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This is probabibly trivial, but I've read on here that a connected compact $n$-manifold $M$ is orientable iff $H_n(M,\mathbb{Z}) \neq 0$, and it's not exactly clear to me why.

Following Hatcher, let's say that a $n$-manifold $M$ is $R$-orientable with coefficients in $R$ iff there is a generator of $H_n(M;R) \cong R$, called a "fundamental class", which is mapped by $i_\ast$ ($i$ being the inclusion $(M, \emptyset) \to (M|x) :=(M, M \setminus \{x\})$) to a generator of $H_n(M|x;R) \cong R$ for all $x$. I want to prove that this is equivalent to $H_n(M;R) \cong R$ for $M$ compact and connected. Clearly Poincaré duality implies that $H_n(M;R) \cong R$ if $M$ is orientable. The converse doesn't seem trivial to me though: from Proposition 3.29 of Hatcher I know that $H_n(M \setminus \{x\}; R ) =0$, and from the long exact sequence of a pair I get $$ 0 \to R \cong H_n(M; R) \xrightarrow{i_\ast} H_n(M|x; R) \cong R \to \dots $$ but this is not enough to say that $i_\ast$ is an isomorphism, unless $H_{n-1}(M\setminus \{x\};R) = 0$, but this doesn't seem to hold in general (take $M = S^1 \times S^1$ and $n=2$). What am I missing?

Move78
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1 Answers1

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As you mentioned, $i_\ast$ takes the fundamental homology class to a generator of the relative group (rather than a multiple of a generator). Therefore $i_\ast$ is onto. Thus, you don't need to compute $H_{n-1}$ at the next stage in the long exact sequence.

Mikhail Katz
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  • Again, it's probably just me, but I can't see exactly why this is true. Isn't this exactly what we want to show? – Move78 Jul 17 '23 at 14:15
  • You wrote that you want to prove that "this is equivalent to $H_n\simeq R$". It certainly is, by the argument I mentioned. One does not need to evaluate $H_{n-1}$ which was your worry. Try to make it clearer what your concern is. – Mikhail Katz Jul 17 '23 at 14:41
  • sorry, I'll try to clarify: I'd like to show that, if $H_n(M) \cong R$, then $i_{\ast}$ maps a generator to a generator. From my understanding, your answer says that $i_\ast$ does exactly that, being onto; but it's not clear to me why it's actually onto. Thanks for your help! – Move78 Jul 17 '23 at 16:23
  • @Move78, Hatcher surely has a description of the generator of the relative group somewhere, and it is simply the fundamental homology class of $M$ viewed as a relative class. Have you tried to look up a description of the generator of the relative group there? – Mikhail Katz Jul 17 '23 at 16:27
  • I think I got it; by Proposition 3.26.b on Hatcher, if $H_n(M;R) \cong R$ but $M$ is not $R$-orientable, then $i_\ast(H_n(M;R))$ is equal the elements in $R$ having order $2$, but this would imply that $R$ has a unit of order $2$, and hence that $M$ is $R$-orientable, which is a contradiction. – Move78 Jul 17 '23 at 21:25