Kerodon

Proof. The existence of the left adjoint $L: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}'$ follows from our assumption that $\operatorname{\mathcal{C}}'$ is a reflective subcategory of $\operatorname{\mathcal{C}}$, and the $(\kappa ,\lambda )$-compactness of $L$ follows from our assumption that $\iota $ is $(\kappa ,\lambda )$-finitary (Example 9.4.2.17). Let $\operatorname{\mathcal{C}}'_0 \subseteq \operatorname{\mathcal{C}}'$ be the full subcategory spanned by objects of the form $L(C)$, where $C \in \operatorname{\mathcal{C}}$ is $(\kappa ,\lambda )$-compact. We will complete the proof by showing that $\operatorname{\mathcal{C}}'$ is an $\operatorname{Ind}_{\kappa }^{\lambda }$-completion of $\operatorname{\mathcal{C}}'_0$. Since the functor $L$ is $(\kappa ,\lambda )$-compact, each object of $\operatorname{\mathcal{C}}'_0$ is $(\kappa ,\lambda )$-compact when viewed as an object of $\operatorname{\mathcal{C}}'$. It will therefore suffice to show that every object $X \in \operatorname{\mathcal{C}}'$ can be the colimit of a $\lambda $-small $\kappa $-filtered diagram in $\operatorname{\mathcal{C}}'_0$. Without loss of generality, we may assume that $X = L(Y)$ for some object $Y \in \operatorname{\mathcal{C}}$. Using Corollary 9.3.4.17, we realize $Y$ as the colimit of a $\lambda $-small $\kappa $-filtered diagram $q: \operatorname{\mathcal{K}}\rightarrow \operatorname{\mathcal{C}}$ carrying each object of $\operatorname{\mathcal{K}}$ to a $(\kappa ,\lambda )$-compact object of $\operatorname{\mathcal{C}}$. It follows that $X = L(Y)$ is the colimit of the diagram $(L \circ q): \operatorname{\mathcal{K}}\rightarrow \operatorname{\mathcal{C}}'_0$. $\square$

Proof. We first show that $(1)$ implies $(2)$. Let $\widetilde{X}$ be an essentially finite Kan complex and let $u: \widetilde{X} \rightarrow X$ be a morphism which exhibits $X$ as an $n$-truncation of $\widetilde{X}$. Our assumption that $\widetilde{X}$ is essentially finite guarantees that there is a weak homotopy equivalence $w: K \rightarrow \widetilde{X}$, where $K$ is a finite simplicial set. Since $u$ is $(n+1)$-connective, the composite map $(u \circ w): K \rightarrow X$ is also $(n+1)$-connective.

We now prove the converse. Suppose there is a finite simplicial set $K$ and an $(n+1)$-connective morphism $f: K \rightarrow X$. Using Proposition 3.1.8.1, we can factor $f$ as a composition $K \xrightarrow {w} \widetilde{X} \xrightarrow {u} X$, where $w$ is anodyne (hence a weak homotopy equivalence) and $u$ is a Kan fibration (so that $\widetilde{X}$) is a Kan complex. Since $f$ is $(n+1)$-connective, $u$ is also $(n+1)$-connective, and therefore exhibits $X$ as an $n$-truncation of $\widetilde{X}$. $\square$

Proof. The implication $(1) \Rightarrow (2)$ follows from Corollary 4.7.6.22, since every essentially finite Kan complex is also essentially finite when regarded as an $\infty $-category (Proposition 9.2.7.4). Conversely, suppose that condition $(2)$ is satisfied. Then there is a categorically $(n+1)$-connective morphism $f: K \rightarrow X$, where $K$ is a finite simplicial set. Applying Remark 4.8.8.5, we deduce that $f$ is $(n+1)$-connective, so that $X$ belongs to $\operatorname{\mathcal{S}}^{\leq n}_{\mathrm{fin}}$. $\square$

Proof. The case $n = -1$ is vacuous, since $\operatorname{\mathcal{QC}}^{\leq -1}_{\mathrm{fin}} = \operatorname{\mathcal{QC}}^{\leq -1}$ (see Example 4.7.6.5). Let us therefore assume that $n \geq 0$. Note that the initial object of $\operatorname{\mathcal{QC}}^{\leq n}$ is the empty $\infty $-category $\emptyset $, which is essentially finite. It will therefore suffice to show that $\operatorname{\mathcal{QC}}^{\leq n}_{\mathrm{fin}}$ is closed under pushouts (Corollary 7.6.2.42). Suppose we are given a pushout square

in the $\infty $-category $\operatorname{\mathcal{QC}}^{\leq n}$, where $\operatorname{\mathcal{C}}$, $\operatorname{\mathcal{C}}_0$, and $\operatorname{\mathcal{C}}_1$ belong to $\operatorname{\mathcal{QC}}^{\leq n}_{\mathrm{fin}}$; we wish to show that $\operatorname{\mathcal{C}}_{01}$ also belongs to $\operatorname{\mathcal{QC}}^{\leq n}_{\mathrm{fin}}$.

Let $L: \operatorname{\mathcal{QC}}\rightarrow \operatorname{\mathcal{QC}}^{\leq n}$ be a left adjoint to the inclusion map, given on objects by the construction $\operatorname{\mathcal{E}}\mapsto \operatorname {h}_{\mathit{\leq {}n}}{\mathit{(\operatorname{\mathcal{E}})}}$ (see Variant 6.2.2.14). Without loss of generality, we may assume that $\operatorname{\mathcal{C}}= L(\operatorname{\mathcal{D}})$, $\operatorname{\mathcal{C}}_0 = L( \operatorname{\mathcal{D}}_0 )$, and $\operatorname{\mathcal{C}}_1 = L( \operatorname{\mathcal{D}}_1 )$ for essentially finite $\infty $-categories $\operatorname{\mathcal{D}}$, $\operatorname{\mathcal{D}}_0$, and $\operatorname{\mathcal{D}}_1$. By virtue of Remark 9.4.3.11, we may further assume that there is a categorical equivalence $K \rightarrow \operatorname{\mathcal{D}}$, where $K$ is a simplicial set of dimension $\leq (n+1)$.

Note that the reflection map $\operatorname{\mathcal{D}}_0 \rightarrow L(\operatorname{\mathcal{D}}_0) = \operatorname{\mathcal{C}}_0$ is categorically $(n+1)$-connective (Example 4.8.5.19). Applying Proposition 4.8.5.25, we conclude that the induced map $\operatorname{Fun}(K, \operatorname{\mathcal{D}}_0) \rightarrow \operatorname{Fun}(K, \operatorname{\mathcal{C}}_0)$ is essentially surjective, and therefore the functor $\operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{D}}_0 ) \rightarrow \operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{C}}_0 ) \simeq \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{C}}_0)$ is also essentially surjective. We may therefore assume without loss of generality that $F_0 = L(G_0)$, for some functor $G_0: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}_0$. Similarly, we can assume that $F_1 = L(G_1)$ for some functor $G_1: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}_1$. Since the functor $L: \operatorname{\mathcal{QC}}\rightarrow \operatorname{\mathcal{QC}}^{\leq n}$ preserves finite colimits, it follows that $\operatorname{\mathcal{C}}_{01}$ is a local $(n-1)$-truncation of an $\infty $-category $\operatorname{\mathcal{D}}_{01}$ which is a colimit of the diagram $\operatorname{\mathcal{D}}_0 \xleftarrow {G_0} \operatorname{\mathcal{D}}\xrightarrow {G_1} \operatorname{\mathcal{D}}_1$ in $\operatorname{\mathcal{QC}}$. To complete the proof, it will suffice to show that $\operatorname{\mathcal{D}}_{01}$ is essentially finite as an $\infty $-category, which is a special case of Corollary 9.2.7.7. $\square$

Proof. Let $\operatorname{\mathcal{QC}}^{\leq n}_{0} \subseteq \operatorname{\mathcal{QC}}^{\leq n}$ be the smallest full subcategory which contains $\Delta ^1$ and is closed under finite colimits; we wish to show that $\operatorname{\mathcal{QC}}^{\leq n}_{\mathrm{fin}}$ is contained in $\operatorname{\mathcal{QC}}^{\leq n}_{0}$ (the reverse inclusion follows from Proposition 9.4.3.14). Let $\operatorname{\mathcal{QC}}_0 \subseteq \operatorname{\mathcal{QC}}$ be the full subcategory spanned by those $\infty $-categories $\operatorname{\mathcal{D}}$ satisfying $\operatorname {h}_{\mathit{\leq {}n}}{\mathit{(\operatorname{\mathcal{D}})}} \in \operatorname{\mathcal{QC}}^{\leq n}_{0}$. Then $\operatorname{\mathcal{QC}}_0$ contains $\Delta ^1$ and is closed under finite colimits, and therefore contains all essentially finite $\infty $-categories (Proposition 9.2.7.8). If $\operatorname{\mathcal{C}}$ belongs to $\operatorname{\mathcal{QC}}^{\leq n}_{\mathrm{fin}}$, then $\operatorname{\mathcal{C}}$ is equivalent to $\operatorname {h}_{\mathit{\leq {}n}}{\mathit{(\operatorname{\mathcal{D}})}}$ for some essentially finite $\infty $-category $\operatorname{\mathcal{D}}$, so that $\operatorname{\mathcal{C}}\in \operatorname{\mathcal{QC}}^{\leq n}_{0}$ as desired. $\square$

Proof. We may assume without loss of generality that $n \geq 0$ (otherwise, $\operatorname{\mathcal{S}}^{\leq n}_{\mathrm{fin}} = \operatorname{\mathcal{S}}^{\leq n}$ and there is nothing to prove). Note that the inclusion functor $\operatorname{\mathcal{S}}^{\leq n} \hookrightarrow \operatorname{\mathcal{QC}}^{\leq n}$ has a right adjoint (given by the construction $\operatorname{\mathcal{C}}\mapsto \operatorname{\mathcal{C}}^{\simeq }$; see Example 6.2.2.25), and therefore preserves small colimits (Corollary 7.1.4.28). The desired result now follows by combining Propositions 9.4.3.14 and 9.4.3.9. $\square$

Proof. We proceed as in the proof of Corollary 9.4.3.15. Let $\operatorname{\mathcal{S}}^{\leq n}_{0} \subseteq \operatorname{\mathcal{S}}^{\leq n}$ be the smallest full subcategory which contains $\Delta ^0$ and is closed under finite colimits; we wish to show that $\operatorname{\mathcal{S}}^{\leq n}_{\mathrm{fin}}$ is contained in $\operatorname{\mathcal{S}}^{\leq n}_{0}$ (the reverse inclusion follows from Corollary 9.4.3.16). Let $\operatorname{\mathcal{S}}_0 \subseteq \operatorname{\mathcal{S}}$ be the full subcategory spanned by those Kan complexes $X$ satisfying $\pi _{\leq n}(X) \in \operatorname{\mathcal{S}}^{\leq n}_{0}$. Then $\operatorname{\mathcal{S}}_0$ contains $\Delta ^0$ and is closed under finite colimits, and therefore contains all essentially finite Kan complexes (Proposition 9.2.6.3). If a Kan complex $Y$ belongs to $\operatorname{\mathcal{S}}^{\leq n}_{\mathrm{fin}}$, then it is homotopy equivalent to $\pi _{\leq n}(X)$ for some essentially finite Kan complex $X$, so that $Y \in \operatorname{\mathcal{S}}^{\leq n}_{0}$ as desired. $\square$

Proof. We will assume that $n \geq -1$ (the case $n \leq -2$ is trivial: see Example 9.4.3.7). It follows from Corollary 5.5.3.21 that the $\infty $-category $\operatorname{\mathcal{QC}}^{\leq n}_{\mathrm{fin}}$ is locally $n$-truncated, and from Proposition 9.4.3.14 that it admits finite colimits. Applying Variant 8.5.4.3, we deduce that $\operatorname{\mathcal{QC}}^{\leq n}_{\mathrm{fin}}$ is idempotent-complete. Since the full subcategory $\operatorname{\mathcal{S}}^{\leq n}_{\mathrm{fin}} \subseteq \operatorname{\mathcal{QC}}^{\leq n}_{\mathrm{fin}}$ is closed under retracts (Remark 8.5.1.19), it is also idempotent-complete (Proposition 8.5.4.10). $\square$

Proof of Proposition 9.4.3.9. Let $n$ be an integer and let $X$ be an $n$-truncated Kan complex. Applying Proposition 9.4.3.4, we see that $X$ is compact as an object of the $\infty $-category $\operatorname{\mathcal{S}}^{\leq n}$ if and only if is a retract of $\pi _{\leq n}(Y)$, for some Kan complex $Y$ which is compact as an object of $\operatorname{\mathcal{S}}$. If this condition is satisfied, then $Y$ is a retract of an essentially finite Kan complex $Z$ (Proposition 9.2.6.7), so that $X$ is also a retract of $\pi _{\leq n}(Z)$ (Remark 8.5.1.6). Since $\pi _{\leq n}(Z)$ belongs to $\operatorname{\mathcal{S}}^{\leq n}_{\mathrm{fin}}$, it follows that $X$ belongs to $\operatorname{\mathcal{S}}^{\leq n}_{\mathrm{fin}}$ (Corollary 9.4.3.18). $\square$