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Why is $H_{i}(M,\mathbb{Z}_{2}) = H^{i}(M,\mathbb{Z}_{2})$ for a closed manifold $M$? (Hatcher states this on p. 249 in his proof of Corollary 3.37.) Thanks.

The Phenotype
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Ashley
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  • Because $\mathbb{Z}_2$ is a field. You can see the implication via universal coefficients or Poincare duality, do you know either one? – Kevin Carlson Apr 14 '14 at 18:39
  • @KevinCarlson I'm perhaps being a bit silly here, but how do you get the isomorphism $H^{n-i}(M, \Bbb Z_2) \cong H^i(M, \Bbb Z_2)$ from Poincare duality? –  Apr 14 '14 at 18:45
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    @Mike, you're right, I don't actually have a way to do that without going through the result we're using here. OP, $H_i(M,\mathbb{Z}_2)$ is a finite-dimensional vector space. $H^i(M,\mathbb{Z}_2)$ is its dual, and finite-dimensional vector spaces over any field are isomorphic to their dual. – Kevin Carlson Apr 14 '14 at 18:56
  • @KevinCarlson I think you should upgrade that to an answer :) –  Apr 14 '14 at 18:58
  • Ah okay, thank you very much. – Ashley Apr 14 '14 at 19:01
  • @user33206 The key here is that $H_i(M, \Bbb Z_2)$ is finite dimensional. Hatcher proves this in full generality in corollaries A.8 and A.9 in the appendix. –  Apr 14 '14 at 19:02

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$H_i(M,\mathbb{Z}_2)$ is a finite-dimensional vector space. $H^i(M,\mathbb{Z}_2)$ is its dual, and finite-dimensional vector spaces over any field are isomorphic to their duals.

Kevin Carlson
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