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 surjective
^{1}.

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