Prove that any map from a surface to another one preserves at least one angle Tissot proved that for any kind of representation of a surface onto another, there exists, at every point of the first surface, two perpendicular tangents, which are unique unless the angles are preserved at that point (i.e. the map is conformal), such that the images of these two tangents are perpendicular on the second surface.
I can't find the paper; and don't really know if this is easy to prove.
The only idea I had so far is picking two perpendicular tangents of the first surface and rotate them and try using something like Bolzano, but it hasn't worked so far and I don't think it will.
Does anyone have a reference/knows the general idea of how to prove that? I would really appreciate that. Thanks!
 A: I'll do it for a linear transformation $T : V \rightarrow W$. Consider the matrix $R_\theta$ that rotates a vector with a counterclockwise angles $\theta$. Then, if $\{e_1,e_2\}$ is an orthonormal basis, $A_\theta = \{R_\theta\ e_1, R_\theta\ e_2\}$ is too, $\forall\ \theta$. 
Let's define $F: [0,2\pi] \rightarrow \mathbb{R}$ by $F(\theta)=<Tv,Tw>$
If $F(\theta) = 0$ for any $\theta$, I won. Suppose $F(0) > 0$, then, by linearity, $F(\pi/2) = - F(0) < 0$. Then, by Bolzano, I won.
[Edit] For the uniqueness part:
Suppose $\{v,w\}, \{x,y\}$ both satisfy what's said in the statement. (of norm=1). Let's prove the transformation is conformal, then.
Let $a,b$ be some vectors. We can write $a=a_1 v + a_2 w$, and $b = b_1 v + b_2 w$.
$<a,b> = a_1 b_1 + a_2 b_2$
$<Ta,Tb> = a_1 b_1 ||Tv||^2+ a_2 b_2 ||T(w)||^2$
Let's see, then, that $||Tv||^2 = ||Tw||^2$.
We can write x,y in function of v,w:
$x=x_1 v + x_2 w$, and $y = y_1 v + y_2 w$
As $<T(v),T(w)> = 0$, we get that $x_1 y_1 = - \frac{x_2 y_2 ||T(w)||^2}{||T(v)||^2}$ (1)
Then, $||Tv||^2 = ||Tw||^2$ iff $x_1 y_1 = -x_2 y_2$, which happens because $x,y$ are orthonormal. (I'm using here that $\{v,w\}, \{x,y\}$ are not the same pair of vectors and hence (1) is not the equation 0 = 0).
The end!
A: You need the Polar decomposition for a linear map. So a (real) linear map is always the composition of a rotation and a symmetric map (thus is diagonalizable). Since a rotation does not alter angles, so we need only to consider the symmetric map. Then we can find an orthonormal basis of the space, with respect to which the map is only a stretch on the base vectors.
