Differential form on $\mathbb R\mathbb P^2$ Let $f:S^2\to\mathbb R\mathbb P^2$ be a mapping that takes the point of the unit sphere into a straight line passing through this point and the center of the sphere. Let $r:S^2\to S^2$ be a mapping that takes every point of the sphere to its diametrically opposite point. Let $w$ be a 2-form on $S^2$. Prove that the existence of a 2-form $\eta$ on $\mathbb R\mathbb P^2$ such that $w=f^*\eta$ is equivalent to $r^*w=w$. ($f^*$ is a pull-back map, in our case $f^*: \Omega^2(\mathbb R\mathbb P^2) \to \Omega^2(S^2)$ ) I don't understand what should I show to prove that 2-form is exist.

The idea is to push the form $\omega$ forward by the local diffeomorphism $f$. By "pushing forward" I really mean pulling back by the inverse. In other words, around any point in the projective plane you can find a small open set $U$ such that there exist two distinct sections $s_1,s_2:U \to S^2$ of the double cover $f$ with $r \circ s_i = s_j$ if $i \
eq j$. Now define $\eta = s_1^* \omega$ on the open set $U$. This yields a well-defined global form because the equivariance condition gives us $$s_2^* \omega = (r \circ s_1)^* \omega = s_1^* (r^* \omega) = s_1^*\omega.$$

So even if you picked two distinct sections on two neighbourhoods of a point, the equivariance condition guarantees that the pulled back form $\eta$ agrees on the intersection of the neighbourhoods.