ScalingStacks

Proof. Without loss of generality we may identify ${\mathcal{A}}$ with its essential image in ${\mathcal{B}}$, so that an arrow $f\colon X\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Y$ is in $T$ if and only if any cofiber $Z$ of $f$ lies in ${\mathcal{A}}$. By [52, 5.5.4.15] it suffices to show that $T\subseteq\overline{S}$, the strongly saturated class of arrows of ${\mathcal{B}}$ generated by $S$ (see [52, 5.5.4.5]). To see this, let $X\overset{f}{\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}}Y\overset{g}{\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}}Z$ be a cofiber sequence in ${\mathcal{B}}$ such that $Z$ is in ${\mathcal{A}}$. Then $Z=\colim_{\alpha}Z_{\alpha}$ is a $\kappa$-filtered colimit of objects $Z_{\alpha}\in{\mathcal{A}}^{\kappa}\subset{\mathcal{B}}^{\kappa}$ and $Y=\colim_{\alpha}Y_{\alpha}$ is a $\kappa$-filtered colimit of objects $Y_{\alpha}=Y\times_{Z}Z_{\alpha}$. Now $Y_{\alpha}$ may not be $\kappa$-compact, so write $Y_{\alpha}=\colim_{\beta}Y_{\alpha\beta}$ for some $Y_{\alpha\beta}\in{\mathcal{B}}^{\kappa}$ and consider the resulting diagram of cofiber sequences

in which the lower right and upper left squares are cartesian, which implies that these two squares are also cocartesian and that the maps $X_{\alpha}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}X$ and $Z_{\alpha\beta}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Z_{\alpha}$ are equivalences. Hence $Z_{\alpha\beta}\in{\mathcal{A}}^{\kappa}\subseteq{\mathcal{B}}^{\kappa}$ and we conclude that $g_{\alpha\beta}$ and therefore $f_{\alpha\beta}$ as well are maps in ${\mathcal{B}}^{\kappa}$; in particular, $f_{\alpha\beta}$ is an arrow in $S$. It follows from [52, 5.5.4.5] that the pushout $f_{\alpha}$ of $f_{\alpha\beta}$ along $X_{\alpha\beta}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}X_{\alpha}\simeq X$ is in $\overline{S}$, and we see from [52, 5.5.4.12] that $f\simeq\colim_{\alpha}f_{\alpha}:X\simeq\colim_{\alpha}X_{\alpha}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\colim Y_{\alpha}\simeq Y$ is then also in $\overline{S}$. ∎