map to the sphere that preserves measure Let $A$ be any Borel set in $\mathcal{B}([0,10]\times[0,10])$. I want to have a map $\phi$ from $\mathbb{R}^2$ to the unit sphere $S^2\subset \mathbb{R}^3$ such that the spherical measure $\sigma(\phi(A))$ can be estimated from below and above by a constant times the Lebesgue measure |A| of $A$. The constant shall, of course, not depend on $A$. Is such a map known?
 A: As $|A|$ is unbounded, the lower constant must be $0$, whatever the transformation (unless you allow non-injective mappings).

A lifting map
$$z=\sqrt{1-x^2-y^2}$$ will map a centered square of size $\dfrac1{\sqrt2}$ to a unit half-sphere with a coefficient between $1$ and twice the area of the projected patch.

In fact you can make the square as small as you want and map with a coefficient $\sim1$. IMO the question is incomplete.
A: Letting $J_\phi$ denote the absolute value of the Jacobian determinant of $\phi\colon \mathbb R^2\to S^2$, the change of variable formula for integrals reads 
$$
\int_{\phi(A)} \, d\sigma = \int_A J_\phi\, dx.$$ 
Thus, you are asking whether there is a map $\phi$ and a constant $C>0$ such that 
$$
\frac{1}{C}\le J_\phi(x) \le C.$$ 
If you take the inverse of the stereographic projection, that is 
$$
\phi(x, y)=\left( \frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2}, \frac{1-x^2-y^2}{1+x^2+y^2}\right), $$ 
then the Jacobian is 
$$
J_\phi(x, y)=\left(\frac{2}{1+x^2+y^2}\right)^2, $$ 
and if you restrict yourself to any bounded domain of $\mathbb R^2$ it is bounded both above and below.
A: Consider as $\varphi$ the inverse of the stereographic projection.
A: If you take the inverse of any equal-area planar map of a spherical Earth model with the map scaled and translated so that $[0,10]\times[0,10]$ is inside the map, the spherical measure $\sigma(\phi(A))$ will be exactly a fixed constant times $\lvert A\rvert$ for any $A$.
The constants for the upper and lower bound then are the same.
Numerous equal-area projections are known.
