An isometry of the upper half plane which is a "bounded distance from the identity" must be the identity? In Farb-Margalit's book:
https://www.maths.ed.ac.uk/~v1ranick/papers/farbmarg.pdf
on page 204, in the proof of Prop 7.7, they say that if $S_g$ is a closed hyperbolic surface of genus $g$, and $\phi$ is an isometry which is isotopic to the identity, then it has a lift to an isometry of the upper half plane (with the standard hyperbolic metric) which is a "bounded distance from the identity". Then, they claim that using the classification of isometries of $\mathbb{H}$, such a lift must in fact be the identity.
What do they mean by "bounded distance"?
Why must there exist a lift which is bounded distance from the identity?
Why is the identity the only isometry which has bounded distance from the identity?
(I suspect that the right answer to the first question will in fact answer the last two, though I thought I'd ask just in case)
 A: Let $H:S\times I\rightarrow S$ be a homotopy such that $H(-,0)=Id_s$ and $H(-,1)=\phi$. Let $\pi:\mathbb H^2\rightarrow S$ be a covering map which is a local isometry. We can assume such $H$ is smooth by Whitney's approximation theorem.
Because of the covering homotopy theorem(see Theorem III.3.4 in Bredon's book) there is a unique homotopy $G:\mathbb H^2\times I\rightarrow \mathbb H^2$ making the diagram 
$\require{AMScd}$
\begin{CD}
\mathbb H^2\times I@>G>> \mathbb H^2\\
@V\pi\times\operatorname{Id}_I VV @VV\pi V\\
S\times I@>>H>S ,
\end{CD}
conmute, and $G\upharpoonright \mathbb H^2\times\{0\}=\operatorname{Id}_{\mathbb H^2}$. Then $\varphi:=G(-,1)$ is a lift of $\phi$. Let us see $\varphi$ has bounded distance to the identity, that is, $d(\varphi(x),x)<M$ for all $x\in\mathbb H^2$. 
The path $t\in I\mapsto H(\pi(x),t)$ of $S$ joins $\pi(x)$ and $\phi(x)=\pi\circ\varphi(x)$. Then $t\mapsto G(x,t)$ is a path joining $x$ and $\varphi(x)$, and thus has the same length as the first path; as $\pi$ is a local isometry. However, we have a continuous function $h:S\rightarrow[0,\infty)$ such that  for each $y\in S$, $h(y)$ is the length of the path $t\mapsto H(y,t)$. Hence as $S$ is compact, this function is bounded above by some $M\geq 0$. Thus for each $x\in\mathbb H^2$ the path $t\mapsto G(x,t)$ has length $\leq M$. Thus $d(x,\varphi(x))\leq M$ for all $x\in\mathbb H^2$.
Now let us prove $\varphi$ is the identity. We can do this by proving that the extension of $\varphi$ to the boundary of $\mathbb H^2$, $\partial \mathbb H^2$, is the identity. But if $\gamma(t)$ is a geodesic half-closed line of $\mathbb H^2$ parametrized by arc-length on $[0,\infty)$, then $\varphi\circ\gamma(t)$ has also the same properties and these two half-closed lines represent the same point of $\partial\mathbb H^2$ as $$\sup_{t\geq 0} d(\gamma(t),\varphi\circ\gamma(t))\leq M<\infty.$$
Therefore $\varphi$ is the identity.
