ScalingStacks

[05UD]

Proof. Let G:Δ⁡[1]→W1G\colon\Delta[1]\rightarrow W_{1} denote the path connecting gg and g′g^{\prime}. Then it suffices to note that a dotted arrow exists in

Δ⁡[0]\displaystyle{{\Delta[0]}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}H\scriptstyle{H}W3\displaystyle{{W_{3}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}Δ⁡[1]\displaystyle{{\Delta[1]}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}(s0​d1​G,G,s0​d0​G)\scriptstyle{(s_{0}d_{1}G,G,s_{0}d_{0}G)}Maps​𝒮⁡(Z⁡(3),W)\displaystyle{{\Map_{s{\operatorname{\mathcal{S}}}}(Z(3),W)}}

where HH is a lift of (s0​d1​g′,g′,s0​d0​g′)=(i​dx′,g′,i​dy′)(s_{0}d_{1}g^{\prime},g^{\prime},s_{0}d_{0}g^{\prime})=(id_{x^{\prime}},g^{\prime},id_{y^{\prime}}) to W3W_{3}, since the right-hand vertical map is a fibration. ∎

Original mathematics by the credited authors. Source collection and HTML conversion remain in progress.

Charles Rezk

Original source: arXiv:math/9811037v3

    Original source page 13

    Original source · math/9811037v3