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Consider a figure-eight geodesic $\delta $( geodesic with exactly one self-intersection point at p ) on a pair of pants Y with three geodesic boundaries $ \gamma_i$ and three perpendiculars between them $ \alpha_i $. Assume $\delta= \delta_1 $ followed by $\delta_2 $, where $\delta_i $ are freely homotopic to the geodesic boundaries $\gamma_i$. $\delta_1, \delta_2 $ intersect at p.I have two ( basic ) questions :

  1. Why does p always lie on $\alpha_3 $?
  2. Why does $\delta $ intersect $\alpha_i $ orthogonally ?

May be we have to consider the self-isometry of Y which "exchanges" the two right-angled hexagons and has the fixed-point set $ \alpha_1 \cup \alpha_2 \cup \alpha_3 $ ?

Thank you !

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You are absolutely right, your method does the trick.

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