The following is a problem from a topology qualifying exam I am studying for:
Show there does not exist a map $S^2\times S^2\to \mathbb{CP}^2$ with odd degree.
I think I am doing something wrong, because I am getting that the degree of any such map must be zero. I argue as follows: By the universal coefficient theorem, the degree is the same whether we compute it on homology or on cohomology. Hence the degree of our map, lets call it $f$, is the same as computing the homomorphism $f^*:H^4(\mathbb{CP}^2)\to H^4(S^2\times S^2)$. We know that the cohomology ring of $\mathbb{CP}^2$ is isomorphic to $\mathbb{Z}[\alpha]$ where $|\alpha|=2$, hence a generator of the top cohomology is $\alpha^2=\alpha\cup \alpha$, hence we need only compute $f^*(\alpha \cup \alpha)$. But the cup product is functorial, so we have $f^*(\alpha\cup \alpha)=f^*\alpha \cup f^*\alpha$.
However, we note that $f^*\alpha$ is an element of $H^2(S^2\times S^2)$, and we claim that the square of any element in this cohomology is zero. Indeed, the projection map $p:S^2\times S^2\to S^2$ by projecting on either coordinate induces the map $p^*$ which is an isomorphism on $H^2$ by considering cellular cohomology. But clearly the square of any element in $H^2(S^2)$ is zero, since it will be an element of $H^4(S^2)=0$. Hence $f^*\alpha \cup f^*\alpha=0$, so the degree is zero.
Am I making some mistake here?