$\pi^* : \Omega^r(P_n(R)) \rightarrow \Omega^r(R^{n+1}\setminus \{0\})$ is injective

Question

Let $\pi : R^{n+1} \setminus \{0\} \rightarrow P_n(R)$ be the canonical projection. Show that the induced map $$\pi^* : \Omega^r(P_n(R)) \rightarrow \Omega^r(R^{n+1}\setminus \{0\})$$ is injective for all $r$ with $0 \le r \le n$.

What I know:

$\Omega^r(P_n(R)) = \{\omega \mid \omega \text{ a differential } r \text{-form on } P_n(R)\}$, where $\omega: P_n(R) \rightarrow \Lambda^rT^*P_n(R)$ as $pr_r \circ \omega = id_{P_n(R)}$ and $pr_r: \Lambda^rT^*P_n(R) \rightarrow P_n(R)$, where $pr_r(p,\alpha)=p$.

And $\Omega^r(R^{n+1} \setminus \{0\}) = \{\omega \mid \omega \text{ a differential } r\text{-form on } R^{n+1} \setminus \{0\}\}$, where $\omega: R^{n+1} \setminus \{ 0\} \rightarrow \Lambda^rT^*R^{n+1} \setminus \{0\}$ as $pr_r \circ \omega=id_{R^{n+1}\setminus\{0\}}$ and $pr_r: \Lambda^rT^*R^{n+1} \setminus\{0\} \rightarrow R^{n+1} \setminus \{0\}$, where $pr_r(p,\alpha)=p$.

I can't figure out how I can connect things to verify injectivity...

Answer 1

I'm assuming that $P_n(R)$ is real projective space and $\pi$ is the usual submersion. The fact that $\pi$ is a differentiable submersion is all you need here.

If $\omega$ is an $r$-form on $P_n(R)$ and $\pi^\ast \omega =0$, then for all $x \in R^{n+1} \setminus \{0\}$ and all $v_1,\dots, v_r$ in $T_xR^{n+1} \setminus \{0\}$ we have $\omega(\mathrm{d} {\pi}_x v_1,\dots,\mathrm{d}{\pi}_x v_r)=0$. But since $\pi$ is a submersion, $\mathrm{d}\pi_x$ is surjective and therefore $\omega_x$ is zero.