Schwarz-Pick lemma I have a strange problem, in a book from which I need to get prepared to give a lecture, it is written lemma, with no proof:
Let F be an analytic function from one simply-connected domain E to another, let us say G. Then $\lambda_E(x,y) \geq \lambda_G(F(x),F(y))$ where $\lambda(x,y)$ is the hyperbolic metric,and if G is a simply-connected region with at least two boudary points, then $\lambda_G(x,y)=\lambda(\phi(x),\phi(y))$ and where $\phi$ is a conformal isomorphism from G to unit disc, which exists due to Riemann theorem.
And it is named Shwarz-Pick lemma. When I looked for the lemma , under that name there is just some lemma that includes derivatives, not this. The problem is that $\geq$ which exists in the lemma, because, if F is a conformal isomorphism, we could easily decompose it into $\phi_1^{-1} \circ f \circ \phi$ and $\phi_1, \phi$ being respectively isomorphisms. But, when you write everything down, definition and property that $f$ doesn't change $\lambda$, because it is a Moebius mapping, you get EQUALITY. But inequality I don't see. Please help.
 A: The book is good.
Bibliography:
Rivard.Un lemme de Schwarz-Pick à points multiples.
Avkhadiev & Wirths.Schwarz-Pick type inequalities.
Goloff & To.Holomorphic mappings, the Schwarz-Pick lemma, and curvature.
Amar & Amar.Sur les théorèmes de Schwarz-Pick et Nevanlinna dans Cn.
Yamashita.The Pick version of the Schwarz lemma and comparison of the Poincaré densities.
Anderson & Vasil'ev.Lower Schwarz-Pick estimates and angular derivatives.
Kaptanoglu.Some refined Schwarz-Pick lemmas.
Beardon.The Schwarz-Pick lemma for derivatives.
Yous should read the Avkhadiev & Wirths.Schwarz-Pick type inequalities book (BirkHauser, frontier in mathematics) for the link between the schwarz pick lemma and the Poincaré metrics.
A: You can conjugate your $F$ with a function $f:\>D\to D$ mapping the unit disk to itself. The statement is then equivalent with
$$\lambda\bigl(f(x),f(y)\bigr)\leq \lambda(x,y)\qquad(x,y\in D)\ .$$
For the proof we may assume $x=f(x)=0$; but then the statement is equivalent with the usual Schwarz' Lemma.
