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?

`$X_n \to X_{n+1}$`

so that for all such CW-inclusions, any element of`$$ [K,X_n] \times_{F(K)} F(L) $$`

has a lift to`$[L,X_{n+1}]$`

. $\endgroup$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. $\endgroup$2more comments