Image (category theory)

General Definition

Given a category C and a morphism f : X → Y in C , the image of f is a monomorphism m : I → Y satisfying the following universal property:

Contents

• General Definition
• Second definition
• Examples
• References

1. There exists a morphism e : X → I such that f = m e .
2. For any object I ′ with a morphism e ′ : X → I ′ and a monomorphism m ′ : I ′ → Y such that f = m ′ e ′ , there exists a unique morphism v : I → I ′ such that m = m ′ v .

Remarks:

1. such a factorization does not necessarily exist
2. e is unique by definition of m monic
3. v is monic.
4. m = m ′ v already implies that m is unique.

The image of f is often denoted by Im f or Im ( f ) .

Proposition: If C has all equalizers then the e in the factorization f = m e of (1) is an epimorphism.

Second definition

In a category C will all finite limits and colimits, the image is defined as the equalizer ( I m , m ) of the so-called cokernel pair ( Y ⊔ X Y , i 1 , i 2 ) .

Remarks:

1. Finite bicompleteness of the category ensures that pushouts and equalizers exist.
2. ( I m , m ) can be called regular image as m is a regular monomorphism, i.e. the equalizer of a pair of morphism. (Recall also that an equalizer is automatically a monomorphism).
3. In an abelian category, the cokernel pair property can be written i 1 f = i 2 f   ⇔   ( i 1 − i 2 ) f = 0 = 0 f and the equalizer condition i 1 m = i 2 m   ⇔   ( i 1 − i 2 ) m = 0 m . Moreover, all monomorphisms are regular.

Examples

In the category of sets the image of a morphism f : X → Y is the inclusion from the ordinary image { f ( x )   |   x ∈ X } to 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 f can be expressed as follows:

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