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On page 250 of Hatcher he says for a closed $R$-orientable $n$-manifold $M$, consider the cup product pairing

$H^k(M;R) \times H^{n-k}(M;R) \rightarrow R;~~ (\varphi, \psi) \mapsto (\varphi \smile \psi)[M]$.

Here $[M]$ means a fundamental class for $M$. My confusing point is the statement $(\varphi \smile \psi)[M]$. I'm not totally sure how $(\varphi \smile \psi)[M]$ is supposed to be evaluated since $\varphi \smile \psi$ is a cohomology class rather than a cochain, and $[M]$ is a homology class rather than a cochain, so $(\varphi \smile \psi)[M]$ doesn't seem to be a well defined statement unless I'm missing something.

My best guess here is that he is being sloppy here and meaning that $\varphi$ and $\psi$ and $[M]$ are representatives of cohomology/homology classes in the above cohomology groups and then the right hand side is business as usual. But then again can't be sure..

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  • $\begingroup$ You mean the product may depend on the choice of rep? It does not, though I cant think of a proof now. Elements of (Co)homology are ultimately (Co)chains. $\endgroup$
    – gary
    Jul 26 '17 at 1:25
  • $\begingroup$ Yes, your guess is correct and this notation is the standard Kronecker pairing between Cohomology and homology. Get used to it. $\endgroup$ Jul 26 '17 at 2:34
  • $\begingroup$ @gary No I was just confused by the notation since it was never explained as far as I can tell, and I've read pretty much all of the book up to that point. Not the first time Hatcher has not defined notation he's using that he shouldn't be assuming we know what it means. $\endgroup$
    – TuoTuo
    Jul 26 '17 at 2:42
  • $\begingroup$ @MoisheCohen Thanks, and it's not at all difficult to get used to it, as long as it's explained once. $\endgroup$
    – TuoTuo
    Jul 26 '17 at 2:44
  • $\begingroup$ @TuoTuo: Sadly, I have to agree with MoisheCohen: Sometimes authors are sloppy. $\endgroup$
    – gary
    Jul 26 '17 at 17:12
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Your guess is correct: this notation means to choose a cochain $\alpha$ representing $\varphi\smile\psi$ and a chain $x$ representing $[M]$ and evaluate $\alpha(x)$. It is easy to check that this is independent of the choice of representatives: if you add a coboundary $\delta \beta$ to $\alpha$ and a boundary $\partial y$ to $x$, you get \begin{align*} (\alpha+\delta\beta)(x+\partial y)&=\alpha(x)+\alpha(\partial y)+\delta\beta(x)+\delta\beta(\partial y) \\ &=\alpha(x)+\delta\alpha(y)+\beta(\partial x)+\beta(\partial\partial y) \\ &=\alpha(x) \end{align*} since $\delta\alpha=0$ and $\partial x=0$.

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