ScalingStacks

Proof. Forming each of the pushouts of $S_{00}$ in $\gaunt_{n}$ yields an equivalence, so $X$ is local with respect to $S_{00}$.

Now let $S_{0}^{\prime}\subseteq S_{0}$ denote the class of morphisms $f:U\to V$ in $S_{0}$ such that $\nu X$ is local with respect to $f$ for any gaunt $X$. We have observed that $S_{0}^{\prime}$ contains $S_{00}$. It is also visibly closed under isomorphism.

We complete the proof by showing that $S^{\prime}_{0}$ is closed under the operation $-\times_{C_{i}}N$ for any $N\in\Upsilon_{n}$. Indeed, suppose $U\to V$ a morphism of $S_{0}^{\prime}$. We claim that for any morphism $V\to C_{k}$ and any functor $N\to C_{k}$, the map

$$\Upsilon_{n}(V\times_{C_{k}}N,X)\to\Upsilon_{n}(U\times_{C_{k}}N,X)$$

is a bijection. For each $W\in\Upsilon_{n}$, we have

	$\displaystyle\gaunt_{n}(W\times_{C_{i}}N,X)$	$\displaystyle\cong(\gaunt_{n}/C_{k})(W\times_{C_{k}}N,X\times C_{k})$
		$\displaystyle\cong(\gaunt_{n}/C_{k})(W,\uHom_{C_{k}}(N,X\times C_{k})),$

where $\uHom_{C_{k}}(N,-)$ denotes the right adjoint of Lemma 5.6. The claim now follows from the observation that as $\uHom_{C_{i}}(N,X\times C_{i})$ is gaunt, it is local with respect to $U\to V$. ∎