44

Tukia , Homeomorphic conjugates of Fuchsian groups

Proof . Let Go be the subgroup of elements geG such that g|S is orientation preserving. Let Eq{A, X\ Eçy{A\ etc. denote the essential intersection number when it is taken with respect to Gq. We will prove first that there is a G-complex X" extending X such that

( 4 ) Eo{AX")=^0,

We prove (4) by induction on Eq{A, X). If Eq{A, X) = 0, there is nothing to prove. So suppose that (4) is true for all multi-axes В and all G-complexes У such that

£о ( Д Ю<£о(Д^)>0.

Let Р(Л, X) = (Q,..., С J be as in (1) and let

Ai=^AnCi and С^ = Сс^. Since

i

there is i such that £o(^i) > 0. If £o(^i) > 0? then G^o = G^, n Go must be non-elementary and hence g|ßCj is orientation preserving for every geG^o by Lemma 5АО. By Theorem 3E, there is an essential axis В of Q such that Nß = N{Ai) and Pß is an endpoint of an axis conjugate to Ai essentially intersecting Ai (with respect to G^ol^Cf) and such that

Eo { B ) <Eo { Ai ) .

Hence , by the inductive assumption, there is a G-complex Y extending X such that Eq{B, У) = 0. This means that if С is a component of D\Y, then Bn С is either empty, a non-essential axis of С or an essential axis of С with Eq{B n C) = 0.

Let hßE Gl be such that Pß is an endpoint of hßAi. Then A and hßA intersect essentially. There is at most one component С of 0\У such that AnC and hßAnC intersect essentially. If there is no such C, then Eo{Y, A)<Eo{A, X) by Lemma 5B1 and we are done by the inductive assumption. If there is one such C, then С is a subset of Cf. Furthermore, BnC is an axis of С such that N{BnC) and N{AnC) are in the same component of dC\L{Gc) ^^d P{BnC) and one endpoint of hßAnC are in another.

Now remember that Ео{Вг\С) = 0. By Lemmas 3E and 5АО, there is an axis B' of С equivalent to ß n С and which is simple with respect to G^ n Go which has index one or two in G^. Hence B' has at most binary intersection with respect to G^ and so we can adjoin B' to У by Lemma 5A1 and obtain a G-complex Yß^ extending X.

Since B' is equivalent to BnQ B' connects the components of dC\L{Gc) containing N{AnC) and an endpoint of hßAnC. Hence, for no component С of D\Yß, do AnC and hßAnC intersect essentially. Thus E{A, Yß.)<E{A, X) by Lemma 5B1. By the inductive assumption we can now find an extension X" of X for which (4) is true.