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6:50 AM
Here's something that feels like it should be true, but for which I have no idea how to tackle. Concretely, I have a K(G,1), where G is purely torsion but infinite. And I have a G-module whose underlying abelian group is finite (so torsion, but much more strongly so: I'd also be happy with a torsion group with finite exponent). Then it seems to me that I'm in the following situation:
I have a space all of whose homotopy groups are in some Serre class (i.e. torsion groups), and local coefficients with fibre in some Serre class, with some extra niceness (this is important!). Then are all the cohomology groups also in that Serre class? I guess the homology groups are, but then I need to use some extra juice to get the cohomology groups there.
Probably something about Ext groups, no?
I'm cautious, because of examples like this: math.stackexchange.com/questions/3254029/… which is where I'm headed, though I will fall into the gap between the standard hypotheses and the counterexample there.
 
 
1 hour later…
7:57 AM
Kind of hard to follow your train of thought, let me try to rephrase: You're asking whether the cohomology groups of K(G,1) with coefficients in M are torsion, where M is finite, and G is an infinite torsion group? Or are you asking something more general about Serre classes?

The first thing is trivially true, without any conditions on G. H^*(X;M) is always annihilated by the order of M.
 
 
2 hours later…
10:14 AM
I suspected the thing in the last comment, but I was wondering if there is a more general/formal version of this in case I need something more widely applicable
I'm definitely assuming more general local coefficients, and many sources I look at chicken out and just give constant coefficients, else I'd not be so worried that I'm missing something.
 
I don't think the general statement you asked for is correct. For example, take a space Y with homology given by Q in degree 2, and endow it with a C_2 action by -1 on H_2(Y). Take X = Y_{hC_2} (or (Y x EC_2)_{C_2}, if you prefer). Then X has the homology of a point, but H_*(X; Z[C_2]) = H_*(Y;Z), which is Q in degree 2. So this is a counterexample for the Serre class of finitely generated abelian groups.
(sorry, of course X doesn't have the homology of a point, but the homology of BC_2. But this is still finitely generated in each degree, so this is still a counterexample)
 
Weird.
Is it clear where it comes from?
The nontorsion I mean?
Oh, sorry, I was worrying first about local coefficient, but you went and gave a general counterexample for a different Serre class.
 
Well, homology with local coefficients encodes info about homology of all covers. But stuff like the existence of acyclic groups tells you that the homology of a space should be bad at controlling the homology of all its covers
 
That's why I was conscious of extra hypotheses on the coefficients (so, in my example, finite, not just a torsion group)
 
Yeah, if you're just thinking about torsion, then the first argument just works (torsion coefficients give torsion cohomology, without any conditions on the space, and no Serre class argument required)
 
10:45 AM
OK, cool, thanks!
I confess I have started doing spectral sequence calculations for the first time in my life, 11 years post-PhD, so there's a lot going on my head at the moment.
Though I suspect it's not quite as simple as you say, in general. The cohomology |N with the discrete topology, with coefficients in Q/Z is not torsion, since one can take a function in H^0 picking out a sequence of elements with unbounded order.
 
11:09 AM
oh, that's a good point, sorry! Yes, I guess torsion coefficients only imply torsion homology, for cohomology it generally fails... Even with a constant coefficient system. For example, if you have an infinite free abelian group as H_n, by UCT you'll get exactly what you said in H^n with Q/Z-coefficients.
 
Though I would imagine that torsion-with-bounded-order will be ok to follow your argument through, and in particular, in the finite case we are all good!
 

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