ScalingStacks

Proof. By Remark 8.1.1, we can assume $\xi$ is outgoing for $Z$. We will show that

(8.1.2)

the isomorphism of Lemma 8.1.2 is compatible with the differentials.

The proposition will follow immediately from (8.1.2).

Let $S^{\prime\prime}$ be a subset of $S$ with $n$ elements, and let $T$ be a finite subset of $M$. Let $a:\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S^{\prime\prime},\{\xi(1),\ldots\xi(n)\})\wedge\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S\setminus S^{\prime\prime},T)\to L^{\bullet}(T,S,e^{n})$ be the map of Lemma 8.1.2. Let $\alpha\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S^{\prime\prime},\{\xi(1),\ldots\xi(n)\})$ and $\beta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S\setminus S^{\prime\prime},T)$. Let $\theta=\alpha\boxtimes\beta=a(\alpha\wedge\beta)$. The statement (8.1.2) will follow from the following property:

(8.1.3)

$$a(d(\alpha\otimes\beta))=d(\theta).$$

We have

$$a(d(\alpha\otimes\beta))=d(\alpha)\boxtimes\beta+\alpha\boxtimes d(\beta)+\sum_{\zeta}a((\operatorname{id}\nolimits\otimes\beta)\cdot g_{S^{\prime\prime},\zeta}(\alpha))$$

where $\zeta$ runs over positive admissible homotopy classes of paths starting in $S^{\prime\prime}$ and ending in $S\setminus S^{\prime\prime}$.

We have

$$D(\theta)/\mathrm{inv}=\bigl(D(\alpha)/\mathrm{inv}\bigr)\sqcup\bigl(D(\beta)/\mathrm{inv}\bigr)\sqcup\coprod_{(s_{1},s_{2})\in S^{\prime\prime}\times(S\setminus S^{\prime\prime})}I(\alpha_{s_{1}},\beta_{s_{2}})\cap D(\theta).$$

Fix $(s_{1},s_{2})\in S^{\prime\prime}\times(S\setminus S^{\prime\prime})$. Let $S^{\prime}=(S^{\prime\prime}\setminus\{s_{1}\})\sqcup\{s_{2}\}$. Let $\zeta$ be a smooth path $s_{1}\to s_{2}$. Let $u^{\prime}=\operatorname{id}\nolimits_{S\setminus(S^{\prime}\sqcup\{s_{1}\})}\boxtimes\zeta$. Write $g_{S^{\prime\prime},\zeta}(\alpha)=v\wedge u$ with $u:S\setminus S^{\prime}\to S\setminus S^{\prime\prime}$ and $v:S^{\prime}\to\{\xi(1),\ldots,\xi(n)\}$. We take $u=0$ and $v=0$ if $g_{S^{\prime\prime},\zeta}(\alpha)=0$. If $g_{S^{\prime\prime},\zeta}(\alpha)\neq 0$, then $u=u^{\prime}$.

Assume $(\beta\cdot u)\neq 0$. Then $\alpha_{s_{1}}\circ\zeta^{-1}$ and $\beta_{s_{2}}\circ\zeta$ are smooth, and $\zeta$ and $\bar{\zeta}$ have opposite orientations, since $\bar{\zeta}$ is negative (it starts in $\xi({\mathbf{Z}}_{\geq 1})$ and ends in $M$). It follows that $\zeta\in L(\theta)$.

Assume $\zeta\in L(\theta)$. Since $\bar{\zeta}$ is negative, it follows that $\zeta$ is positive, then $(\theta^{\zeta})_{s}=\theta_{s}$ for $s{\not\in}\{s_{1},s_{2}\}$, while $(\theta^{\zeta})_{s_{1}}=\beta_{s_{2}}\circ\zeta$ and $(\theta^{\zeta})_{s_{2}}=\alpha_{s_{1}}\circ\zeta^{-1}$. We deduce that $(\beta\cdot u)\boxtimes v=\theta^{\zeta}$ if $\beta\cdot u\neq 0$. So, the assertion (8.1.3) is a consequence of the following:

(8.1.4)

$$\text{given }\zeta\in L(\theta)\text{ positive, we have }\beta\cdot u\neq 0\text{ if and only if }\zeta\in D(\theta).$$

We will prove that statement by reduction to the non-singular case. Let $f:\hat{Z}\to Z$ be a non-singular cover. The morphism $\xi:{\mathbf{R}}_{>0}\to Z$ lifts uniquely to a morphism of curves $\hat{\xi}:{\mathbf{R}}_{>0}\to\hat{Z}$. Let $\hat{M}=f^{-1}(M)$ and let $\hat{\alpha}:\hat{S}^{\prime\prime}\to\{\hat{\xi}(1),\ldots\hat{\xi}(n)\}$ and $\hat{\zeta}$ be the unique lifts of $\alpha$ and $\zeta$ to $\hat{Z}$. There exist subsets $\hat{S},\hat{T}$ of $\hat{M}$ and a lift $\hat{\beta}:\hat{S}\setminus\hat{S}^{\prime\prime}\to\hat{T}$ of $\beta$ such that $\hat{\zeta}\in L(\hat{\theta})$, where $\hat{\theta}=\hat{\alpha}\boxtimes\hat{\beta}$ (Lemma 7.4.28). We have $\zeta\in D(\theta)$ if and only if $\hat{\zeta}\in D(\hat{\theta})$ (Lemma 7.4.28).

Write $g_{\hat{S}^{\prime\prime},\hat{\zeta}}(\hat{\alpha})=\hat{v}\wedge\hat{u}$ as above. We have $f(\hat{u})=u$ and $f(\hat{v})=v$. We have $g_{S^{\prime\prime},\zeta}(\alpha)\neq 0$ if and only if $g_{\hat{S}^{\prime\prime},\hat{\zeta}}(\hat{\alpha})\neq 0$. Finally, $\beta\cdot u\neq 0$ if and only if $\hat{\beta}\cdot\hat{u}\neq 0$. This completes the reduction of (8.1.4) to the case of $\hat{Z}$.

So, we now prove (8.1.4) assuming $Z$ is smooth. Note that $Z(\xi)$ is isomorphic (as a $1$-dimensional space) to an interval of ${\mathbf{R}}$. We consider $\zeta:s_{1}\to s_{2}$ in $L(\theta)$ positive with $s_{1}\in S^{\prime\prime}$ and $s_{2}\in S\setminus S^{\prime\prime}$.

Remark 7.4.11 shows that $\beta\cdot u^{\prime}\neq 0$ if and only if $i(\beta_{s},\beta_{s_{2}}\circ\zeta)=i(\beta_{s},\beta_{s_{2}})+i(\operatorname{id}\nolimits_{s},\zeta)$ for all $s\in S\setminus(S^{\prime}\sqcup\{s_{1}\})$. That equality is always satisfied unless there are $\zeta^{\prime\prime}:s_{1}\to s$ and $\zeta^{\prime}:s\to s_{2}$ positive. In that case, $\bar{\zeta}^{\prime\prime}$ is negative and the equality is satisfied if and only if $\bar{\zeta}^{\prime}$ is positive.

We have $u\neq 0$ if and only if given $\zeta^{\prime\prime}:s_{1}\to s$ and $\zeta^{\prime}:s\to s_{2}$ positive with $s\in S^{\prime}\setminus\{s_{2}\}$, then $\bar{\zeta}^{\prime\prime}=\alpha_{s_{2}}\circ\zeta^{\prime\prime}\circ\alpha_{s_{1}}^{-1}$ is positive.

We deduce that $\zeta\in D(\theta)$ if and only if $\beta\cdot u\neq 0$. The proposition follows. ∎