I'm trying to show that $H^2(\mathbb{P}^2_\mathbb{R})=0$ by pulling back closed 2-forms $\omega$ on $\mathbb{P}^2_\mathbb{R}$ to $S^2$ using the fact that $\pi^*\omega$ is exact (where $\pi:S^2\longrightarrow\mathbb{P}^2_\mathbb{R}$ is the canonical quotient map). If we write $\pi^*\omega=d\eta$ for $\eta\in\Omega^1(S^2)$, I would like to "average" $\eta$ to produce a 1-form $\kappa\in\Omega^1(S^2)$ that descends to a 1-form $\tilde{\kappa}$ on $\mathbb{P}^2_\mathbb{R}$, akin to my approach in 1. However, I'm not entirely sure what conditions I need to check to prove that this is the case.
In showing the analogous statement for 1-forms, I was able to just define a smooth function $g:S^2\longrightarrow S^2$ by $$g(p)=\frac{1}{2}\left(f(p)+f(-p)\right)=\frac{1}{2}\left(f(p)+(\alpha^*f)(p)\right)$$ if $\pi^*\omega=df$ for $f\in \Omega^0(S^2)$, and checking that $g(p)=g(-p)$.
Returning to the 2-form case, I first tried to find a 1-form $\kappa$ on $S^2$ such that $\kappa_p(v)=\kappa_{-p}(v)$, but I'm not even sure this statement makes sense: here, we'd want to regard $v$ as the "same" tangent vector in both $T_pS^2$ and $T_{-p}S^2$. Anyway, I wasn't able to come up with a $\kappa$ that works, assuming this is the "correct" thing to check.
Are there general conditions would I have to check to ensure this descent works (and similarly, a natural definition for $\kappa$)? I'm hoping to establish that $H^2(\mathbb{P}^2_\mathbb{R})=0$ using this approach. Thank you for any help.
References:
The canonical quotient $\pi:S^2\longrightarrow \mathbb{P}^2_\mathbb{R}$ forms a (double-sheeted) cover of $\mathbb{P}^2_\mathbb{R}$ with deck group $\operatorname{Aut}(\pi)\cong\mathbb{Z}/2\mathbb{Z}$. Because $\pi$ is a smooth covering, it can be shown that $G=\operatorname{Aut}(\pi)$ acts smoothly, freely, and properly on $S^2$. Thus we can write $\mathbb{P}^2_\mathbb{R}$ as the orbit space $S^2/G$.
Fix $p\in S^2$, and observe that the orbit $G\cdot p=\pi^{-1}([p])$ is an embedded submanifold of $S^2$. This gives that $T_p(G\cdot p)=\operatorname{ker}d\pi_p$, and since $d\pi_p$ is a surjection, the isomorphism $$T_p\mathbb{P}^2_\mathbb{R}\cong \frac{T_pS^2}{T_p(G\cdot p)}.$$ As a consequence, we can regard $T_p\mathbb{P}^2_\mathbb{R}$ as the collection of pairs $\{(p,v),(-p,-v)\}$ such that $p\cdot v=0$. In particular, to check that a 2-form $\tau\in\Omega^2(S^2)$ descends to $\mathbb{P}^2_\mathbb{R}$, it suffices to show that $\tau$ is $\alpha^*$-invariant.
Let $$\kappa=\frac{1}{2}(\eta+\alpha^*\eta).$$ It is immediate that $\alpha^*\kappa=\kappa$, so $\kappa$ furnishes to a 2-form $\tilde{\kappa}$ such that $\pi^*\tilde{\kappa}=\kappa$. Moreover, \begin{align*} d\kappa_p=\frac{1}{2}(d\eta_p+\alpha^*d\eta_p)&=\frac{1}{2}(\pi^*\omega_p+\alpha^*\pi^*\omega_p)\\ &=\frac{1}{2}(\pi^*\omega_p+(\pi\circ \alpha)^*\omega_p)\\ &=\pi^*\omega_p. \end{align*} The naturality of $d$ implies $d\kappa_p=\pi^*d\tilde{\kappa}_p$, so the injectivity of $\pi^*$ gives the desired result.