Editing Homomorphism (section)

===Epimorphism===

In [[algebra]], '''epimorphisms''' are often defined as [[surjective]] homomorphisms.<ref name="Birkhoff.1967"/>{{rp|134}}<ref name="Burris.Sankappanavar.2012" />{{rp|43}} On the other hand, in [[category theory]], [[epimorphism]]s are defined as '''right cancelable''' [[morphism]]s.<ref name=workmath/> This means that a (homo)morphism <math>f: A \to B</math> is an epimorphism if, for any pair <math>g</math>, <math>h</math> of morphisms from <math>B</math> to any other object <math>C</math>, the equality <math>g \circ f = h \circ f</math> implies <math>g = h</math>.

A surjective homomorphism is always right cancelable, but the converse is not always true for algebraic structures. However, the two definitions of ''epimorphism'' are equivalent for [[set (mathematics)|sets]], [[vector space]]s, [[abelian group]]s, [[module (mathematics)|modules]] (see below for a proof), and [[group (mathematics)|groups]].<ref>{{cite journal |last=Linderholm |first=C. E. |year=1970 |title=A group epimorphism is surjective |journal=The American Mathematical Monthly |volume=77 |issue=2 |pages=176–177|doi=10.1080/00029890.1970.11992448 }}</ref> 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 [[semigroup]]s and [[ring (mathematics)|rings]]. The most basic example is the inclusion of [[integer]]s into [[rational number]]s, 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.<ref name=workmath/><ref>{{cite book | page=363 | title=Hopf Algebra: An Introduction | zbl=0962.16026 | series=Pure and Applied Mathematics | volume=235 | location=New York City | publisher=Marcel Dekker | first1=Sorin | last1=Dăscălescu | first2=Constantin | last2=Năstăsescu | first3=Șerban | last3=Raianu | year=2001 | isbn=0824704819 }}</ref>

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 [[inverse function#Left and right inverses|right inverse]] and thus it is itself a left inverse of that other homomorphism. That is, a homomorphism <math>f\colon A \to B</math> is a split epimorphism if there exists a homomorphism <math>g\colon B \to A</math> such that <math>f\circ g = \operatorname{Id}_B.</math> 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 

<math display="block">\text {split epimorphism} \implies \text{epimorphism (surjective)}\implies \text {epimorphism (right cancelable)};</math>

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.
{{collapse top|Equivalence of the two definitions of epimorphism}}
Let <math>f\colon A \to B</math> be a homomorphism. We want to prove that if it is not surjective, it is not right cancelable.

In the case of sets, let <math>b</math> be an element of <math>B</math> that not belongs to <math>f(A)</math>, and define <math>g, h\colon B \to B</math> such that <math>g</math> is the [[identity function]], and that <math>h(x) = x</math> for every <math>x \in B,</math> except that <math>h(b)</math> is any other element of <math>B</math>. Clearly <math>f</math> is not right cancelable, as <math>g \neq h</math> and <math>g \circ f = h \circ f.</math>

In the case of vector spaces, abelian groups and modules, the proof relies on the existence of [[cokernel]]s and on the fact that the [[zero map]]s are homomorphisms: let <math>C</math> be the cokernel of <math>f</math>, and <math>g\colon B \to C</math> be the canonical map, such that <math>g(f(A)) = 0</math>. Let <math>h\colon B\to C</math> be the zero map. If <math>f</math> is not surjective, <math>C \neq 0</math>, and thus <math>g \neq h</math> (one is a zero map, while the other is not). Thus <math>f</math> is not cancelable, as <math>g \circ f = h \circ f</math> (both are the zero map from <math>A</math> to <math>C</math>).
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