Suppose that $g$ is a $C^{1,1}$ (i.e., continuously differentiable with locally Lipschitz first derivative) Riemannian metric on a smooth manifold $M$. It seems to be known that locally the exponential map $\exp\_p$ corresponding to $g$ at some point $p$ of the manifold is still a local homeomorphism (and Lipschitz). I am looking for a proof of this result (if it is true). The standard proof for $C^2$ metrics uses the fact that the tangent map of $exp\_p$ at $0\in T\_pM$ is the identity, so that the inverse function theorem gives that $exp\_p$ is a local diffeomorphism. However, for a $C^{1,1}$ metric, $\exp\_p$ is only (locally) Lipschitz, so it is not clear that there even exists a tangent map of $exp\_p$ at $0$ and the inverse function theorem does not apply. Is there some argument that can replace the use of the inverse function theorem in this situation? 



Is the exponential map of a $C^{1,1}$ Riemannian metric a local homeomorphism?
Note that $C^{1,1}$ metric admits an approximation by $C^2$-metrics with uniformly bounded $C^2$-norm.
In particular the curvature is bounded, hence we get a bounds for the distortion of the exponential map depending on the size of the neighborhood of $0$.
It remains to pass to the limit.


