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Let $C$ be a Serre class which satisfies the additional axioms about $\otimes, \mathrm{Tor}, K(A,1)$'s.

It is then easy to check that if $F\to X\to B$ is a Serre fibration with $\pi_1(B)$ acting trivially on the homology of $F$, and the (positive) homologies of $B,F$ are in $C$, then the same is true of $X$.

(For that we don't even need $K(A,1)$'s)

Indeed it follows easily from the Serre spectral sequence and the universal coefficient theorem applied on $E^2$.

I'm wondering if this holds when $\pi_1(B)$ acts nontrivially.

Is it true that under the hypotheses (minus the hypothesis on the action) $H_p(B, H_q(F)) \in C$ ? (Here it's therefore homology with local coefficients) If not, does $H_p(X) \in C$ still hold for some other reason ?

(I'm only interested in homology in strictly positive degrees)

EDIT : more specifically, here are all the axioms for $C$:

1) if $0\to M\to N\to L\to 0$ is a short exact sequence, $N\in C \iff M,L \in C$

2) If $A,B\in C, A\otimes B, \mathrm{Tor}(A,B) \in C$

3) If $A\in C$, for all $k>0, H_k(K(A,1))\in C$

Note that if there is a counterexample where $C$ doesn't satisfy 3), I'm also interested

2nd Edit : William answered the question I asked, but I'm wondering if anyone has an example with a connected fiber, so I'll leave this as unaccepted for a couple of days to see if one can find an example with connected fiber.

Maxime Ramzi
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1 Answers1

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What about the bundle $\mathbb{Z}/2 \to S^n \to \mathbb{R}P^n$ where $n>0$ is even? Then for all $k>0$ the groups $H_k(\mathbb{Z}/2)$ and $H_k(\mathbb{R}P^n)$ are all in the Serre class of finite groups, but $H_n(S^n)\cong \mathbb{Z}$.

William
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  • Well now I'm a bit annoyed that I didn't think of that. I think I was fixating on having a connected fiber.. Before I accept, would you know an example where all 3 spaces are path connected ? – Maxime Ramzi Jun 08 '19 at 09:59
  • @Max I can't come up with something off the top of my head, but I will keep thinking about it. – William Jun 08 '19 at 17:13
  • Do you then mind if I wait a bit before accepting your answer ? (Like a couple of days, to see if anyone can come up with something connected) – Maxime Ramzi Jun 08 '19 at 18:59
  • Sure, it's up to you. I'm wondering if it's possible to come up with an example where you just replace the $\mathbb{Z}/2$ fibres in this principal bundle with a connected space $F$ whose positive homology groups are finite and which has an appropriate $\mathbb{Z}/2$ action. Specifically I'm wondering about things like $F = \mathbb{R}P^2 \times \mathbb{R}P^2$ where the action switches the fibres, or something like $F = K(A,1)$ where $A$ is a finite abelian group and the $\mathbb{Z}/2$ action is induced by the inversion homomorphism. I haven't managed to work anything out yet. – William Jun 09 '19 at 01:48