2012-08-16 19:33:17 8 Comments

In many places (on MO, elsewhere on the Internet, and perhaps even in some textbooks) one finds a statement of the classical Brown representability theorem that looks something like this:

If $F$ is a contravariant functor from the (weak) homotopy category of topological spaces $\mathrm{Ho}(\mathrm{Top}_\ast)$ to the category $\mathrm{Set}_\ast$ of pointed sets which sends coproducts (i.e. wedges) to products and weak/homotopy pushouts (or CW triads) to weak pullbacks, then $F$ is repesentable.

However, I have been unable to find a convincing proof of this statement. All the proofs I have found fall into one of these classes:

- Proofs of an analogous statement where $\mathrm{Ho}(\mathrm{Top}_*)$ is replaced by its subcategory of pointed
*connected*spaces. - Proofs which purport to prove the above statment, but which actually implicitly assume all spaces are connected at some point. (It's sometimes hard to tell whether a given proof falls into this class or the previous one, since some classical algebraic topologists seem to use "space" to mean "connected pointed space".)
- Proofs that all cohomology theories are representable by a spectrum. Here the suspension axiom implies that the behavior on connected spaces determines the whole theory.

The issue is that the collection of pointed spheres $\{S^n | n\ge 0\}$ is not a strongly generating set for all of $\mathrm{Ho}(\mathrm{Top}_*)$: a weak equivalence of pointed spaces is still required to induce isomorphisms of homotopy groups with *all* basepoints, whereas mapping out of pointed spheres detects only homotopy groups at the *specified* basepoint (and hence at any other basepoint in the same component).

Is the above version of the theorem true, without connectedness hypotheses? If so, where can I find a proof?

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## 1 comments

## @Karol Szumiło 2012-08-19 18:18:54

I was thinking more about this question and found another paper by Heller which offers an answer (unfortunately a negative one). The paper is

Freyd, Peter; Heller, Alex

Splitting homotopy idempotents. II.J. Pure Appl. Algebra 89 (1993), no. 1-2, 93–106.This paper introduces a notion of conjugacy idempotent. It is a triple $(G, g, b)$ consisting of a group $G$, an endomorphism $g \colon G \to G$ and an element $b \in G$ such that for all $x \in G$ we have $g^2(x) = b^{-1} g(x) b$. The theory of conjugacy idempotents can be axiomatized by equations, so there is an initial conjugacy idempotent $(F, f, a)$. The Main Theorem of the paper says (among other things) that $f$ does not split in the quotient of the category of groups by the conjugacy congruence.

Now $f$ induces an endomorphism $B f \colon B F \to B F$ which is an idempotent in $\mathrm{Ho} \mathrm{Top}$ and it follows (by the Main Lemma of the paper) that it doesn't split. It is then easily concluded that $(B f)_{+} \colon (B F)_{+} \to (B F)_{+}$ doesn't split in $\mathrm{Ho} \mathrm{Top}_{*}$.

The map $(B f)_{+}$ induces an idempotent of the representable functor $[-, (B F)_{+}]_{*}$ which does split since this is a $\mathrm{Set}$ valued functor. Let $H \colon \mathrm{Ho} \mathrm{Top}_{*}^\mathrm{op} \to \mathrm{Set}$ be the resulting retract of $[-, (B F)_{+}]_{*}$. It is half-exact (i.e. satisfies the hypotheses of Brown's Representability) as a retract of a half-exact functor. However, it is not representable since a representation would provide a splitting for $(B f)_{+}$.

The same argument with $B f$ in place of $(B f)_{+}$ shows the failure of Brown's Representability in the unbased case.

## @Tyler Lawson 2012-08-20 02:59:21

Excellent! Thanks for pointing out my mistake and the references. At the very least, I've failed to account for the difference between representing a functor on all CW-complexes and representing it on finite CW-complexes.

## @Karol Szumiło 2012-08-20 05:40:27

I guess many people, including myself, are guilty of making this mistake. At some point I automatically assumed that finite CW-complexes strongly generate $\mathrm{Ho} \mathrm{Top}$, but then I realized that I cannot prove it, which is what led me to the paper I mentioned in the comment. The most striking outcome of those arguments is that both $\mathrm{Ho} \mathrm{Top}$ and $\mathrm{Ho} \mathrm{Top}_*$ fail to have strongly generating sets at all. It just seems that the unstable homotopy category is much weirder than we tend to assume.

## @Mike Shulman 2012-08-20 18:11:07

Or, alternatively, that the stable homotopy category is nicer than we have a right to expect. (-: I see that Tyler posted a comment at the unpointed question mathoverflow.net/questions/11456/… , but I think Karol should post this as an answer there, so that it could be accepted.

## @Karol Szumiło 2012-08-21 13:13:52

I reposted my answer.