 Cohomological obstruction to a section of the twistor bundle

If the displayed fiber bundle had a section $s$, its projection $p$ would satisfy $s^{\ast }{p}^{\ast }=1$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^4)$ with $|h|=2$. For a generator $u\in H^4(S^4;\mathbb{Z})$, write $p^{\ast }u=m{h}^2$. But $s^{\ast }h=0$ since $H^2(S^4;\mathbb{Z})=0$, giving $u=s^{\ast }{p}^{\ast }u=0$. This contradiction proves that no section exists, without needing the value of $m$.