The crucial step in the proof for purposes of this discussion (carried out in the lemmas on pages 38 and 39) is to consider the situation that $\delta$ is an ideal triangulation, $h$ is an oriented ideal arc whose interior intersects $\delta$ transversely and efficiently ($h$ does not double back across the same arc in a triangle of $\delta$), $x_0$ is the first point where $h$ crosses $\delta$ transversely, and $\alpha$ is the edge of $\delta$ containing $x_0$. In this situation one wants to prove that $\alpha$ belongs to two distinct triangles of $\delta$. \delta$and so can be flipped; this is basically the inductive step for proving connectivity. The proof of this step is to consider the possibility that$\alpha$belongs to only a single triangle$T$of$\delta$--- meaning that$\alpha$is the ideal arc obtained by identifying 2 sides of$T$--- and to derive a contradiction. In the article this step is carried out only in the orientable category, where the result of gluing the 2 sides of$T$must be a disc (depicted in the diagrams of that paper as a puncture piece''). However, the conclusion of this argument remains true in the nonorientable category, and the proof requires just one more case to be considered, namely, when the result of gluing the 2 sides of$T$is a Mobius band. In this situation, the efficient intersection condition would imply that$h$is trapped in the interior of the Mobius band, winding more and more closely around its core and crossing$\alpha$infinitely often, contradicting that the number of intersections must be finite. With this consideration, the whole proof should go through otherwise unscathed; orientability is not otherwise used. 1 My article "Tiling the measured foliation space of a punctured surface", Trans. Math. 306 no. 1 (1988) contains a proof of this fact in the case of oriented surfaces. It is essentially the same as Hatcher's proof of contractibility, but focussing solely on the issue of connectivity, which introduces some simplifications. The proof works, with a little bit more effort, in the nonorientable case as well. The crucial step in the proof for purposes of this discussion (carried out in the lemmas on pages 38 and 39) is to consider the situation that$\delta$is an ideal triangulation,$h$is an oriented ideal arc whose interior intersects$\delta$transversely and efficiently ($h$does not double back across the same arc in a triangle of$\delta$),$x_0$is the first point where$h$crosses$\delta$transversely, and$\alpha$is the edge of$\delta$containing$x_0$. In this situation one wants to prove that$\alpha$belongs to two distinct triangles of$\delta$. The proof of this step is to consider the possibility that$\alpha$belongs to only a single triangle$T$of$\delta$--- meaning that$\alpha$is the ideal arc obtained by identifying 2 sides of$T$--- and to derive a contradiction. In the article this step is carried out only in the orientable category, where the result of gluing the 2 sides of$T$must be a disc (depicted in the diagrams of that paper as a puncture piece''). However, the conclusion of this argument remains true in the nonorientable category, and the proof requires just one more case to be considered, namely, when the result of gluing the 2 sides of$T$is a Mobius band. In this situation, the efficient intersection condition would imply that$h$is trapped in the interior of the Mobius band, winding more and more closely around its core and crossing$\alpha\$ infinitely often, contradicting that the number of intersections must be finite.