Fix a pair of pants in the plane (i.e. a closed disk minus two smaller open disks). Name the boundary components $\alpha\_1,\alpha\_2$ and $\alpha\_3$ and fix three positive lengths $\ell\_1,\ell\_2$ and $\ell\_3$.


Let $M$ be the set of smooth Riemannian metrics on the pair of pants which are hyperbolic and such that all three boundary components are geodesics with lengths $\ell\_1,\ell\_2$ and $\ell\_3$ respectively ($M$ is endowed with the $C^\infty$ topology). 


There is a unique conformal metric $g\_0 \in M$ and one knows that given $g \in M$ there is an isotopy from the identity $f\_t, t \in [0,1]$ such that $f\_t(\alpha\_i) \subset \alpha\_i$ for $i =1,2,3$, $f\_0$ is the identity map, and the pushforward metric $f\_1^\*g$ is equal to $g\_0$.


**My question:** Given a continuous one parameter family of metrics $g\_s \in M$ can one choose isotopies $f\_{s,t}$ taking each one to $g\_0$ in a continuous way?


**Edit:** I was wrong. Once the pair of pants is fixed there is a unique choice of $\ell\_1,\ell\_2$ and $\ell\_3$ such that there is a conformal hyperbolic metric with boundary geodesics of those lengths. For other values of the $\ell\_i$ the solutions to the Beltrami equation in Misha's answer do not (and cannot) leave the given pair of pants invariant.



Continuous choice of isotopies between hyperbolic metrics on the pair of pants
First, the metric $g= Edx^2 + 2Fdxdy + Gdy^2$ defines the Beltrami differential
$$
\mu =\frac{E − G + 2iF}{E + G + 2\sqrt{EG − F^2}}
$$
For every $\mu$ one has the canonical family of Beltrami differentials
$t\mu$, $0\le t\le 1$. If $g=g\_s$ depends smoothly on another parameter $s$, we also get$t\mu\_s$. Solution of the corresponding Beltrami equation depends smoothly on 
the pair of parameters $(s,t)$. 


