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

  • 3
    $\begingroup$ 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: mathoverflow.net/questions/104866/… $\endgroup$ Aug 20, 2012 at 14:58
  • $\begingroup$ It's also worth emphasizing that the statement above of "the classical Brown Representability Theorem" is not correct. The classical statement restricts to connected spaces, and the example of Heller's that Szumilo describes below shows that the statement is false without this assumption. $\endgroup$ Nov 15, 2017 at 14:56

2 Answers 2


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:

  • Peter Freyd and Alex Heller, Splitting homotopy idempotents. II. J. Pure Appl. Algebra 89 no. 1-2 (1993) pp 93–106, doi:10.1016/0022-4049(93)90088-B.

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.


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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