Do curvature differences obstruct a.e orientation-preserving isometries? Is there an example of a pair $M,N$ of connected, oriented equidimensional Riemannian manifolds with the following properties:


*

*$M$ is everywhere non-flat, $N$ is flat.

*There exist a map $f:M \to N$ which is differentiable almost everywhere (a.e), and $df$ is an orientation-preserving isometry a.e.
(An easier goal: Find a pair of manifolds which are not locally isometric, but which admit a map as in 2. We should probably restrict here to manifolds without boundary, since otherwise $M=[0,1],N=\mathbb{R},f(x)=x$ is an example. )
Context:
The point is to see whether curvature differences obstruct existence of a.e orientation-preserving isometries.
If we omit the requirement on the orientation, then there is a lot of flexibility;
Gromov showed that for any metric $g$ on the unit $d$-dimensional disk $\mathbb{D}^d$ there is an a.e isometry $f:(\mathbb{D}^d,g) \to (\mathbb{R}^d,e)$. ($e$ is the Euclidean metric).

Further comments:


*

*Gromov's a.e isometry cannot be orientation-preserving: 


It's $1$-Lipschitz, and hence in $W^{1,\infty}(M,N)$, and every map $f \in W^{1,\infty}(M,N)$ satisfying $df \in \text{SO}$ a.e is  smooth. (Thus Gromov's map cannot be orientation-preserving or orientation-reversing on any open subset of the domain. It must "switch" orientations in an"infinite rate").


*An a.e orientation preserving isometry does not need to be smooth:


For an example take $M=[0,1],N=[0,2],f(x)=c(x)+x$ where $c$ is the Cantor function. Then $f'=1$ a.e. 
This example can be used to show that there is an a.e orientation-preserving isometry from a circle of radius $1$ into a circle of radius $2$. (Of course, there is no smooth local isometry from the former into the latter).
 A: There is a discontinuous map $f\colon\mathbb{S}^2\to\mathbb{R}^2$ such that $d_xf$ is defined and isometric for almost all $x$. (If you want a continuous one then I am sure the answer is "no")
To construct such $f$ do the following:


*

*Start with a sequence of finer and finer subdivision $(K_n)$ of $\mathbb{S}^2$ into polygons; say next subdivision divedes each polygon in 4 nearly equal pieces. 

*Construct a maximal tree $T_n$ of cuts in in the 1-skeleton of $K_n$, in such a way that $T_n$ is obtained from $T_{n-1}$ by adding minimal length of cuts. 

*Finally consider the development of these polygons in the plane and pass to the limit. 

There are few estimates in the construction you have to take care of, but I am sure everything should work.
A: I will add an explanation of Anton Petrunin's comment
Question : There is no continuous map $$f :B\rightarrow
\mathbb{E}^2$$ where $B$ is a geodesic ball in $S^2(1)$ of radius
$\varepsilon$ s.t. $df$ is
isometric a.e.
Proof : Assume that $f$ is a continuous map.
 Note that ${\rm length}\ f(\partial B)={\rm length}\ \partial B$. In further, $f(B)$ is
 enclosed by $f(\partial B)$. Since $f$ is a volume preserving map, by isoperimetric inequality, it is a
 contradiction.
