Let $\mathcal C$ be a category. Say that a class of objects $\mathcal S \subseteq \mathcal C$ is *weakly cogenerating* if the functors $Hom_{\mathcal C}(-,S)$ are jointly conservative, for $S \in \mathcal S$. That is, a map $X \to Y$ in $\mathcal C$ is an isomorphism if and only if it induces bijections $Hom_C(Y,S) \to Hom_C(X,S)$ for every $S \in \mathcal S$.

Of course, every category $C$ admits a weakly cogenerating class -- namely, take $\mathcal S = \mathcal C$. But it's frequently important to have a cogenerating *set* -- i.e. to require that $\mathcal S$ is small.

**Question:** Does the homotopy category (of spaces) admit a weak cogenerating set?

It's clear that the homotopy category of simply-connected spaces admits a weak cogenerating set -- we can take $\mathcal S = \{K(\mathbb Z, n) \mid n \geq 2\}$ or alternatively $\mathcal S = \{K(k,n) \mid n \geq 2, k \in \{\mathbb Q, \mathbb F_p\}\}$ in this case by the cohomology Whitehead theorem. But I'm pessimistic about the chances of doing something similar with arbitrary spaces.

Relatedly, I wonder whether the category of groups admits a weak cogenerating set.

I also wonder whether the class of

*truncated*spaces -- those spaces $S$ for which $\pi_k(S) = 0$ for $k$ sufficiently large -- is a cogenerating class for the homotopy category. What about the class of Eilenberg-MacLane spaces?

stronglycogenerating. $\endgroup$ – Alexander Campbell Dec 16 '19 at 0:29don'task that $Hom_C(-,S)$ be jointlyfaithful, so a weakly cogenerating class in the above sense is not even cogenerating! $\endgroup$ – Tim Campion Dec 16 '19 at 0:30In fact the above definition of a strong generator is far from ideal unless $\mathcal{B}_0$ admits finite limits; but we have chosen it for simplicity since this is usually the case in applications." Of course, this is not the case in your application. $\endgroup$ – Alexander Campbell Dec 16 '19 at 0:34