proof that the schwarz map defined as ratios of gauss hypergeometric functions is univalent The ratio of two linearly independent solutions of the Guass hypergeometric differential equation defines a map from the upper half plane to a Schwarz triangle. Everything I read tells me that this map is injective, but I cannot find a proof. Is there a simple proof?
Also, can we similarly prove a similar result for the map 
$$\sigma:(x,y) \to (\frac{G(x,y)}{F(x,y)},\frac{H(x,y)}{F(x,y)})$$
where $F$, $G$, and $H$ are solutions of Appell's $F_1$ system?
 A: "Everything you read" has it wrong: this map is not necessarily injective.
For example, there exists a Schwarz equation whose solution is a ratio of two
solutions of a Gauss equation, and the solution is $z^{10}$, which is not
injective in the upper half-plane. The "triangle" in question has angles
$10\pi,\pi,10\pi$. It is indeed a "triangle" in some sense, and you can even make
a paper model. But it does not fit into the plane.
The correct result is the following. Suppose that you have an (honest) circular triangle
in the plane. That is the sides are arcs of circles, and the angles are at most $2\pi$. 
Then (by the Riemann mapping theorem) there exists a conformal (injective) map of the upper
half-plane onto this triangle. This map satisfies a Schwarz equation. 
All other solutions
of this Schwarz equation also map the upper half-plane injectively onto triangles (obtained by a fractional-linear transformation of the original one).
But not all Schwarz equations occur in this way. Parameters of the Schwarz equations
have to satisfy certain (simple) conditions for the solution to be injective in the upper half-plane.
There is one Complex Analysis textbook (in English) where this question is treated in great detail: C. Caratheodory, vol. II. For those who read German or Russian, also Hurwitz-Courant.
