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Consider a smooth, closed, compact finite-dim manifold. We have Poincare Duality to relate the cocycles and cycles.

I would like to know where I can find a reference for a proof that the cup product of the Cohomology Ring is given by the intersection of the corresponding cycles.

Griffiths and Harris talk about intersection number, and discuss this result in chapter 0, Hatcher's book doesn't mention this explicitly as far as I can tell, Katz' little book on enumerative geometry alludes to this, Fulton's book on Young Tableaux dodges this, etc.

I am preparing to give a talk on Schubert Cells and Schubert calculus, and I realized that I have not checked the details of this proof.

Thanks in advance!

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The tag Schubert-varieties is not quite right, but the tag schubert-cells does not exist. –  B. Bischof Apr 4 '10 at 22:27
It's a chapter in Bredon's "Geometry and topology". You can also prove it readily from what's in Hatcher, but I don't think it's assembled that way. You need the Thom isomorphism and its relation to Poincare duality to get anywhere. –  Ryan Budney Apr 4 '10 at 22:31
Michael Hutchings has some lecture notes on this for cohomology classes of smooth closed submanifolds; key point is to use Thom class. One can also ask for analogue in etale cohomology without smoothness or properness hypotheses, for which one has to modify the technique to make a rigorous proof in such generality. –  BCnrd Apr 4 '10 at 22:52
Here are the notes that Brian Conrad refers to: math.berkeley.edu/~hutching/teach/215b-2005/cup.pdf –  Kevin H. Lin Apr 4 '10 at 23:05
For de Rham cohomology, I think you can find this in Bott-Tu. (Indeed it's almost immediate from the explicit tubular-neighborhood construction of forms $\omega_M$ representing $\text{PD}(M)$: if $M$ and $N$ are transverse, then $\omega_M\wedge \omega_N$ is equal to $\omega_{M\cap N}$ if you choose your coordinates consistently.) –  Tom Church Apr 5 '10 at 0:04
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1 Answer

Bott and Tu do this completely, in the de Rham theoretic setting of course.

Here's an alternate proof I plan to use in singular theory next time I teach this material, which I find slightly more direct than using Thom classes (which require the tubular neighborhood theorem, etc):

Definition: Given a collection $S = \{W_i\}$ of submanifolds of a manifold $X$, define the smooth chain complex transverse to $S$, denoted ${C^S}_*(X)$, by using the subgroups of the singular chain groups in which the basis chains $\Delta^n \to X$ are smooth and transverse to all of the $W_i$.

Lemma: The inclusion ${C^S}_*(X) \to C_*(X)$ is a quasi-isomorophism, for any such collection $S$.

Now if $W \in S$ then "count of intersection with $W$" gives a perfectly well-defined element $\tau_W$ of ${\rm Hom}(C^S_*(X), A)$ and thus by this quasi-isomorphism a well-defined cocycle if the $W$ is proper and has no boundary. It is immediate that this cocycle evaluates on cycles which are represented by closed submanifolds through intersection count. It is also not hard (but takes a bit to work out all the details) to show that the cup product of these cochains (when the submanifolds intersect transversally) is given by the intersection class of their intersection - we compute on the chains which intersect all of $W$, $V$ and $W \cap V$ transversally and reduce to linear settings. Consider for example $W$ the $x$-axis in the plane, $V$ with $y$-axis, and then various $2$-simplices can contain the origin (or not) and have various faces which intersect the axes (or not) all consistent with the formula for cup product.

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