ScalingStacks

Proof. We have

$$\sigma=(R_{\xi_{1}^{-}}\circ\textrm{mult})\circ(R_{\xi_{1}^{-}}\otimes R_{\xi_{2}^{-}}\otimes\varepsilon_{L_{\xi_{1}^{+}},R_{\xi_{1}^{-}}})\circ(R_{\xi_{1}^{-}}\otimes\lambda\otimes R_{\xi_{1}^{-}})\circ(\eta_{L_{\xi_{1}^{+}},R_{\xi_{1}^{-}}}\otimes\operatorname{id}\nolimits).$$

We have $\alpha=(\operatorname{id}\nolimits\boxtimes\alpha_{\xi_{2}^{-}(-1)})\cdot(\alpha_{|U}\boxtimes\operatorname{id}\nolimits)$, hence $\alpha\otimes\beta=(\operatorname{id}\nolimits\boxtimes\alpha_{\xi_{2}^{-}(-1)})\otimes(\alpha_{|U}\cdot\beta)$. As a consequence, it is enough to prove the first statement of the lemma assuming that $\alpha_{|U}=\operatorname{id}\nolimits_{U}$. In that case, the composition above is given by

	$\displaystyle\alpha\otimes\beta$	$\displaystyle\mapsto\sum_{x\in\tilde{\xi}_{1}^{-1}(T)}(\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}_{1}(x)\}}\boxtimes\tilde{\xi}_{1}([-1\to x]))\otimes(\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}_{1}(x)\}}\boxtimes\tilde{\xi}_{1}([x\to 1]))\otimes\alpha\otimes\beta$
		$\displaystyle\mapsto\sum_{x\in\tilde{\xi}_{1}^{-1}(T)}(\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}_{1}(x)\}}\boxtimes\tilde{\xi}_{1}([-1\to x]))\otimes(\alpha_{\xi_{2}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([x\to 1]))\otimes\beta$
		$\displaystyle\mapsto(\operatorname{id}\nolimits\boxtimes\beta_{\xi_{1}^{-}(-1)})\otimes(\alpha_{\xi_{2}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes\beta_{|S}$
		$\displaystyle\mapsto(\operatorname{id}\nolimits\boxtimes\beta_{\xi_{1}^{-}(-1)})\otimes(\alpha_{\xi_{2}^{-}(-1)}\boxtimes\beta_{|S}).$

It is immediate to check that the formula for $\sigma^{-1}$ does produce an inverse. ∎

Proof. Assume first $\alpha_{|U\setminus\{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))\}}=\operatorname{id}\nolimits$ and $\beta_{|S}=\operatorname{id}\nolimits$. We have

	$\displaystyle\rho(\alpha\otimes\beta)$	$\displaystyle=\varepsilon_{1}R_{\xi_{1}^{-}}L_{\xi_{1}^{+}}\circ L_{\xi_{1}^{+}}\tau L_{\xi_{1}^{+}}(\alpha\otimes\beta\otimes\eta_{1}(\operatorname{id}\nolimits_{S}))$
		$\displaystyle=\varepsilon_{1}R_{\xi_{1}^{-}}L_{\xi_{1}^{+}}\biggl(\sum_{x\in\tilde{\xi}_{1}^{-1}(S)}\alpha\otimes\tau\Bigl((\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr)\Bigr)\otimes\bigl(\tilde{\xi}_{1}([x\to 1])\boxtimes\operatorname{id}\nolimits\bigr)\biggr)$
		$\displaystyle=\varepsilon_{1}R_{\xi_{1}^{-}}L_{\xi_{1}^{+}}\biggl(\sum_{x\in I}\alpha\otimes\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr)\otimes(\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes\bigl(\tilde{\xi}_{1}([x\to 1])\boxtimes\operatorname{id}\nolimits\bigr)\biggr)$
		$\displaystyle=\delta_{1}(\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes(\alpha_{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))}\boxtimes\operatorname{id}\nolimits)$

where $I=\{x\in\tilde{\xi}_{1}^{-1}(S)\ |\ (\tilde{\xi}_{1}([-1\to x])\boxtimes(\beta_{\xi_{1}^{-}(-1)}\circ[\xi_{1}^{-}(-2)\to\xi_{1}^{-}(-1)])\cdot\tau\neq 0\}$.

Since $\rho$ is a morphism of $({\mathcal{S}}_{M}(Z),{\mathcal{S}}_{M}(Z))$-bimodules, the general result follows using the decompositions $\alpha=(\alpha_{|U\setminus\{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))\}}\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{+}(1)})\cdot(\operatorname{id}\nolimits\boxtimes\alpha_{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))})$ and $\beta=(\operatorname{id}\nolimits\boxtimes\beta_{\xi_{1}^{-}(-1)})\cdot(\beta_{|S}\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})$. ∎

Proof. It is immediate that $d(f_{1})=0$. For the second equality, consider $\alpha\in R^{\bullet}_{\xi_{2}^{-}}(T,U)$ and $\beta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z_{\xi})}(S,U)$. Since $\alpha\otimes\beta=(\operatorname{id}\nolimits\boxtimes\alpha_{\xi_{2}^{-}(-1)})\otimes(\alpha_{|U}\cdot\beta)$, we can assume that $\alpha_{|U}=\operatorname{id}\nolimits$. We have

$$d(f_{2})(\alpha\otimes\beta)=\alpha\cdot\bigl(d(\beta\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])+d(\beta)\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]\bigr).$$

We have

	$\displaystyle d(\beta\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])$	$\displaystyle=d\bigl((\operatorname{id}\nolimits\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})\bigr)$
		$\displaystyle=(\operatorname{id}\nolimits\boxtimes[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\cdot d(\operatorname{id}\nolimits_{U}\boxtimes[\xi_{1}^{-}(-1)\to\xi_{1}^{+}(1)])\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})$
		$\displaystyle\ \ \ \ +(\operatorname{id}\nolimits\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])\cdot(d(\beta)\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})$
		$\displaystyle=(\operatorname{id}\nolimits_{U}\boxtimes[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\sum_{x\in\tilde{\xi}_{1}^{-1}(U)}\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([x\to 1])\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr)$
		$\displaystyle\ \ \ \ \cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})+d(\beta)\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]$

hence

$$d(f_{2})(\alpha\otimes\beta)=\alpha\cdot\sum_{x\in\tilde{\xi}_{1}^{-1}(U)}\Bigl(\operatorname{id}\nolimits\boxtimes\bigl([\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\tilde{\xi}_{1}([x\to 1])\bigr)\boxtimes\tilde{\xi}_{1}([-1\to x])\Bigr)\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})$$

		$\displaystyle=\sum_{x\in\tilde{\xi}_{1}^{-1}(U)}\Bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\boxtimes\bigl(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\tilde{\xi}_{1}([x\to 1])\bigr)\Bigr)\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})$
		$\displaystyle=f_{1}\circ u(\alpha\otimes\beta).$

The lemma follows. ∎

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.