ScalingStacks

Proof. The counit of the adjunction is $\varepsilon=\kappa_{1}\circ\mathrm{mult}$. Let $\gamma\in R_{\xi^{-}}^{\bullet}(T,S)$. Let $\eta$ be the map defined in the lemma. We have

$$\eta(\operatorname{id}\nolimits_{T})=\sum_{x\in\tilde{\xi}^{-1}(T)}(\tilde{\xi}([-1\to x])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}})\otimes(\tilde{\xi}([x\to 1])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}}),$$

hence

$$(\operatorname{id}\nolimits\otimes\varepsilon)\circ(\eta\otimes\operatorname{id}\nolimits)(\gamma)=\sum_{x\in\tilde{\xi}^{-1}(T)}(\tilde{\xi}([-1\to x])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}})\cdot\kappa_{1}\bigl((\tilde{\xi}([x\to 1])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}})\cdot\gamma\bigr)$$

Let $x$ be the unique element of $\tilde{\xi}^{-1}(\chi(\gamma)(\xi^{-}(-1)))$. We have $\gamma_{\xi^{-}(-1)}=\tilde{\xi}([-1\to x])$ and $\kappa_{1}\bigl((\tilde{\xi}([x\to 1])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}})\cdot\gamma\bigr)=\gamma_{|S}$, hence

$$(\operatorname{id}\nolimits\otimes\varepsilon)\circ(\eta\otimes\operatorname{id}\nolimits)(\gamma)=(\gamma_{\xi^{-}(-1)}\boxtimes\operatorname{id}\nolimits_{T\setminus\{\chi(\gamma)(\xi^{-}(-1))\}})\otimes\gamma_{|S}$$

We deduce that

$$\mathrm{mult}\circ(\operatorname{id}\nolimits\otimes\varepsilon)\circ(\eta\otimes\operatorname{id}\nolimits)(\gamma)=\gamma$$

and the lemma follows. ∎