Monomorphisms and epimorphisms

The classes of monomorphisms, epimorphisms, bimorphisms, split monomorphisms, and split epimorphisms are closed under composition and contain all isomorphisms. To show these classes are closed under composition, consider two morphisms $f:A\to B$ and $g:B\to C$. If $f$ and $g$ are monomorphisms then for $a,a'\in\operatorname{hom}(X,A)$,

and so $gf$ is a monomorphism. If $f$ and $g$ are epimorphisms then for $c,c'\in\operatorname{hom}(C,X)$,

and so $gf$ is an epimorphism. If $f$ and $g$ are split monomorphisms then there exists $F:B\to A$ and $G:C\to B$ such that $Ff=1_A$ and $Gg=1_B$. So $gf$ is a split monomorphism since $(FG)(gf)=1_A$. Similarly, if $g$ and $f$ are split epimorphisms then $(gf)(FG)=1_C$ and so $gf$ is a split epimorphism.

Let $f:A\to B$ be an isomorphism then, $f$ has inverse $g:B\to A$ and so for every pair of morphisms $a,a'\in\operatorname{hom}(X,A)$ and $b,b'\in\operatorname{hom}(B,X)$ we have that

and

Hence, every isomorphism is a monomorphism and an epimorphism and so it is a bimorphism. Also, every isomorphism is a split monomorphism and a split epimorphism since it has a unique inverse.

Let $f:A\to B$ and $g:B\to C$ be two morphisms:

• If the composition $gf$ is a monomorphism then for $a,a'\in\operatorname{hom}(X,A)$, $fa=fa'\implies gfa=gfa'\implies a=a'$ and so $f$ is a monomorphism.
• If the composition $gf$ is an epimorphism then for $c,c'\in\operatorname{hom}(C,X)$, $cg=c'g\implies cgf=c'gf\implies c=c'$ and so $g$ is an epimorphism.
• If the composition $gf$ is a split monomorphism then there exists a morphism $h:C\to A$ such that $h(gf)=1_A$, so $(hg)f=1_A$ and hence, $f$ is a split monomorphism.
• If the composition $gf$ is a split epimorphism then there exists a morphism $h:C\to A$ such that $(gf)h=1_C$, so $g(fh)=1_C$ and hence, $g$ is a split epimorphism.
• If $f:A\to B$ is a split monomorphism then there is a morphism $g:B\to A$ with $gf=1_A$ and so $(fg)f=f(gf)=f1_A=f=1_Bf$ and since $f$ is an epimorphism, we have that $fg=1_B$. Hence, $f$ is an isomorphism.

The axiom of choice can be stated as follows.

For every surjective map of sets $f:A\to B$, there exists a map $g:B\to A$ with $fg=1_B$ ¹.

Using this, and the fact that every epimorphism in $\\textbf{Sets}$ is a surjective map of sets, it is easy to see that the axiom of choice can be formulated as

Every epimorphism in $\\textbf{Sets}$ is split ¹.

The inclusion map from the additive monoid of natural numbers to the additive group of integers (considered as a monoid) is a bimorphism in $\\textbf{Mon}$. Let $\mathbb{N}=(\mathbb{N},+,0)$ denote the additive monoid of natural numbers and $\mathbb{Z}=(\mathbb{Z},+,0)$ denote the additive group of integers considered as a monoid. Then for every object $X$ in $\\textbf{Mon}$ and for every pair of morphisms $a,a'\in\operatorname{hom}(X,\mathbb{N})$ we have that if $ia=ia'$ then for every $x\in X$, $ia(x)=ia'(x)$ and so $i(a(x))=i(a'(x))\implies a(x)=a'(x)$. Hence, $i$ is a monomorphism. Additionally, for every pair of morphisms $b,b'\in\operatorname{hom}(\mathbb{Z},X)$ and $n\in\mathbb{N}$, $bi(n)=b'i(n)\implies b(\alpha)=b'(\alpha)$ for all $\alpha\in\mathbb{Z}^+$ with $\alpha = i(n)$. But this means $b(0)=b'(0)$ and so $b(\alpha)+b^{-1}(\alpha)=b(\alpha -\alpha)=b(0)=b'(0)=b'(\alpha -\alpha) = b'(\alpha)+(b')^{-1}(\alpha)$ since $b(0)=b'(0)=0$. Thus, we have that $b(-\alpha) = b^{-1}(\alpha)=(b')^{-1}(\alpha) = b'(-\alpha)$ and so $b=b'$. Hence, $i$ is an epimorphism and so it is a bimorphism.