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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.

Use the standard complex orientations and the product orientation. Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ satisfy $a^2=b^2=0$ and $⟨ab,[S^2\times S^2]⟩=1$, and let $x\in H^2({\mathbb{CP}}^2;\mathbb{Z})$ satisfy $⟨x^2,[{\mathbb{CP}}^2]⟩=1$. These are the cohomology rings of the product of two spheres and the Complex projective plane.

For a map from the Complex projective plane, write $f^{\ast }a=rx$ and $f^{\ast }b=sx$. Naturality of the cup product gives $r^2{x}^2=f^{\ast }(a^2)=0$ and $s^2{x}^2=f^{\ast }(b^2)=0$. Since $x^2$ has infinite order, $r=s=0$. It follows that $f^{\ast }(ab)=0$, so every such map has degree $0$.

In the reverse direction, write $g^{\ast }x=ra+sb$. ThenEvery even mapping degree really occurs. To construct it, identify $S^2$ with ${\mathbb{CP}}^1$ and use the Segre embeddingThe generator $x_3\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ pulls back to $a+b$: restricting to either factor gives a projective line and hence coefficient $1$. By the cellular approximation theorem, $\sigma$ is homotopic to a cellular map. Its four-dimensional domain then maps into the four-skeleton ${\mathbb{CP}}^2$ of ${\mathbb{CP}}^3$. Denote that map by $g_0$; restriction of $x_3$ to ${\mathbb{CP}}^2$ is $x$, so $g_0^{\ast }x=a+b$ and $\mathrm{deg}{g}_0=2$.

For any integer $m$, choose a map $d_m:S^2\to S^2$ of mapping degree $m$. For $m>0$ one may use $z\mapsto z^m$ on the Riemann sphere, for $m<0$ use $z\mapsto {\overline{z}}^{|m|}$, and for $m=0$ use a constant map. Then $g_m=g_0\circ (d_m\times \mathrm{id})$ satisfies $g_m^{\ast }x=ma+b$, so the possible degrees are exactly all even integers, with $g_m$ realizing $2m$. The cellular approximation here deforms this particular map into the skeleton; it does not require a retraction of ${\mathbb{CP}}^3$ onto ${\mathbb{CP}}^2$.

Finally, take the connected sum of oriented manifolds with both summands carrying their standard complex orientations. Its degree-two cohomology has generators $u,v$ withThis follows by choosing the generators supported away from the two balls used to form the connected sum, so mixed cup products vanish while each square gives the common orientation class. WriteIf $\mathrm{deg}h=n$, the three cup product relations sayTaking determinants gives $-(\det P{)}^2=n^2$, which forces $n=0$. Therefore the only possible degree to the connected sum is $0$, realized by a constant map. This is a degree constraint from intersection forms: the indefinite intersection form of the sphere product cannot pull back a definite form with a nonzero degree multiplier.