Let \(A \in \Ab(T_{\leq 0})\) and \(n \geq 1\). Then for any object \(X \in T\) there is an equivalence
\[\Hom_T(X,\bB^{n+1}A) \quad\simeq\quad \Gerb_n^A(T_{/X}), \qquad \qquad f \mapsto \fib_0(f).\]
Proof
First, note that the functor is well-defined. By Example 3.42, the map \(0\colon *\to \bB^{n+1}A\) is an \(n\)-gerbe. Its fiber is \(\bB^nA\), so Lemma 3.49 supplies it with a canonical band by \(A\), which is inherited by every pullback. The claim is that this is the universal\(n\)-gerbe banded by \(A\).To see this, consider an \(n\)-gerbe \(\widetilde{X} \to X\) banded by \(A\). There is a unique cartesian square of the form Here the cartesian square is required to identify the given band on \(\widetilde X\) with the band pulled back from the universal gerbe. More formally, the homotopy fiber of the functor in the statement over \(\widetilde X\to X\) is the subanima of
spanned by the band-compatible cartesian squares. We show that this subanima is contractible.Step 1: Suppose first that \(\widetilde{X} \to X\) admits a section \(s\colon X \to \widetilde{X}\). In the slice topos \(T_{/X}\), the section makes \(\widetilde X\) into a pointed \(n\)-gerbe. By Lemma 3.49, there is a unique band-preserving pointed isomorphism \(\widetilde X\iso X\times\bB^nA\).A band-compatible cartesian square with the universal gerbe is now the same as a map \(u\colon X\to\bB^{n+1}A\) together with a trivialization of its fiber \(P_u:=X\times_{\bB^{n+1}A}*\to X\). Such a trivialization is equivalently a section of \(P_u\to X\), or equivalently a nullhomotopy of \(u\). The anima of pairs consisting of a map \(u\colon X\to\bB^{n+1}A\) and a nullhomotopy of \(u\) is \(\Hom_T(X,*)\), which is contractible. The band compatibility is built into the unique pointed identification supplied by Lemma 3.49. This proves the claim in the pointed case.Step 2: We now prove the claim in general. Since \(n\geq1\), the map \(\widetilde X\to X\) is an effective epimorphism. We may therefore take \(U=\widetilde X\); the pullback \(\widetilde X\times_XU\to U\) admits the diagonal as a section. Let \(\check{C}(U/X)\) denote the Čech nerve, and write \(U_k:=\check{C}(U/X)_k\) and \(\widetilde U_k:=\widetilde X\times_XU_k\). Each map \(\widetilde U_k\to U_k\) is an \(n\)-gerbe banded by \(A\) admitting a section. Step 1 therefore shows that the anima of band-compatible cartesian squares
is contractible for every \(k\). Objects, isomorphisms, and identifications of their bands satisfy descent along effective epimorphisms. Hence the anima of band-compatible cartesian squares from \(\widetilde X\to X\) to the universal gerbe is the limit of these contractible animae, and is therefore contractible. Thus every homotopy fiber of the functor in the statement is contractible, proving that it is an equivalence.
