Let $(M,g)$ be a compact non-orientable Riemannian manifold and consider its orientable double cover $(\tilde{M},\tilde{g})$ with the natural projection $\pi:\tilde{M}\to M$. Then $\pi$ is a local isometry. Is is true that there exists a small enough uniform radius $r$ (i.e., not depending on $x$) such that for all $x\in \tilde{M}$, the restriction of $\pi$ to the ball $B(x,r)$ of radius $r$ centered at $x$ in $\tilde{M}$ is an isometry?
I cannot come up with a proof of this, neither a counter example.
Let $M$ be a compact Riemannian manifold, there exists $r>0$ be the convexity radius of $M$; this number is less than the injectivity radius of $M$. Let $\pi: \tilde M\to M$ be any covering map with $\tilde M$ connected ($\pi$ does not have to be 2-fold) that is a Riemannian isometry (pull-back of the Riemannian metric on $M$ is the Riemannian metric on $\tilde M$). Then $r$ is again less than the convexity radius of $\tilde M$, as $\pi$ sends geodesics to geodesics and preserves arclength. Since $r$ is less than injectivity radius of $\tilde M$ and of $M$, for every $x\in \tilde M$ the restriction of $\pi$ to the geodesic ball $B(x,r)$ is 1-1.
Now, an isometry in the sense of metric geometry is a map which preserves distances. Distances are computed via minimizing geodesics. For each metric ball $B=B(x,r)\subset \tilde M$ ($r$ is as above), for all $y,z\in B(x,r)$, $d(y,z)=L(yz)$ where $L$ is the length and $xy\subset B$ is the unique geodesic connecting $x$ to $y$, while $d$ is the distance function on $\tilde M$. Since $\pi(B)= B(\pi(x), r)$, $\pi(yz)\subset B$ is a geodesic. Hence, $d(\pi(y), \pi(z))= L(\pi(yz))= d(y, z)$. qed