Lemma 13.9 (0MKH) · ScalingStacks

Lemma 13.9. If $T^{(b)}_{\Theta_{n}}$ coincides with $T_{\Theta_{n}}$ , then so does the class $T^{(a)}_{\Theta_{n}}$ .

Proof. First note that as $\Theta_{n}$ is dense in $\pre(\Theta_{n})$ , and $T_{\Theta_{n}}$ is strongly saturated, it is enough to consider $H\in\Theta_{n}$ representable. Let $f\colon U\to V$ be a morphism in $T_{\Theta_{n}}$ , let $V\to C_{i}$ be given, and let $H\to C_{i}$ be arbitrary. There exists a unique factorization $H\to C_{k}\hookrightarrow C_{i}$ , with $H\to C_{k}$ nondegenerate. Consider the following diagram of pullbacks in $\pre(\Theta_{n})$ :

Since $T^{(c)}_{\Theta_{n}}=T_{\Theta_{n}}$ , we have $[U^{\prime}\to V^{\prime}]\in T_{\Theta_{n}}$ , and if $T^{(b)}_{\Theta_{n}}=T_{\Theta_{n}}$ , then we also have $[U^{\prime\prime}\to V^{\prime\prime}]\in T_{\Theta_{n}}$ , as desired. ∎