Let $M$ be a connected, non-compact, non-orientable topological manifold of dimension $n$.
Question: Is the top singular cohomology group $H^n(M,\mathbb Z)$ zero?
This naïve question does not seem to be answered in the standard algebraic topology treatises, like those by Bredon, Dold, Hatcher, Massey, Spanier, tom Dieck, Switzer,...
Some remarks.
a) Since $H_n(M,\mathbb Z)=0$ (Bredon, 7.12 corollary) we deduce by the universal coefficient theorem: $$ H^n(M,\mathbb Z) =\operatorname {Ext}(H_{n-1}(M,\mathbb Z), \mathbb Z)\oplus \operatorname {Hom} (H_n(M,\mathbb Z),\mathbb Z)=\operatorname {Ext}(H_{n-1}(M,\mathbb Z),\mathbb Z )$$
But since $H_{n-1}(M,\mathbb Z)$ need not be finitely generated I see no reason why $\operatorname {Ext}(H_{n-1}(M,\mathbb Z),\mathbb Z)$ should be zero.
b) Of course the weaker statement $H^n(M,\mathbb R) =0$ is true by the universal coefficient theorem, or by De Rham theory if $M$ admits of a differentiable structure.
c) This question was asked on this site more than 8 years ago but the accepted answer is unsubstanciated since it misquotes Bredon.
Indeed, Bredon states in (7.14, page 347) that $H^n(M,\mathbb Z)\neq0$ for $M$ compact, orientable or not, but says nothing in the non-compact case, contrary to what the answerer claims.
2 Answers
I believe you can deduce this from the corresponding statement in the orientable case. Let $\tilde M$ be the oriented double cover. Make an exact sequence of cochain complexes $$ 0 \to C^\bullet(M;\mathbb Z^t)\to C^\bullet(\tilde M;\mathbb Z)\xrightarrow{p_!} C^\bullet(M;\mathbb Z)\to 0, $$ where $\mathbb Z^t$ is the local system on $M$ corresponding to the non-orientability of $M$. (The map $p_!$ is dual to the transfer map taking each singular simplex of $M$ to the sum of its two lifts to $\tilde M$.) Then there is an exact sequence $$ H^n(\tilde M;\mathbb Z)\to H^n(M;\mathbb Z)\to H^{n+1}(M;\mathbb Z^t) $$ with the first and third groups trivial.
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$\begingroup$ Thank you for this answer, Tom. What exactly are the elements of the group $C^k (M;\mathbb Z^t)$? $\endgroup$ Commented May 27, 2021 at 18:14
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$\begingroup$ The local system that I am calling $\mathbb Z^t$ provides a group $\mathbb Z^t(x)$ at each point $x\in M$, namely the infinite cyclic group whose two generators are the two orientations of $M$ at $x$. An element of $C^k(M;\mathbb Z^t)$ is a function that assigns to each singular $k$-simplex $\sigma$ of $M$ an element of the group associated with a point of $\sigma$ (say, the number $0$ vertex). $\endgroup$ Commented May 28, 2021 at 0:19
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$\begingroup$ It should be possible to find a short exact sequence of sheaves on $M$ which recovers the exact sequence above. The first should be the orientation sheaf and the third should be constant $\mathbb Z$. What is their extension? I am a bit confused. $\endgroup$– Z. MCommented May 28, 2021 at 6:36
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$\begingroup$ Dear Tom, thank you very much again for quickly (and very satisfyingly!) answering the question in my comment. If I may be so bold as to abuse your patience once more: I "knew" that orientable non-compact manifolds satisfy $H^n(M,\mathbb Z)=0$, a fact indeed applied in your answer to $\tilde M$. Do you have an easy explanation or a reference for that statement, since I must confess that I don't remember why it is true! $\endgroup$ Commented May 28, 2021 at 7:23
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$\begingroup$ Sorry for being stupid. This sequence is essentially $0\to\mathbb Z\to\mathbb Z[C_2]\to\mathbb Z\to0$ of $C_2$-groups, which is indeed non-split. $\endgroup$– Z. MCommented May 28, 2021 at 15:57
I think it's worth pointing out that at least for smooth or PL $n$-manifolds $M$ that are connected but not compact, something much stronger holds than the $n$th homology vanishing -- the manifold is actually homotopy equivalent to an $(n-1)$-dimensional simplicial complex! This is a theorem of Whitehead from
J. H. C. Whitehead, The immersion of an open 3-manifold in euclidean 3-space, Proc. London Math. Soc. (3) 11 (1961), 81–90.
I gave a modern treatment of it in my note here. In that note, I say that the manifold is smooth, but really all the proof uses is PL (I should fix this sometime).
I don't know if this holds if the manifold is not smooth or PL, but I suspect it does.
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$\begingroup$ Conversely, it's also true that vanishing of $H^{\ge n}(X)$ for all local coefficient systems implies that $X$ is homotopy equivalent to a cell complex of dimension $<n$ (except maybe for very small $n$). $\endgroup$ Commented May 27, 2021 at 3:07
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$\begingroup$ Also, I believe that $H^p(M;G)$ is isomorphic to $H_{n-p}(M;G\otimes \mathbb Z^t)^{lf}$(locally finite homology twisted by orientation), and that $H_0(M)^{lf}$ with any twisted or untwisted coefficients is pretty clearly zero for a connected non-compact manifold $M$. $\endgroup$ Commented May 27, 2021 at 16:34
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1$\begingroup$ @TomGoodwillie: I believe that you're right on both counts, which gives an alternate approach to Whitehead's theorem except maybe in very low dimensions. If I remember correctly, the result you're referring to in your first comment is due to Stallings, right? I remember seeing the paper long ago, but never got around to reading it. In any case, one nice thing about Whitehead's proof is that it gives a very explicit and visually appealing deformation retract onto an (n-1)-dimensional spine. $\endgroup$ Commented May 27, 2021 at 17:19
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$\begingroup$ Thank you for the amazing information contained in Whitehead's theorem, and for your note, Andy. $\endgroup$ Commented May 27, 2021 at 18:21
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1$\begingroup$ Andy, you're probably right that it was Stallings. I learned this fact from John Klein. Sketch proof: Suppose $H^{\>d}(X;M)=0$ for every $\mathbb Z[\pi_1(X)]$-module $M$. Make a $(d-1)$ dimensional complex $K$ with $(d-1)$-connected map $f:K\to X$. The relative homology $H_\ast(f;\mathbb Z[\pi_1(X)])$ vanishes below degree $d$. Because of the cohomology hypothesis, $H_d(f;\mathbb Z[\pi_1(X)])$ is the only nontrivial group and is a projective module. Attach enough $d$-cells to $K$ (trivially) to make that module free. A basis for it tells you how to attach $d$-cells to finish the job. $\endgroup$ Commented May 28, 2021 at 0:30