Image (Category Theory)

Right. You want the Wikipedia article on the image of a morphism in category theory, but… enhanced. More detail, more… flavor. As if the original was some bland, beige sketch that needed a proper artist. Fine. Let’s see what we can scrape from the dust.

Image of a Morphism

In the rather bleak landscape of category theory, a field of mathematics that often feels like staring into a void, the concept of the image of a morphism serves as a rather grim generalization of the image of a function. It’s not about making things pretty; it’s about structure, about the inherent decomposition of a process.

General Definition

Let’s be precise, though precision can be a tedious affair. Consider a category, let’s call it C. Within this C, we have a morphism, let’s denote it as f, which maps from an object X to an object Y. We write this as f: X → Y.

The image of this f, denoted by Im f, is essentially a monomorphism – a morphism that is injective in spirit, if not always in literal form. Let this monomorphism be m: I → Y. This m isn't just some arbitrary arrow; it’s defined by a rather stringent universal property.

This property dictates two crucial conditions:

Factorization: There must exist a morphism, let’s call it e, that maps from X to I. This e must be such that when you compose it with m, you get back the original morphism f. In other words, f = m ∘ e. This tells us that f can be broken down, or factored, into two parts: first, this e that maps from the source of f to the image object I, and then m that maps from I to the target of f.
Uniqueness and Universality: Now, this is where the "universal property" really bites. Imagine you have any other way to factor f like this. Suppose there’s another object, I', and a morphism e': X → I', and a monomorphism m': I' → Y, such that f = m' ∘ e'. If this other factorization exists, then there must be a unique morphism, let’s call it v, mapping from our original image object I to this new object I'. This v must satisfy the condition that when you compose m' with v, you get back our original image monomorphism m. So, m = m' ∘ v.

This is the grim elegance of it. The image I is the "most efficient" or "minimal" object in Y that f can factor through, in a way that preserves the essential structure of f’s mapping.

Remarks on the Grim Details:

This factorization, this f = m ∘ e, doesn’t always exist. Sometimes, the universe of the category is just too uncooperative.
The morphism e itself, once m is fixed, is unique. This is a direct consequence of m being a monomorphism. If e and e'' both worked such that f = m ∘ e and f = m ∘ e'', then m ∘ e = m ∘ e''. Since m is monic, it’s “one-to-one” enough to imply e = e''.
The relationships m' ∘ e' = f = m ∘ e and m = m' ∘ v are crucial. They lead to e' = v ∘ e. This is because f = m' ∘ e' and f = m ∘ e = (m' ∘ v) ∘ e = m' ∘ (v ∘ e). Again, the monic nature of m' ensures that e' = v ∘ e.
And yes, v itself is also a monomorphism. If v ∘ x = v ∘ y, then m' ∘ v ∘ x = m' ∘ v ∘ y, which means m' ∘ x = m' ∘ y. Since m' is monic, x = y.
The uniqueness of v is already guaranteed by the condition m = m' ∘ v. If there were another v'' such that m = m' ∘ v'', then m' ∘ v = m' ∘ v''. Because m' is monic, v = v''.

The image of f is commonly referred to as Im f or Im(f). It’s a mark left, a trace of the morphism’s journey.

A Proposition and its Unpleasant Proof:

If our category C is generous enough to possess all equalizers, then the morphism e in the factorization f = m ∘ e is actually an epimorphism.

Proof: Let’s assume we have two morphisms, α and β, such that α ∘ e = β ∘ e. Our goal, the grim objective, is to prove that α = β. Since the equalizer of (α, β) exists – remember, our category is blessedly equipped with them – e can be factored as e = q ∘ e', where q is a monomorphism. Now, our original factorization becomes f = m ∘ e = m ∘ (q ∘ e') = (m ∘ q) ∘ e'. Notice that m ∘ q is also a monomorphism, as it’s a composition of monomorphisms. By the universal property of the image (the one we discussed earlier), there must exist a unique morphism, let’s call it v, mapping from I to the equalizer of (α, β), denoted Eq(α, β). This v must satisfy m = (m ∘ q) ∘ v. Since m is a monomorphism, this implies id_I = q ∘ v, where id_I is the identity morphism on I. Furthermore, we have m ∘ q = (m ∘ q ∘ v) ∘ q. Since m ∘ q is a monomorphism, we can cancel it out to get id_Eq(α, β) = v ∘ q. This establishes an equivalence, an isomorphism, between I and Eq(α, β). Thus, id_I = q ∘ v effectively equalizes (α, β). This means that α and β must be the same: α = β. It’s a rather convoluted way of saying that e “covers” everything it needs to for α and β to be identical.

Second Definition

Let’s consider a category C that is particularly well-endowed, possessing all finite limits and colimits. In such a category, the image of a morphism f: X → Y can be defined in a different, perhaps more constructivist, manner.

It is defined as the equalizer of a specific pair of morphisms, often called a "cokernel pair." This pair arises from the cocartesian square formed by taking the coproduct (or pushout) of Y with itself over X. More formally, it’s the equalizer (Im, m) of the morphisms i_1 and i_2 from Y to Y ⊔_X Y.

The diagram looks something like this:

      f
X --------> Y
|           |
|           | i_1
|           |
Y ⊔_X Y <----- Y
      i_2

Here, Y ⊔_X Y is the pushout of f with itself. The morphisms i_1 and i_2 are the canonical inclusions into this pushout. The equalizer (Im, m) is the pair such that i_1 ∘ m = i_2 ∘ m, and m: Im → Y is the universal morphism with this property.

Remarks on the Second Definition:

If a category is bicomplete (meaning it has all finite limits and colimits), then the pushouts and equalizers we need are guaranteed to exist. It’s a matter of having the right tools.
This (Im, m) can be termed the "regular image" because m is a regular monomorphism. A regular monomorphism is, by definition, an equalizer of some pair of morphisms. And as a reminder, all equalizers are automatically monomorphisms.
In an abelian category, this cokernel pair condition takes on a slightly neater form. For a morphism f, the condition i_1 ∘ f = i_2 ∘ f is equivalent to (i_1 - i_2) ∘ f = 0. Similarly, for the equalizer m, i_1 ∘ m = i_2 ∘ m is equivalent to (i_1 - i_2) ∘ m = 0. In such categories, all monomorphisms are regular, which simplifies things considerably.

Theorem: If every morphism f in a category can always be factored through regular monomorphisms, then the two definitions of the image – the one based on the universal property and this one based on equalizers – coincide. They become indistinguishable.

Proof:

First Definition Implies the Second: Let’s assume the first definition holds, where f = m ∘ e, and m is a regular monomorphism.
- Equalization: We need to show that i_1 ∘ m = i_2 ∘ m. Since (i_1, i_2) form the cokernel pair of f, we know that i_1 ∘ f = i_2 ∘ f. And from our earlier proposition (assuming the category has all equalizers), e must be an epimorphism. Therefore, i_1 ∘ f = i_2 ∘ f implies i_1 ∘ m = i_2 ∘ m.
- Universality: In a category with all colimits (or at least all pushouts), m itself will have a cokernel pair, say (Y ⊔_I Y, c_1, c_2). As a regular monomorphism, (I, m) is the equalizer of some pair (b_1, b_2): Y → B. We claim it’s also the equalizer of (c_1, c_2): Y → Y ⊔_I Y. Because b_1 ∘ m = b_2 ∘ m (since m is the equalizer of b_1, b_2), the diagram for the cokernel pair of m provides a unique morphism u': Y ⊔_I Y → B such that b_1 = u' ∘ c_1 and b_2 = u' ∘ c_2. Now, consider any morphism m': I' → Y that equalizes (c_1, c_2). This means c_1 ∘ m' = c_2 ∘ m'. Then, b_1 ∘ m' = (u' ∘ c_1) ∘ m' = u' ∘ (c_1 ∘ m') = u' ∘ (c_2 ∘ m') = (u' ∘ c_2) ∘ m' = b_2 ∘ m'. Because (I, m) is the equalizer of (b_1, b_2), there must be a unique morphism h': I' → I such that m' = m ∘ h'. This is precisely the universality condition for (I, m) being the equalizer of (c_1, c_2). Finally, we use the cokernel pair diagram of f. There exists a unique morphism u: Y ⊔_X Y → Y ⊔_I Y such that c_1 = u ∘ i_1 and c_2 = u ∘ i_2. Therefore, any morphism g that equalizes (i_1, i_2) will also equalize (c_1, c_2) (because c_1 ∘ g = u ∘ i_1 ∘ g and c_2 ∘ g = u ∘ i_2 ∘ g, and if i_1 ∘ g = i_2 ∘ g, then c_1 ∘ g = c_2 ∘ g). This means g must factor uniquely through m, i.e., g = m ∘ h' for some h'. This confirms that (I, m) is indeed the equalizer of (i_1, i_2).
Second Definition Implies the First:
- Factorization: If we take the second definition, (Im, m) is the equalizer of (i_1, i_2). Let’s consider the original morphism f: X → Y. The diagram for the cokernel pair of f gives us i_1 ∘ f = i_2 ∘ f. Since m is the universal equalizer for (i_1, i_2), there must be a unique morphism, let’s call it h, such that f = m ∘ h. This is our factorization.
- Universality: Suppose we have another factorization of f where m': I' → Y is a regular monomorphism, so f = m' ∘ e'. Since m' is a regular monomorphism, it’s the equalizer of some pair (d_1, d_2): Y → D. This means d_1 ∘ m' = d_2 ∘ m'. Now, d_1 ∘ f = d_1 ∘ (m' ∘ e') = (d_1 ∘ m') ∘ e'. Similarly, d_2 ∘ f = (d_2 ∘ m') ∘ e'. Since d_1 ∘ m' = d_2 ∘ m', it follows that d_1 ∘ f = d_2 ∘ f. Looking at the cokernel pair diagram of f again, i_1 ∘ f = i_2 ∘ f. If we also have d_1 ∘ f = d_2 ∘ f, then by the universality of the equalizer of (d_1, d_2) (applied to f instead of m'), there exists a unique morphism u'': Y ⊔_X Y → D such that d_1 = u'' ∘ i_1 and d_2 = u'' ∘ i_2. We know i_1 ∘ m = i_2 ∘ m because m is the equalizer of (i_1, i_2). Applying u'' to this, we get u'' ∘ i_1 ∘ m = u'' ∘ i_2 ∘ m, which simplifies to d_1 ∘ m = d_2 ∘ m. Now, consider the fact that m' is the equalizer of (d_1, d_2). Since d_1 ∘ m = d_2 ∘ m, by the universality property of the equalizer m', there must exist a unique morphism v: I → I' such that m = m' ∘ v. This is precisely the condition required by the universal property of the first definition.

Examples

Let’s ground this in something less abstract, shall we?

Category of Sets: In the most fundamental category of sets, the image of a morphism f: X → Y is the inclusion map from the actual set-theoretic image {f(x) | x ∈ X} into Y. It’s the subset of Y containing precisely the elements that f actually maps to. Simple, direct.
Concrete Categories: For many other concrete categories – think of groups, abelian groups, or modules – the image of a morphism is defined analogously to the image in the category of sets. It’s the image of the underlying set-theoretic function, endowed with the structure of the category.
Normal Categories: In any normal category that has a zero object and is equipped with kernels and cokernels for every morphism, the image of a morphism f can be expressed with a certain brutal economy: im f = ker(coker f). You take the cokernel, and then the kernel of that. It's a chain reaction of algebraic operations.
Abelian Categories: If we're in an abelian category (which is a special kind of normal category), and f happens to be a monomorphism, then f = ker(coker f), and consequently, f = im f. The image is the morphism itself. A neat, self-contained loop.

Essential Image

There’s a related, though somewhat more esoteric, notion called the "essential image." It’s less about direct factorization and more about encompassing subcategories.

A subcategory C within a larger category B is called "replete" if, for any object x in C, and any isomorphism ι: x → y where ι and y are also within B, both ι and y must belong to C. It’s a requirement of completeness, of not leaving any isomorphic copies of your objects out in the cold.

Given a functor F: A → B between categories, the "essential image" of F is the smallest replete subcategory of the target category B that contains the image of A under F. It’s the smallest possible "container" that includes all that F maps from A into B, while also adhering to the rules of repleteness. It's like a meticulously curated collection, ensuring that if you have something, you have all its equivalent forms too.