Say I have two closed hyperbolic surfaces $X,Y$ and a smooth, $(1+\epsilon)$-bilipschitz map $f : X \to Y$ for some small $\epsilon$. Pick a simple closed curve $c \subset X$, and let $X',Y'$ be the results of `length t conformal grafting' along $c$ and $f(c)$, respectively. (Namely, we cut $X$ along $c$, patch in a Euclidean annulus of length $t$, and then uniformize to get another hyperbolic surface, and we do something similar on $Y$ to get $Y'$.) Then $X',Y'$ are also metrically close, since you can use $f$ to construct a quasiconformal map from one to the other that has small dilatation, and nearly conformal hyperbolic surfaces are nearly isometric. (Essentially since you can take limits and use that conformal maps $D \to D$ are hyperbolic isometries.)
So what if I now have infinite type surfaces? Let's say $(X,x),(Y,y)$ are two pointed hyperbolic surfaces (possibly infinite type) that are close, meaning that for some large $R$, there are subsets $A\subset X$ and $B\subset Y$ containing the $R$-balls around $x$ and $y$, and a $(1+\epsilon)$-bilipschitz map $f : A \to B$ with $f(x)=y$, for some small $\epsilon$. Say I do length $t$ conformal grafting on some curve $c$ that's close to $x$, and on its $f$-image in $Y$. Are the resulting surfaces $X',Y'$ close, in the sense that large hyperbolic neighborhoods around their basepoints are close? You can construct quasiconformal maps between $A \cup \text{ (annulus)}$ and $B \cup \text{(annulus)}$ using $f$, just like before, but the problem I have is saying that the these sets contain large radius hyperbolic balls around the basepoints after uniformization.
Any thoughts on this from the grafting gurus? Thanks in advance.
