identical. Thus there is an isometric bijection $\varphi:S'_n\rightarrow S^+_{n+1} $ which lifts to an isometry $\tilde{\varphi}:\tilde{S}'_n\rightarrow
\tilde{S}^+_{n+1}$. Since $\tilde{S}'_n$ and $\tilde{S}^+_{n+1}$ are subsets of hyperbolic space, $\tilde{\varphi}$ must be the restriction of a Möbius transformation and $\Gamma_n=\tilde{\varphi}^{-1}G^+_{n+1}\tilde{\varphi}.$ It follows also that $\Gamma_{n+1}=\tilde{\varphi}^{-1}G_{n+1}\tilde{\varphi} $ is a Fuchsian group representing $S_{n+1} $ and $\Gamma_{n+1}\supset\Gamma_n.$

Note that $\varphi$ maps the geodesic $\alpha_0$ on $S'_n$ to its counterpart on $S^+_{n+1} $ and identifies the corresponding copies of $A^-$. Thus $\tilde{\varphi}$ will map $I$ to a geodesic in $\tilde{S}^+_{n+1}$ that covers $\alpha_0$ and takes the left half-plane $H^-$ in ${\Bbb{H}}^2$ to a corresponding hyperbolic half-plane in $\tilde{S}^+_{n+1}$ that covers $A^-.$ It follows that for all integers $n\geq 0$, $H^-$ is precisely invariant in $\Gamma_n$ under $\Gamma_0$; that is, $h(H^-)\cap H^- \not =\emptyset$ if and only if $h\in \Gamma_0.$ Thus, $H^-$ projects to $A^-$ and $I$ projects to $\alpha_0$ on each $S_n.$

We are now in a position to define the limiting surface $S_{\infty}.$ First, define $\d{\Gamma_{\infty}=\cup_{n=0}^{l}\Gamma_n}$, where $2< l\leq\infty$ is the number of components in the complement of $C$ on $\alpha_0.$ None of the $\Gamma_n$ contain elliptic elements, and therefore the same must be true of $\Gamma_{\infty}$. Since, for $n> 0$ $\Gamma_n$ is non-elementary, it 