$\newcommand{\R}{\mathbf R}$ Let $p:(E, e)\to (X, x)$ be a covering projection map. We know that for any path $\gamma:I\to X$ such that $\gamma(0)=x$, there is a unique lift $\Gamma:I\to E$ such that $\Gamma(0)=e$.
Question. Are there any known generalizations of this. More precisely, if $f:(Y, y)\to (X, x)$ is any continuous map, then are there some nice conditions on $Y$ which guarantee a lift of $f$ to a map $F:(Y, y)\to (E, e)$?
Why I am interested in the above is because of the following.
Suppose we have a continuous map $\varphi:\R^n\to G_k(\R^m)$, where $G_k(\R^m)$ denotes the Grassmannian manifold of $k$-planes in $\R^m$.
Can we left this map to a map $\Phi:\R^n\to F_k(\R^m)$, where $F_k(\R^m)$ is the manifold of $k$-frames in $\R^m$?
The question is meaningful when interpretted geometrically.
EDIT: My second question can be tackled in the following way. For each $\mathbf p\in \R^n$, let $\gamma_{\mathbf p}:[0, 1]\to \R^n$ be the path which takes $0$ to $\mathbf 0$ and $1$ to $\mathbf p$ in a linear fasion.
Let $\sigma_0$ be a $k$-frame in $\R^m$ such that $\pi(\sigma_0)=f(\mathbf 0)$, where $\pi:F_k(\R^m)\to G_k(\R^m)$ is the natural map. Then for each $\mathbf p\in \R^n$, we get a unique $k$-frame $\sigma_{\mathbf p}$ by lifting the path $f\circ \gamma_{\mathbf p}:[0, 1]\to G_k(\R^m)$. Now I think the assignment $\mathbf p\to \sigma_{\mathbf p}$ would be a continuous lifting of $f$. The fact that it is a lifting is clear. Only the continuity in question.
In answer to your first question, from page 61 of Hatcher's book, you have that
Now, if $Y$ is a simply connected space (as is the case for $\mathbb{R}^n$) then all the condition for the existence of a lift is automatically satisfied and you always have a lift along a covering map.