Example 10.2.0.1. Let $\mathbf{Q}$ denote the field of rational numbers, let $\mathbf{Z} \subseteq \mathbf{Q}$ denote the ring of integers, and let $f: \mathbf{Z} \hookrightarrow \mathbf{Q}$ denote the inclusion map. Then $f$ is both a monomorphism and an epimorphism in the category of commutative rings. Consequently, the ring homomorphism $f$ admits (at least) two factorizations as an epimorphism followed by a monomorphism, given by the diagrams

## 10.2 Regular $\infty $-Categories

Let $X$ and $Y$ be sets and let $f: X \rightarrow Y$ be a function. Recall that the *image* of $f$ is the subset $\operatorname{im}(f) \subseteq Y$ consisting of those elements $y \in Y$ satisfying $y = f(x)$ for some element $x \in X$. Writing $i$ for the inclusion of $\operatorname{im}(f)$ into $Y$, the function $f$ then factors as a composition

This factorization admits a more abstract characterization: it is determined (up to unique isomorphism) by the requirements that $i$ is injective (that is, it is a monomorphism in the category of sets) and that $f_0$ is surjective (that is, it is an epimorphism in the category of sets).

The construction $f \mapsto \operatorname{im}(f)$ has counterparts in many other categories. For example, every homomorphism of commutative rings $f: R \rightarrow S$ has a tautological factorization

which is again characterized (up to unique isomorphism) by the requirements that that $i$ is injective and that $f_0$ is surjective. Here the first demand is equivalent to the condition that $i$ is a monomorphism in the category of commutative rings. However, the second demand is more subtle. Every surjective ring homomorphism is an epimorphism in the category of commutative rings, but the converse is false in general.

To address the phenomenon described in Example 10.2.0.1, it is convenient to modify the definition of epimorphism.

Definition 10.2.0.2. Let $\operatorname{\mathcal{C}}$ be a category which admits fiber products. We will say that a morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ is a *regular epimorphism* if it exhibits $Y$ as a coequalizer of the pair of projection maps $\pi _0, \pi _1: X \times _{Y} X \rightarrow X$.

Remark 10.2.0.3. Let $\operatorname{\mathcal{C}}$ be a category which admits fiber products and let $f: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$. Then $f$ is an epimorphism if and only if, for every object $Z \in \operatorname{\mathcal{C}}$, the function

is injective. The condition that $f$ is a regular epimorphism is (in general) stronger: it requires also that the image of $\theta _{Z}$ is the collection of morphisms $h: X \rightarrow Z$ which satisfy the identity $h \circ \pi _0 = h \circ \pi _1$; here $\pi _0$ and $\pi _1$ denote the projection maps from $X \times _{Y} X$ to $X$.

Example 10.2.0.4. Let $\operatorname{\mathcal{C}}= \operatorname{Set}$ be the category of sets and let $f: X \twoheadrightarrow Y$ be an epimorphism in $\operatorname{\mathcal{C}}$: that is, a surjective function. Then $f$ is a regular epimorphism: that is, it exhibits $Y$ as a quotient of the equivalence relation $\equiv _{f}$, defined by the requirement

Exercise 10.2.0.5. Let $f: R \rightarrow S$ be a homomorphism of commutative rings. Show that $f$ is a regular epimorphism (in the category of commutative rings) if and only if it is surjective (as a map of sets). In particular, the inclusion map $\mathbf{Z} \hookrightarrow \mathbf{Q}$ of Example 10.2.0.1 is an epimorphism in the category of commutative rings which is not regular.

Let $\operatorname{\mathcal{C}}$ be a category which admits fiber products and let $f: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$. We will say that an object $Y_0 \in \operatorname{\mathcal{C}}$ is an *image of $f$* if the morphism $f$ factors as a composition

where $i$ is a monomorphism and $f_0$ is a regular epimorphism. It is not difficult to show that if such a factorization exists, then it is uniquely determined up to (canonical) isomorphism: for example, the object $Y_0$ can be recovered as the coequalizer of the pair of projection maps $X \times _ Y X \rightrightarrows X$. To emphasize the uniqueness, we will typically denote the object $Y_0$ by $\operatorname{im}(f)$ and refer to it as *the* image of $f$. This motivates the following:

Definition 10.2.0.6. Let $\operatorname{\mathcal{C}}$ be a category. We say that $\operatorname{\mathcal{C}}$ is *regular* if it satisfies the following conditions:

- $(1)$
The category $\operatorname{\mathcal{C}}$ admits finite limits (in particular, it admits fiber products).

- $(2)$
Every morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ has an image: that is, we can write $f$ as a composition

\[ X \stackrel{f_0}{\twoheadrightarrow } Y_0 \stackrel{i}{\hookrightarrow } Y \]where $i$ is a monomorphism and $f_0$ is a regular epimorphism.

- $(3)$
The collection of regular epimorphisms is stable under the formation of pullbacks. That is, for every pullback diagram

\[ \xymatrix@R =50pt@C=50pt{ X' \ar [d]^{f'} \ar [r] & X \ar [d]^{f} \\ Y' \ar [r] & Y } \]in the category $\operatorname{\mathcal{C}}$, if $f$ is a regular epimorphism, then $f'$ is also a regular epimorphism.

Example 10.2.0.7. The axioms of Definition 10.2.0.6 tend to be satisfied by any category $\operatorname{\mathcal{C}}$ whose objects can be described as “sets with algebraic structure.” For example:

The category of sets is regular.

The category of groups is regular.

The category of abelian groups is regular.

The category of associative rings is regular.

The category of commutative rings is regular.

Example 10.2.0.8. Let $\operatorname{\mathcal{C}}$ be the category of partially ordered sets (with morphisms given by nondecreasing functions). Then $\operatorname{\mathcal{C}}$ is not regular: it satisfies conditions $(1)$ and $(2)$ of Definition 10.2.0.6, but does not satisfy condition $(3)$ (see Exercise 10.2.2.16).

Our goal in this section is to extend Definition 10.2.0.6 to the setting of $\infty $-categories. The first step is to find an appropriate $\infty $-categorical counterpart for the notion of regular epimorphism. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category which admits fiber products. Then, to every morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$, one can associate a diagram

If $\operatorname{\mathcal{C}}$ is (the nerve of) an ordinary category, then $f$ is a regular epimorphism if and only if (10.18) is a coequalizer diagram. In the $\infty $-categorical setting, this condition is almost never satisfied (even if $f$ is an isomorphism). To guarantee that a morphism $g: X \rightarrow Z$ factors (up to homotopy) through $f$, it is typically not enough to know that $g \circ \pi _0$ is homotopic to $g \circ \pi _1$: one needs a homotopy satisfying further coherence conditions, whose formalization involves iterated fiber products $X \times _{Y} X \times _ Y \cdots \times _{Y} X$. Recall that the collection of all such fiber products can be organized into an augmented simplicial object $\operatorname{\check{C}}_{\bullet }(X/Y)$ called the *Čech nerve* of $f$ (Definition 10.1.5.4), which we display informally as

We will say that $f$ is a *quotient morphism* if $\operatorname{\check{C}}_{\bullet }(X/Y)$ is a colimit diagram in the $\infty $-category $\operatorname{\mathcal{C}}$.

Remark 10.2.0.9. In §10.2.2, we adopt a slightly different definition of quotient morphism (Definition 10.2.2.1), which makes sense in any $\infty $-category $\operatorname{\mathcal{C}}$ (that is, we do not need to assume that $\operatorname{\mathcal{C}}$ admits fiber products). Our definition is formulated using the language of *sieves*, which we review in §10.2.1. When $\operatorname{\mathcal{C}}$ admits fiber products, the sieve-theoretic definition reduces to the requirement that $\operatorname{\check{C}}_{\bullet }(X/Y)$ is a colimit diagram (see Proposition 10.2.2.4).

Warning 10.2.0.10. Let $\operatorname{\mathcal{C}}= \operatorname{N}_{\bullet }( \operatorname{\mathcal{C}}_0 )$ be the nerve of an ordinary category $\operatorname{\mathcal{C}}_0$, and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. Then $f$ is a quotient morphism if and only if it is a regular epimorphism in $\operatorname{\mathcal{C}}_0$, in the sense of Definition 10.2.0.2 (see Corollary 10.2.2.7). In particular, every quotient morphism in $\operatorname{\mathcal{C}}$ is an epimorphism. Beware that, if $\operatorname{\mathcal{C}}$ is not assumed to be the nerve of an ordinary category, then the analogous statement is false: quotient morphisms in $\operatorname{\mathcal{C}}$ are usually not epimorphisms (that is, they are not monomorphisms when viewed as morphisms in the opposite $\infty $-category $\operatorname{\mathcal{C}}^{\operatorname{op}}$). See Warning 10.2.2.10).

Let $\operatorname{\mathcal{C}}$ be an $\infty $-category containing a morphism $f: X \rightarrow Y$. We will say that an object $Y_0 \in \operatorname{\mathcal{C}}$ is an *image of $f$* if there exists a diagram

where $f_0$ is a quotient morphism and $i$ is a monomorphism (Definition 10.2.3.1). In §10.2.3, we will show that if such a diagram exists, then it is unique up to isomorphism (in fact, up to a contractible space of choices: see Proposition 10.2.3.14). Following our discussion of the classical case, we will typically denote the object $Y_0$ by $\operatorname{im}(f)$ and refer to it as the *image* of the morphism $f$ (Notation 10.2.3.12).

Let $\operatorname{\mathcal{C}}$ be an $\infty $-category which admits fiber products. Beware that, in general, the collection of quotient morphisms in $\operatorname{\mathcal{C}}$ is not closed under pullback (see Exercise 10.2.2.16). We say that a morphism $f: X \rightarrow Y$ if $\operatorname{\mathcal{C}}$ is a *universal quotient morphism* if, for every pullback diagram

the morphism $f'$ is a quotient morphism. In §10.2.4, we extend this definition to the setting of an $\infty $-category $\operatorname{\mathcal{C}}$ which does not necessarily admit pullbacks (see Definition 10.2.4.1 and Corollary 10.2.4.7) and study its properties. The notion of universal quotient morphism is in some respects better behaved than the notion of quotient morphism: for example, the collection of universal quotient morphisms is always closed under composition (Proposition 10.2.4.12), while the collection of quotient morphisms need not be (Exercise 10.2.2.15).

Armed with a good theory of quotient morphisms, we can formulate an $\infty $-categorical analogue of Definition 10.2.0.6. We say that an $\infty $-category $\operatorname{\mathcal{C}}$ is *regular* if it satisfies the following axioms (Definition 10.2.5.1):

- $(1)$
The $\infty $-category $\operatorname{\mathcal{C}}$ admits finite limits.

- $(2)$
Every morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ has an image: that is, we can write $f$ as a composition of a quotient morphism $X \twoheadrightarrow Y_0$ with a monomorphism $Y_0 \hookrightarrow Y$.

- $(3)$
The collection of quotient morphisms in $\operatorname{\mathcal{C}}$ is stable under pullback: that is, every quotient morphism is a universal quotient morphism.

In §10.2.5, we discuss various formulations of this definition and give some examples of regular $\infty $-categories. In particular, we show that the $\infty $-category of spaces $\operatorname{\mathcal{S}}$ is regular (Corollary 10.2.5.6) and that the collection of regular $\infty $-categories is closed under the formation of slice constructions (Proposition 10.2.5.9) and left exact localization (Proposition 10.2.5.19).