Contents

category theory

# Contents

## Idea

In category theory, the concept of epimorphism is a generalization or strengthening of the concept of surjective functions between sets (example below).

The formally dual concept is that of monomorphism, similarly related to the concept of injective function.

Common jargon includes “is a mono” or “is monic” for “is a monomorphism”, and “is an epi” or “is epic” for “is an epimorphism”, and “is an iso” for “is an isomorphism”.

## Definition

A morphism $f \colon X \to Y$ in some category is called an epimorphism (sometimes abbrieviated to epi, or described as being epic), if for every other object $Z$ and every pair of parallel morphisms $g_1,g_2 \colon Y \to Z$ then

$\left( g_1\circ f \,=\, g_2 \circ f \right) \;\Rightarrow\; \left( g_1 = g_2 \right) \,.$

Stated more abstractly, this says that $f$ is an epimorphism precisely if for every object $Z$ the hom-functor $Hom(-,Z)$ sends it to an injective function

$Hom(Y,Z) \xhookrightarrow{\;\; f^* \;\;} Hom(X,Z)$

between hom-sets. Since injective functions are the monomorphisms in Set (example below) this means that $f$ is an epimorphism precisely if $Hom(f,Z)$ is a monomorphism for all $Z$.

Finally, this means that $f$ is an epimorphism in a category $\mathcal{C}$ precisely if it is a monomorphism in the opposite category $\mathcal{C}^{op}$.

## Examples

### General

###### Example

Every isomorphism is both an epimorphism and a monomorphism.

###### Example

(epimorphisms of sets)

The epimorphisms in the category Set of sets are precisely the surjective functions.

Thus the concept of epimorphism may be thought of as a category-theoretic generalization of the concept of surjection.

But beware that in categories of sets with extra structure, epimorphisms need not be surjective (in contrast to monomorphisms, which are usually injective).

###### Example

(epimorphisms of sheaves)

A morphism in the category of sheaves of sets over a topological space $X$ is epimorphism iff the induced map on the level of each stalk is surjective. See also math.stackexchange here.

###### Example

(epimorphisms of rings)

In the categories Ring or CRing of (commutative) rings and ring homomorphisms between them, a morphism $f: R\to S$ is epi iff the canonical inclusion of the right module $S_R$ into its extension of scalars along $f$, $S_R\to S_R\otimes_R S$ is a bijection (hence isomorphism), or equivalently, the multiplication map $S\otimes S\to S$ factors to an isomorphism $S\otimes_R S\to S$ of $R$-bimodules.

It follows that the multiplication induces natural isomorphism of endofunctors with components $M\mapsto M\otimes_R S\otimes_R S\cong M\otimes_R S$. While the extension of scalars followed by restriction of scalars is a monad (by adjunction), this implies that it is an idempotent monad, hence the extension of scalars is a localization functor.

In both categories, every surjective ring homomorphisms is an epimorphism, however there are many epimorphisms which are not surjective maps. For example, every flat localization of rings (e.g. Ore localization) is the same as a flat epimorphism; (nontrivial) localizations are not surjective.

Any faithfully flat epimorphism of rings is an isomorphism.

An epimorphism of commutative rings is surjective iff it is a finite morphism. For epimorphisms of commutative rings see Stacks Project, 10.106 Epimorphisms of rings and MathOverflow what-do-epimorphisms-of-commutative-rings-look-like.

Often, though, the surjections correspond to a stronger notion of epimorphism.

But beware that the converse fails:

### Examples of monos that are epi but not iso

The following lists some examples of morphisms that are both monomorphisms and epimorphisms, but not necessarily isomorphisms.

###### Example

In the category of Hausdorff topological spaces, the inclusion $A \hookrightarrow X$ of a dense subspace is an epimorphism.

See this Prop. for proof.

###### Example

In unital Rings, the canonical inclusion $\mathbb{Z} \overset{i}{\hookrightarrow} \mathbb{Q}$ of the integers into the rational numbers is an epimorphism.

See this Prop. for proof.

## Properties

###### Proposition

The following are equivalent:

• $f : x \to y$ is an epimorphism in $C$;

• $f$ is a monomorphism in the opposite category $C^{op}$;

• precomposition with $f$ is a monomorphism in Set: that is, for all $c \in C$, $- \circ f : Hom(y,c) \to Hom(x,c)$ is an injection;

• $\array{ x &\stackrel{f}{\to}& y \\ {}^{\mathllap{f}}\downarrow && \downarrow^{\mathrlap{Id}} \\ y &\underset{Id}{\to}& y }$

is a pushout diagram.

###### Proposition

If $f \colon x \to y$ and $g \colon y \to z$ are epimorphisms, so is their composite $g f$. If $g f$ is an epimorphism, so is $g$.

###### Proposition

Every coequalizer $x \to y$

$z \stackrel{\to}{\to} x \to y$

is an epimorphism.

The converse of the above proposition fails, and an epimorphism is called a regular epimorphism if it is the coequalizer of some pair of morphisms.

###### Proposition

Epimorphisms are preserved by pushout: if $f \colon x \to y$ is an epimorphism and

$\array{ x &\to& a \\ {}^{\mathllap{f}}\downarrow && \downarrow^{g} \\ y &\to& b }$

is a pushout diagram, then also $g$ is an epimorphism.

###### Proof

Let $h_1,h_2 : b \to c$ be two morphisms such that $\stackrel{g}{\to} \stackrel{h_1}{\to} = \stackrel{g}{\to} \stackrel{h_2}{\to}$. Then by the commutativity of the diagram also $x \to y \to b \stackrel{h_1}{\to} c$ equals $x \to y \to b \stackrel{h_2}{\to} c$. Since $x \to y$ is assumed to be epi, it follows that $y \to b \stackrel{h_1}{\to} c$ equals $y \to b \stackrel{h_2}{\to} c$. But this means that $h_1$ and $h_2$ define the same cocone. By the universality of the pushout $b$ there is a unique map of cocones from $b$ to $c$. Hence $h_1$ must equal $h_2$. Therefore $g$ is epi.

###### Proposition

Epimorphisms are preserved by any left adjoint functor, or more generally any functor that preserves pushouts: if $F \colon C \to D$ is a functor that preserves pushouts and $f \in Mor(C)$ an epimorphism then $F(f) \in Mor(D)$ is an epimorphism.

###### Proof

If $F : C \to D$ is a left adjoint we can argue this way: by the adjunction natural isomorphism we have for all $d \in Obj(D)$

$Hom_D(L(f),d) \simeq Hom_C(f,R(d)) \,.$

The right hand is a monomorphism by assumption, hence so is the left hand, hence $L(f)$ is epi.

More generally, if $F$ preserves pushouts we can use the fact that $f$ is epic iff

$\array{ x &\stackrel{f}{\to}& y \\ {}^{\mathllap{f}}\downarrow && \downarrow^{\mathrlap{Id}} \\ y &\underset{Id}{\to}& y }$

is a pushout diagram.

###### Proposition

Epimorphisms are reflected by faithful functors.

###### Proof

Let $F \colon \mathcal{C}\longrightarrow \mathcal{D}$ be a faithful functor. Consider $f \colon x \longrightarrow y$ a morphism in $\mathcal{C}$ such that $F(f) \colon F(x)\longrightarrow F(y)$ is an epimorphism in $\mathcal{D}$. We need to show that then $f$ itself is an epimorphism.

So consider morphisms $g,h \colon y \longrightarrow z$ such that $g \circ f = h \circ f$. We need to show that this implies that already $g = h$ (injectivity of $Hom(f,z)$). But functoriality implies that $F(g)\circ F(f) = F(h) \circ F(f)$, and since $F(f)$ is epic this implies that $F(g) = F(h)$. Now the statement follows with the assumption that $F$ is faithful, hence injective on morphisms.

Epimorphisms get along with colimits in a number of ways, some of which we have seen above. Here is another:

###### Proposition
1. Any morphism to an initial object is an epimorphism.

2. The coproduct of epimorphisms is an epimorphism.

###### Proof

For the first suppose $0 \in C$ is initial and $f : x \to 0$. Given morphisms $g,h: 0 \to y$ with $g \circ f = h \circ f$ we have $g = h$ simply because $0$ is initial.

For the second suppose $f_1 : x_1 \to y_1$ and $f_2 : x_2 \to y_2$ are epimorphisms; we wish to show that $f_1 + f_2 : x_1+x_2 \to y_1 + y_2$ is an epimorphism. Suppose we have morphisms $g, h: y_1+y_2 \to z$ with $g \circ (f_1+f_2) = h \circ (f_1 + f_2)$. Then $g \circ i_1 \circ f_1 = h \circ i_1 \circ f_1$ where $i_1 : y_1 \to y_1 + y_2$ is the canonical map into the coproduct. Since $f_1$ is epic we conclude $g \circ i_1 = h \circ i_1$. Similarly we have $g \circ i_2 = h \circ i_2$. It follows that $g = h$.

Epimorphisms do not get along quite as well with limits. For example, the projections from a Cartesian product onto its factors, e.g. $p_1 \colon x_1 \times x_2 \to x_1$, are not always epimorphisms (even in $Set$: take $x_2$ to be empty).

## Variations

There are a sequence of variations on the concept of epimorphism, which conveniently arrange themselves in a total order. In order from strongest to weakest, we have:

In the category of sets, every epimorphism is effective descent (and even split if you believe the axiom of choice). Thus, it can be hard to know, when generalising concepts from Set to other categories, what kind of epimorphism to use. The following discussion may be helpful in this regard.

First we note:

• Descent and effective descent morphisms are only defined in a category with pullbacks. The other notions can be defined in any category, although of course for an effective epimorphism one must in general assert the existence of the kernel pair.

Moreover, if the category has finite limits, then the picture becomes much simpler:

• If a strict epimorphism has a kernel pair, then it is effective and hence also regular. Thus, in a category with pullbacks, effective = regular = strict. Probably for this reason, there is substantial variation among authors in their use of these words; some use “effective epi” or “regular epi” to mean what we have called a “strict epi”.

• Likewise, in a category with pullbacks, every extremal epimorphism is strong, since monomorphisms are always pullback-stable.

• Moreover, in a category with equalizers, strong and extremal epimorphisms do not need to explicitly be asserted to be epic; that follows from the other condition in their definition.

Also worth noting are:

• In a regular category, every extremal epimorphism is a descent morphism (i.e. a pullback-stable regular epimorphism). Thus in this case there remain only four types of epimorphism: split, effective descent, regular, and plain.

• In an exact category, or a category that has pullback-stable reflexive coequalizers (which implies that it is regular), any regular epimorphism is effective descent. Thus in this case we have only three types: split, regular, and plain.

• In a pretopos (hence also in a topos), every epimorphism is regular, leaving only two types: split and plain. The collapsing of these two types into one is called the axiom of choice for that category.

Thus, in general, the two serious distinctions come

• Between split epimorphisms and regular ones: in very few categories are all regular epimorphisms split. Splitting of even regular epimorphisms is a form of the axiom of choice, which may be valid in Set (if you believe it) but very often fails internally.

• Between extremal epimorphisms and “plain” epimorphisms: in many categories, the plain epimorphisms are oddly behaved, but the extremal ones are what we would expect. For instance, the inclusion $\mathbb{Z}\hookrightarrow\mathbb{Q}$ is an epimorphism of rings, but the extremal epimorphisms of rings are just the surjective ring homomorphisms. More generally, in all algebraic categories (categories of algebra for a Lawvere theory), which are regular, the regular epimorphisms are the morphisms whose underlying function is surjective.

Moreover, even in non-regular categories, there seems to be a strong tendency for strong/extremal epimorphisms to coincide with regular/strict ones. For example, this is the case in Top, where both are the class of quotient maps. (The plain epimorphisms are the surjective continuous functions.)

However, the distinction is real. For instance, in the category generated by the following graph:

$\array{ &&&& C\\ &&& ^f\nearrow\\ A& \underoverset{h}{k}{\rightrightarrows} & B \\ &&& _g\searrow\\ &&&& D}$

subject to the equations $f h = f k$ and $g h = g k$, both $f$ and $g$ are strong, but not strict, epimorphisms.

Textbook accounts: