# Image (category theory) – Wikipedia

In category theory, a branch of mathematics, the image of a morphism is a generalization of the image of a function.

## General definition

Given a category

${displaystyle C}$

and a morphism

${displaystyle fcolon Xto Y}$

in

${displaystyle C}$

, the image[1]
of

${displaystyle f}$

is a monomorphism

${displaystyle mcolon Ito Y}$

satisfying the following universal property:

1. There exists a morphism
${displaystyle ecolon Xto I}$

such that

${displaystyle f=m,e}$

.

2. For any object
${displaystyle I’}$

with a morphism

${displaystyle e’colon Xto I’}$

and a monomorphism

${displaystyle m’colon I’to Y}$

such that

${displaystyle f=m’,e’}$

, there exists a unique morphism

${displaystyle vcolon Ito I’}$

such that

${displaystyle m=m’,v}$

.

Remarks:

1. such a factorization does not necessarily exist.

2. ${displaystyle e}$

is unique by definition of

${displaystyle m}$

monic.

3. ${displaystyle m’e’=f=me=m’ve}$

, therefore

${displaystyle e’=ve}$

by

${displaystyle m’}$

monic.

4. ${displaystyle v}$

is monic.

5. ${displaystyle m=m’,v}$

${displaystyle v}$

is unique.

The image of

${displaystyle f}$

is often denoted by

${displaystyle {text{Im}}f}$

or

${displaystyle {text{Im}}(f)}$

.

Proposition: If

${displaystyle C}$

has all equalizers then the

${displaystyle e}$

in the factorization

${displaystyle f=m,e}$

of (1) is an epimorphism.[2]

Proof

Let

${displaystyle alpha ,,beta }$

be such that

${displaystyle alpha ,e=beta ,e}$

, one needs to show that

${displaystyle alpha =beta }$

. Since the equalizer of

${displaystyle (alpha ,beta )}$

exists,

${displaystyle e}$

factorizes as

${displaystyle e=q,e’}$

with

${displaystyle q}$

monic. But then

${displaystyle f=(m,q),e’}$

is a factorization of

${displaystyle f}$

with

${displaystyle (m,q)}$

monomorphism. Hence by the universal property of the image there exists a unique arrow

${displaystyle v:Ito Eq_{alpha ,beta }}$

such that

${displaystyle m=m,q,v}$

and since

${displaystyle m}$

is monic

${displaystyle {text{id}}_{I}=q,v}$

. Furthermore, one has

${displaystyle m,q=(mqv),q}$

and by the monomorphism property of

${displaystyle mq}$

one obtains

${displaystyle {text{id}}_{Eq_{alpha ,beta }}=v,q}$

.

This means that

${displaystyle Iequiv Eq_{alpha ,beta }}$

and thus that

${displaystyle {text{id}}_{I}=q,v}$

equalizes

${displaystyle (alpha ,beta )}$

, whence

${displaystyle alpha =beta }$

.

## Second definition

In a category

${displaystyle C}$

with all finite limits and colimits, the image is defined as the equalizer

${displaystyle (Im,m)}$

of the so-called cokernel pair

${displaystyle (Ysqcup _{X}Y,i_{1},i_{2})}$

.[3]

[clarification needed]

Remarks:

1. Finite bicompleteness of the category ensures that pushouts and equalizers exist.

2. ${displaystyle (Im,m)}$

can be called regular image as

${displaystyle m}$

is a regular monomorphism, i.e. the equalizer of a pair of morphisms. (Recall also that an equalizer is automatically a monomorphism).

3. In an abelian category, the cokernel pair property can be written
${displaystyle i_{1},f=i_{2},f Leftrightarrow (i_{1}-i_{2}),f=0=0,f}$

and the equalizer condition

${displaystyle i_{1},m=i_{2},m Leftrightarrow (i_{1}-i_{2}),m=0,m}$

. Moreover, all monomorphisms are regular.

Theorem — If

${displaystyle f}$

always factorizes through regular monomorphisms, then the two definitions coincide.

Proof

First definition implies the second: Assume that (1) holds with

${displaystyle m}$

regular monomorphism.

• Equalization: one needs to show that
${displaystyle i_{1},m=i_{2},m}$

. As the cokernel pair of

${displaystyle f, i_{1},f=i_{2},f}$

and by previous proposition, since

${displaystyle C}$

has all equalizers, the arrow

${displaystyle e}$

in the factorization

${displaystyle f=m,e}$

is an epimorphism, hence

${displaystyle i_{1},f=i_{2},f Rightarrow i_{1},m=i_{2},m}$

.

• Universality: in a category with all colimits (or at least all pushouts)
${displaystyle m}$

itself admits a cokernel pair

${displaystyle (Ysqcup _{I}Y,c_{1},c_{2})}$

Moreover, as a regular monomorphism,

${displaystyle (I,m)}$

is the equalizer of a pair of morphisms

${displaystyle b_{1},b_{2}:Ylongrightarrow B}$

but we claim here that it is also the equalizer of

${displaystyle c_{1},c_{2}:Ylongrightarrow Ysqcup _{I}Y}$

.

Indeed, by construction

${displaystyle b_{1},m=b_{2},m}$

thus the “cokernel pair” diagram for

${displaystyle m}$

yields a unique morphism

${displaystyle u’:Ysqcup _{I}Ylongrightarrow B}$

such that

${displaystyle b_{1}=u’,c_{1}, b_{2}=u’,c_{2}}$

. Now, a map

${displaystyle m’:I’longrightarrow Y}$

which equalizes

${displaystyle (c_{1},c_{2})}$

also satisfies

${displaystyle b_{1},m’=u’,c_{1},m’=u’,c_{2},m’=b_{2},m’}$

, hence by the equalizer diagram for

${displaystyle (b_{1},b_{2})}$

, there exists a unique map

${displaystyle h’:I’to I}$

such that

${displaystyle m’=m,h’}$

.

Finally, use the cokernel pair diagram (of

${displaystyle f}$

) with

${displaystyle j_{1}:=c_{1}, j_{2}:=c_{2}, Z:=Ysqcup _{I}Y}$

: there exists a unique

${displaystyle u:Ysqcup _{X}Ylongrightarrow Ysqcup _{I}Y}$

such that

${displaystyle c_{1}=u,i_{1}, c_{2}=u,i_{2}}$

. Therefore, any map

${displaystyle g}$

which equalizes

${displaystyle (i_{1},i_{2})}$

also equalizes

${displaystyle (c_{1},c_{2})}$

and thus uniquely factorizes as

${displaystyle g=m,h’}$

. This exactly means that

${displaystyle (I,m)}$

is the equalizer of

${displaystyle (i_{1},i_{2})}$

.

Second definition implies the first:

Then

${displaystyle d_{1},m’=d_{2},m’ Rightarrow d_{1},f=d_{1},m’,e=d_{2},m’,e=d_{2},f}$

so that by the “cokernel pair” diagram (of

${displaystyle f}$

), with

${displaystyle j_{1}:=d_{1}, j_{2}:=d_{2}, Z:=D}$

, there exists a unique

${displaystyle u”:Ysqcup _{X}Ylongrightarrow D}$

such that

${displaystyle d_{1}=u”,i_{1}, d_{2}=u”,i_{2}}$

.

Now, from

${displaystyle i_{1},m=i_{2},m}$

(m from the equalizer of (i1, i2) diagram), one obtains

${displaystyle d_{1},m=u”,i_{1},m=u”,i_{2},m=d_{2},m}$

, hence by the universality in the (equalizer of (d1, d2) diagram, with f replaced by m), there exists a unique

${displaystyle v:Imlongrightarrow I’}$

such that

${displaystyle m=m’,v}$

.

## Examples

In the category of sets the image of a morphism

${displaystyle fcolon Xto Y}$

is the inclusion from the ordinary image

${displaystyle {f(x)~|~xin X}}$

to

${displaystyle Y}$

. In many concrete categories such as groups, abelian groups and (left- or right) modules, the image of a morphism is the image of the correspondent morphism in the category of sets.

In any normal category with a zero object and kernels and cokernels for every morphism, the image of a morphism

${displaystyle f}$

can be expressed as follows:

im f = ker coker f

In an abelian category (which is in particular binormal), if f is a monomorphism then f = ker coker f, and so f = im f.