# Do spaces admit a weak cogenerating set?

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?

• It amuses me that, in my native language, such a class of objects would be called strongly cogenerating. – Alexander Campbell Dec 16 '19 at 0:29
• @AlexanderCampbell Ah, but I don't ask that $Hom_C(-,S)$ be jointly faithful, so a weakly cogenerating class in the above sense is not even cogenerating! – Tim Campion Dec 16 '19 at 0:30
• Good point! Indeed, after Kelly, in his book (where my usage comes from), defines a strong generator, he says "In 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. – Alexander Campbell Dec 16 '19 at 0:34

For any infinite set $$X$$ let $$S_X$$ be the group of bijections $$\sigma \colon X\to X$$ such that $$\{x : \sigma(x)\neq x\}$$ is finite. This still has signature homomorphism, and the alternating subgroup $$A_X$$ is simple, and has the same cardinality as $$X$$. Now let $$\mathcal{G}$$ be a set of groups, and put $$\kappa = \max \{|G|:G\in\mathcal{G}\}$$. Then $$\text{Hom}(A_X,G)$$ will be a singleton for all $$G\in\mathcal{G}$$ and $$X$$ with $$|X|>\kappa$$ (because the kernel of any homomorphism is nontrivial by cardinality, and so is the whole of $$A_X$$ by simplicity). So $$\mathcal{G}$$ is not a weak cogenerating set.
EDIT To summarize the discussion in the comments, we can indeed deduce the corresponding result for the homotopy category with a little more work. Choose an acyclic simple group $$G$$ bigger than the fundamental group of any space in $$\mathcal S$$. Then any map $$f: BG \to S$$ for $$S \in \mathcal S$$ is trivial on $$\pi_1$$ by simplicity, so it lifts to the universal cover $$\tau_{\geq 2} S$$. By acyclicity, the composite map $$BG \to \tau_{\geq 2} S \to K(\pi_2(S),2)$$ is trivial so $$f$$ lifts through the 2-connected cover $$\tau_{\geq 3} S$$. Continue in this manner, lifting through the Whitehead tower to see that $$f$$ is nullhomotopic. Thus $$\mathcal S$$ does not distinguish $$BG$$ from a point, and is not weakly cogenerating.
• Can we adapt this to the homotopy category like this? Given a set $S$ of spaces, make a group $G$ bigger than all fundamental groups of all elements of $S$, such that $G$ is the fundamental group of an acyclic space. Now if $X\in S$ then every map $F\to X$ is homotopic to a constant because we can first lift to a universal cover using cardinality and then lift all the way up the Postnikov tower using acyclicity. So the map $F\to \ast$ induces a bijection $Hom(\ast,X)\to Hom(F,X)$. – Tom Goodwillie Dec 16 '19 at 2:43
• @TomGoodwillie Thanks, I'd accept this as an answer! To clarify, you claim there exist simple groups $G$ of arbitrarily large cardinality such that $G = \pi_1(F)$ for an acyclic space $F$. Choosing such $F$ sufficiently large, and given $f: F \to X$ for $X \in \mathcal S$, we inductively find lifts $f_k: F \to \tau_{\geq k+1} X$ through the Whitehead tower of $X$ since the composite map $F \xrightarrow{f_k} \tau_{\geq k} X \to K(\pi_k X, k)$ is null -- by simplicity when $k=1$ and by acyclicity from there on. So $f$ factors through the limit of the Whitehead tower, which is contractible. – Tim Campion Dec 16 '19 at 14:45
• Concerning the existence of such $G,F$, I think it's easy -- let $G$ be a sufficiently large simple group (e.g. the alternating group as in Neil's answer). Note that since $G$ is simple, its abelianization is trivial, i.e. $H_1(BG) = 0$. Start with $BG$ and then just keep gluing in cells of dimension 3 and higher to kill all the homology to obtain $F$. This doesn't affect the fundamental group. – Tim Campion Dec 16 '19 at 14:46
• You can't kill H_2 by attaching 3-cells if pi_2 is trivial. But I'm sure it's easy. If G is simple (and nonabelian) and H_2G is trivial then the homotopy fiber of $BG\to BG^+$ is acyclic and has $\pi_1=G$. – Tom Goodwillie Dec 16 '19 at 18:48