Question: Given a continuous map $f:\mathbb{R}P^n\rightarrow\mathbb{R}P^n$, is there automatically a continuous map $g:S^n\rightarrow S^n$ such that $f,g$ commute with the covering map $S^n\rightarrow\mathbb{R}P^n$? More generally, if $X$ is a path-connected, locally simply-connected space, and $U$ is its universal cover, with covering map $\pi:U\rightarrow X$, does every map $f:X\rightarrow X$ lift to a map $g:U\rightarrow U$ such that $f\pi = \pi g$?
Thoughts: On intuitive, straight-from-the-definition grounds, it seems to me this should be true. However, I feel that if it is true there ought to be an elegant and short proof based on something like the homotopy lifting property of covering maps, though I don't see how to make this work. If this is true, I'd appreciate a hint about how to do it that way.
Here's my intuitive reasoning: for a given $x\in X$, pick $y\in \pi^{-1}(x)$ and $z\in \pi^{-1}(f(x))$; we can assign $g(y)=z$ and then in a small enough neighborhood of $y$, $g$ is determined by the homeomorphisms that send small neighborhoods of $x,f(x)$ to neigborhoods of $y,z$. Then we let $\pi_1(X)$, identified with the group of deck transformations of $U$ over $X$, act on $\pi^{-1}(x)$, and this tells us how to define $g$ on all other preimages of $x$. We have an initial choice (indexed by $\pi_1(X)$) of which $z\in\pi^{-1}(f(x))$ should be the image of $y$, but after that, everything is determined at least in a small neighborhood of $\pi^{-1}(x)$. Now given $x'\in X$, take a path from $x$ to $x'$ and extend the definition of $g$ to $\pi^{-1}(x')$ by extending it bit by bit along the path, using the same procedure above, with the only difference that now the choice of $g(y)$ for each $y$ has already been made, since we can take a sequence of overlapping neighborhoods along the path and the choice in the previous neighborhood determines the next one. This feels a little hand-wavy to me, especially at the end.
There is Proposition 1.33 in Hatcher's Algebraic Topology which says that a lifting of $f: Y \to X$ with respect to covering map $p: \tilde{X} \to X$ for $Y$ path-connected and locally path-connected exists if and only if $f_*(π_1(Y)) ⊆ p_*(π_1(\tilde{X}))$ where $f_*, p_*$ are homomorphisms between fundamental groups induced by $f, p$ respectively. So the proposition of your question would hold if your $X$ was simply connected and locally path-connected. And it seems that for $p: S^n \to \mathbb{R}P^n$ and $f = id_{\mathbb{R}P^n}$ it doesn't hold for $n = 1$ when $\mathbb{R}P^n = S^1$ which is not simply connected. $p_*: \mathbb{Z} \to \mathbb{Z}$ is then multiplication by $2$ and $f_*$ is identity on $\mathbb{Z}$.