Every smooth map $S^4 \to\mathbb{CP}^2$ has degree zero

Question

I'm trying to show that every smooth map $f: S^4 \to \mathbb{CP}^2$ has degree zero.

I'm not sure how to go about this, but I know that the oriented intersection number between any two closed 2-dimensional submanifolds of $S^4$ is zero. Maybe there's a way of showing that the inverse image of a regular value of any such $f$ is the oriented intersection of two such submanifolds in $S^4$?

Answer 1

Note that $f$ induces a homomorphism $f^* : H^4 (\mathbb{CP}^2; \mathbb{Z}) \to H^4(S^4; \mathbb{Z})$. As $\mathbb{CP}^2$ and $S^4$ are orientable and closed, both cohomology groups are isomorphic to $\mathbb{Z}$ and therefore $f^*$ is given by multiplication by an integer; that integer is the degree of $f$.

There is $\omega \in H^2(\mathbb{CP}^2; \mathbb{Z})$ such that $\omega^2$ generates $H^4(\mathbb{CP}^2; \mathbb{Z})$. Now note that $f^*(\omega^2) = (f^*\omega)^2$, but $f^*\omega = 0$ as $H^2(S^4; \mathbb{Z}) = 0$. Therefore $f^*(\omega^2) = 0$ and so $f$ has degree zero.

Note that the smoothness hypothesis was not needed, so the conclusion actually holds for any continuous map $S^4 \to \mathbb{CP}^2$.

Answer 2

Consider the 5-dimensional Hopf fibration $S^1 \to S^5 \to \Bbb CP^2$. The long exact sequence of homotopy groups contains the sequence $\pi_4(S^1) \to \pi_4(S^5) \to \pi_4(\Bbb CP^2)\to \pi_3(S^1)$. As all three of the corresponding homotopy groups of spheres vanish trivially, $\pi_4(\Bbb CP^2)=0$. As degree is an invariant of homotopy and the above calculation shows any map $S^4 \to \Bbb CP^2$ is nullhomotopic, we have any such map has degree $0$. One can probably dissect this proof as just repeated applications of the homotopy lifting property.