An epimorphism in a category $C$ is a morphism $f : X \to Y$ such that every contravariant hom-functor $Hom(-,Z)$ sends it to an injection
In more elementary terms, $f: X \to Y$ is an epimorphism if, given any $g, h: Y \to Z$, such that the composites $\stackrel{f}{\to} \stackrel{g}{\to} = \stackrel{f}{\to} \stackrel{h}{\to}$ are equal, then already $g = h$:
The epimorphisms in Set are the surjective functions; thus epimorphisms can be thought of as a categorical notion of surjection. However, this is frequently not quite right: in categories of sets with extra structure, epimorphisms need not be surjective (unlike the case for monomorphisms, which are usually injective). Often, though, the surjections correspond to a stronger notion of epimorphism.
The following are equivalent
$f : x \to y$ is an epimorphism in $C$;
$f$ is a monomorphism in the opposite category $C^{op}$;
for all $d$, $Hom(f,d)$ is a monomorphism in Set – an injection;
is a pushout diagram.
Epimorphisms are preserved by pushout: if $f : x \to y$ is an epimorphism and
is a pushout diagram, then also $g$ is an epimorphism.
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.
Epimorphisms are preserved by left adjoint functors:
if $F : C \to D$ is a functor which is left adjoint then for $f \in Mor(C)$ an epimorphism also $F(f) \in Mor(D)$ is an epimorphism
One argument is this:
By the adjunction natural isomorphism we have for all $d \in Obj(D)$
The right hand is a monomorphism by assumption, hence so is the left hand, hence $L(f)$ is epi.
Another argument is this: use that by the above $f$ is epi precisely if
is a pushout diagram and observe that left adjoint functors preserve pushouts (and of course identities).
Epimorphisms are reflected by faithful functors.
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 epi this implies that $F(g) = F(h)$. Now the statement follows with the assumption that $F$ is faithful, hence injective on morphisms.
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:
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:
subject to the equations $f h = f k$ and $g h = g k$, both $f$ and $g$ are strong, but not strict, epimorphisms.