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The classical Brown Representability Theorem states: Denote $hCW_*$ the homotopy category of pointed CW-complexes. Let $F : hCW_* \to Set_*$ be a contravariant functor. Then $F$ is representable if and only if

  • $F$ respects coproducts, i.e. $F(\vee_{i \in I} X_i) = \prod_{i \in I} F(X_i)$ for all families $X_i$ of pointed CW-complexes.
  • $F$ satisfies a sort of mayer-vietoris-axiom: If $X$ is a pointed CW-complex which is the union of two pointed subcomplexes $A,B$, then the canonical map $F(X) \to F(A) \times_{F(A \cap B)} F(B)$ is surjective1.

What about omitting the base points? So let $F :hCW \to Set$ be a contravariant functor that satisfies the analogous properties as above (replace the wedge-sum by the disjoint union). Is then $F$ representable? I'm not sure if we just can copy the proof of the pointed case (which can be found, e.g., in Switzer's book "Algebraic Topology - Homology and Homotopy", Representability Theorems). For example, $F(pt)$ can be anything (in contrast to the pointed case), it will be the set of path components in the classifying space. Besides, the proof uses homotopy groups and in particular the famous theorem of Whitehead, which deal with pointed CW-complexes. Nevertheless, I hope that $F$ is representable ... what do you think?

As a first step, we may define for every $i \in F(pt)$ the subfunctor $F_i$ of $F$ by $F_i(Y) = \{f \in F(Y) : \forall y : pt \to Y : f|_{y} = i \in F(pt)\}$, which should be thought as the connected component associated to $i$. Then it's not hard to show that $F_i$ satisfies the same properties as $F$ and that $F_i = [-,X_i]$ implies $F = [-,\coprod_i X_i]$. In other words, we may assume that $F(pt)=pt$ (so that the classifying space will be connected).

1 You can't expect it to be bijective, cf. question about categorical homotopy colimits

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The answer to this question turns out to be negative. The reader who comes across this question should see the following question and the answers by Karol Szumiło:… – Tyler Lawson Aug 20 '12 at 14:58
up vote 9 down vote accepted

This is a copy of my answer to Brown representability for non-connected spaces which I repost here per request in the comment.

A negative answer to the question can be concluded from this paper:

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.

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Yes, Brown representability holds for such functors. There are not really any material differences between this and the proof of Brown representability in the pointed case.

EDIT: My previous version of this was not rigorous enough. I was trying to be clever and get away with just simple cell attachments, which only work if you already know that the functor is represented by a space. Sorry for the delay in reworking, but this particular proof has enough details that it takes time to write up.

As you say, you begin by decomposing such functors so without loss of generality $F(pt)$ is a single point.

Start with $X_{-1}$ as a point. Assume you've inductively constructed an $(n-1)$-dimensional complex $X_{n-1}$ with an element $x_{n-1} \in F(X_{n-1})$ so that, for all CW-inclusions $Z \to Y$ of finite CW complexes with $Y$ formed by attaching a $k$-cell for $k < n$, the map $$ [Y,X_{n-1}] \to [Z,X_{n-1}] \times_{F(Z)} F(Y) $$is surjective.

Now, define a "problem" of dimension $n$ to be a CW-inclusion $Z \to Y$ where $Y$ is a subspace of $\mathbb{R}^\infty$ formed by attaching a single $n$-cell to $Z$, together with an element of $$ Map(Z,X_{n-1}) \times_{F(Z)} F(Y). $$ The fact that $Y$ has a fixed embedding in $\mathbb{R}^\infty$ means that there is a set of problems $S$, whose elements are tuples $(Z_s,Y_s,f_s,y_s)$ with $f_s$ a map $Z_s \to X_{n-1}$ and $y_s$ is a compatible element of $F(Y)$.

Let $X_n$ be the pushout of the diagram $$ X_{n-1} \leftarrow \coprod_{s \in S} Z_s \rightarrow \coprod_{s \in S} Y_s $$ where the lefthand maps are defined by the maps $f_s$ and the righthand maps are the given $CW$-inclusions. This is a relative $CW$-inclusion formed by attaching a collection of $n$-cells; therefore, $X_n$ still has the extension property for relative cell inclusions of dimension less than $n$.

The space $X_n$ is homotopy equivalent to the homotopy pushout of the given diagram, which is formed by gluing together mapping cylinders. Specifically, $X_n$ is weakly equivalent to the space $$ X_{n-1} \times \{0\} \cup (\coprod_S Z_s \times [0,1]) \cup (\coprod_S Y_s \times \{1\}) $$ which decomposes into two CW-subcomplexes: $$ A = X_{n-1} \times \{0\} \cup (\coprod Z \times [0,1/2]) $$ which deformation retracts to $X_{n-1}$, and $$ B = (\coprod Z_s \times [0,1/2]) \cup (\coprod Y_s \times \{1\}) $$ which deformation retracts to $\coprod Y_s$ with intersection $A \cap B \cong \coprod Z_s$. The Mayer-Vietoris property and the coproduct axiom then imply that there is an element $x_n \in F(X_n)$ whose restriction to $A$ is $x_{n-1}$ and whose restriction to $B$ is $\prod y_s$.

Taking colimits, you have a CW-complex $X$ with an element $x \in F(X)$ (constructed using a mapping telescope + Mayer-Vietoris argument) so that, for all CW-inclusions $Z \to Y$ obtained by attaching a single cell, the map $$ [Y,X] \to [Z,X] \times_{F(Z)} F(Y) $$ is surjective.

Now you need to show that for any finite CW-complex $K$, $[K,X] \to F(X)$ is a bijection.

First, surjectivity is straightforward by induction on the skeleta of $K$. More specifically, for any $K$ with subcomplex $L$, element of $F(K)$, and map $L \to X$ realizing the restriction to $F(L)$, you induct on the cells of $K\setminus L$. Then, injectivity: if you have two elements $K \to X$ with the same images in $F(K)$, you use the above-proven stronger surjectivity property to the inclusion $K \times \{0,1\} \to K \times [0,1]$ to show that there is a homotopy between said maps.

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$Y_n$ is the disjoint union of $X_{n-1}$ and spheres, right? – Martin Brandenburg Jan 12 '10 at 15:28
No, it's the wedge because of the simplifying assumption F(pt) = pt. If you take a disjoint union then $Y_n$ no longer agrees with $X_{n-1}$ on low spheres. – Tyler Lawson Jan 12 '10 at 15:41
hmm, how do you construct elements in $F(Y_n)$ when there is no wedge-axiom? I think the disjoint union works here since $[S^k,-]$ commutes with disjoint unions (since $S^k$ is path-connected) and $[S^k,S^n]$ has exactly one element for $k<n$. – Martin Brandenburg Jan 12 '10 at 15:50
You cut each sphere you're attaching into two discs, one of which is joined to $X_{n-1}$ at the center. Then $Y_n$ is a union of a coproduct of discs, along a coproduct of equators, with something that retracts down to $X_{n-1}$. So far as the disjoint union, the set $[S^k, X_{n-1} \coprod S^n]$ is a disjoint union of a point and $[S^k, X_{n-1}]$ for $k < n$ and in particular does not agree with it. – Tyler Lawson Jan 12 '10 at 16:01
I don't see why this should work because $F(S^n) \to F(D^n_+) \times F(D^n_-)$ is not injective; then there is no reason why the naive preimages of elements of $F(S^n)$ are preimages. – Martin Brandenburg Jan 13 '10 at 15:59

An answer to this is started in Boardman's Stable Operations in Generalized Cohomology (Handbook of Algebraic Topology, also available via Steve Wilson's homepage (Wilson and Boardman - together with Johnson - cowrote the follow-up paper in the same volume). Theorem 3.6 states:

Let $h(-)$ be an ungraded cohomology theory as above [same conditions as in the question except that it lands in abelian groups]. Then:
(a) $h(-)$ is represented in $\operatorname{Ho}$ [homotopy category of spaces homotopy equivalent to CW-complexes] by an $H$-space $H$, with a universal class $\iota \in h(H,o) \subset h(H)$ that induces an isomorphism $\operatorname{Ho}(X,H) \cong h(X)$ of abelian groups by $f \mapsto h(f)i$ for all $X$;
(b) For any cohomology theory $k(-)$, operations $\theta : h(-) \to k(-)$ correspond to elements $\theta \iota \in k(H)$.

The proof given depends on references, which is why I say that the answer is "started" in this paper. The argument goes:

  1. Brown representability gives a based connected space representing $h(-,o)$ on based connected spaces.
  2. West shows that $h(-,o)$ is represented on all based spaces (i.e. drops the connected assumption).
  3. Then the "disjoint basepoint" trick represents $h(-)$ on all spaces.

Now that I look at it, the essence of the "disjoint basepoint" trick probably does need the additional assumption of the functor landing in abelian groups, since it uses the split short exact sequence

$$ 0 \to h(X,o) \to h(X) \to h(o) \to 0 $$

to relate relative and absolute cohomology. Thus $h(X) \cong h(X^+,o)$.

So, for abelian groups then you are fine. Is it necessary that you land in Set?

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