$f_n \to f$ then $\tilde{f_n}\to \tilde{f}$ where $\tilde{f_n}$ and $\tilde{f}$ are the lifting of $f_n$ and $f$.

Question

Let $p:(\mathbb{R}^2,\tilde{u})\to (\mathbb{R}^2\setminus\{0\},u_0) $ be a covering map and let $f_n:(I^2,u)\to (\mathbb{R}^2\setminus\{0\},u_0)$ be a sequence of function which converges uniformly to $f$ on every compact subset of $I^2=[0,1]\times [0,1]$. Let $\tilde{f_n}$ and $\tilde{f}$ be unique lifting of $f_n$ and $f$. Then is it true that $\tilde{f_n}\to \tilde{f}$ uniformly on each compact subset of $I^2$?

Answer 1

First of all, $I^2$ is itself compact, so uniform convergence on compact subsets is just uniform convergence in this case. Otherwise this is true by standard lifting and compactness arguments. By compactness, the image $f(I^2)$ has some positive distance $r>0$ to $0$, so local inverses of $p$ exist in $r$-neighborhoods of every $y \in f(I^2)$. Label these local inverses by points $x \in I$ such that $p_x^{-1}$ is the local inverse of $p$ in the $r$-neighborhood of $f(x)$ which is encountered by path lifting $f$ along the straight line from $u$ to $x$. (By simple connectedness this is really independent of the path.) This means that $\tilde{f}(x) = p_x^{-1} (f(x))$ for all $x \in I^2$. We also have that $p_x^{-1} = p_y^{-1}$ if $|x-y|$ is small enough, in the intersection of the $r$-neighborhoods abour $f(x)$ and $f(y)$. And by compactness we can use finitely many of these inverse branches to construct the lifting, which shows that these are uniformly continuous in some neighborhood of $f(I^2)$. Then you can use these same inverse branches of $p$ to lift $\tilde{f_n}$ for $n$ large enough and get uniform convergence.