0PB3 Lemma 7.4.15. Let ζ,ζ′,ζ′′∈L(θ)\zeta,\zeta^{\prime},\zeta^{\prime\prime}\in L(\theta) such that ζ=ζ′∘ζ′′\zeta=\zeta^{\prime}\circ\zeta^{\prime\prime}. If ζ′(0)∈Zo\zeta^{\prime}(0)\in Z_{o} and θζ′(0)=id\theta_{\zeta^{\prime}(0)}=\operatorname{id}\nolimits, then ζ′\zeta^{\prime} and ζ′′\zeta^{\prime\prime} have opposite orientations.
0PB4 Proof. Let z=ζ′(0)=ζ′′(1)z=\zeta^{\prime}(0)=\zeta^{\prime\prime}(1). We have ζ′∈I(idz,θζ(1))\zeta^{\prime}\in I(\operatorname{id}\nolimits_{z},\theta_{\zeta(1)}). Since ζ¯′=θζ(1)∘ζ′\bar{\zeta}^{\prime}=\theta_{\zeta(1)}\circ\zeta^{\prime} is smooth and has opposite orientation to ζ′\zeta^{\prime}, it follows that ζ′(0+)∈ι(C(z)+)\zeta^{\prime}(0+)\in\iota(C(z)^{+}). Similarly, ζ′′(1−)∈ι(C(z)+)\zeta^{\prime\prime}(1-)\in\iota(C(z)^{+}). We deduce that ζ′\zeta^{\prime} and ζ′′\zeta^{\prime\prime} have opposite orientations. ∎