TITLE: Flips of triangulations on non-orientable surfaces
QUESTION [11 upvotes]: Let $N_{k,r}$ be a non-orientable surface of genus $k$ (i.e the connected sum of $k$ projective planes) and with $p\geq 1$ punctures. 
I'm looking at ideal triangulations of the surface, namely maximal collection of pairwise disjoint ideal arcs (joining punctures) on the surface. 
Given a triangulation and an arc $\alpha$ in that triangulation, we can perform a flip along $\alpha$ : the arc belongs to two triangles forming a quadrilateral, the flip along $\alpha$ gives the new triangulation where the arc has been replaced by the other diagonal of the quadrilateral, and all the other arcs remains unchanged. (When the arc belongs to only one triangle, we cannot do the flip.)
A classical result in the orientable case states that given any two triangulations of an orientable surface, there is a sequence of flips that transform one into the other (Hatcher, On triangulations of surfaces). So the so-called triangulation graph, whose vertices are triangulations and edges are flips, is connected.
I am looking for the similar result for non-orientable surfaces. Does anyone have a reference ? My guess is that the result should also hold but I did not find anything on the litterature.

REPLY [10 votes]: 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 $\delta$ 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 unscathed; orientability is not otherwise used.