TITLE: Is the exponential map of a $C^{1,1}$ Riemannian metric a local homeomorphism?
QUESTION [10 upvotes]: 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?

REPLY [5 votes]: More generally the exponential map of any Lipschitz connection or spray is a Lipeomorphism (bi-Lipschitz homeomorphism) in a neighborhood of the origin. The proof uses Leach's      inverse function theorem which holds under strong differentiability. The non-trivial part is to show that the exponential map is strongly differentiable at the origin. Details are given here http://arxiv.org/abs/1308.6675