ScalingStacks

Proof. If $S$ is of small generation then $S^{-1}{\mathcal{C}}$ is presentable and corepresents the functor $\mathrm{Fun}^{\mathrm{L}}_{S}({\mathcal{C}},-)$ by [52, 5.5.4.14, 5.5.4.20]. Conversely, if this functor is corepresentable by ${\mathcal{C}}^{\prime}$ then the identity ${\mathcal{C}}^{\prime}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{C}}^{\prime}$ determines a colimit-preserving functor ${\mathcal{C}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{C}}^{\prime}$. Let $T$ be the class of arrows in ${\mathcal{C}}$ which become invertible in ${\mathcal{C}}^{\prime}$, and note that $S\subseteq T$, $T$ is strongly saturated [52, 5.5.4.10], and $T$ is of small generation [52, 5.5.4.16] (the last claim uses the fact that the equivalences in ${\mathcal{C}}^{\prime}$ is the strongly saturated class generated by the identity of the initial object of ${\mathcal{C}}^{\prime}$, which follows from [52, 5.5.4.5, 5.5.4.6]). Thus $T^{-1}{\mathcal{C}}\simeq{\mathcal{C}}^{\prime}$, so ${\mathcal{C}}^{\prime}$ also corepresents the functor $\mathrm{Fun}^{\mathrm{L}}_{T}({\mathcal{C}},-)$, showing that a colimit-preserving functor ${\mathcal{C}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{D}}$ inverts the arrows of $S$ if and only if it inverts the arrows of $T$. Since $S$ is strongly saturated, we conclude that $S=T$. ∎