ScalingStacks

Proof. The proper retracts of $C_{n}$ are precisely the cells $C_{k}$ for $0\leq k<n$, and, as before, each $C_{k}$ is distinguished as the unique such retract such that, up to isomorphism, there exists precisely $k$ other objects which occur as further proper retracts (the cells $C_{s}$ for $s<k$). Thus it is enough to show that $F(C_{n})\cong C_{n}$ for the $n$-cell alone, as this implies the analogous statement $F(C_{k})\cong C_{k}$ for all $0\leq k\leq n$.

Recall that a generator of a category $\mathcal{C}$ is an object $X$ such that the corepresentable functor $\mathcal{C}(X,-):\mathcal{C}\to\set$ is faithful [30, pg. 127]. The collection of generators is preserved under any autoequivalence. The $n$-cell $C_{n}$ is a generator for $\cat_{n}$ and hence also $\gaunt_{n}$, however the $k$-cells $C_{k}$ for $k<n$ are not generators. Thus no proper retract of $C_{n}$ is a generator. We claim that in fact $C_{n}$ is the unique generator such that every proper retract is not a generator. If this characterization holds, then any autoequivalence necessarily preserves the $n$-cell up to automorphism and the lemma follows.

In fact we will prove a stronger statement: we claim that the $n$-cell $C_{n}$ is a retract of every generator of $\gaunt_{n}$. Now consider the gaunt $n$-category $\partial C_{n+1}$. This may be written as

$$\partial C_{n+1}=C_{n}\cup^{\partial C_{n}}C_{n}.$$

There are exactly two non-identity $n$-morphisms in $\partial C_{n+1}$; call them $a,b:C_{n}\to\partial C_{n+1}$. Observe that these two functors differ only on the unique nontrivial $n$-morphism of $C_{n}$. The unique non-trivial $n$-morphism of $C_{n}$, viewed as a map $C_{n}\to C_{n}$, corresponds to the element $\id\in\gaunt_{n}(C_{n},C_{n})$.

Suppose now that $X$ is a generator of $\gaunt_{n}$. Then there must exist a functor $X\to C_{n}$ such that the induced map $X_{n}\to\gaunt_{n}(C_{n},C_{n})$ contains the element $\id\in\gaunt_{n}(C_{n},C_{n})$ in its image, for otherwise $\mathcal{C}(X,-)$ would not be able to distinguish $a$ and $b$, contradicting the fact that $X$ is a generator. Thus there exists an $n$-morphism $f$ of $X$ which maps via this functor to $\id\in\gaunt_{n}(C_{n},C_{n})$. Corresponding to $f$ is a section $C_{n}\to X$ that carries the unique nontrivial $n$-morphism of $C_{n}$ to $f$. This exhibits $C_{n}$ as a retract of $X$, as desired. ∎