ScalingStacks

Proof. Note that the second assumption on $I^{\prime}$ shows that the full subquiver of $\Gamma(\theta)$ with vertex set $I^{\prime}$ is a disjoint union of oriented lines and oriented circles.

Let $i_{1}\neq i_{2}\in I$. If $i_{1},i_{2}\in I\setminus I^{\prime}$, then $u(i_{1})\neq u(i_{2})$. Assume now $i_{1}\in I^{\prime}$ and $i_{2}\in I\setminus I^{\prime}$. Since $i_{1}\to i_{2}$ is not an arrow of the quiver, we have $u(i_{1})\neq i_{2}$, hence $u(i_{1})\neq u(i_{2})$. Finally if $i_{1},i_{2}\in I^{\prime}$, then $u(i_{1})\neq u(i_{2})$. We have shown that $u$ is a braid.

Note that there is a (unique) decomposition $\theta=\theta^{u}\circ u$ with $|\theta|=|\theta^{u}|+|u|$. In order to show that $\theta^{u}\cdot u\neq 0$, we can replace $\theta$ by $\theta_{|I\setminus I_{0}}$ and $M$ by $M\setminus I_{0}$, thanks to Lemma 7.4.25. So, we assume now that $I_{0}=\emptyset$.

Let $q:\tilde{Z}\to Z$ be a non-singular cover of $Z$. Let $\tilde{M}=q^{-1}(M)$. Let $\tilde{\theta}:\tilde{I}\to\tilde{J}$ be the unique lift of $\theta$ to $\tilde{Z}$. We have a decomposition $\tilde{\theta}_{i}=\tilde{\beta}^{i}\cdot\tilde{\alpha}^{i}$ for $i\in\tilde{I}$ and $q(\tilde{\alpha}^{i})=\alpha^{q(i)}$.

Let $\tilde{I}^{\prime}=q^{-1}(I^{\prime})\cap\tilde{I}$. Note that $q$ induces a morphism of quivers $\Gamma(\tilde{\theta})\to\Gamma(\theta)$, hence $\tilde{I}^{\prime}$ satisfies the assumptions of the lemma and we have a decomposition $\tilde{\theta}=\tilde{\theta}^{\tilde{u}}\circ\tilde{u}$. Since $q(\tilde{u})=u$, it follows that if the lemma holds for $\tilde{\theta}$, then it holds for $\theta$.

We assume now that $Z$ is non-singular. If the lemma holds for connected components of $Z$, it will hold for $Z$, hence it is enough to prove the lemma for $Z$ connected. Assume now $Z$ is connected. There is an injective morphism of curves $f:Z\to S^{1}$, where $S^{1}$ is unoriented. It the lemma holds for $S^{1}$, it holds for $Z$.

We assume finally that $Z=S^{1}$ unoriented. Let $i_{1}\neq i_{2}\in I^{\prime}$ such that $i(u_{i_{1}},u_{i_{2}})\neq 0$. Note that $u_{i_{1}}$ and $u_{i_{2}}$ have opposite directions and $i(u_{i_{1}},u_{i_{2}})=1$. Furthermore, $\theta_{i_{r}}$ has the same direction as $u_{i_{r}}$, hence $i(\theta_{i_{1}},\theta_{i_{2}})=i((\theta^{u})_{i_{1}},(\theta^{u})_{i_{2}})+1$. Given $i_{1}\neq i_{2}\in I$ with $i_{1}{\not\in}I^{\prime}$, we have $i(u_{i_{1}},u_{i_{2}})=0$. It follows from Remark 7.4.11 that $\theta^{u}\cdot u\neq 0$. This completes the proof of the lemma. ∎