A question on nomenclature of Image and Coimages in categories with zero-morphisms.

Question

I am currently doing some reading around categories with (a consistent family of) zero-morphisms and results about kernels and cokernels, including learning about abelian categories, but also not just limited to that particular case. In such categories and for a fixed morphism $f:X\to Y$ there exists a definition (up-to isomorphism) of an image of $f$ as the kernel of the cokernal of $f$ and dually the coimage defined by the cokernel of the kernel of $f$ provided, necessarily, that such limiting and colimiting objects exist. More-generally however, there exists a more widely applicable definition of image and coinage given by that of an initial mono-factorisation of $f$ and a terminal epi-factorisation of $f$ respectively.

Looking at both definitions side-by-side I notice I notice ways in which the two definitions are lacking in cohesion with one another: more specifically, I find that the notion of image in the general sense lines up (not exactly but) more precisely with the notion of coimage in the categories with zero-morphisms case (and likewise coimage in the former with image in the latter).

There must be a pragmatic reason as to why the theory is framed this way in all the introductory texts I have come across, and my question is why is this so?

I will give formal definitions for the constructs I have talked about in the above paragraphs so my question can be made precise, the reader can understand may nomenclature, (and so one can call me out if I have conceptual errors with the definitions!)

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Fix a category $\mathscr C$ and morphism $f: X \to Y$. We make the following definitions:

Mono-Factorisation

The triple $\mathbb M := (A,a,b)$ is a mono-factorisation of $f$ if $A$ is an object of $\mathscr C$ and $a: X \to A$, $b: A \to Y$ satisfy $f = b \circ a$ with $b$ being monic. For any other mono-factorisation $\mathbb M := (A^\prime,a^\prime,b^\prime)$ we say that a map $u: A \to A^\prime$ is a map from $\mathbb M$ to $\mathbb M^\prime$ if $u \circ a = a^\prime$ and $b^\prime \circ u = b$.

One may, of course, form a category out mono-factorisations and and maps between them, but this isn't relevant. Dually one can form the notion of an epi-factorisation explicitly with a definition like that of the above, or just appeal to formal duality.

Image (Most General Category-Theoretic Case)

The triple $\mathbb I :=(I,e,m)$ is an image factorisation if it is a mono-factorisation and for any other mono-factorisation $\mathbb I^\prime := (I^\prime,e^\prime,m^\prime)$ there exists a unique mediating morphism $u: \mathbb I \to \mathbb I^\prime$.

Image (Categories with Zero-morphisms)

Suppose that $f$ has a cokernel $\mathbb D := (D,d)$ and the cokernel map $d$ has a kernel $\mathbb L := (L,l)$ then there exists a unique map $l^\prime: X \to L$ satisfying $l \circ l^\prime =f$. Set the image to be any such factorisation $(L,l^\prime,l)$.

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For the remainder of the text we will assume that $f$ has kernel $\mathbb K := (K,k)$ and said map $k$ has cokernel $\mathbb C := (C,c)$. My evidence for the cokernel-kernel (CK) factorisation being most suited to be the image factorisation comes from the following lemma:

Lemma 1: A factorisation property of the cokernel of the kernel of a morphism.

There exists a unique map $c^\prime: C \to Y$ satisfying $c^\prime \circ c = 0_{X,Y}$. If $\mathbb I^\prime := (I^\prime,e^\prime,m^\prime)$ is any mono-factorisation of $f$ there exists a unique map $u:C \to I^\prime$ such that $u \circ c = e^\prime$ and $m^\prime \circ u = c^\prime$.

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From the above lemma we see that any mono-factorisation of $f$ factors uniquely through the CK-factorisation $(C,c,c^\prime)$ and thus $\mathbb C$ is an image factorisation if and only if $c^\prime$ is monic. One sees then that $(C,c,c^\prime)$ functions already very closely to an image map, and various sets axioms that imply $c^\prime$ is monic are common in the literature.

Now also assume that $f$ has a cokernel $\mathbb D := (D,d)$ and and that $d: Y \to D$ has a kernel $\mathbb L := (L,l)$. then $L$ represents a coimage in the 'categories with zero-mmrohpsism' sense. Using the above lemma we can find a unique ma $\mu: C \to L$ that is the usual `coimage to image map' of the theory.

Corollary 1: CK to KC (coimage to image) map.

There exists a unique map $\mu: C \to L$ satisfying $\mu \circ c = l^\prime$ and $l \circ \mu = c^\prime$.

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Just looking from a super zoomed-out point of view then one observes that $C$ has a mapping property concerning maps out of the object, which is just like the image in the most general sense (being a colimit or equivalently a initial object.) This observation can, of course be dualized to concern $L$ and the coimage.

It seems to be the case that this inconsistency can be easily addressed if instead of introducing the notion of an image of $f$ as the kernel of the cokernal of $f$, to instead make the reader aware of such factorisation and then prove that this forms an image (& coimage) factorisation in the case that the unique morphism $\mu: C \to L$ is an isomorphism i.e. replace the image-definition for categories with zero morphisms to that of a theorem.

In a certain sense this terminology is completely understandable in the abelian case, even if there is some possibility for confusion, as it is a standard result that everything works out (read: $\mu$ is an isomorphism). However, I have been doing some extra reading around the concepts and I see that even outside the abelian case the same nomenclature is adopted. For example, in this introductory text on semi-abelian categories, which deals with cases where $\mu$ is epic and monic yet not an isomorphism.

In the case one has that $(C,c,c^\prime)$ is an image map and $(L,l^\prime,l)$ a coimage map in the categoric sense but the former a coimage and the later an image in the categories with zero-morphisms sense. I.e. the two definitians are diametrically opposite to one another! Moreover, in such a case it is not necesrry that $\mu$ is an isomorphism, so then the definitions really do result in different objects/factorisations.

My question in essence boils down to the following: if these two notions of image do not coincide in general, why was the decision made to define the image as the KC factorisation and coimage as CK factorisation when the converse assignment would have lead to a definition that holds true to the more wildly applicable one on a larger collection of maps $f: X \to Y$? My guess is that the kernel-cokernel factorisation has more useful properties in practise that generalise the notion of image then the cokernel-kernel factorisation for standard applications (e.g. chain complexes), though at my level of knowledge on the subject I can only see that the fact that the map $l$ is guaranteed to be monic whenever $\mathbb L$ is defined as being such a property.

Does anyone who has knowledge (either historical or practical) about this theory have an enlightening argument as to why the nomenclature is as it is?

Answer 1

One reason why the image is called the image is because in a category of modules $R\mathbf{Mod}$ over a ring $R$, if $f:M \to N$ is a morphism of modules then $\operatorname{im}(f)$ can be chosen to be quite literally the image of the map $f$, i.e., $\operatorname{im}(f) = \lbrace f(m) \; | \; m \in M \rbrace$. The coimage is isomorphic to the module $M/\operatorname{Ker}(f)$ and the statement that the image is isomorphic to the coimage in $R\mathbf{Mod}$ comes down to the familiar First Isomorphism Theorem $$ \operatorname{im}(f) \cong \frac{M}{\operatorname{Ker}(f)}. $$ Because a lot of homological algebra and the theory of Abelian categories was developed to study modules (and their chain complexes, homology, cohomology, etc.) in a systematic way, the terminology was developed to keep the classical module theory and group theory in mind.

Another reason people like to use images to factor morphisms is the fact that when given your morphism $f:X \to Y$ in your category $\mathscr{C}$, the factorization $X \xrightarrow{e_f} \operatorname{im}(f) \xrightarrow{m_f} Y$, the morphism $m$ is monic. In this way we can think of the image as a subobject of $Y$, which is helpful for porting the module-theoretic intuition of images to arbitrary Abelian categories.

Answer 2

Notation gives frequently rise to discussion or confusion. In many cases authors do not use standardized terminology, but use the same expression for slightly different things.

I think the concept of "image" in category theory is such an example. The historical origin is the classical set theoretic image $f(X)$ of a morphism $f : X \to Y$ in a number of "basic categories". This "naive" concept has been in use long before the framework of category theory was developped. Examples of morphisms having such images are

• functions between sets

• homomorphisms between groups

• linear maps between vector spaces (or more generally, $R$-modules)

and many more. In these cases the set theoretic image $f(X)$ inherits the structure of an object of the appropriate category from the range object $Y$ (which makes $f(X)$ a subobject of $Y$), and the inclusion function $i : f(X) \to Y$ turns out to be a morphism in the category which has a certain universal property. See Attempt to define the notion of subobjects and my answer to it.

The problem is that this naive approach is based on the elements of the sets which underly the objects and thus does not generalize to abstract categories $\mathbf C$.

A possible general approach is what you describe as Image (Most General Category-Theoretic Case). In fact, this agrees in most basic cases with the naive (element-based) concept. However, in my opinion it is inadequate because in general the monomorphisms form a class which is too big to cover properly the concept of a subobject (which images should be). For example, in the category $\mathbf{Top}$ of topological spaces and continuous maps the monomorphisms are precisely the injective maps $m : B \to Y$ - and these are in general no embeddings in the topological sense (homeomorphisms onto the subspace $m(B) \subset Y$).

An interesting modification dealing with this problem is the concept of an $M$-image (see e.g. here):

Let $M$ be a subclass of the monomorphisms in $\mathbf C$. The elements of $M$ may be regarded as the class of all subjects in $\mathbf C$. Given a morphism $f : X \to Y$ in $\mathbf C$, the $M$-image of $f$ is the smallest subobject $im(f) : B \to Y$ through which $f$ factors (if it exists).

Examples:

1. Let $M$ be the class of all regular monomorphisms (i.e. equalizers) in $\mathbf C$. Then the $M$-images are denoted as regular images.

2. If $\mathbf C$ is a category with zero-morphisms, we may take for $M$ the class of normal monomorphisms (i.e. kernels) in $\mathbf C$. Note that normal monomorphisms are regular monomorphisms. The $M$-images are denoted as normal images. For the sake of simplicity let us assume that all morphisms of $\mathbf C$ have kernels and cokernels. Then it is a nice exercise to prove that the morphism $\ker(\operatorname{coker}(f))$ is a normal image of $f : X \to Y$.

In abelian categories the classes of monomorphisms, regular monomorphisms and normal monomorphisms agree. Thus images (= $Mono$-images in the sense of your Most General Category-Theoretic Case), regular images and normal images are the same.

The category $\mathbf {Grp}$ of groups (which has zero-morphisms) is an example where this is not the case: The image of a homomorphism $f : X \to Y$ is the inclusion $f(X) \to Y$, the normal image is the inclusion $[f(X)] \to Y$, where $[-]$ denotes the normal closure of a subgroup. You see that it depends on what you want to achieve with an image-concept; in general there is more than one such concept.

Coimages are in general not related to images. Dually to the above definition we may start with a subclass $E$ of the epimorphisms in $\mathbf C$. The elements of $E$ may be regarded as the class of all quotient objects in $\mathbf C$. Given a morphism $f : X \to Y$ in $\mathbf C$, the $E$-coimage of $f$ is the biggest quotient object $coim(f) : X \to A$ through which $f$ factors (if it exists).

One cannot expect that in the coimage-factorization $f = f^c \circ coim(f)$ one has a monomorphism $f^c$; thus this $f^c$ is in general not even a candidate for an image in any sense. However, in abelian categories image- and coimage-factorizations agree, and this may mislead to believe that it is true in general.

In a category with zero-morphisms, kernels and cokernels we have both normal images and normal coimages (a normal epimorphism is a cokernel). The normal coimage is given as $\operatorname{coker}(\ker(f))$. They are related as in your question, but the morphism $\mu$ is in general no isomorphism. In the normal-image-factorization $f = im(f) \circ f^i$ the morphism $f^i$ is in general no epimorphism (e.g. in $\mathbf {Grp}$) and in the normal-coimage-factorization $f = f^c \circ coim(f)$ the morphism $f^c$ is in general no monomorphism (e.g. in the dual category of $\mathbf {Grp}$).