ScalingStacks

Proof. Note that Lemma 6.2.9 shows that $d$ is homogeneous of degree $1$. The compatibility of $d$ with $F_{I}$ follows from Lemma 3.2.4.

Consider now $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{H}}_{n}}(I,J)$ non-zero. There exists $\tau\in\operatorname{Hom}\nolimits_{{\mathcal{H}}_{n}}(J,I)$ with $\ell(\tau)=0$. We have $d(\tau\circ\sigma)=\tau\circ d(\sigma)$, hence $d^{2}(\tau\circ\sigma)=\tau\circ d^{2}(\sigma)$. The compatibility of $F_{I}$ with $d$ shows that $d^{2}(\tau\circ\sigma)=0$. Since $\tau$ is invertible, we deduce that $d^{2}(\sigma)=0$.

Consider finally $\sigma^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{H}}_{n}}(J,K)$ and fix $\tau^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{H}}_{n}}(K,J)$ with $\ell(\tau^{\prime})=0$. We have $d(\tau^{\prime}\circ\sigma^{\prime}\circ\sigma\circ\tau)=\tau^{\prime}\circ d(\sigma^{\prime}\circ\sigma)\circ\tau$ and it follows from the compatibility of $F_{J}$ with $d$ that

	$\displaystyle d(\tau^{\prime}\circ\sigma^{\prime}\circ\sigma\circ\tau)$	$\displaystyle=F_{J}\bigl(d(F_{J}^{-1}(\tau^{\prime}\circ\sigma^{\prime}\circ\sigma\circ\tau))\bigr)=F_{J}\bigl(d(F_{J}^{-1}(\tau^{\prime}\circ\sigma^{\prime})\circ F_{J}^{-1}(\sigma\circ\tau))\bigr)$
		$\displaystyle=F_{J}\bigl(d(F_{J}^{-1}(\tau^{\prime}\circ\sigma^{\prime}))\circ F_{J}^{-1}(\sigma\circ\tau)\bigr)+F_{J}\bigl(F_{J}^{-1}(\tau^{\prime}\circ\sigma^{\prime}))\circ d(F_{J}^{-1}(\sigma\circ\tau)\bigr)$
		$\displaystyle=d(\tau^{\prime}\circ\sigma^{\prime})\circ\sigma\circ\tau+\tau^{\prime}\circ\sigma^{\prime}\circ d(\sigma\circ\tau).$

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