### 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 $\mathcal{H}$ denote the homotopy category of pointed connected CW complexes, and let $F:\mathcal{H}^{op}\to Set_*$ 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 $\mathcal{C}$ be cocomplete, locally small, and well-copowered, with a small generating set, and let $F:\mathcal{C}^{op}\to Set$ be a functor that takes colimits in $\mathcal{C}$ 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 $\mathcal{H}$ 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 $\mathcal{K}$ be a category with coproducts, weak pushouts, and weak sequential colimits, which admits a strongly generating set $\mathcal{G}$, closed under finite coproducts and weak pushouts, and such that for $X\in\mathcal{G}$ the hom-functor $\mathcal{K}(X,-)$ takes the weak sequential colimits to actual colimits. Then if $F:\mathcal{K}^{op}\to Set$ takes coproducts to products and weak pushouts to weak pullbacks, it is representable.

Recall that $\mathcal{G}$ is a *generating set* if the hom-functors $\mathcal{K}(X,-)$ 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 $\{S^n | n \ge 1 \}$ 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 $(S^n)_+$ 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 $(\infty,1)$-category of unpointed spaces *does* have a strong generator in the $(\infty,1)$-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 $\phi:G\to G$ and an element $b\in G$ such that $\phi(\phi(x)) = b \cdot \phi(x)\cdot b^{-1}$ 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 $\phi$ induces an endomorphism $B \phi :B G \to B G$, and since free homotopies of continuous maps between classifying spaces of groups correspond to conjugations, $\phi$ 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, $\phi$ induces an idempotent of the representable functor $[-,B G]$, which *does* split (since idempotents split in $Set$); 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 $\phi$ does not split.

Finally, $B \phi _+ :B G_+ \to B G_+$ 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 $H^n(X) \cong H^{n+1}(\Sigma X)$ (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 $\{S^n | n \in \mathbb{Z}\}$ 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.)

## Re: Brown representability

…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.)