epimorphic (or, more precisely, extremal-epimorphic — so “extremally generating” would be an even better name — but strong and extremal epimorphisms are the same in any category with pullbacks, so people are often sloppy about the difference, and I’ve decided not to try to fight that battle). However, some people say “separating set” and “generating set” where I say “generating set” and “strongly generating set”, and I don’t know whether all of these people require a generating set (in their terminology) to also be separating. (In a category with equalizers, the implication is automatic — but of course the homotopy category lacks equalizers.)

It would be nice to have a word for morphisms which satisfy the property that makes an epimorphism strong, but are not necessarily themselves epic, and similarly for the sort of “strongly not-generating sets” that we see here.

a counterexample in the non-based case which is even more striking since it is a half-exact functor which is not even a retract of a representable functor. … it seems that the problem is that the fundamental group of the representing space would have to be proper class.

Thanks for pointing that out here! Although I’m not sure whether I find it more striking to have a functor that fails to be representable because an idempotent fails to split, or a functor that fails to be representable for cardinality reasons. I think both of them seem like rather “boring” reasons. (-: But I suppose the second example implies that Brown representability also fails in the Cauchy completion of the unbased homotopy category, which is not obvious if you’ve only seen the first example.

Posted by: Mike Shulman on August 26, 2012 4:51 AM | Permalink | Reply to this

### Re: Brown representability

It may be worth mentioning at this point, in case some reader is unaware, that there is no unanimity in the literature on the meaning of “generating set”.

There is one such notion that is especially relevant to this discussion, namely generating sets in triangulated categories. A set of objects $𝒢$ of a triangulated category $𝒯$ is generating if given an object $X$ such that for all $G\in 𝒢$ and $m\in ℤ$ we have $𝒯\left({\Sigma }^{m}G,X\right)=0$, then it follows that $X=0$. In the presence of coproducts this is (somewhat non-trivially) equivalent to $𝒢$ generating all objects $𝒯$ under shifts, cones and coproducts.

For example in the stable homotopy category the sphere spectrum $𝕊$ is generating even though $\left\{{\Sigma }^{m}𝕊\right\}$ is not a generating set in the sense of your post because of the existence of non-trivial stable cohomology operations.

The relevance is that if $𝒯$ has coproducts and a set of compact generators, then every cohomological functor ${𝒯}^{\mathrm{op}}\to \mathrm{Ab}$ is representable. The proof is a variation on the usual proof, but it works even more neatly because of the stability.

Posted by: Karol Szumiło on August 27, 2012 11:08 AM | Permalink | Reply to this

### Re: Brown representability

Isn’t that the same as this sort of “strong non-generation”, though, because a morphism in a triangulated category is an isomorphism iff its cone is zero, and hom-functors are exact?

Posted by: Mike Shulman on August 27, 2012 6:24 PM | Permalink | Reply to this

### Re: Brown representability

You are right of course. I noticed that this material is not very well covered in nLab. Maybe I will try to add something.

Posted by: Karol Szumiło on August 28, 2012 10:25 AM | Permalink | Reply to this

### Re: Brown representability

Maybe it’s worth adding another “issue” that disappears when you go stable, say the smallness of the generators. I learned that from the paper “Representability theorems for presheaves of spectra” by Jardine. It contains a version of Brown representability for any compactly generated model category (an object $x$ is compact in $C$ if ${\mathrm{colim}}_{i}\left[x,{y}_{i}\right]\to \left[x,{\mathrm{colim}}_{i}{y}_{i}\right]$ is a bijection in $\mathrm{Ho}\left(C\right)$ for every (countable) sequential colimit). However, Brown representability seems to be unknown if the category is $\alpha$-compactly generated for a regular cardinal $\alpha$ (change sequential colimits for $\alpha$-sequential colimits). In the case $C$ is stable, its homotopy category will be well generated and you can proof Brown representability even without any model structure (as done in Neeman’s book). Jardine also shows that this problem disappears if we “enrich” the problem, i.e. if we start with a simplicial $\alpha$-compactly generated (cofibrantly generated pointed) model category $C$ and a simplicial functor $F:{C}^{\mathrm{op}}\to {\mathrm{sSet}}_{*}$ (sending w.e. to w.e and homotopy colimits to homotopy limits), then $F\left(-\right)\simeq \mathrm{Map}\left(-,X\right)$.
Posted by: Oriol Raventós on September 5, 2012 4:59 PM | Permalink | Reply to this

### Re: Brown representability

I’d just like to explicitly cite Heller’s paper “On the representability of homotopy functors” J. London Math. Soc. (2) 23 (1981), no. 3, 551–562, which Karol already mentioned on math overflow, as his Theorem 1.3 is the most general positive result that I know, but I don’t think the paper is as well known as it should be.

I should also mention that the representability of homology functors (which isn’t representability in the sense of category theory, since it involves a symmetric monoidal structure), is much more subtle.

Posted by: Dan Christensen on September 26, 2012 5:18 PM | Permalink | Reply to this

### Re: Brown representability

Alex Heller’s papers from that time were full of insights that perhaps now are better appreciated, yet you rarely see references to them. He did a lot of work on homotopy Kan extensions, for instance.

Posted by: Tim Porter on September 26, 2012 8:32 PM | Permalink | Reply to this

### Re: Brown representability

Indeed they are! I’ve been discovering that his monograph “Homotopy theories” and later paper on stable homotopy theories are full of useful stuff about what we now call derivators.

Posted by: Mike Shulman on September 26, 2012 10:36 PM | Permalink | Reply to this

### Re: Brown representability

Yes, in that memoir Alex had some comment about Grothendieck doing something but it seems to be going in a different direction! I seem to remember telling both of them about the others work way back in 1984 or shortly after, but I do not know if the message was understood. They thought of themselves doing different things.

Jean-Marc Cordier and I worked on trying to extend his proofs (including that of Brown representability) to the simplicially enriched context (as we called it then). That part of our results were never published but some of the more basic stuff did appear in the paper (Homotopy Coherent Category Theory, Trans. Amer. Math. Soc. 349 (1997) 1-54.).

Posted by: Tim Porter on September 27, 2012 6:45 AM | Permalink | Reply to this

### Re: Brown representability

How about a link to `Alex Heller’ if there is such an entry?
And if it’s there, does it due justice to his work?

Posted by: jim stasheff on September 27, 2012 1:37 PM | Permalink | Reply to this

### Re: Brown representability

There is now, but it does not do justice to his contribution to the subject. I have made some links but not enough.

Posted by: Tim Porter on September 28, 2012 7:56 AM | Permalink | Reply to this

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