ScalingStacks

Assume $\xi^{-1}(M)$ has no maximum. Fix an increasing sequence $m_{0},m_{1},\ldots$ of points of $(0,1)$ with $\xi(m_{i})\in M$ for all $i$ and with $\lim_{i}m_{i}>t$ for all $t\in\xi^{-1}(M)$.

Fix $n\geq 0$ and define the braid $\beta_{r}:\{m_{r},\ldots,m_{r+n-1}\}\to\{1,\ldots,n\}$ of ${\mathbf{R}}_{>0}$ by $(\beta_{r})_{m_{r+i}}=[m_{r+i}\to i+1]$.

Let $S$ and $T$ be two finite subsets of $M$. Consider $r$ such that $m_{r}>\xi^{-1}(t)$ for all $t\in T\cap\xi({\mathbf{R}}_{>0})$. There is an isomorphism

It follows that there are isomorphisms functorial in $S$ and $T$

Here, the colimit is taken over the invertible maps $\xi(\theta_{r})$, where $\theta_{r}:\{m_{r},\ldots,m_{r+n-1}\}\to\{m_{r+1},\ldots,m_{r+n}\}$ is the braid in ${\mathbf{R}}_{>0}$ given by $(\theta_{r})_{m_{s}}=[m_{s}\to m_{s+1}]$.

We deduce that $T\mapsto(S\mapsto L^{\bullet}(T,S,e^{n}))$ is isomorphic to the functor