Let $M,N$ be Riemannian manifolds and $f: M \to N$ a local isometry. If $M$ is complete ,$f$ is a smooth covering.
Let's take $U=B(x,r)$ a normal neighbourhood of $x \in N$. I managed to prove that $f^{-1}(U)=\bigcup_{y \in f^{-1}(x)}B(y,r)$ abd that $f: B(y,r) \to B(x,r)$ is a diffeomorphism. I just can't see why these open balls should be disjoint.
First, we show that there exists $c>0$ and $c<r$ such that for every $y,z\in f^{-1}(x), d(y,z)>c$. Suppose that for every $n>0,$ there exists $x_n,y_n\in f^{-1}(x)$ such that $d(x_n,y_n)<{1\over n}$. Let $n$ such that ${1\over n}<r$, $d(x_n,y_n)<{1\over n}<r$ implies that $y_n\in B(x_n,r)$ this is impossible since $f(x_n)=f(y_n)$ and the restriction of $f$ to $B(x_n,r)$ is injective.
$f^{-1}(B(x,{c\over 4}))=\bigcup_{y\in f^{-1}(x)}B(y,{c\over 4})$. Let $y,z\in f^{-1}(x)$ and $u\in B(y,{c\over 4})\cap B(z,{c\over 4})$, $d(y,z)<d(y,u)+d(u,z)<{c\over 4}+{c\over 4}<c<r$ contradiction, since the restriction of $f$ to $B(y,r)$ is injective.