Lemma 3.49.
Let \(n\geq1\). The trivial \(n\)-gerbe \(\bB^n A\) is banded by \(A\). Moreover, the anima of pointed \(n\)-gerbes banded by \(A\) and band-preserving pointed isomorphisms is contractible, with distinguished object \(\bB^nA\).
Proof
For the first claim, consider the two canonical maps \(\bB^nA \to (\bB^nA)^{S^n}\) and \(\Omega^n\bB^nA \to (\bB^nA)^{S^n}\). Using that \(\bB^nA \in \CGrp(T)\), this induces a map
\[\bB^nA \times A \simeq \bB^nA \times \Omega^n\bB^nA \to (\bB^nA)^{S^n}\]
in \(T_{/\bB^nA}\). We need to show that this induces an isomorphism on \(0\)-truncations. Since \(n\geq1\), the map \(* \twoheadrightarrow \bB^nA\) is an effective epimorphism. Pullback along it is conservative by Lemma 2.29, so the claim may be checked after this base change. There the map becomes the identity on \(\Omega^n\bB^nA\).For the second claim, apply the equivalence \(\bOmega^n\colon \mathrm{EM}_n(T)\iso E_n\Grp(T_{\leq0})\) from Corollary 3.45. Under this equivalence, a band on \(X\) is precisely an isomorphism \(\bOmega^nX\cong A\). The anima of pairs consisting of an object \(A'\) and an isomorphism \(A'\cong A\) is contractible. Transporting this statement across the equivalence proves the claim, including uniqueness of the band-preserving pointed isomorphism.