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I heard that every topological $n$-manifold $M$ is $\mathbb{F}_2$-orientable, but then for $M=\mathbb{R}^2$ is must be $H_2(\mathbb{R}^2;\mathbb{F}_2)\neq 0$?

In lecture we had the lemma: Let $M$ is a connected n-manifold, then: $M$ is $R$-orientable iff $H_n(M;R)\cong R$.

Something must be wrong, I always thought that $H_2(\mathbb{R}^2;\mathbb{F}_2)=0$ because $M$ is contractible. Or is the lemma only true for compact manifolds?

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    $\begingroup$ There is a statement for non-compact manifolds, but you will need to use something like compactly supported cohomology. I believe this is also in Hatcher. $\endgroup$
    – Thomas Rot
    Mar 25 '16 at 11:37
  • $\begingroup$ good to know, thanks $\endgroup$
    – user325096
    Mar 25 '16 at 18:21
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Yes, this is true only for compact manifolds (I assume that your "manifolds" are not allowed to have boundary). In fact, $H_n(M;R)\cong R$ iff $M$ is $R$-orientable and compact. You should be able to find a proof in any text that covers Poincaré duality. For instance, this follows from Theorem 3.26 and Proposition 3.29 of Hatcher's Algebraic Topology (note the hypothesis that $M$ is closed in Theorem 3.26).

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