A continuous a.e orientation-preserving isometry is locally injective? Let $M,N$ be $d$-dimensional Riemannian manifolds. Let $f:M \to N$, and suppose $f$ is continuous, differentiable almost everywhere (a.e
) and that $df$ is an orientation-preserving isometry a.e. 
Question: Is it true that there exist a ball $B_{\epsilon}(p) \subseteq M$ such that $f|_{B_{\epsilon}(p)}$ is injective?

I want to reply to your comment in
https://mathoverflow.net/questions/264873/do-curvature-differences-obstruct-a-e-orientation-preserving-isometries/267163#267163
Question : Assume that $$f :B\rightarrow
\mathbb{E}^2$$ is a map s.t. 
(1) $B$ is a geodesic ball in $S^2(1)$ of radius $\varepsilon$
(2) $df$ is isometric a.e.
(3) a.e. orientation preserving
Then $f$ is not continuous 
EXE : Gromov's map $S^2\rightarrow \mathbb{E}^2$ is not
orientation preserving, since it can view as a limit of piecewise
distance preserving maps that are not an a.e.-orientation-preserving
maps.
Proof of Question : Assume that $f$ is continuous. Hence by considering following EXE, we conclude that it is volume preserving 
Consider a triangulation $T_i$ of $B_\epsilon (p)$ where each $T_i$ is 2-dimensional geodesic triangle. Since $f$ is orientation preserving then $
f(T_i)$ are not overlapping except measure $0$-set. 
Hence we have that $\gamma_i$ is a curve going to $\partial
B_\epsilon(p)$ and $f\circ \gamma_i$ goes to $\partial f(B_\epsilon
(p))$ s.t. $$\lim_i\ {\rm length}\ \gamma_i={\rm
length}\ \partial B_\epsilon (p) \geq \lim_i\ {\rm length}\ f\circ
\gamma_i $$ since $f$ is short.
Hence by isoperimetric inequality, it is a contradiction.
EXE : Consider a continuous map $ f: [0,1]^2\rightarrow \mathbb{E}^2$ s.t.
assume that $c(t)=f(t,1)$ is a continuous curve whose image has
2-dimensional.
Hence note that curve $f(t,1-\varepsilon_n)$ where
$\varepsilon_n\rightarrow 0$ has an arbitrary large length $l_n$
with $\lim_n\ l_n=\infty$ Hence $f$ is not isometric on
$[0,1]\times [1-\varepsilon_n,1)$
