ScalingStacks

Proof. Consider $\alpha_{i}\in R_{\xi_{i}^{-}}^{\bullet}(T,U)$ and $\beta_{i}\in R_{\xi_{i}^{-}}^{\bullet}(U,S)$. In order to prove that the equality of the lemma holds when applied to $((\alpha_{2},\alpha_{1})\otimes(\beta_{2},\beta_{1}))$, we can assume that $(\alpha_{i})_{|T}=\operatorname{id}\nolimits$ and $(\beta_{i})_{|S}=\operatorname{id}\nolimits$, since the morphisms involved in the equality commute with the right action of ${\mathcal{S}}_{M}(Z)$.

We have

$$\tau\circ(f_{i}\otimes f_{j})(\alpha_{i}\otimes\beta_{j})=$$

		$\displaystyle=\tau\Bigl(\bigl((\alpha_{i})_{\xi_{i}^{-}(-1)}\cdot[\xi_{1}^{-}(-1)\to\xi_{i}^{-}(-1)]\boxtimes\operatorname{id}\nolimits\bigr)\otimes\bigl(\operatorname{id}\nolimits\boxtimes(\beta_{j})_{\xi_{j}^{-}(-1)}\cdot[\xi_{1}^{-}(-1)\to\xi_{j}^{-}(-1)]\bigr)\Bigr)$
		$\displaystyle=d_{i,j}\bigl((\beta_{j})_{\xi_{j}^{-}(-1)}\cdot[\xi_{1}^{-}(-1)\to\xi_{j}^{-}(-1)]\boxtimes\operatorname{id}\nolimits\bigr)\otimes\bigl(\operatorname{id}\nolimits\boxtimes(\alpha_{i})_{\xi_{i}^{-}(-1)}\cdot[\xi_{1}^{-}(-1)\to\xi_{i}^{-}(-1)]\bigr)$

where $d_{2,1}=0$, $d_{1,2}=1$ and $d_{i,i}=1$ if $\bigl((\alpha_{i})_{\xi_{i}^{-}(-1)}\cdot[\xi_{i}^{-}(-2)\to\xi_{i}^{-}(-1)]\boxtimes(\beta_{i})_{\xi_{i}^{-}(-1)}\bigr)\cdot\tau\neq 0$ and $d_{i,i}=0$ otherwise. We deduce that the lemma holds when applied to $((\alpha_{2},\alpha_{1})\otimes(\beta_{2},\beta_{1}))$, hence it holds in general. ∎