ScalingStacks

Proof. By Proposition 4.7 the restriction of $E$ to $\mathbb{G}_{n}$ is necessarily of the form $\rho(g)$ for some $g\in(\mathbb{Z}/2)^{n}$. It suffices to prove that $E\circ\rho(g)\simeq\rho(g)\circ\rho(g)$, which is in turn simply the identity, so without loss of generality we may assume $\rho(g)=\id$, that is, that $E$ restricts to the identity functor on $\mathbb{G}_{n}$.

Now in this case, the isomorphisms

$$\gaunt_{n}(C_{k},X)\cong\gaunt_{n}(F(C_{k}),F(X))=\gaunt_{n}(C_{k},F(X))$$

are natural in both $C_{k}$ and $X$, whence one obtains a natural isomorphism $\gamma\colon U\cong U\circ F$, where $U\colon\gaunt_{n}\to\Fun(\mathbb{G}_{n}^{\mathrm{op}},\set)$ is the forgetful functor from gaunt $n$-categories to globular sets. It thus remains to show that this natural isomorphism is compatible with the compositions $\ast_{j}$.

For this, consider

$$w_{j}^{k}\colon C_{k}\to C_{k}\cup^{C_{j-1}}C_{k},$$

the morphism that corepresents the composition law $\ast_{j}$, as above. Since $E$ preserves compositional pushouts, one obtains a commutative diagram

Hence for any gaunt $n$-category $X$, we obtain a commutative diagram

in which the top and bottom morphisms are exactly the composition functors. Hence the natural isomorphism $\gamma$ is compatible with compositions, whence it lifts to a natural isomorphism $\id\cong E$, as desired. ∎