ScalingStacks

Proof. Let $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(I,J)$. We have $L(F_{\alpha}(\sigma))=(\alpha\times\alpha)(L(\sigma))$, hence $\ell(F_{\alpha}(\sigma))=\ell(\sigma)$. We have $R_{\alpha}(\llbracket\sigma\rrbracket)=\llbracket F_{\alpha}(\sigma)\rrbracket$, hence $L_{\alpha}(m(\sigma))=m(F_{\alpha}(\sigma))$. We deduce that $\Gamma_{\sigma}(\deg(\sigma))=\deg(F_{\alpha}(\sigma))$.

We have $D(F_{\alpha}(\sigma))=(\alpha\times\alpha)(D(\sigma))$ and $F_{\alpha}(s_{i_{1},i_{2}})=s_{\alpha(i_{1}),\alpha(i_{2})}$ for $i_{1},i_{2}\in\tilde{I}$ with $i_{1}-i_{2}{\not\in}n{\mathbf{Z}}$, hence $F_{\alpha}$ is compatible with $d$. ∎