ScalingStacks

Proof. Note first that all the maps of the diagram are functorial with respect to $S\in{\mathcal{S}}_{M}(Z_{\xi})$.

$\bullet\$Let $\gamma\in R_{\xi_{2}^{-}}(U,S)$, $\beta\in L_{\xi_{1}^{+}}(V,U)$ and $\alpha\in R_{\xi_{2}^{-}}(T,V)$. We will show that

(8.3.2)

$$\mathrm{action}\circ(\Xi\otimes f_{2})(\alpha\otimes\beta\otimes\gamma)=f_{2}\circ R_{\xi_{2}^{-}}(\mathrm{mult}\circ\Xi)\circ\tau L_{\xi_{1}^{+}}\circ R_{\xi_{2}^{-}}\lambda(\alpha\otimes\beta\otimes\gamma).$$

Since $\gamma=(\operatorname{id}\nolimits\boxtimes\gamma_{\xi_{2}^{-}(-1)})\cdot\gamma_{|S}$ and since $\mathrm{action}\circ(\Xi\otimes f_{2})$ and $f_{2}\circ R_{\xi_{2}^{-}}(\mathrm{mult}\circ\Xi)\circ\tau L_{\xi_{1}^{+}}\circ R_{\xi_{2}^{-}}\lambda$ are morphisms of ${\mathcal{S}}_{M}(Z)^{\operatorname{opp}\nolimits}$-modules, we can assume $\gamma_{|S}=\operatorname{id}\nolimits$. We have $\alpha\otimes\beta=(\operatorname{id}\nolimits\boxtimes\alpha_{\xi_{2}^{-}(-1)})\otimes(\alpha_{|V}\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{+}(1)}\cdot\beta)$, hence we can assume $\alpha_{|V}=\operatorname{id}\nolimits$. We can also assume that $\beta\otimes\gamma\neq 0$.

We have

$$\mathrm{action}\circ(\Xi\otimes f_{2})(\alpha\otimes\beta\otimes\gamma)=\bigl(\operatorname{id}\nolimits_{V}\boxtimes(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\bigr)\cdot\beta$$

$$\cdot\bigl(\operatorname{id}\nolimits_{S}\boxtimes(\gamma_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])\bigr)$$

$$=\delta_{1}\beta_{|S\setminus\chi(\beta)^{-1}(\xi_{1}^{+}(1))}\boxtimes(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\beta_{\chi(\beta)^{-1}(\xi_{1}^{+}(1))})$$

$$\boxtimes\bigl((\beta\circ\gamma)_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]\bigr)$$

where $\delta_{1}=\delta_{i(\alpha_{\xi_{2}^{-}(-1)},(\beta\circ\gamma)_{\xi_{2}^{-}(-1)})=0}$. On the other hand, we have

$$f_{2}\circ R_{\xi_{2}^{-}}(\mathrm{mult}\circ\Xi)\circ\tau L_{\xi_{1}^{+}}\circ R_{\xi_{2}^{-}}\lambda(\alpha\otimes\beta\otimes\gamma)=$$

		$\displaystyle=f_{2}\circ R_{\xi_{2}^{-}}(\mathrm{mult}\circ\Xi)\circ\tau L_{\xi_{1}^{+}}\Bigl(\alpha\otimes\bigl((\beta_{\chi(\gamma)(\xi_{2}^{-}(-1))}\cdot\gamma_{\xi_{2}^{-}(-1)})\boxtimes\operatorname{id}\nolimits\bigr)\otimes\beta_{|S}\Bigr)$
		$\displaystyle=\delta_{1}f_{2}\circ R_{\xi_{2}^{-}}(\mathrm{mult}\circ\Xi)\Bigl(\bigl(\operatorname{id}\nolimits\boxtimes(\beta_{\chi(\gamma)(\xi_{2}^{-}(-1))}\cdot\gamma_{\xi_{2}^{-}(-1)})\bigr)\otimes(\operatorname{id}\nolimits\boxtimes\alpha_{\xi_{2}^{-}(-1)})\otimes\beta_{|S}\Bigr)$
		$\displaystyle=\delta_{1}f_{2}\biggl(\bigl(\operatorname{id}\nolimits\boxtimes(\beta_{\chi(\gamma)(\xi_{2}^{-}(-1))}\cdot\gamma_{\xi_{2}^{-}(-1)})\bigr)\otimes\Bigl(\bigl(\operatorname{id}\nolimits\boxtimes(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\bigr)\cdot\beta_{|S}\Bigr)\biggr)$
		$\displaystyle=\delta_{1}\bigl((\beta_{\chi(\gamma)(\xi_{2}^{-}(-1))}\cdot\gamma_{\xi_{2}^{-}(-1)})\cdot[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]\bigr)\boxtimes\Bigl(\bigl(\operatorname{id}\nolimits\boxtimes(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\bigr)\cdot\beta_{|S}\Bigr)$
		$\displaystyle=\mathrm{action}\circ(\Xi\otimes f_{2})(\alpha\otimes\beta\otimes\gamma).$

We deduce that (8.3.2) holds.

$\bullet\$Let $\gamma\in R_{\xi_{1}^{-}}(U,S)$, $\beta\in L_{\xi_{1}^{+}}(V,U)$ and $\alpha\in R_{\xi_{2}^{-}}(T,V)$. We will show that

(8.3.3)

$$\mathrm{action}\circ(\Xi\otimes f_{1})(\alpha\otimes\beta\otimes\gamma)=\bigl(f_{1}\circ R_{\xi_{1}^{-}}(\mathrm{mult}\circ\Xi)\circ\sigma L_{\xi_{1}^{+}}\circ R_{\xi_{2}^{-}}\rho+f_{2}\circ R_{\xi_{2}^{-}}\varepsilon\bigr)(\alpha\otimes\beta\otimes\gamma).$$

As before, we can assume $\gamma_{|S}=\operatorname{id}\nolimits$, $\alpha_{|V}=\operatorname{id}\nolimits$ and $\beta\otimes\gamma\neq 0$. We put $u_{1}=\chi(\gamma)(\xi_{1}^{-}(-1))$ and $u_{2}=\chi(\beta)^{-1}(\xi_{1}^{+}(1))$.

We have

$$\mathrm{action}\circ(\Xi\otimes f_{1})(\alpha\otimes\beta\otimes\gamma)=\bigl(\operatorname{id}\nolimits_{V}\boxtimes(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\bigr)\cdot\beta\cdot\bigl(\operatorname{id}\nolimits_{S}\boxtimes\gamma_{\xi_{1}^{-}(-1)}\bigr)$$

$$=\begin{cases}\delta_{2}(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{-}(1)\to\xi_{2}^{-}(-1)])\boxtimes\beta_{|S}&\text{ if }u_{1}=u_{2}\\ \delta_{3}(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\beta_{u_{2}})\boxtimes(\beta_{u_{1}}\circ\gamma_{\xi_{1}^{-}(-1)})\boxtimes\beta_{|S\setminus\{u_{2}\}}&\text{ otherwise}\end{cases}$$

where

• •

  $\delta_{2}=1$ if $\gamma_{\xi_{1}^{-}(-1)}(1-)=\iota(\beta_{u_{2}}(0+))$ and $\delta_{2}=0$ otherwise

• •

  $\delta_{3}=1$ if $\beta_{|U}\cdot(\operatorname{id}\nolimits_{S}\boxtimes\gamma_{\xi_{1}^{-}(-1)})\neq 0$ and $\delta_{3}=0$ otherwise.

We have

$$f_{2}\circ R_{\xi_{2}^{-}}\varepsilon(\alpha\otimes\beta\otimes\gamma)=\delta_{2}f_{2}(\alpha\otimes\beta_{|S})=\delta_{2}(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{-}(1)\to\xi_{2}^{-}(-1)])\boxtimes\beta_{|S}.$$

We have

$$f_{1}\circ R_{\xi_{1}^{-}}(\mathrm{mult}\circ\Xi)\circ\sigma L_{\xi_{1}^{+}}\circ R_{\xi_{2}^{-}}\rho(\alpha\otimes\beta\otimes\gamma)=$$

		$\displaystyle=\delta^{\prime}_{3}f_{1}\circ R_{\xi_{1}^{-}}(\mathrm{mult}\circ\Xi)\circ\sigma L_{\xi_{1}^{+}}\Bigl(\alpha\otimes\bigr(\beta_{|U\setminus\{u_{2}\}}\cdot(\gamma_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\bigl)\otimes(\beta_{u_{2}}\boxtimes\operatorname{id}\nolimits)\Bigr)$
		$\displaystyle=\delta^{\prime}_{3}\delta^{\prime\prime}_{3}f_{1}\circ R_{\xi_{1}^{-}}(\mathrm{mult}\circ\Xi)\circ\sigma L_{\xi_{1}^{+}}\Bigl(\alpha\otimes\bigr(\beta_{|S\setminus\{u_{2}\}}\boxtimes(\beta_{u_{1}}\circ\gamma_{\xi_{1}^{-}(-1)})\bigl)\otimes(\beta_{u_{2}}\boxtimes\operatorname{id}\nolimits)\Bigr)$
		$\displaystyle=\delta^{\prime}_{3}\delta^{\prime\prime}_{3}f_{1}\circ R_{\xi_{1}^{-}}(\mathrm{mult}\circ\Xi)\Bigl(\bigl((\beta_{u_{1}}\circ\gamma_{\xi_{1}^{-}(-1)})\boxtimes\operatorname{id}\nolimits\bigr)\otimes(\alpha_{\xi_{2}^{-}(-1)}\boxtimes\beta_{|S\setminus\{u_{2}\}})\otimes(\beta_{u_{2}}\boxtimes\operatorname{id}\nolimits)\Bigr)$
		$\displaystyle=\delta^{\prime}_{3}\delta^{\prime\prime}_{3}(\beta_{u_{1}}\circ\gamma_{\xi_{1}^{-}(-1)})\boxtimes(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\beta_{u_{2}})\boxtimes\beta_{|S\setminus\{u_{2}\}}$

where

• •

  $\delta^{\prime}_{3}=1$ if $u_{1}\neq u_{2}$ and $(\operatorname{id}\nolimits_{u_{1}}\boxtimes\beta_{u_{2}})\cdot(\gamma_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits_{u_{2}})\neq 0$ and $\delta^{\prime}_{3}=0$ otherwise

• •

  $\delta^{\prime\prime}_{3}=1$ if $\beta_{|U\setminus\{u_{2}\}}\cdot(\operatorname{id}\nolimits_{S\setminus\{u_{2}\}}\boxtimes\gamma_{\xi_{1}^{-}(-1)})\neq 0$ and $\delta^{\prime\prime}_{3}=0$ otherwise.

Since $\delta_{3}=\delta^{\prime}_{3}\delta^{\prime\prime}_{3}$, we deduce that (8.3.3) holds and the lemma follows. ∎