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Let $X$ be a topological manifold of dimension $n$ (assuming perhaps that there is a countable basis of open sets). Do NOT assume that $X$ is compact, or oriented, or triangulable (so do not assume it to be smooth).

  1. Can we still conclude that the homology groups $H_i(X, \mathbb{Z})$ are finitely-generated?
  2. Can we still conclude that $H_i(X, \mathbb{Z})=0$ for $i > n$ ?

The second point is related to the first, as one can show that $H_i(X, \mathbb{Q})=0$ and $H_i(X, \mathbb{F}_p) = 0$ for all primes $p$, when $i>n$. If $H_i(X, \mathbb{Z})$ were known to be finitely-generated, we would conclude by the universal coefficients theorem.

[To see the claim: if $X$ is $k$-oriented, it is part of Poincaré duality that $H_i(X, k)=0$ for $i>n$, and $X$ is always $\mathbb{F_2}$-oriented. Next, consider $Y \to X$ the canonical 2-sheeted cover with $Y$ oriented, and apply the homology Serre spectral sequence to the fibration $Y \to X \to B\mathbb{Z}/2$; this gives $H_i(X, k)=0$ when $i>n$ for each ring $k$ in which 2 is invertible.]

thanks! Pierre

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    $\begingroup$ A countably infinite discrete set of points violates condition $1$. $\endgroup$
    – user2520938
    Commented May 13, 2020 at 13:19
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    $\begingroup$ Point 1 is false, see @user2520938's comment (for a connected example, take an infinite connected sum of genus 1 surfaces). Point 2 is Proposition 3.29 in Hatcher if $X$ is noncompact, and Theorem 3.26 if $X$ is compact. Also you have to be a bit careful with your spectral sequence argument since the base of your fibration is not connected. $\endgroup$
    – Najib Idrissi
    Commented May 13, 2020 at 13:47
  • $\begingroup$ I'm pretty sure that answer on question 2 depends a lot on the choice of homology theory. And it seems to me that "tubular neigborhood" of 2-dimensional Hawaiian earring would have nontrivial homology in degrees higher that 2 (Barrat&Milnor, An example of anomalous singular theory). $\endgroup$
    – Denis T
    Commented May 13, 2020 at 14:52
  • $\begingroup$ @Najib: thanks for the reference! as for the spectral sequence, $BC_2$ is certainly connected, but you probably worry about simple-connectedness; however, page 2 of the sequence is just $H_*(C_2, H_*(Y))$, the homology of the group $C_2$ with nontrivial coefficients, and this vanishes in positive degrees if multiplication by 2 is an isomorphism on $H_*(Y)$. $\endgroup$
    – Pierre
    Commented May 13, 2020 at 15:29
  • $\begingroup$ @Denis: the example in Barratt-Milnor is not a manifold; are you saying that one could produce a manifold example by taking a tubular neighbourhood? This would violate both Hatcher's result and my spectral sequence argument (since they use rational coefficients in this paper anyway). $\endgroup$
    – Pierre
    Commented May 13, 2020 at 15:34

1 Answer 1

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It's already been noted in the comments that 1) is false: take for example an infinite genus surface.

2) is true. For an oriented manifold you have $H_i(M)\cong H^{n-i}_c(M)$, the cohomology with compact supports (see Hatcher Theorem 3.35). For $i>n$, the right side is $0$ trivially. If $M$ is not oriented, you have essentially the same theorem but with twisted coefficients in the cohomology. See page 207 of the Springer edition of Bredon's Sheaf Theory.

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  • $\begingroup$ I actually prefer the reference to 3.29 from Hatcher's book, pointed out by Najib in the comments. $\endgroup$
    – Pierre
    Commented May 15, 2020 at 21:02

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