Let $g:\mathbb R\mathbb P^{2n}\longrightarrow \mathbb R\mathbb P^{2n}$ be a continuous map where $\mathbb R\mathbb P^{2n}=\mathbb S^{2n}/\{\pm x\}$. Is it true there exists $f:\mathbb S^{2n}\longrightarrow \mathbb S^{2n}$, possibly continuous, such that the diagram $$\require{AMScd}\begin{CD} S^{2n} @>{f}>> S^{2n}\\ @VVV @VVV \\ \mathbb R\mathbb P^{2n} @>{g}>> \mathbb R\mathbb P^{2n} \end{CD}$$ is commutative? In the vertical arrow are the natural projections $q:S^{2n}\longrightarrow \mathbb R\mathbb P^{2n}$.
Maybe the even dimension is not needed.
Important Consequence: Once this problem has been solved we can conclude every continuous map $f:\mathbb R\mathbb P^{2n}\longrightarrow \mathbb R\mathbb P^{2n}$ has a fixed point. This is done as follows:
Using degree theory, or only homotopy theoric arguments, we are able to show that for every continuous map $f:\mathbb S^{2n}\longrightarrow \mathbb S^{2n}$ there exists $x_0\in\mathbb S^{2n}$ such that $f(x_0)=x_0$ or $f(x_0)=-f(x_0)$. Consequently, if there exists the function $f$ that solves my problem then taking such $x_0$ gives $g([x_0])=[x_0]$, that is, $[x_0]$ is a fixed point of $g$.
Yes, such a continuous map $f$ follows from the lifting property of the universal cover.