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$\newcommand{\R}{\mathbf R}$ 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.

caffeinemachine
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  • Are you still assuming that $E\to X$ is a covering map? It seems that $F_k(\mathbb R^m) \to G_k(\mathbb R^n)$ is not a covering map. –  Oct 10 '15 at 09:07
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    Anyway, $F_k(\mathbb R^m) \to G_k(\mathbb R^m)$ is a fiber bundle and if you have a map $\mathbb R^m \to G_k(\mathbb R^m)$, you can "pullbcak" the fiber bundle to $\mathbb R^n$. As $\mathbb R^n$ is contractible, the pullback bundle is trivial and so has section. This is the same as lifting the map to $F_k(\mathbb R^m)$. There's a related discussion here: http://math.stackexchange.com/questions/186145/a-fiber-bundle-over-euclidean-space-is-trivial –  Oct 10 '15 at 09:12
  • @JohnMa Oh. I was under the impression that $F_k(\R^m)\to G_k(\R^m)$ is a covering map. Never tried to prove it. My reason for thinking this was that $S^n\to \R P^n$ is a covering map. Anyway. Thanks for the insights. Need some time to think. Perhaps you could expand this to an answer. Thanks. – caffeinemachine Oct 10 '15 at 09:45

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In answer to your first question, from page 61 of Hatcher's book, you have that

Given covering map $p:(\tilde{X},\tilde{x}_0)\to(X,x_0)$ and a map $f:(Y,y_0)\to (X,x_0)$, with $Y$ path connected and semi-locally simply connected. Then a lift $\tilde{F}:(Y,y_0)\to(\tilde{X},\tilde{x}_0)$ exists if and only if $f_\star(\pi_1(Y,y_0))\subset p_\star(\pi_1(\tilde{X},\tilde{x}_0))$

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.

ChesterX
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