ScalingStacks

Proof. Consider $J,J^{\prime}\subset{\mathbf{Z}}/n$. We have an injective map $f:\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(J,J^{\prime})\to{\mathbf{Z}}^{J},\ \sigma\mapsto(\sigma(j)-j))_{b}$, where $j\in\{1,\ldots,n\}$ and $b=j+n{\mathbf{Z}}$. The image of that map is the set of those $c\in{\mathbf{Z}}^{J}$ such that $\{c_{b}+b\}_{b}=J^{\prime}$ and we obtain a bijection

	$\displaystyle\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(J,J^{\prime})$	$\displaystyle\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{{\mathcal{P}}^{\bullet}_{M}(\vec{S}^{1})}(\{a_{j}\}_{j\in\tilde{J}\cap[1,n]},\{a_{j^{\prime}}\}_{j^{\prime}\in\tilde{J}^{\prime}\cap[1,n]})$
	$\displaystyle\sigma$	$\displaystyle\mapsto\bigl(F_{j^{\prime},j}(f(\sigma)_{j+n{\mathbf{Z}}})\bigr)_{j\in\tilde{J}\cap[1,n],\ j^{\prime}\in\tilde{J}^{\prime}\cap[1,n],\ \sigma(j)-j^{\prime}\in n{\mathbf{Z}}}.$

We deduce that $F$ induces a bijection on pointed $\operatorname{Hom}\nolimits$-sets. Consider now $\sigma:J\to J^{\prime}$ and $\sigma^{\prime}:J^{\prime}\to J^{\prime\prime}$ two maps in ${\mathcal{S}}_{n}$. Given $j\in\tilde{J}\cap[1,n]$, we have

$$F(\sigma^{\prime}\sigma)_{a_{j}}=F_{j}(\sigma^{\prime}\sigma(j)-j)=F_{j}(\sigma^{\prime}(\sigma(j))-\sigma(j)+\sigma(j)-j)=F_{\sigma(j)}(\sigma^{\prime})_{a_{\sigma(j)}}\circ F_{j}(\sigma)_{a_{j}}.$$

We deduce that $F$ is a functor and the first statement of the proposition follows.

Consider now $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(J,J^{\prime})$.

The map $F(\sigma)$ is in ${\mathcal{P}}^{\bullet}_{M}(\dot{S}^{1})$ if and only if $\sigma(j)\geq 0$ for all $j\in[1,n]\cap\tilde{J}$, hence if and only if $\sigma$ is in ${\mathcal{S}}_{n}^{+}$.

The map $F(\sigma)$ is in ${\mathcal{P}}^{\bullet}_{M}(\vec{S}^{1})$ if and only if $\sigma(j)-j\geq 0$ for all $j\in\tilde{J}$, hence if and only if $\sigma$ is in ${\mathcal{S}}_{n}^{++}$.

The map $F(\sigma)$ is in ${\mathcal{P}}^{\bullet}_{M}(I)$ if and only if $\sigma(j)\in[1,n]$ for all $j\in\tilde{J}\cap[1,n]$, hence if and only if $\sigma$ is in ${\mathcal{S}}_{n}^{f}$.

The proposition follows. ∎