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In category theory, an epimorphism (also called an epic morphism or, colloquially, an epi) is a morphism ''f'' : ''X'' → ''Y'' that is right-cancellative in the sense that, for all morphisms , : Epimorphisms are categorical analogues of surjective functions (and in the category of sets the concept corresponds to the surjective functions), but it may not exactly coincide in all contexts; for example, the inclusion is a ring-epimorphism. The dual of an epimorphism is a monomorphism (i.e. an epimorphism in a category ''C'' is a monomorphism in the dual category ''C''op). Many authors in abstract algebra and universal algebra define an epimorphism simply as an ''onto'' or surjective homomorphism. Every epimorphism in this algebraic sense is an epimorphism in the sense of category theory, but the converse is not true in all categories. In this article, the term "epimorphism" will be used in the sense of category theory given above. For more on this, see the section on Terminology below. ==Examples== Every morphism in a concrete category whose underlying function is surjective is an epimorphism. In many concrete categories of interest the converse is also true. For example, in the following categories, the epimorphisms are exactly those morphisms which are surjective on the underlying sets: *Set, sets and functions. To prove that every epimorphism ''f'': ''X'' → ''Y'' in Set is surjective, we compose it with both the characteristic function ''g''1: ''Y'' → of the image ''f''(''X'') and the map ''g''2: ''Y'' → that is constant 1. *Rel, sets with binary relations and relation preserving functions. Here we can use the same proof as for Set, equipping with the full relation ×. *Pos, partially ordered sets and monotone functions. If ''f'' : (''X'',≤) → (''Y'',≤) is not surjective, pick ''y''0 in ''Y'' \ ''f''(''X'') and let ''g''1 : ''Y'' → be the characteristic function of and ''g''2 : ''Y'' → the characteristic function of . These maps are monotone if is given the standard ordering 0 < 1. *Grp, groups and group homomorphisms. The result that every epimorphism in Grp is surjective is due to Otto Schreier (he actually proved more, showing that every subgroup is an equalizer using the free product with one amalgamated subgroup); an elementary proof can be found in (Linderholm 1970). *FinGrp, finite groups and group homomorphisms. Also due to Schreier; the proof given in (Linderholm 1970) establishes this case as well. *Ab, abelian groups and group homomorphisms. *''K''-Vect, vector spaces over a field ''K'' and ''K''-linear transformations. *Mod-''R'', right modules over a ring ''R'' and module homomorphisms. This generalizes the two previous examples; to prove that every epimorphism ''f'': ''X'' → ''Y'' in Mod-''R'' is surjective, we compose it with both the canonical quotient map ''g'' 1: ''Y'' → ''Y''/''f''(''X'') and the zero map ''g''2: ''Y'' → ''Y''/''f''(''X''). *Top, topological spaces and continuous functions. To prove that every epimorphism in Top is surjective, we proceed exactly as in Set, giving the indiscrete topology which ensures that all considered maps are continuous. *HComp, compact Hausdorff spaces and continuous functions. If ''f'': ''X'' → ''Y'' is not surjective, let ''y'' in ''Y''-''fX''. Since ''fX'' is closed, by Urysohn's Lemma there is a continuous function ''g''1:''Y'' → () such that ''g''1 is 0 on ''fX'' and 1 on ''y''. We compose ''f'' with both ''g''1 and the zero function ''g''2: ''Y'' → (). However there are also many concrete categories of interest where epimorphisms fail to be surjective. A few examples are: *In the category of monoids, Mon, the inclusion map N → Z is a non-surjective epimorphism. To see this, suppose that ''g''1 and ''g''2 are two distinct maps from Z to some monoid ''M''. Then for some ''n'' in Z, ''g''1(''n'') ≠ ''g''2(''n''), so ''g''1(''-n'') ≠ ''g''2(''-n''). Either ''n'' or ''-n'' is in N, so the restrictions of ''g''1 and ''g''2 to N are unequal. *In the category of algebras over commutative ring R, take R() → R(), where R() is the group ring of the group G and the morphism is induced by the inclusion N → Z as in the previous example. This follows from the observation that 1 generates the algebra R() (note that the unit in R() is given by 0 of Z), and the inverse of the element represented by n in Z is just the element represented by -n. Thus any homomorphism from R() is uniquely determined by its value on the element represented by 1 of Z. *In the category of rings, Ring, the inclusion map Z → Q is a non-surjective epimorphism; to see this, note that any ring homomorphism on Q is determined entirely by its action on Z, similar to the previous example. A similar argument shows that the natural ring homomorphism from any commutative ring ''R'' to any one of its localizations is an epimorphism. *In the category of commutative rings, a finitely generated homomorphism of rings ''f'' : ''R'' → ''S'' is an epimorphism if and only if for all prime ideals ''P'' of ''R'', the ideal ''Q'' generated by ''f''(''P'') is either ''S'' or is prime, and if ''Q'' is not ''S'', the induced map Frac(''R''/''P'') → Frac(''S''/''Q'') is an isomorphism (EGA IV 17.2.6). *In the category of Hausdorff spaces, Haus, the epimorphisms are precisely the continuous functions with dense images. For example, the inclusion map Q → R, is a non-surjective epimorphism. The above differs from the case of monomorphisms where it is more frequently true that monomorphisms are precisely those whose underlying functions are injective. As to examples of epimorphisms in non-concrete categories: * If a monoid or ring is considered as a category with a single object (composition of morphisms given by multiplication), then the epimorphisms are precisely the right-cancellable elements. * If a directed graph is considered as a category (objects are the vertices, morphisms are the paths, composition of morphisms is the concatenation of paths), then ''every'' morphism is an epimorphism. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「epimorphism」の詳細全文を読む スポンサード リンク
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