ScalingStacks

Proof. We need to check that $f^{\#}$ is compatible with composition. This is clear if $Z$ and $Z^{\prime}$ are non-singular. In general, consider two maps $\zeta^{\prime}_{1}$ and $\zeta^{\prime}_{2}$ in ${\mathcal{S}}(Z^{\prime},1)$ such that $f^{\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})\neq 0$. Let $\hat{f}:\hat{Z}\to\hat{Z}^{\prime}$ be the map corresponding to $f$ between non-singular covers $q:\hat{Z}\to Z$ and $q^{\prime}:\hat{Z}^{\prime}\to Z^{\prime}$.

We have

	$\displaystyle q^{\#}f^{\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})$	$\displaystyle=\hat{f}^{\#}q^{\prime\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})=\hat{f}^{\#}\bigl(q^{\prime\#}(\zeta^{\prime}_{2})\circ q^{\prime\#}(\zeta^{\prime}_{1})\bigr)=\bigl(\hat{f}^{\#}q^{\prime\#}(\zeta^{\prime}_{2})\bigr)\circ\bigl(\hat{f}^{\#}q^{\prime\#}(\zeta^{\prime}_{1})\bigr)$
		$\displaystyle=q^{\#}f^{\#}(\zeta^{\prime}_{2})\circ q^{\#}f^{\#}(\zeta^{\prime}_{1}),$

hence $f^{\#}(\zeta^{\prime}_{1})\neq 0$ and $f^{\#}(\zeta^{\prime}_{2})\neq 0$ since $q^{\#}f^{\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})\neq 0$ by Lemma 7.3.17. It follows that $f^{\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})=f^{\#}(\zeta^{\prime}_{2})\circ f^{\#}(\zeta^{\prime}_{1})$. ∎