Kerodon

Proof. The equivalence of $(1) \Leftrightarrow (2)$ is a special case of Corollary 10.3.4.8, and the implication $(2) \Rightarrow (3)$ is immediate from the definitions. We next show that $(3)$ implies $(4)$. Fix a diagram of the form (10.27). If condition $(3)$ is satisfied, then we can choose a diagram

in the $\infty $-category $\operatorname{\mathcal{C}}$ where $q$ is a universal quotient morphism, $i$ is a monomorphism, and the lower vertical composition coincides with $f$. Since $\operatorname{\mathcal{C}}$ admits finite limits, this diagram admits a right Kan extension

so that the right square and outer rectangle are pullback diagrams. By construction, the inverse image $u^{-1}( \operatorname{im}(f) )$ is the isomorphism class of the fiber product $Y' \times _{Y} Y_0$ (regarded as an object of the $\infty $-category $\operatorname{\mathcal{C}}_{/Y'}$ via the morphism $i'$). On the other hand, the uniqueness of limits guarantees that $Y' \times _{Y} X$ is isomorphic to $X'$ as an object of $\operatorname{\mathcal{C}}_{/Y'}$, so the image of $f'$ coincides with the image of the composite morphism $i' \circ q'$. To prove that this image coincides with $[Y' \times _{Y} Y_0]$, it suffices to show that $q'$ is a quotient morphism in $\operatorname{\mathcal{C}}$. This follows from Corollary 10.3.4.7, since the left half of (10.28) is also a pullback square (Proposition 7.6.2.28).

We now complete the proof by showing that $(4)$ implies $(1)$. Suppose we are given a pullback square (10.27), where $f$ is a quotient morphism; we wish to show that $f'$ is also a quotient morphism. By virtue of Proposition 10.3.3.6, the assumption that $f$ is a quotient morphism guarantees that $\operatorname{im}(f) = [Y]$ is the largest element of $\operatorname{Sub}(Y)$, and we wish to show that $\operatorname{im}(f') = [Y']$ is the largest element of $\operatorname{Sub}(Y')$. This follows immediately from $(4)$, since the inverse image construction $u^{-1}: \operatorname{Sub}(Y) \rightarrow \operatorname{Sub}(Y')$ preserves largest elements (see Construction 9.3.4.32). $\square$

Proof. Corollary 7.4.1.3 guarantees that $\operatorname{\mathcal{S}}$ admits finite limits. By virtue of Corollary 10.3.5.5, it will suffice to show that every map of Kan complexes $f: X \rightarrow Y$ factors as a composition $i \circ q$, where $q: X \twoheadrightarrow Y_0$ is a universal quotient morphism in $\operatorname{\mathcal{S}}$ and $i: Y_0 \hookrightarrow Y$ is a monomorphism in $\operatorname{\mathcal{S}}$. For this, we can take $i: Y_0 \hookrightarrow Y$ to be the inclusion of the essential image of $f$ (which is a monomorphism by Example 9.3.4.10), and $q: X \rightarrow Y_0$ to be the restriction of $f$ (which is a universal quotient morphism by Proposition 10.3.4.17. $\square$

Proof. It follows from Remark 7.1.3.11 that the $\infty $-category $\operatorname{\mathcal{C}}_{/Z}$ admits finite limits. By virtue of Corollary 10.3.5.5, it will suffice to show that every morphism $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ can be realized as the composition of a universal quotient morphism $\widetilde{X} \twoheadrightarrow \widetilde{Y}_0$ with a monomorphism $\widetilde{Y}_0 \hookrightarrow \widetilde{Y}$. Let $f: X \rightarrow Y$ denote the image of $\widetilde{f}$ in the $\infty $-category $\operatorname{\mathcal{C}}$. Since $\operatorname{\mathcal{C}}$ is regular, we can choose a $2$-simplex $\sigma :$

of $\operatorname{\mathcal{C}}$, where $q$ is a universal quotient morphism and $i$ is a monomorphism. The inclusion map $\operatorname{N}_{\bullet }( \{ 0 < 2 \} ) \hookrightarrow \Delta ^2$ is right anodyne (Lemma 4.3.7.8), so we can lift $\sigma $ to a $2$-simplex

in the $\infty $-category $\operatorname{\mathcal{C}}_{/Z}$. We conclude by observing that $\widetilde{i}$ is a monomorphism (Remark 9.3.4.24) and $\widetilde{q}$ is a universal quotient morphism (Proposition 10.3.4.15). $\square$

Proof. The implication $(1) \Rightarrow (2)$ follows from the observation that $F$ preserves monomorphisms (Proposition 9.3.4.21), and the implication $(2) \Rightarrow (3)$ is immediate from the definitions. We will complete the proof by showing that $(3)$ implies $(1)$. Let $f: X \twoheadrightarrow Y$ be a quotient morphism in $\operatorname{\mathcal{C}}$; we wish to show that $F(f)$ is a quotient morphism in $\operatorname{\mathcal{D}}$. By virtue of Proposition 10.3.3.6, our hypothesis can be reformulated as an equality $\operatorname{im}(f) = [Y]$ in the partially ordered set $\operatorname{Sub}(Y)$, and we wish to prove an equality $\operatorname{im}( F(f) ) = [ F(Y) ]$ in the partially ordered set $\operatorname{Sub}( F(Y) )$. This is clear, since the map $\operatorname{Sub}(Y) \rightarrow \operatorname{Sub}( F(Y) )$ preserves largest elements (see Remark 10.3.5.11). $\square$

Proof. Let $f: X \twoheadrightarrow Y$ be a quotient morphism in $\operatorname{\mathcal{C}}$; we wish to that $F(f)$ is quotient morphism in $\operatorname{\mathcal{D}}$. Let $\operatorname{\check{C}}_{\bullet }(X/Y): \operatorname{N}_{\bullet }( \operatorname{{\bf \Delta }}_{+}^{\operatorname{op}} ) \rightarrow \operatorname{\mathcal{C}}$ be a Čechnerve of $f$ (Notation 10.2.5.5. Since $f$ is a quotient morphism (Remark 10.3.4.2), $\operatorname{\check{C}}_{\bullet }(X/Y)$ is a colimit diagram in $\operatorname{\mathcal{C}}$ (Proposition 10.3.2.4). Our assumption that $F$ preserves geometric realizations guarantees that $F \circ \operatorname{\check{C}}_{\bullet }(X/Y)$ is a colimit diagram in the $\infty $-category $\operatorname{\mathcal{D}}$. Since $F$ preserves finite limits, $F \circ \operatorname{\check{C}}_{\bullet }(X/Y)$ is a Čechnerve of the morphism $F(f): F(X) \rightarrow F(Y)$. Applying Proposition 10.3.2.4 again, we deduce that $F(f)$ is a quotient morphism in $\operatorname{\mathcal{D}}$. $\square$

Proof. Assume that $q$ is a quotient morphism in $\operatorname{\mathcal{C}}$; we will show that it is also a quotient morphism in $\operatorname{\mathcal{C}}_0$ (the analogous assertion for universal quotient morphisms then follows from the criterion of Corollary 10.3.4.7). Since $\operatorname{\mathcal{C}}$ admits pullbacks and $\operatorname{\mathcal{C}}_0$ is stable under the formation of pullbacks, it follows that $\operatorname{\mathcal{C}}_0$ also admits pullbacks. Applying Proposition 10.2.5.6, we deduce that $q$ admits a Čechnerve $\operatorname{\check{C}}_{\bullet }(X/Y): \operatorname{N}_{\bullet }( \operatorname{{\bf \Delta }}_{+}^{\operatorname{op}} ) \rightarrow \operatorname{\mathcal{C}}_0$, which is also a Čechnerve of $q$ in the $\infty $-category $\operatorname{\mathcal{C}}$. Since $q$ is a quotient morphism, $\operatorname{\check{C}}_{\bullet }(Z/Y)$ is a colimit diagram in $\operatorname{\mathcal{C}}$ (Proposition 10.3.2.4). It follows that $\operatorname{\check{C}}_{\bullet }(X/Y)$ is also a colimit diagram in $\operatorname{\mathcal{C}}_0$, so that $q$ is a quotient morphism in $\operatorname{\mathcal{C}}_0$. $\square$

Proof. Let $f: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$. Since $\operatorname{\mathcal{C}}$ is regular, $f$ can be factored as the composition of a universal quotient morphism $q: X \twoheadrightarrow Y_0$ with a monomorphism $i: Y_0 \hookrightarrow Y$. If $X$ and $Y$ belong to $\operatorname{\mathcal{C}}_0$, then our assumption guarantees that we can arrange that $Y_0$ is also contained in $\operatorname{\mathcal{C}}_0$. In this case, $i$ is also a monomorphism in the $\infty $-category $\operatorname{\mathcal{C}}_0$, and Proposition 10.3.5.17 guarantees that $q$ is a universal quotient morphism in the subcategory $\operatorname{\mathcal{C}}_0$. Allowing $f$ to vary and invoking Corollary 10.3.5.5, we conclude that $\operatorname{\mathcal{C}}_0$ is regular. $\square$

Proof. Since the functor $F$ has a right adjoint, it preserves geometric realizations of simplicial objects (Corollary 7.1.4.22). Applying Proposition 10.3.5.16, we deduce that the functor $F$ carries quotient morphisms in $\operatorname{\mathcal{C}}$ to quotient morphisms in $\operatorname{\mathcal{D}}$. It will therefore suffice to show that $\operatorname{\mathcal{D}}$ is regular.

It follows from Corollary 7.1.4.27 (together with Corollary 6.2.2.19) that the $\infty $-category $\operatorname{\mathcal{D}}$ admits finite limits. We next show that every morphism $v: D \rightarrow D'$ in $\operatorname{\mathcal{D}}$ has an image. Let $\epsilon : (F \circ G) \rightarrow \operatorname{id}_{\operatorname{\mathcal{D}}}$ be the counit of an adjunction between $F$ and $G$. Since $G$ is fully faithful, the natural transformation $\epsilon $ is an isomorphism. We can therefore replace $v$ by the morphism $(F \circ G)(v)$, and thereby reduce to the case where $v = F(u)$ for some morphism $u: C \rightarrow C'$ in $\operatorname{\mathcal{C}}$. In this case, our assumption that $\operatorname{\mathcal{C}}$ has images guarantees that we can factor $u$ as a composition $C \stackrel{q}{\twoheadrightarrow } C'_0 \stackrel{i}{\hookrightarrow } C'$, where $q$ is a quotient morphism and $i$ is a monomorphism (in the $\infty $-category $\operatorname{\mathcal{C}}$). It follows that $v$ can be written as the composition of $F(q)$ (which is a quotient morphism in $\operatorname{\mathcal{D}}$, as noted above) with $F(i)$ (which is a monomorphism in $\operatorname{\mathcal{D}}$ by virtue of Proposition 9.3.4.21). In particular, the object $F'(C_0)$ is an image of $v$.

We now complete the proof by showing that if

is a pullback diagram in $\operatorname{\mathcal{C}}$ where $f$ is a quotient morphism, then $f'$ is also a quotient morphism. Since $\operatorname{\mathcal{C}}$ is regular, we can choose a $2$-simplex

which exhibits $Y$ as an image of $G(f)$. It follows that $F(\sigma )$ exhibits $F(Y)$ as an image of the morphism $(F \circ G)(f)$ in the $\infty $-category $\operatorname{\mathcal{D}}$. Note that $(F \circ G)(f)$ is isomorphic to $f$ (as an object of $\operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{D}})$), and is therefore a quotient morphism (Corollary 10.3.2.12). Applying Proposition 10.3.3.6, we conclude that $F(i)$ is an isomorphism in $\operatorname{\mathcal{D}}$.

Amalgamating the $2$-simplex $\sigma $ with $G(g)$, we obtain a diagram

in the $\infty $-category $\operatorname{\mathcal{C}}$. Since $\operatorname{\mathcal{C}}$ admits finite limits, this diagram admits a right Kan extension

so that the square on the right and the outer rectangle are pullback squares. Note that, after applying the functor $F$, the outer rectangle of this diagram is isomorphic to (10.29). We are therefore reduced to showing that the functor $F$ carries the upper horizontal composition in (10.30) to a quotient morphism in $\operatorname{\mathcal{D}}$. Since $F$ preserves pullback squares, $F(i')$ is a pullback of $F(i)$ and is therefore an isomorphism (Corollary 7.6.2.27). Using Corollary 10.3.2.12, we are reduced to showing that $F(q')$ is a quotient morphism in $\operatorname{\mathcal{D}}$. In fact, we claim that $q'$ is a pullback morphism of $\operatorname{\mathcal{C}}$. This follows from our assumption that $\operatorname{\mathcal{C}}$ is regular, since $q$ is a quotient morphism by construction and the left half the diagram (10.30) is a pullback square (Proposition 7.6.2.28). $\square$