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?

  • 1
    $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.