Several kinds of homomorphisms have a specific name, which is also defined for general
morphisms. s, labelled as "Inner").
Isomorphism An
isomorphism between
algebraic structures of the same type is commonly defined as a
bijective homomorphism. In the more general context of
category theory, an isomorphism is defined as a
morphism that has an
inverse that is also a morphism. In the specific case of algebraic structures, the two definitions are equivalent, although they may differ for non-algebraic structures, which have an underlying set. More precisely, if f: A\to B is a (homo)morphism, it has an inverse if there exists a homomorphism g: B\to A such that f\circ g = \operatorname{Id}_B \qquad \text{and} \qquad g\circ f = \operatorname{Id}_A. If A and B have underlying sets, and f: A \to B has an inverse g, then f is bijective. In fact, f is
injective, as f(x) = f(y) implies x = g(f(x)) = g(f(y)) = y, and f is
surjective, as, for any x in B, one has x = f(g(x)), and x is the image of an element of A. Conversely, if f: A \to B is a bijective homomorphism between algebraic structures, let g: B \to A be the map such that g(y) is the unique element x of A such that f(x) = y. One has f \circ g = \operatorname{Id}_B \text{ and } g \circ f = \operatorname{Id}_A, and it remains only to show that is a homomorphism. If * is a binary operation of the structure, for every pair x, y of elements of B, one has g(x*_B y) = g(f(g(x))*_Bf(g(y))) = g(f(g(x)*_A g(y))) = g(x)*_A g(y), and g is thus compatible with *. As the proof is similar for any
arity, this shows that g is a homomorphism. This proof does not work for non-algebraic structures. For example, for
topological spaces, a morphism is a
continuous map, and the inverse of a bijective continuous map is not necessarily continuous. An isomorphism of topological spaces, called
homeomorphism or bicontinuous map, is thus a bijective continuous map, whose inverse is also continuous.
Endomorphism An
endomorphism is a homomorphism whose
domain equals the
codomain, or, more generally, a
morphism whose source is equal to its target. This means that a (homo)morphism f:A \to B is a monomorphism if, for any pair g, h of morphisms from any other object C to A, then f \circ g = f \circ h implies g = h. These two definitions of
monomorphism are equivalent for all common algebraic structures. More precisely, they are equivalent for
fields, for which every homomorphism is a monomorphism, and for
varieties of
universal algebra, that is algebraic structures for which operations and axioms (identities) are defined without any restriction (the fields do not form a variety, as the
multiplicative inverse is defined either as a
unary operation or as a property of the multiplication, which are, in both cases, defined only for nonzero elements). In particular, the two definitions of a monomorphism are equivalent for
sets,
magmas,
semigroups,
monoids,
groups,
rings,
fields,
vector spaces and
modules. A
split monomorphism is a homomorphism that has a
left inverse and thus it is itself a right inverse of that other homomorphism. That is, a homomorphism f\colon A \to B is a split monomorphism if there exists a homomorphism g\colon B \to A such that g \circ f = \operatorname{Id}_A. A split monomorphism is always a monomorphism, for both meanings of
monomorphism. For sets and vector spaces, every monomorphism is a split monomorphism, but this property does not hold for most common algebraic structures.
An injective homomorphism is left cancelable: If f\circ g = f\circ h, one has f(g(x))=f(h(x)) for every x in C, the common source of g and h. If f is injective, then g(x) = h(x), and thus g = h. This proof works not only for algebraic structures, but also for any
category whose objects are sets and arrows are maps between these sets. For example, an injective continuous map is a monomorphism in the category of
topological spaces. For proving that, conversely, a left cancelable homomorphism is injective, it is useful to consider a
free object on x. Given a
variety of algebraic structures a free object on x is a pair consisting of an algebraic structure L of this variety and an element x of L satisfying the following
universal property: for every structure S of the variety, and every element s of S, there is a unique homomorphism f: L\to S such that f(x) = s. For example, for sets, the free object on x is simply \{x\}; for
semigroups, the free object on x is \{x, x^2, \ldots, x^n, \ldots\}, which, as, a semigroup, is isomorphic to the additive semigroup of the positive integers; for
monoids, the free object on x is \{1, x, x^2, \ldots, x^n, \ldots\}, which, as, a monoid, is isomorphic to the additive monoid of the nonnegative integers; for
groups, the free object on x is the
infinite cyclic group \{\ldots, x^{-n}, \ldots, x^{-1}, 1, x, x^2, \ldots, x^n, \ldots\}, which, as, a group, is isomorphic to the additive group of the integers; for
rings, the free object on x is the
polynomial ring \mathbb{Z}[x]; for
vector spaces or
modules, the free object on x is the vector space or free module that has x as a basis.
If a free object over x exists, then every left cancelable homomorphism is injective: let f\colon A \to B be a left cancelable homomorphism, and a and b be two elements of A such f(a) = f(b). By definition of the free object F, there exist homomorphisms g and h from F to A such that g(x) = a and h(x) = b. As f(g(x)) = f(h(x)), one has f \circ g = f \circ h, by the uniqueness in the definition of a universal property. As f is left cancelable, one has g = h, and thus a = b. Therefore, f is injective.
Existence of a free object on x for a variety (see also ): For building a free object over x, consider the set W of the
well-formed formulas built up from x and the operations of the structure. Two such formulas are said equivalent if one may pass from one to the other by applying the axioms (
identities of the structure). This defines an
equivalence relation, if the identities are not subject to conditions, that is if one works with a variety. Then the operations of the variety are well defined on the set of
equivalence classes of W for this relation. It is straightforward to show that the resulting object is a free object on x.
Epimorphism In
algebra,
epimorphisms are often defined as
surjective homomorphisms. The importance of these structures in all mathematics, especially in
linear algebra and
homological algebra, may explain the coexistence of two non-equivalent definitions. Algebraic structures for which there exist non-surjective epimorphisms include
semigroups and
rings. The most basic example is the inclusion of
integers into
rational numbers, which is a homomorphism of rings and of multiplicative semigroups. For both structures it is a monomorphism and a non-surjective epimorphism, but not an isomorphism. A wide generalization of this example is the
localization of a ring by a multiplicative set. Every localization is a ring epimorphism, which is not, in general, surjective. As localizations are fundamental in
commutative algebra and
algebraic geometry, this may explain why in these areas, the definition of epimorphisms as right cancelable homomorphisms is generally preferred. A
split epimorphism is a homomorphism that has a
right inverse and thus it is itself a left inverse of that other homomorphism. That is, a homomorphism f\colon A \to B is a split epimorphism if there exists a homomorphism g\colon B \to A such that f\circ g = \operatorname{Id}_B. A split epimorphism is always an epimorphism, for both meanings of
epimorphism. For sets and vector spaces, every epimorphism is a split epimorphism, but this property does not hold for most common algebraic structures. In summary, one has \text {split epimorphism} \implies \text{epimorphism (surjective)}\implies \text {epimorphism (right cancelable)}; the last implication is an equivalence for sets, vector spaces, modules, abelian groups, and groups; the first implication is an equivalence for sets and vector spaces. Let f\colon A \to B be a homomorphism. We want to prove that if it is not surjective, it is not right cancelable. In the case of sets, let b be an element of B that not belongs to f(A), and define g, h\colon B \to B such that g is the
identity function, and that h(x) = x for every x \in B, except that h(b) is any other element of B. Clearly f is not right cancelable, as g \neq h and g \circ f = h \circ f. In the case of vector spaces, abelian groups and modules, the proof relies on the existence of
cokernels and on the fact that the
zero maps are homomorphisms: let C be the cokernel of f, and g\colon B \to C be the canonical map, such that g(f(A)) = 0. Let h\colon B\to C be the zero map. If f is not surjective, C \neq 0, and thus g \neq h (one is a zero map, while the other is not). Thus f is not cancelable, as g \circ f = h \circ f (both are the zero map from A to C). == Kernel ==