Covering projection $S^{2k} \to \mathbb{P}_\mathbb{R}^{2k}$ induces $0$ in (integral) (co)homology?

Question

For any $k$, I have the following questions.

1. Does the covering projection $S^{2k} \to \mathbb{P}_\mathbb{R}^{2k}$ induce $0$ in (integral) homology?
2. Does the covering projection $S^{2k} \to \mathbb{P}_\mathbb{R}^{2k}$ induce $0$ in (integral) cohomology?

Answer 1

Note that $H^*(\mathbb{R}P^{2n};\mathbb{Z})=\mathbb{Z}[a]/(2a, a^{k+1})$ for $a\in H^2$, so we have that $p^*:H^*(\mathbb{R}P^{2n};\mathbb{Z})\to H^*(S^n;\mathbb{Z})$ is zero since the source is torsion and the right is free for $*>0$. For homology, note that the only interesting dimension is the top-one, and it is clear that this map vanishes as well since its degree is $2$.