9
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ It amuses me that, in my native language, such a class of objects would be called strongly cogenerating. $\endgroup$ – Alexander Campbell Dec 16 '19 at 0:29
  • $\begingroup$ @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! $\endgroup$ – Tim Campion Dec 16 '19 at 0:30
  • $\begingroup$ 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. $\endgroup$ – Alexander Campbell Dec 16 '19 at 0:34
4
$\begingroup$

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.

It doesn't seem to be straightforward to deduce the corresponding result for the homotopy category.

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $Hom(\ast,X)\to Hom(F,X)$. $\endgroup$ – Tom Goodwillie Dec 16 '19 at 2:43
  • 1
    $\begingroup$ @TomGoodwillie Yes, I think that works. I don't remember the construction of large acyclic groups but I think it is in Kan-Thurston. $\endgroup$ – Neil Strickland Dec 16 '19 at 8:09
  • $\begingroup$ @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. $\endgroup$ – Tim Campion Dec 16 '19 at 14:45
  • $\begingroup$ 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. $\endgroup$ – Tim Campion Dec 16 '19 at 14:46
  • 1
    $\begingroup$ 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 $\pi_1=G$. $\endgroup$ – Tom Goodwillie Dec 16 '19 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.