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By the Kunneth Theorem, we know that the degree two cohomology of $\mathbb{P}^{1} \times \mathbb{P}^{1}$ with integral coefficients is generated by two hyperplane classes $H_{1}$ and $H_{2}$ which I think we can take to be the Poincare duals of the fundamental classes of the two $\mathbb{P}^{1}$'s. I guess more correctly, the images of the two fundamental classes under the obvious embeddings.

$$[\mathbb{P}^{1}]_{1} = \text{PD}(H_{1})\,\,\,\,\,\,\text{and}\,\,\,\,\,\,\,[\mathbb{P}^{1}]_{2} = \text{PD}(H_{2})$$

I want to compute the following intersection product which I know intuitively should be one,

$$[\mathbb{P}^{1}]_{1} \cdot [\mathbb{P}^{1}]_{2} = \int_{\mathbb{P}^{1} \times \mathbb{P}^{1}} H_{1} \smile H_{2}$$

Now I think what I want to do from here is either use the inclusions or projections to pushforward (or pullback) the integrals to either $\mathbb{P}^{1}$. My first question is, can I use either one and get the correct result? I tried computations both ways, and each way there's something that I can't get to make complete sense. I was hoping someone could lay out the cleanest way to do a computation like this for me please!

My second question, is that I believe I'm implicitly using the de Rham Theorem above. Specifically, I think I'm using that

$$H_{\text{dR}}(X) \cong H^{*}_{\text{sing}}(X, \mathbb{R})$$

such that wedge products are sent to cup products. That roughly justifies my use of $\smile$ above. But in intersection theory, particularly above, we usually want to consider integral singular cohomology, but the de Rham isomorphism is for singular cohomology with real coefficients. Is there something I'm overlooking here that allows me to use either?

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    $\begingroup$ You're working with real differential forms that represent integral cohomology classes. For your first computation, why not use Fubini's Theorem to compute the integral of the wedge product of the two Kähler forms? $\endgroup$ – Ted Shifrin Apr 30 '17 at 22:36
  • $\begingroup$ Thanks a lot :) Looking at this (math.stackexchange.com/questions/29797/…) question, it looks like there's a canonical map from integral cohomology to de Rham cohomology. So I guess what you mean is that I'm taking my integral classes and using their images under this map as my differential forms? I guess it's fair to say that not all differential forms can be represented by integral classes. $\endgroup$ – Benighted May 1 '17 at 0:07
  • $\begingroup$ You might also take a look at the discussion at this question. $\endgroup$ – Ted Shifrin May 1 '17 at 0:31

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