<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Consider the map $\varphi: {\bf D}\rightarrow {\bf D}$ given by
$\varphi(z) = z^2$.  Calculate the pullback of the hyperbolic
element of arc-length from ${\bf D}$ using $\varphi$.  (That is,
define a new element of arc-length on ${\bf D}$ by setting ${\rm
length}(f) = {\rm length}_{\bf D}(\varphi\circ f)$ for a path $f:
[a,b]\rightarrow {\bf D}$.)  Is the pullback of the hyperbolic
element of arc-length by $\varphi$ the hyperbolic element of
arc-length on ${\bf D}$?
\medskip

{\bf Answer}

The pull back of the standard hyperbolic metric on ${\bf{D}}$:

$$\phi:{\bf{D}} \longrightarrow {\bf{D}}\ \ \ f:[a,b]
\longrightarrow {\bf{D}}\ {\rm{a\ path}}$$

\begin{eqnarray*}
{\rm{length}}(f) &amp; = &amp; {\rm{length}}_{\bf{D}}(\phi \circ f)\\ &amp; =
&amp; \ds\int_{\phi \circ f} \ds\frac{1}{1-|z|^2} |dz|\\ &amp; = &amp;
\ds\int_a^b \ds\frac{2}{1-|\phi(f(t))|^2}|\phi'(f(t))||f'(t)|
\,dt\\ &amp; = &amp; \ds\int_f \ds\frac{2}{1-|\phi(z)|^2}|\phi'(z)||\,dz|
\end{eqnarray*}

\newpage
\un{with $\phi(z)=z^2$}:

\begin{eqnarray*} \ds\frac{2}{1-|\phi(z)|^2} &amp; = &amp;
\ds\frac{2}{1-|z|^4}2|z|\\ &amp; = &amp; \ds\frac{4|z|}{1-|z|^4}
\end{eqnarray*}

Since $\ds\frac{2}{1-|z|^2}=\ds\frac{4|z|}{1-|z|^4}$, we do not
recover the hyperbolic metric on ${\bf{D}}$ via the pull back.

\end{document}
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