 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 implyThe two factor classes give the source matrix $H=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, while the target matrix is $I_2$. If $P$ is the integral matrix for the pullback in these bases, thenTaking determinants gives $-(\det P{)}^2=d^2$, whose left side is nonpositive and whose right side is nonnegative. Thus $d=0$. 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.