Local diffeomorphisms, covering maps and smooth path lifting

Question

Let $f: M\to N$ be a surjective local diffeomorphism of noncompact smooth manifolds. Suppose that every smooth path is liftable, that is, for any smooth path $\gamma: [0,1]\to N$ and any point $p\in f^{-1}\gamma(0)$ there exists $\tilde \gamma: [0,1]\to M$, such that $\tilde \gamma(0)=p$ and $f\circ \tilde \gamma=\gamma$. By smooth path here I mean a map $[0,1]\to N$ that extends to a smooth map $(0-\varepsilon,1+\varepsilon)$.

Does it imply that $f$ is a covering?

In our situation being a covering is equivalent to the lifting property for continuous or rectifiable paths but the proofs that I am aware of (https://www.math.ucdavis.edu/~kapovich/EPR/cov.pdf) apparently do not go through when we know how to lift only smooth paths. The bottleneck is lifting homotopies.

Actually I am interested in cases where the class of liftable paths is even more restricted, though still abundant, like the class of piecewise linear paths.

In general, even if we can approximate uniformly a given path: $\gamma_i\to \gamma$ as ${i\to\infty}$, with $\gamma_i(0)=\gamma(0)$, $\gamma_i(1)=\gamma(1)$ and $lenght(\gamma_i)\to lenght(\gamma)$ by paths that lift to $\tilde\gamma_i$ with $\tilde\gamma_i(0)=p\in f^{-1} \gamma (0)$ and $\gamma$ is liftable on $[0,1)$, it does not allow us to conclude that $\gamma$ is liftable on the whole $[0,1]$. So, even if the answer to the question is 'yes', it likely doesn't follow for free by an approximation argument.

Answer 1

To show that $f$ is a covering map, pick any point $n∈N$. Choose an open neighborhood $U$ of $n$ that is diffeomorphic to ${\bf R}^n$. It suffices to show that the restriction $g\colon V\to U$ of the map $f$ to any connected component $V$ of $f^{-1}(U)$ is a diffeomorphism.

The map $g$ is surjective because otherwise the smooth path lifting property is violated for a smooth path that connects a point in the image of $g$ with a point not in the image of $g$.

The map $g$ is injective: if $g(x)=g(y)$, choose a smooth path $p\colon[0,1]→V$ that connects $x$ and $y$. Choose a smooth homotopy $h\colon[0,1]⨯[0,1]→U$ that contracts $gp$ to a point.

Take the supremum $S$ of $Q⊂[0,1]$ consisting of those $q∈[0,1]$ such that $h$ restricted to $[0,q]⨯[0,1]$ has a (necessarily unique) smooth lifting along $g$. We claim that $S∈Q$. Indeed, consider the smooth path $h(S,-)$, which has a smooth lifting along $g$. By compactness of $[0,1]$, this lifting extends to a smooth lifting of $h(t,-)$, where $t$ belongs to a sufficiently small open neighborhood $W$ of $S$. In particular, for $t<S$ this smooth lifting must coincide with the previously constructed smooth lifting along $g$ of the restriction of $h$ to $[0,S)⨯[0,1]$. Thus $S∈Q$. This argument also shows that $Q$ is an open subset of $[0,1]$, so $Q=[0,1]$, which completes the proof.