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

1. There exists a morphism $e\colon X\to I$ such that $f=m\,e$.
2. For any object $I'$ with a morphism $e'\colon X\to I'$ and a monomorphism $m'\colon I'\to Y$ such that $f=m'\,e'$, there exists a unique morphism $v\colon I\to I'$ such that $m=m'\,v$.

Remarks:

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

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 $f=m\,e$ 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 $e=q\,e'$ with $q$ monic. But then $f=(m\,q)\,e'$ is a factorization of $f$ with $(m\,q)$ monomorphism. Hence by the universal property of the image there exists a unique arrow $v:I\to Eq_{\alpha ,\beta }$ such that $m=m\,q\,v$ and since $m$ is monic ${\text{id}}_{I}=q\,v$. Furthermore, one has Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle m\,q=(mqv)\,q} and by the monomorphism property of $mq$ one obtains ${\text{id}}_{Eq_{\alpha ,\beta }}=v\,q$.

This means that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle I\equiv Eq_{\alpha ,\beta }} and thus that ${\text{id}}_{I}=q\,v$ 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (Im,m)} of the so-called cokernel pair $(Y\sqcup _{X}Y,i_{1},i_{2})$, which is the cocartesian of a morphism with itself over its domain, which will result in a pair of morphisms $i_{1},i_{2}:Y\to Y\sqcup _{X}Y$, on which the equalizer is taken, i.e. the first of the following diagrams is cocartesian, and the second equalizing.

Remarks:

1. Finite bicompleteness of the category ensures that pushouts and equalizers exist.
2. Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (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 $i_{1}\,f=i_{2}\,f\ \Leftrightarrow \ (i_{1}-i_{2})\,f=0=0\,f$ and the equalizer condition $i_{1}\,m=i_{2}\,m\ \Leftrightarrow \ (i_{1}-i_{2})\,m=0\,m$. 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 $f,\ i_{1}\,f=i_{2}\,f$ and by previous proposition, since $C$ has all equalizers, the arrow $e$ in the factorization $f=m\,e$ is an epimorphism, hence $i_{1}\,f=i_{2}\,f\ \Rightarrow \ i_{1}\,m=i_{2}\,m$.
• Universality: in a category with all colimits (or at least all pushouts) $m$ itself admits a cokernel pair $(Y\sqcup _{I}Y,c_{1},c_{2})$

Moreover, as a regular monomorphism, $(I,m)$ is the equalizer of a pair of morphisms $b_{1},b_{2}:Y\longrightarrow B$ but we claim here that it is also the equalizer of $c_{1},c_{2}:Y\longrightarrow Y\sqcup _{I}Y$.

Indeed, by construction $b_{1}\,m=b_{2}\,m$ thus the "cokernel pair" diagram for $m$ yields a unique morphism Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u':Y\sqcup _{I}Y\longrightarrow B} such that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle b_{1}=u'\,c_{1},\ b_{2}=u'\,c_{2}} . Now, a map $m':I'\longrightarrow Y$ which equalizes Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (c_{1},c_{2})} also satisfies Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle b_{1}\,m'=u'\,c_{1}\,m'=u'\,c_{2}\,m'=b_{2}\,m'} , hence by the equalizer diagram for $(b_{1},b_{2})$, there exists a unique map Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h':I'\to I} such that $m'=m\,h'$.

Finally, use the cokernel pair diagram (of $f$) with $j_{1}:=c_{1},\ j_{2}:=c_{2},\ Z:=Y\sqcup _{I}Y$ : there exists a unique $u:Y\sqcup _{X}Y\longrightarrow Y\sqcup _{I}Y$ such that $c_{1}=u\,i_{1},\ c_{2}=u\,i_{2}$. Therefore, any map $g$ which equalizes $(i_{1},i_{2})$ also equalizes $(c_{1},c_{2})$ and thus uniquely factorizes as Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g=m\,h'} . This exactly means that $(I,m)$ is the equalizer of $(i_{1},i_{2})$.

Second definition implies the first:

• Factorization: taking $m':=f$ in the equalizer diagram ($m'$ corresponds to $g$), one obtains the factorization Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f=m\,h} .
• Universality: let $f=m'\,e'$ be a factorization with $m'$ regular monomorphism, i.e. the equalizer of some pair $(d_{1},d_{2})$.

Then $d_{1}\,m'=d_{2}\,m'\ \Rightarrow \ d_{1}\,f=d_{1}\,m'\,e=d_{2}\,m'\,e=d_{2}\,f$ so that by the "cokernel pair" diagram (of $f$), with $j_{1}:=d_{1},\ j_{2}:=d_{2},\ Z:=D$, there exists a unique Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u'':Y\sqcup _{X}Y\longrightarrow D} such that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle d_{1}=u''\,i_{1},\ d_{2}=u''\,i_{2}} .

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 $v:Im\longrightarrow I'$ such that $m=m'\,v$.

Examples

In the category of sets the image of a morphism $f\colon X\to Y$ 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.

See also

• Subobject
• Coimage
• Image (mathematics)