ScalingStacks

Proof. Let $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z_{\xi})}(I,J)$. Let $n=\mu(\theta)$. We show by induction on $n$ that there exists $\alpha\in F_{n}\cap C_{n}$ such that $q(\alpha)=\theta$.

Assume $n=1$. Let $s\in I$ such that $\mu(\theta_{s})=1$. There is a decomposition $\theta_{s}=\theta_{s}^{r-}\cdot\theta_{s}^{r}$ as in §7.4.6. We define $\alpha\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z)}(I\sqcup\{-1\},J\sqcup\{1\})$ by $\alpha_{s^{\prime}}=\theta_{s^{\prime}}$ for $s^{\prime}\neq s$, $\alpha_{s}=[0\to 1]\cdot\theta_{s}^{r}$ and $\alpha_{-1}=\theta_{s}^{r-}\cdot[-1\to 0]$. We have $\alpha\in A_{1}=F_{1}\cap C_{1}$ and $q(\alpha)=\theta$.

Assume now $n>1$. Consider a decomposition $\theta=r^{\prime}(\theta)\cdot r(\theta)$ as in Lemma 7.4.27. There exists $\alpha\in A_{1}$ and $\beta\in F_{n-1}\cap C_{n-1}$ such that $q(\alpha)=r(\theta)$ and $q(\beta)=r^{\prime}(\theta)$. Let $\gamma=\beta\ast\alpha\in C_{n}$. We have $q(\gamma)=\theta$.

Let $s=\gamma^{-1}(n)=\alpha^{-1}(1)$. We have $\mu(r(\theta)_{s})=1$. Let $i\in(1,n-1)$ and $s^{\prime}=\gamma^{-1}(i)$. If $s^{\prime}\in(-n,-1)$, then $I(\gamma_{|\{s^{\prime},s\}})=\emptyset$. Assume $s^{\prime}{\not\in}(-n,-1)$. We have $\theta_{s^{\prime}}^{r}=[i\to 0]\cdot\gamma_{s^{\prime}}$. Since $\mathrm{supp}(\theta_{s}^{r})\subset\mathrm{supp}(\theta_{s^{\prime}}^{r})$, it follows that $I(\gamma_{|\{s^{\prime},s\}})=\emptyset$. Since $\beta\in F_{n-1}$, we deduce that $\gamma\in F_{n}$.

The case of $E\cap C$ follows from that of $F\cap C$ applied to $Z^{\operatorname{opp}\nolimits}$, cf Remark 8.2.6. ∎