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 $C$ and a morphism $f:X\to Y$ in $C$, the image of $f$ is a monomorphism $m:I\to Y$ satisfying the following universal property:

1. There exists a morphism $e:X\to I$ such that $f=me$.
2. For any object $I^{\prime }$ with a morphism $e^{\prime }:X\to I^{\prime }$ and a monomorphism $m^{\prime }:I^{\prime }\to Y$ such that $f=m^{\prime }{e}^{\prime }$, there exists a unique morphism $v:I\to I^{\prime }$ such that $m=m^{\prime }v$.

Remarks:

1. such a factorization does not necessarily exist.
2. $e$ is unique by definition of $m$ monic.
3. $m^{\prime }{e}^{\prime }=f=me=m^{\prime }ve$, therefore $e^{\prime }=ve$ by $m^{\prime }$ monic.
4. $v$ is monic.
5. $m=m^{\prime }v$ already implies that $v$ is unique.

File:Image Theorie des catégories.png

The image of $f$ is often denoted by $\text{Im}f$ or $\text{Im}(f)$. Proposition: If $C$ has all equalizers then the $e$ in the factorization $f=me$ of (1) is an epimorphism.

Proof

Let $\alpha ,\beta$ be such that $\alpha e=\beta e$, one needs to show that $\alpha =\beta$. Since the equalizer of $(\alpha ,\beta )$ exists, $e$ factorizes as $e=q{e}^{\prime }$ with $q$ monic. But then $f=(mq)e^{\prime }$ is a factorization of $f$ with $(mq)$ monomorphism. Hence by the universal property of the image there exists a unique arrow $v:I\to E{q}_{\alpha ,\beta }$ such that $m=mqv$ and since $m$ is monic ${\text{id}}_I=qv$. Furthermore, one has $mq=(mqv)q$ and by the monomorphism property of $mq$ one obtains ${\text{id}}_{E{q}_{\alpha ,\beta }}=vq$.

File:E epimorphism.png

This means that $I\equiv E{q}_{\alpha ,\beta }$ and thus that ${\text{id}}_I=qv$ equalizes $(\alpha ,\beta )$, whence $\alpha =\beta$.

Second definition

In a category $C$ with all finite limits and colimits, the image is defined as the equalizer $(Im,m)$ of the so-called cokernel pair $(Y{\bigsqcup }_{X}Y,i_1,i_2)$, which is the cocartesian of a morphism with itself over its domain, which will result in a pair of morphisms $i_1,i_2:Y\to Y{\bigsqcup }_{X}Y$, on which the equalizer is taken, i.e. the first of the following diagrams is cocartesian, and the second equalizing.

File:Cokernel pair.png

File:Equalizer of the cokernel pair, diagram.png

Remarks:

1. Finite bicompleteness of the category ensures that pushouts and equalizers exist.
2. $(Im,m)$ can be called regular image as $m$ 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 $$\begin{gathered} i_{1}f=i_{2}f \\ \iff \\ (i_1-i_2)f=0=0f \end{gathered}$$ and the equalizer condition $$\begin{gathered} i_{1}m=i_{2}m \\ \iff \\ (i_1-i_2)m=0m \end{gathered}$$. Moreover, all monomorphisms are regular.

Theorem — If $f$ always factorizes through regular monomorphisms, then the two definitions coincide.

Proof

First definition implies the second: Assume that (1) holds with $m$ regular monomorphism.

• Equalization: one needs to show that $i_{1}m=i_{2}m$ . As the cokernel pair of $$\begin{gathered} f, \\ {i}_{1}f=i_{2}f \end{gathered}$$ and by previous proposition, since $C$ has all equalizers, the arrow $e$ in the factorization $f=me$ is an epimorphism, hence $$\begin{gathered} i_{1}f=i_{2}f \\ \Rightarrow \\ {i}_{1}m=i_{2}m \end{gathered}$$.
• Universality: in a category with all colimits (or at least all pushouts) $m$ itself admits a cokernel pair $(Y{\bigsqcup }_{I}Y,c_1,c_2)$

File:Cokernel pair m.png

Moreover, as a regular monomorphism, $(I,m)$ is the equalizer of a pair of morphisms $b_1,b_2:Y⟶B$ but we claim here that it is also the equalizer of $c_1,c_2:Y⟶Y{\bigsqcup }_{I}Y$.

Indeed, by construction $b_{1}m=b_{2}m$ thus the "cokernel pair" diagram for $m$ yields a unique morphism $u^{\prime }:Y{\bigsqcup }_{I}Y⟶B$ such that $$\begin{gathered} b_1=u^{\prime }{c}_1, \\ {b}_2=u^{\prime }{c}_2 \end{gathered}$$. Now, a map $m^{\prime }:I^{\prime }⟶Y$ which equalizes $(c_1,c_2)$ also satisfies $b_1{m}^{\prime }=u^{\prime }{c}_1{m}^{\prime }=u^{\prime }{c}_2{m}^{\prime }=b_2{m}^{\prime }$, hence by the equalizer diagram for $(b_1,b_2)$, there exists a unique map $h^{\prime }:I^{\prime }\to I$ such that $m^{\prime }=m{h}^{\prime }$.

Finally, use the cokernel pair diagram (of $f$) with $$\begin{gathered} j_1:=c_1, \\ {j}_2:=c_2, \\ Z:=Y{\bigsqcup }_{I}Y \end{gathered}$$ : there exists a unique $u:Y{\bigsqcup }_{X}Y⟶Y{\bigsqcup }_{I}Y$ such that $$\begin{gathered} c_1=u{i}_1, \\ {c}_2=u{i}_2 \end{gathered}$$. Therefore, any map $g$ which equalizes $(i_1,i_2)$ also equalizes $(c_1,c_2)$ and thus uniquely factorizes as $g=m{h}^{\prime }$. This exactly means that $(I,m)$ is the equalizer of $(i_1,i_2)$.

Second definition implies the first:

• Factorization: taking $m^{\prime }:=f$ in the equalizer diagram ($m^{\prime }$ corresponds to $g$), one obtains the factorization $f=mh$.
• Universality: let $f=m^{\prime }{e}^{\prime }$ be a factorization with $m^{\prime }$ regular monomorphism, i.e. the equalizer of some pair $(d_1,d_2)$.

File:Equalizerd1d2.png

Then $$\begin{gathered} d_1{m}^{\prime }=d_2{m}^{\prime } \\ \Rightarrow \\ {d}_{1}f=d_1{m}^{\prime }e=d_2{m}^{\prime }e=d_{2}f \end{gathered}$$ so that by the "cokernel pair" diagram (of $f$), with $$\begin{gathered} j_1:=d_1, \\ {j}_2:=d_2, \\ Z:=D \end{gathered}$$, there exists a unique $u^{″}:Y{\bigsqcup }_{X}Y⟶D$ such that $$\begin{gathered} d_1=u^{″}{i}_1, \\ {d}_2=u^{″}{i}_2 \end{gathered}$$.

Now, from $i_{1}m=i_{2}m$ (m from the equalizer of (i₁, i₂) diagram), one obtains $d_{1}m=u^{″}{i}_{1}m=u^{″}{i}_{2}m=d_{2}m$, hence by the universality in the (equalizer of (d₁, d₂) diagram, with f replaced by m), there exists a unique $v:Im⟶I^{\prime }$ such that $m=m^{\prime }v$.

Examples

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

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.

Essential Image

A related notion to image is essential image. A subcategory $C\subset B$ of a (strict) category is said to be replete if for every $x\in C$, and for every isomorphism $\iota :x\to y$, both $\iota$ and $y$ belong to C. Given a functor $F:A\to B$ between categories, the smallest replete subcategory of the target n-category B containing the image of A under F.

See also

• Subobject
• Coimage
• Image (mathematics)