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Let's take $X$ and $Y$ K3 surfaces and $Z\subset X\times Y$ an algebraic cycle of dimension 2. I know that the Poincarè dual of $Z$, namely $[Z]$, is in $H^4(X\times Y,\mathbb{Z})$ and by Kunneth formula $[Z]\in Hom(H^0(X,\mathbb{Z}),H^0(Y,\mathbb{Z}))\oplus Hom(H^2(X,\mathbb{Z}),H^2(Y,\mathbb{Z}))\oplus Hom(H^4(X,\mathbb{Z}),H^4(Y,\mathbb{Z}))$.

So $[Z]$ gives a morphism between the graduated rings $H^*(X,\mathbb{Z})$ and $H^*(Y,\mathbb{Z})$. Let $p_1$ and $p_2$ be the projections, then this morphism is given by

$\mathscr{P}{p_2}_*\mathscr{P}^{-1}(p_1^*(\sigma)\wedge [Z]^*)\in H^l(Y,\mathbb{Z})$ for $\sigma\in H^l(X,\mathbb{Z})$ and $\mathscr{P}$ is the Poincarè duality.

What is not clear to me is how $[Z]$ actually acts on $H^0(X,\mathbb{Z}), H^2(X,\mathbb{Z})$ and $H^4(X,\mathbb{Z})$. I think i can visualize $\mathscr{P}^{-1}(p_1^*(\sigma)\wedge [Z]^*)$ as $\mathscr{P}^{-1}(p_1^*(\sigma))\cap Z$, but this isn't helping so much.

I'm reading from this article at page 16 it says for example that if $Z=X\times p'$ then for $x\in H^0(X,\mathbb{Z})\oplus H^2(X,\mathbb{Z})\oplus H^4(X,\mathbb{Z})$ it is $[Z]_*(x)=deg_4(x).\alpha'$, where $\alpha'$ is the positive generator of $H^4(Y,\mathbb{Z})$. Why is that? actually i don't know what that dot product exactly means, i think that is the intersection, but why should it be with $\alpha'$?

Thank you

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