Let $M$ be a smooth manifold and assume that for each $v\in M$, I have a vector $v(m)\in\mathbb{R}\setminus\{0\}$ with some property and this vector is unique up to multiple: further, it depends smoothly on $m$. I would like to define a smooth map $F: M\to S^{n-1}$ that sends $m$ to a representative unit vector with the required property. Under which conditions can this be done?
I understand that it is a lifting problem: I want to lift a smooth map $f: M\to\mathbb{RP}^{n-1}$ to a map $F: M\to S^{n-1}$. If necessary, I can assume that $\mathrm{dim}\, M<n-1$ but I'm not sure if it is sufficient. Then $f(M)$ has "lower dimension" then $n-1$ and I can assume that it aviods a point, but I can hardly assume that it aviods the whole $\mathbb{RP}^{n-2}$ (in which case I could define a global section on $\mathbb{RP}^{n-1}\setminus\mathbb{RP}^{n-2}\to S^{n-1}$)..
Any hint please?
The problem's all in $\pi_1$. Consider the identity map from $\mathbb {RP}^2$ to itself; does this lift to a map to $S^2$? No. Why not? Because at the level of $\pi_1$, it's the identity map. Since $\pi_1(S^2)$ is trivial, the nontrivial identity map on $\mathbb Z / \mathbb {2Z}$ cannot factor through it.
What if (in your question) the map from $\pi_1(M) \to \pi_1(\mathbb {RP}^n)$ is trivial? Then the homotopy lifting property tells you that the map you're looking for exists.