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Let $M$ be a compact connected $3$-manifold with boundary $\partial M$. If $M$ is nonorientable and $\partial M$ is empty, then how do I see that $H_1(M, \mathbb{Z})$ is infinite?

Stefan Hamcke
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  • You need $\partial M$ non empty, otherwise it is false. The proof is explain in every textbook on 3-manifolds, Hempel, Jaco for instance. just write the exact sequence relating of homology of the manifold, homology relative to the boundary and homology of the boundary, and look at the term conatining $H_1$ – Thomas Nov 25 '15 at 08:05
  • @Thomas can you provide a counter example? – Anubhav Mukherjee Nov 25 '15 at 08:09
  • These object are studied as "homology spheres"https://en.wikipedia.org/wiki/Homology_sphere#Constructions_and_examples – Thomas Nov 25 '15 at 08:14
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    @Thomas: Homology spheres are orientable. – Eric Wofsey Nov 25 '15 at 08:25
  • @EricWofsey, sorry I did not read the question correctly... – Thomas Nov 25 '15 at 09:18
  • Counter example for non-orientable with boundary is $\mathbb RP^2\times I$. Its fundamental group is $\mathbb Z_2$. – mmm Jul 13 '18 at 10:22

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Every odd-dimensional manifold has vanishing Euler characteristic, so $$0 = \chi(M) = b_0 - b_1 + b_2 - b_3.$$We have $b_0 = 1$, and $b_3 = 0$ since $M$ is nonorientable. Hence, $b_1 > 0$ and thus $H_1(M, \mathbb{Z})$ is infinite.

Dario
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user149792
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