the degree of a map from $S^2$ to $S^2$

Question

Is the degree of $\pi\circ p$ $0$?

$\pi\circ p:S^{2}\xrightarrow{p}\mathbb{R}P^2=S^{1}\bigcup_{2}e^{2}\xrightarrow{\pi}S^{1}\bigcup_{2}e^{2}/S^{1}=S^2$,Where $e^2$ is a unit disk.If it is true,can anyone give me a brief proof ,thanks!

Answer 1

If $\pi\circ p$ were non-trivial, say of degree say $d\neq 0$, then it would induce multiplication by $d$ on $H_2S^2$. But since $H_2\mathbb{R}P^2=0$ we have

$(\pi\circ p)_*=\pi_*\circ p_*:H_2S^2\rightarrow H_2\mathbb{R}P^2=0\rightarrow H_2S^2$

factoring through the trivial group. Therefore $(\pi\circ p)_*$ is multiplication by $0$, so we must have $d=0$ and $\pi\circ p\simeq \ast$.