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 $S=T$. ∎

Proof. Let ${\mathcal{C}}$ be a presentable stable $\infty$-category, and note that a colimit-preserving functor ${\mathcal{B}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{C}}$ sends the arrows in $S$ to equivalences in ${\mathcal{C}}$ if and only if its restriction to ${\mathcal{A}}$ is trivial. We therefore may identify

with the full subcategory spanned by those colimit-preserving functors ${\mathcal{B}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{C}}$ which send the arrows in $S$ to equivalences in ${\mathcal{C}}$. It follows from lemma 5.5 that ${\mathcal{B}}/{\mathcal{A}}\simeq T^{-1}{\mathcal{B}}$, where $T$ is the strongly saturated class of arrows of ${\mathcal{B}}$ which become equivalences in ${\mathcal{B}}/{\mathcal{A}}$.

We now show that $S$ is strongly saturated, so that $S=T$. First, suppose given a cofiber sequence $X\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Y\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Z$ in ${\mathcal{B}}$ such that $Z$ lies in the essential image of ${\mathcal{A}}$, and let $X\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}X^{\prime}$ be any map. Then the cofiber of $X^{\prime}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}X^{\prime}\coprod_{X}Y$ is equivalence to $Z$, so is also in the essential image of ${\mathcal{A}}$. Second, given a diagram $f_{\alpha}\colon X_{\alpha}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Y_{\alpha}$ in $\mathrm{Fun}(\Delta^{1},{\mathcal{B}})$ with colimit $f\colon X\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Y$, and suppose that the cofibers $Z_{\alpha}$ of each $f_{\alpha}$ lies in the essential image of ${\mathcal{A}}$. Commuting colimits implies that the cofiber $Z$ of $f$ is computed as the colimit of the $Z_{\alpha}$, and this lies in the essential image of ${\mathcal{A}}$ since ${\mathcal{A}}$ is closed under colimits and the functor ${\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}$ preserves colimits. Lastly, suppose $h=g\circ f$ is a composite of $f\colon X\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Y$ followed by $g\colon Y\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Z$, and write $Y/X$, $Z/Y$, and $Z/X$ for the cofibers of $f$, $g$, and $h$, respectively. Then we have a cofiber sequence $Y/X\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Z/X\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}Z/Y$, so if any two lie in the essential image of ${\mathcal{A}}$ then so does the third. ∎

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 $\overline{S}$. ∎

Proof. By construction, ${\mathcal{B}}/{\mathcal{A}}\subseteq{\mathcal{B}}$ is the full subcategory on those objects $b$ such that $\map(a,b)\simeq*$ for all objects $a$ in the image of ${\mathcal{A}}$. This shows that, as full subcategories of $\Ho({\mathcal{B}})$, $\Ho({\mathcal{B}}/{\mathcal{A}})\subseteq\Ho({\mathcal{B}})/\Ho({\mathcal{A}})$. Conversely, if $b$ is in $\Ho({\mathcal{B}})/\Ho({\mathcal{A}})$, then $\pi_{0}\map(a,b)\simeq\ast$ for each object $a$ in the image of ${\mathcal{A}}$, and we claim that in fact $\map(a,b)\simeq*$. Indeed, ${\mathcal{A}}$ is a stable subcategory of ${\mathcal{B}}$, so that $\pi_{n}\map(a,b)\simeq\pi_{0}\map(\Sigma^{n}a,b)\simeq\ast$. Hence $\Ho({\mathcal{B}})/\Ho({\mathcal{A}})\subseteq\Ho({\mathcal{B}}/{\mathcal{A}})$ as well. ∎

Proof. The fully faithful inclusions ${\mathcal{A}}\subset\Ind_{\kappa}({\mathcal{A}})$ and ${\mathcal{B}}\subset\Ind_{\kappa}({\mathcal{B}})$ show that ${\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}$ is fully faithful if and only if $\Ind_{\kappa}({\mathcal{A}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ind_{\kappa}({\mathcal{B}})$ is fully faithful (for the reverse direction, this follows from the definition of the mapping spaces in $\Ind_{\kappa}(-)$). Thus it remains to check that ${\mathcal{B}}/{\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{C}}$ is an equivalence upon idempotent completion if and only if $\Ind_{\kappa}({\mathcal{B}})/\Ind_{\kappa}({\mathcal{A}})\simeq\Ind_{\kappa}({\mathcal{B}}/{\mathcal{A}})$. Since $\Ind_{\kappa}$ preserves cofibers, it is enough to check that the equivalence $\Ind_{\kappa}({\mathcal{B}})/\Ind_{\kappa}({\mathcal{A}})\simeq\Ind_{\kappa}({\mathcal{B}}/{\mathcal{A}})$ implies the equivalence ${\mathcal{B}}/{\mathcal{A}}\simeq{\mathcal{C}}$ whenever the latter are idempotent complete. Thus, given a $\kappa$-cocomplete small stable $\infty$-category ${\mathcal{D}}$ (which we assume is idempotent complete if $\kappa=\omega$), we must show that

is a fiber sequence of $\infty$-categories. Since ${\mathcal{D}}\simeq(\Ind_{\kappa}({\mathcal{D}}))^{\kappa}$, by adjunction this is equivalent to the sequence

which is a fiber sequence by assumption. ∎