# Intersection in the cartesian product and dominant projection

i must say in advance that i'm not very familiar with algebraic cycles and intersection theory, so i hope my question is not too trivial.

Let $X$ and $Y$ be two K3 surfaces. Let $Z=\sum_{i=1}^{k}n_iZ_i$ be an algebraic cycle in $X\times Y$ such that for all $i$ it is $dim_{\mathbb{C}}Z_i=2$ and $n_i$ are positive integers. I have found this statement:

If for a generic $p\in X$ it is $\sum_{i=1}^{k}n_i(\{p\}\times Y)\cdot Z_i=1$ then there exists a unique $i_0$ for which the projection $Z_{i_0}\rightarrow X$ is dominant and furthermore $n_{i_0}=1$.

I have trouble formalizing the proof of this statement. I mean, if for a certain $p\in X$ it is $\sum_{i=1}^{k}n_i(\{p\}\times Y)\cdot Z_i=1$, then $\{p\}\times Y$ must intersect only one irreducible component of $Z$, call it $Z_{i_p}$, and it must be $n_{i_p}=1$. But for another point $q\in X$ such that $\sum_{i=1}^{k}n_i(\{q\}\times Y)\cdot Z_i=1$ why couldn't there be another irreducible component, call it $Z_{i_q}$, which is the only one intersected by $\{q\}\times Y$ and $n_{i_q}=1$? I think this is connected with the fact that $Y$ and the irreducible components $Z_i$ have the same dimension.

If i can prove that there is only one $Z_{i_0}$ intersecting $\{p\}\times Y$ for all $p\in X$ then i think that the fact that the projection $Z_{i_0}\rightarrow X$ is dominant is straightforward.

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