= Solution
Let $f:S^2\times S^2\to\mathbb{CP}^2\#\mathbb{CP}^2$ have <mapping degree> $d$. Naturality of the <cup product> and evaluation on the fundamental class imply
$$
Q_{S^2\times S^2}(f^*u,f^*v)=d\,Q_{\mathbb{CP}^2\#\mathbb{CP}^2}(u,v).
$$
The two factor classes give the source matrix $H=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, while the target matrix is $I_2$. If $P$ is the integral matrix for the pullback in these bases, then
$$
P^{\mathsf T}HP=dI_2.
$$
Taking determinants gives $-(\det P)^2=d^2$, whose left side is nonpositive and whose right side is nonnegative. Thus $d=0$. \b[There is no map of nonzero degree], including negative degree. This <degree constraint from intersection forms> is a cohomological argument and does not assume that $f$ is smooth.
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