Lemma 7.4.35 (0PC1) · ScalingStacks

0PC1

Lemma 7.4.35. Consider braids $\theta^{\prime\prime}:I\to J$ and $\theta^{\prime}:J\to K$ and assume $\theta=\theta^{\prime}\cdot\theta^{\prime\prime}$ is non-zero. Let $\zeta\in D(\theta)\setminus(D(\theta)\cap D(\theta^{\prime\prime}))$ . Assume $\zeta$ and $\zeta^{-1}$ are oriented.

Define $\alpha^{\prime\prime}:I\to J$ by

\alpha^{\prime\prime}_{s}=\begin{cases}\theta^{\prime\prime}_{\zeta(1)}\circ\zeta&\text{ if }s=\zeta(0)\\ \theta^{\prime\prime}_{\zeta(0)}\circ\zeta^{-1}&\text{ if }s=\zeta(1)\\ \theta^{\prime\prime}_{s}&\text{ otherwise.}\end{cases}

Let $\zeta^{\prime}=\theta^{\prime\prime}_{\zeta(1)}\circ\zeta\circ(\theta^{\prime\prime}_{\zeta(0)})^{-1}$ and $\alpha^{\prime}=(\theta^{\prime})^{\zeta^{\prime}}$ . Then $\alpha^{\prime\prime}$ and $\alpha^{\prime}$ are braids and $\theta=\alpha^{\prime}\cdot\alpha^{\prime\prime}$ .

0PC2

Proof. Since $\zeta$ and $\zeta^{-1}$ are oriented, it follows that $\alpha^{\prime\prime}_{s}$ is oriented for all $s$ . Also, it follows from Lemma 7.4.31 that $\alpha^{\prime}$ is a braid.

Consider first the case where $Z=S^{1}$ unoriented. In that case, the lemma follows from Proposition 7.4.33 and Lemmas 7.4.20 and 6.2.10.

Assume now $Z$ is smooth and connected. There is an injective morphism of curves $f:Z\to S^{1}$ , where $S^{1}$ is unoriented. Since the lemma holds for $S^{1}$ , we deduce that it holds for $Z$ .

When $Z$ is only assumed to be smooth, the lemma follows from the case of the connected component containing $\zeta$ .

Consider now the general case. Let $f:\tilde{Z}\to Z$ be a smooth cover. Let $\tilde{\theta}$ be a braid lifting $\theta$ . There are unique braids $\tilde{\theta}^{\prime}$ and $\tilde{\theta}^{\prime\prime}$ in $\tilde{Z}$ with $\tilde{\theta}=\tilde{\theta}^{\prime}\cdot\tilde{\theta}^{\prime\prime}$ and $f(\tilde{\theta}^{\prime})=\theta^{\prime}$ , $f(\tilde{\theta}^{\prime\prime})=\theta^{\prime\prime}$ . There is a unique $\tilde{\zeta}\in D(\tilde{\theta})$ with $f(\tilde{\zeta})=\zeta$ (Lemma 7.4.28). We have $\tilde{\zeta}{\not\in}D(\tilde{\theta}^{\prime\prime})$ (Lemma 7.4.28). Since the lemma holds for $\tilde{Z}$ , we deduce it holds for $Z$ . ∎