Editing Monomorphism (section)

==Examples==
Every morphism in a [[concrete category]] whose underlying [[function (mathematics)|function]] is injective is a monomorphism; in other words, if morphisms are actually functions between sets, then any morphism which is a one-to-one function will necessarily be a monomorphism in the categorical sense.  In the [[category of sets]] the converse also holds, so the monomorphisms are exactly the [[injective]] morphisms.  The converse also holds in most naturally occurring categories of algebras because of the existence of a [[free object]] on one generator.  In particular, it is true in the categories of all groups, of all [[ring (mathematics)|ring]]s, and in any [[abelian category]].

It is not true in general, however, that all monomorphisms must be injective in other categories; that is, there are settings in which the morphisms are functions between sets, but one can have a function that is not injective and yet is a monomorphism in the categorical sense. For example, in the category '''Div''' of [[divisible group|divisible]] [[abelian group|(abelian) group]]s and [[group homomorphism]]s between them there are monomorphisms that are not injective: consider, for example, the quotient map {{nowrap|''q'' : '''Q''' → '''Q'''/'''Z'''}}, where '''Q''' is the rationals under addition, '''Z''' the integers (also considered a group under addition), and '''Q'''/'''Z''' is the corresponding [[quotient group]].  This is not an injective map, as for example every integer is mapped to 0. Nevertheless, it is a monomorphism in this category.  This follows from the implication {{nowrap|1=''q'' ∘ ''h'' = 0 ⇒ ''h'' = 0}}, which we will now prove.  If {{nowrap|''h'' : ''G'' → '''Q'''}}, where ''G'' is some divisible group, and {{nowrap|1=''q'' ∘ ''h'' = 0}}, then {{nowrap|''h''(''x'') ∈ '''Z''', ∀ ''x'' ∈ ''G''}}.  Now fix some {{nowrap|''x'' ∈ ''G''}}.  Without loss of generality, we may assume that {{nowrap|''h''(''x'') ≥ 0}} (otherwise, choose −''x'' instead).  Then, letting {{nowrap|1=''n'' = ''h''(''x'') + 1}}, since ''G'' is a divisible group, there exists some {{nowrap|''y'' ∈ ''G''}} such that {{nowrap|1=''x'' = ''ny''}}, so {{nowrap|1=''h''(''x'') = ''n'' ''h''(''y'')}}.  From this, and {{nowrap|1=0 ≤ ''h''(''x'') < ''h''(''x'') + 1 = ''n''}}, it follows that

:<math>0 \leq \frac{h(x)}{h(x) + 1} = h(y) < 1 </math>

Since {{nowrap|''h''(''y'') ∈ '''Z'''}}, it follows that {{nowrap|1=''h''(''y'') = 0}}, and thus {{nowrap|1=''h''(''x'') = 0 = ''h''(−''x''), ∀ ''x'' ∈ ''G''}}.  This says that {{nowrap|1=''h'' = 0}}, as desired.  

To go from that implication to the fact that ''q'' is a monomorphism, assume that {{nowrap|1=''q''  ∘ ''f'' = ''q'' ∘ ''g''}} for some morphisms {{nowrap|''f'', ''g'' : ''G'' → '''Q'''}}, where ''G'' is some divisible group. Then {{nowrap|1=''q'' ∘ (''f'' − ''g'') = 0}}, where {{nowrap|(''f'' − ''g'') : ''x'' ↦ ''f''(''x'') − ''g''(''x'')}}.  (Since {{nowrap|1=(''f'' − ''g'')(0) = 0}}, and {{nowrap|1=(''f'' − ''g'')(''x'' + ''y'') = (''f'' − ''g'')(''x'') + (''f'' − ''g'')(''y'')}}, it follows that {{nowrap|(''f'' − ''g'') ∈ Hom(''G'', '''Q''')}}).  From the implication just proved, {{nowrap|1=''q''  ∘ (''f'' − ''g'') = 0  ⇒ ''f'' − ''g'' = 0  ⇔ ∀ ''x'' ∈ ''G'', ''f''(''x'') = ''g''(''x'') ⇔ ''f'' = ''g''}}.  Hence ''q'' is a monomorphism, as claimed.