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Let $M$ be an $n$-dimensional closed manifold. Choose $x \in M$. Using long exact sequence of pairs $(M,M - x)$, we have $$H_k(M - x, \mathbb{Z}) \cong H_k(M, \mathbb{Z})$$ for $k<n-1$. For $k=n-1$, I see this is an isomorphism with $\mathbb{Q}$-coefficient. I am curious if there is an example that they are not isomorphic with integer coefficient.

[edit] I was assuming orientability. In the long exact sequence with $\mathbb{Z}$-coefficient $$H_n(M) \rightarrow H_n(M, M-x) \rightarrow H_{n-1}(M-x) \rightarrow H_{n-1}(M) \rightarrow 0,$$ the first map is an isomorphism so the third is an isomorphism. Sorry for stupid question.

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    $\begingroup$ Try $\mathbb{RP}^2$. $\endgroup$ Commented May 10, 2014 at 12:39

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This is also an isomorphism with $\mathbb{Z}$-coefficients. One way to see it is to use the "open-closed" exact sequence $$H^0(M,\mathbb{Z})\rightarrow H^0(\{x\} ,\mathbb{Z})\rightarrow H^1_c(M-x,\mathbb{Z})\rightarrow H^1(M,\mathbb{Z})\rightarrow 0$$ and Poincaré duality which identifies the last (nontrivial) arrow with $H_{n-1}(M-x,\mathbb{Z})\rightarrow H_{n-1}(M,\mathbb{Z})$.

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    $\begingroup$ This answer assumes the manifold is orientable. In general the map in questions is not an isomorphism as the example of the projective plane shows. $\endgroup$ Commented May 10, 2014 at 12:54
  • $\begingroup$ Right, I was indeed assuming $M$ orientable. $\endgroup$
    – abx
    Commented May 10, 2014 at 13:03
  • $\begingroup$ Thank you. I was assuming $M$ orientable, too. Could you let me know the reference of the exact sequence? $\endgroup$
    – Hwang
    Commented May 10, 2014 at 13:33
  • $\begingroup$ There must be many; one I know (in the more general context of sheaf theory) is in Dimca's book Sheaves in Topology (Universitext, Springer), Remark 2.4.5. $\endgroup$
    – abx
    Commented May 10, 2014 at 13:57

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