Image (category Theory)
Get Image Category Theory essential facts below. View Videos or join the Image Category Theory discussion. Add Image Category Theory to your PopFlock.com topic list for future reference or share this resource on social media.
Image Category Theory

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 ${\displaystyle f\colon X\to Y}$ in ${\displaystyle C}$, the image[1] of ${\displaystyle f}$ is a monomorphism ${\displaystyle m\colon I\to Y}$ satisfying the following universal property:

1. There exists a morphism ${\displaystyle e\colon X\to I}$ such that ${\displaystyle f=m\,e}$.
2. For any object ${\displaystyle I'}$ with a morphism ${\displaystyle e'\colon X\to I'}$ and a monomorphism ${\displaystyle m'\colon I'\to Y}$ such that ${\displaystyle f=m'\,e'}$, there exists a unique morphism ${\displaystyle v\colon I\to I'}$ 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. ${\displaystyle m'e'=f=me=m've\implies e'=ve}$ by ${\displaystyle m'}$ monic.
4. ${\displaystyle v}$ is monic.
5. ${\displaystyle m=m'\,v}$ already implies that ${\displaystyle v}$ 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 ${\displaystyle v:I\to Eq_{\alpha ,\beta }}$ 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 ${\displaystyle I\equiv Eq_{\alpha ,\beta }}$ 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 ${\displaystyle (Y\sqcup _{X}Y,i_{1},i_{2})}$.[3]

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 morphism. (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 ${\displaystyle (Y\sqcup _{I}Y,c_{1},c_{2})}$
Moreover, as a regular monomorphism, ${\displaystyle (I,m)}$ is the equalizer of a pair of morphisms ${\displaystyle b_{1},b_{2}:Y\longrightarrow B}$ but we claim here that it is also the equalizer of ${\displaystyle c_{1},c_{2}:Y\longrightarrow Y\sqcup _{I}Y}$.
Indeed, by construction ${\displaystyle b_{1}\,m=b_{2}\,m}$ thus the "cokernel pair" diagram for ${\displaystyle m}$ yields a unique morphism ${\displaystyle u':Y\sqcup _{I}Y\longrightarrow B}$ 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 ${\displaystyle j_{1}:=c_{1},\ j_{2}:=c_{2},\ Z:=Y\sqcup _{I}Y}$ : there exists a unique ${\displaystyle u:Y\sqcup _{X}Y\longrightarrow Y\sqcup _{I}Y}$ 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:

• Factorization: taking ${\displaystyle m':=f}$ in the equalizer diagram (${\displaystyle m'}$ corresponds to ${\displaystyle g}$), one obtains the factorization ${\displaystyle f=m\,h}$.
• Universality: let ${\displaystyle f=m'\,e'}$ be a factorization with ${\displaystyle m'}$ regular monomorphism, i.e. the equalizer of some pair ${\displaystyle (d_{1},d_{2})}$.
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 ${\displaystyle u'':Y\sqcup _{X}Y\longrightarrow D}$ 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 ${\displaystyle v:Im\longrightarrow I'}$ such that ${\displaystyle m=m'\,v}$.

## Examples

In the category of sets the image of a morphism ${\displaystyle f\colon X\to Y}$ is the inclusion from the ordinary image ${\displaystyle \{f(x)~|~x\in X\}}$ 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.

## References

1. ^ Mitchell, Barry (1965), Theory of categories, Pure and applied mathematics, 17, Academic Press, ISBN 978-0-124-99250-4, MR 0202787 Section I.10 p.12
2. ^ Mitchell, Barry (1965), Theory of categories, Pure and applied mathematics, 17, Academic Press, ISBN 978-0-124-99250-4, MR 0202787 Proposition 10.1 p.12
3. ^ Kashiwara, Masaki; Schapira, Pierre (2006), "Categories and Sheaves", Grundlehren der Mathematischen Wissenschaften, 332, Berlin Heidelberg: Springer, pp. 113-114 Definition 5.1.1