$C \times y_1, C \times y_2$ are numerically equivalent in $C \times Y$ for curve $C$.

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This question comes from my former question.

Let me formulate the question more precisely.

Let $C$ be a projective smooth curve over an algebraically closed field. Let $Y$ be any variety. Why for any closed point $y_1,y_2 \in Y$, $C \times y_1, C \times y_2$ are numerically equivalent in $C\times Y$ ? Here, numerically equivalence means they have same intersection number for any curve.

One can reduce the above case to the case where dimensional of $Y$ is $1$.