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?
 A: This is false. Let $M$ be the interval $(0,1)$. Let $\mu$ be a finite, purely singular measure (i.e., singular with respect to the Lebesgue measure) on $M$ that has no atoms and satisfies $\mu(I)>0$ for every nontrivial subinterval $I\subset (0,1)$ (an example of such $\mu$). Define 
$$f(x) = x - \mu((0,x))$$
This is a continuous function because $\mu$ is nonatomic. It has  $f'=1$ almost everywhere, because a singular measure has zero derivative a.e. (stated here, with a reference to Folland's Real Analysis).
For every interval $I\subset (0,1)$, by virtue of $\mu(I)>0$ and $\mu$ being singular, there exist subintervals $J_1,J_2\subset I$ such that $\mu(J_1)>|J_1|$ and $\mu(J_2) <|J_2|$ where $|\cdot |$ stands for the length of intervals. This can be proved in general, but is easier to see for the specific example of $\mu$ linked above: $I$ contains a dyadic subinterval $D$, and after $n$ dyadic subdivisions of $D$ its leftmost child $D_n^{-}$ and rightmost child $D_n^+$ satisfy $\mu(D_n^+) = p^n\mu(D)$ and $\mu(D_n^-) = (1-p)^n \mu(D)$. Since both have length $|D|/2^n$, this yields the desired subintervals when $n$ is large enough.
Conclusion: $f$ is not monotone on $I$, and therefore is not injective on any subinterval on $M$. 
This example can be promoted to higher dimensions by letting $F(x_1,x_2,\dots,x_d) = (f(x_1), x_2,\dots, x_d)$.

To get a positive result, you may want to assume that $f$ is Lipschitz continuous; then $f$ is a Sobolev map, and the theory of quasiregular maps implies that $f$ is discrete and open, hence a local homeomorphism outside of a branch set of topological dimension $\le d-2$. References: 


*

*Space Mappings with Bounded Distortion by Reshetnyak - contains the proof that Sobolev maps with appropriate constraints on the derivative $Df$ are discrete and open. 

*Quasiregular Mappings by Rickman - covers the same topic from a more topological point of view.

*Discrete open mappings on manifolds by J. Väisälä, Ann. Acad. Sci. Fenn. Ser. A I Math. 1966, no. 392 - a proof that a discrete open mapping has a branch set of topological dimension at most $d-2$ (proved earlier by Chernavskii in the Euclidean setting).

*A new characterization of the mappings of bounded length distortion by  Hajłasz and  Malekzadeh - since your assumption on $Df$ is more restrictive than is required for quasiregular maps, it may be easier to use this source and the Martio-Väisälä article it refers to. If $f$ in your problem is Lipschitz, then it immediately qualifies as a bounded length distortion (BLD) map under the analytic definition, which is known to be equivalent to the geometric definition (see the beginning of this article).

A: 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
$[0,1]\times [1-\varepsilon_n,1)$
