Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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:

  1. Proofs of an analogous statement where $\mathrm{Ho}(\mathrm{Top}_*)$ is replaced by its subcategory of pointed connected spaces.
  2. 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".)
  3. 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?

share|improve this question
    
I think it's true without connectedness hypotheses. I made a hash of trying to write up the unbased version here: mathoverflow.net/questions/11456/… –  Tyler Lawson Aug 16 '12 at 19:48
1  
(I also once made a hash of trying to assign it as a homework problem! math.umn.edu/~tlawson/old/18.906/ps5old.pdf) –  Tyler Lawson Aug 16 '12 at 19:52
    
Is the version that's currently at mathoverflow.net/questions/11456/… still a hash, or do you believe it's correct now? Because I don't really follow it yet.... –  Mike Shulman Aug 16 '12 at 20:56
    
I'm not really happy with it, the logical layout is bad, and it works harder than necessary because it wants to construct a CW-object instead of a cell one. The core is just some kind of small object argument: cut down the number of based CW-inclusions $K \to L$ to a set, and construct a sequence of cell inclusions $X_n \to X_{n+1}$ so that for all such CW-inclusions, any element of $$ [K,X_n] \times_{F(K)} F(L) $$ has a lift to $[L,X_{n+1}]$. –  Tyler Lawson Aug 16 '12 at 21:29
5  
I don't know an answer to this question, but I want to point out that contrary to Tyler Lawson's answer linked above Brown's Representability does not hold in the unbased case. A counterexample is discussed in Proposition 2.1 of Heller, Alex On the representability of homotopy functors. J. London Math. Soc. (2) 23 (1981), no. 3, 551–562. Perhaps this example can be also adapted to disprove the based not necessarily connected case, but I'm not so sure about it. –  Karol Szumiło Aug 18 '12 at 8:39

1 Answer 1

up vote 17 down vote accepted

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.

share|improve this answer
2  
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. –  Tyler Lawson Aug 20 '12 at 2:59
3  
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. –  Karol Szumiło Aug 20 '12 at 5:40
    
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. –  Mike Shulman Aug 20 '12 at 18:11
    
I reposted my answer. –  Karol Szumiło Aug 21 '12 at 13:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.