Normal Deligne–ribbon cover and its deck group

Apply the orbit-cover theorem. The label $(1,J)$ is an object: $W_{J}1=1W_J$, and the identity has minimum length in $W_J$. Thus $\mathcal{D}$ is nonempty. By [F2], every two of its objects are joined by a sequence of simple crossings, which compose to an arrow by [F1]. Hence $\mathcal{D}$ is connected. It is small by its construction in [F1]. The action in [F3] is therefore an action by automorphisms on a connected small groupoid, free on objects. By [F4], $\mathcal{Q}$ is a normal covering and $n\mapsto (C\mapsto nC,a\mapsto na)$ is an isomorphism from $N$ to its deck group.

Transport the covering. By [F5], $\mathcal{G}={\mathcal{G}}^{\mathrm{\prime }}\circ \mathcal{Q}$ and ${\mathcal{G}}^{\mathrm{\prime }}$ is an isomorphism. For an arrow $r$ of $\mathcal{R}$ starting at $\mathcal{G}(C)$, its unique preimage under ${\mathcal{G}}^{\mathrm{\prime }}$ starts at $\mathcal{Q}(C)$; the unique lift of that preimage from $C$ under $\mathcal{Q}$ is precisely the unique lift of $r$ from $C$ under $\mathcal{G}$. Thus $\mathcal{G}$ is a covering. The identity functor upstairs and ${\mathcal{G}}^{\mathrm{\prime }}$ downstairs form the asserted isomorphism of coverings.

Transport deck transformations and normality. For an automorphism $d$ of $\mathcal{D}$, the equality $\mathcal{G}d=\mathcal{G}$ is equivalent, after composing with $({\mathcal{G}}^{\mathrm{\prime }}{)}^{-1}$, to $\mathcal{Q}d=\mathcal{Q}$. Hence the two deck groups are the same subgroup of automorphisms of $\mathcal{D}$. Since ${\mathcal{G}}^{\mathrm{\prime }}$ is injective on objects, the fibres of $\mathcal{G}$ and $\mathcal{Q}$ coincide. The deck group therefore acts transitively on a fibre of $\mathcal{G}$ because it does so for $\mathcal{Q}$. Consequently $\mathcal{G}$ is normal and its deck group is identified with $N$ by the action described in 1.1.