Let $\pi: \mathbb S^2 \to \mathbb {RP}^2$ be the projection map. I understand $\pi$ is smooth. Ideally I should use the tangent space, but this only tells me for $x \in \mathbb{S}^2$ that $$\pi_{*,x}:T_x \mathbb{S}^2 \to T_{\pi(x)} \mathbb{R}P^2 $$ is an isomorphism, but I dont know how to use this to construct an inverse for $\pi$.
Edit: Im sorry,The original question was too long for the title, here is a more precise statement
Let $\pi$ be the projection map $\pi:\mathbb{S}^2 \to \mathbb{R}P^2$, does there exist $s: \mathbb{R}P^2 \to \mathbb{S}^2$ such that $\pi \circ s = \text{id}_{\mathbb{R}P^2}$?
Hint: Try to argue by contradiction. If a smooth map $\sigma$ with the required properties would exist, then it cannot be surjective. Hence $\sigma(\mathbb RP^2)\subset S^2$ would be a non-empty, proper subset. But from the construction, it is easy to deduce that this subset is both open and closed in $S^2$ which contradicts the fact that $S^2$ is connected.