ScalingStacks

Proof. Assume first $f$ is a non-singular cover of $Z^{\prime}$.

Let $\zeta^{\prime}\in L(\theta^{\prime})$. There are $i^{\prime}_{1}\neq i^{\prime}_{2}\in f(I)$ such that $\zeta^{\prime}\in I(\theta^{\prime}_{i^{\prime}_{i}},\theta^{\prime}_{i^{\prime}_{2}})$. By Lemma 7.3.24, there are elements $\zeta_{r}\in f^{-1}(\theta^{\prime}_{i_{r}})$ and $\zeta\in I(\zeta_{1},\zeta_{2})$ such $\zeta^{\prime}=f(\zeta)$. We define $\theta\in f^{-1}(\theta^{\prime})$ by setting $\theta_{\zeta_{r}(0)}=\zeta_{r}$ and by setting $\theta_{i}$ to be any lift of $\theta^{\prime}_{f(i)}$ for all $f(i)\notin\{i^{\prime}_{1},i^{\prime}_{2}\}$. This shows the surjectivity part of the first statement of the lemma.

Consider now $\theta$ and $\hat{\theta}$ maps in ${\mathcal{P}}_{f}^{\bullet}(Z)$ such that $f(\theta)=f(\hat{\theta})=\theta^{\prime}$. Let $\zeta\in L(\theta)$ and $\hat{\zeta}\in L(\hat{\theta})$ such that $f(\zeta)=f(\hat{\zeta})=\zeta^{\prime}$. There are $i^{\prime}_{1}\neq i^{\prime}_{2}\in f(I)$ such that $\zeta^{\prime}\in I(\theta^{\prime}_{i^{\prime}_{1}},\theta^{\prime}_{i^{\prime}_{2}})$. We have $\hat{\theta}_{\hat{\zeta}(t)},\theta_{\zeta(t)}\in f^{-1}(\theta^{\prime}_{i^{\prime}_{r}})$ for $t\in\{0,1\}$. It follows from Lemma 7.3.24 that $\zeta=\hat{\zeta}$. So, the first statement of the lemma holds.

Assume now $f$ is an open embedding. The injectivity of the first map of the lemma is clear, while the surjectivity follows from Lemma 7.3.25.

We deduce the first part of the lemma for $Z$ and $Z^{\prime}$ non-singular and the general case follows now by taking non-singular covers of $Z$ and $Z^{\prime}$ and the lift of $f$.

Let us prove now the second statement of the lemma about $D(\theta)$.

Consider $\theta\in f^{-1}(\theta^{\prime})$ and $\zeta\in L(\theta)$. It is clear that if $f(\zeta)\in D(\theta^{\prime})$, then $\zeta\in D(\theta)$.

Assume now $\zeta\in D(\theta)$. Fix $i_{1}\neq i_{2}\in I$ so that $\zeta\in I(\theta_{i_{1}},\theta_{i_{2}})$.

Let $\zeta^{\prime},\zeta^{\prime\prime}\in L(\theta^{\prime})$ such that $f(\zeta)=\zeta^{\prime}\circ\zeta^{\prime\prime}$. Let $z=\zeta^{\prime}(0)=\zeta^{\prime\prime}(1)$. If $|f^{-1}(\theta^{\prime}_{z})|>1$, then $z\in Z_{o}$ and $\theta^{\prime}_{z}=\operatorname{id}\nolimits$, hence $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations by Lemma 7.4.15. Assume now $\theta^{\prime}_{z}$ has a unique lift. Let $\hat{\zeta}^{\prime}$ and $\hat{\zeta}^{\prime\prime}$ be the unique lifts of $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ (first part of the lemma). By unicity of lifts, we have $\zeta=\hat{\zeta}^{\prime}\circ\hat{\zeta}^{\prime\prime}$. We have $\hat{\zeta}^{\prime},\hat{\zeta}^{\prime\prime}\in L(\theta)$, hence $\hat{\zeta}^{\prime}$ and $\hat{\zeta}^{\prime\prime}$ have opposite orientations. It follows that $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations as well.

Consider now $\zeta^{\prime}:f(i_{1})\to f(i_{2})$ a smooth homotopy class of paths such that $f(\zeta)\circ\zeta^{\prime-1}$ and $\zeta^{\prime-1}\circ f(\zeta)$ are smooth and have the same orientation as $f(\zeta)$ and $\zeta^{\prime}$. Let $\hat{\zeta}^{\prime}$ be the unique lift of $\zeta^{\prime}$. Since $f(\zeta)\circ\zeta^{\prime-1}$ is smooth, it follows that $\hat{\zeta}^{\prime}(0)=i_{1}$ and $\zeta\circ\hat{\zeta}^{\prime-1}$ is smooth and has the same orientation as $\zeta$. Similarly, $\hat{\zeta}^{\prime}(1)=i_{2}$ and $\hat{\zeta}^{\prime-1}\circ\zeta$ is smooth and has the same orientation as $\zeta$. A similar statement holds for $\zeta$ replaced by $\bar{\zeta}$. We deduce that $f(\zeta)\in D(\theta^{\prime})$. ∎

Proof. The statement is true for $Z=S^{1}$ unoriented by Lemmas 3.2.3, 7.4.19 and 7.4.20. It follows from Lemmas 7.4.28 and 7.3.22 that it holds for any connected non-singular $Z$, by embedding it in $S^{1}$. So, the lemma holds for any non-singular $Z$. By realizing an arbitrary $Z$ as a quotient of its non-singular cover, we deduce from Lemmas 7.4.28 and 7.3.22 that the lemma holds for any $Z$. ∎

Proof. Let us show the first statement. We can assume $\theta_{\zeta(0)}^{\zeta}\neq\operatorname{id}\nolimits$.

Assume $\theta_{\zeta(0)}^{\zeta}$ has the same orientation as $\zeta^{-1}$. We have $\theta_{\zeta(1)}=\theta_{\zeta(0)}^{\zeta}\circ\zeta^{-1}$. If $\gamma$ and $\gamma^{\prime}$ are minimal paths in $\theta_{\zeta(0)}^{\zeta}$ and $\zeta^{-1}$, then $\gamma\circ\gamma^{\prime}$ is a minimal path in $\theta_{\zeta(1)}$. Since $\gamma\circ\gamma^{\prime}$ is admissible, it follows that $\gamma$ is admissible, hence $\theta_{\zeta(0)}^{\zeta}$ is admissible.

Otherwise, $\theta_{\zeta(0)}^{\zeta}$ has the same orientation as $\bar{\zeta}^{-1}$ and $\theta_{\zeta(0)}=\bar{\zeta}^{-1}\circ\theta_{\zeta(0)}^{\zeta}$, hence we deduce as above that $\theta_{\zeta(0)}^{\zeta}$ is admissible.

Similarly, $\theta_{\zeta(1)}^{\zeta}$ is admissible and we deduce that $\theta^{\zeta}$ is a braid.

Let us prove the second part of the lemma. When $Z=S^{1}$ unoriented, this holds by Lemmas 7.4.20, 6.2.9 and 7.4.19 and Proposition 7.4.18. We deduce that the lemma holds when $Z$ is a connected non-singular curve, by embedding $Z$ in $S^{1}$. So, it holds when $Z$ is a non-singular curve (since $\operatorname{supp}\nolimits(\zeta)$ is contained in a connected component of $Z$).

Consider now a general $Z$ and the non-singular cover $q:\hat{Z}\to Z$. There is a braid $\hat{\theta}$ in $\hat{Z}$ with $q(\hat{\theta})=\theta$ (Lemma 7.4.5) and there is $\hat{\zeta}\in L(\hat{\theta})$ such that $\zeta=q(\hat{\zeta})$ (Lemma 7.4.28). The considerations above show that $\hat{\theta}^{\hat{\zeta}}$ is a braid in $\hat{Z}$, hence $\theta^{\zeta}=q(\hat{\theta}^{\hat{\zeta}})$ is a braid in $Z$. The statement on degrees follows from Lemmas 7.4.28 and 7.4.12. ∎

Proof. Lemma 7.4.28 shows that $d(f^{\#}(\theta^{\prime}))=f^{\#}(d(\theta^{\prime}))$ for any $\theta^{\prime}$ and that $d(f(\theta))=f(d(\theta))$ if $|f^{-1}f(\theta)|=1$.

Assume $Z=S^{1}$ (unoriented) and consider a finite subset $M$ of $Z$ as in §7.4.3. We use the notations of that section. It follows from Lemma 7.4.19 that the isomorphism $F$ of Proposition 7.4.18 induces an isomorphism of ${\mathbf{F}}_{2}$-linear categories $F:{\mathbf{F}}_{2}[{\mathcal{H}}_{n}]\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}_{M}(Z)$. It follows now from Lemma 7.4.20 that this isomorphism commutes with $d$. In particular, $d$ is a differential on ${\mathcal{S}}_{M}(Z)$. Since this holds for any finite subset $M$ of $Z$, we deduce that $d$ is a differential on ${\mathcal{S}}(Z)$.

Consider now a non-singular connected $Z$ and an injective morphism of curves $f:Z\hookrightarrow S^{1}$. Since $f$ induces a faithful ${\mathbf{F}}_{2}$-linear functor ${\mathcal{S}}(Z)\to{\mathcal{S}}(S^{1})$ commuting with $d$, we deduce that $d$ is a differential on ${\mathcal{S}}(Z)$.

The decomposition (7.4.3) is compatible with $d$, hence $d$ is a differential on ${\mathcal{S}}(Z)$ for any non-singular $Z$.

Consider now a general $Z$ and $q:\hat{Z}\to Z$ its non-singular cover. Since the additive ${\mathbf{F}}_{2}$-linear functor $q^{\#}$ commutes with $d$, it follows that $d$ is a differential on ${\mathcal{S}}(Z)$.

The last statement of the theorem follows from Lemma 7.3.17. ∎