Description Morphisms from Category theory in the language of set theory

Question

I have a good understanding of set theory and axiomatic set theory and I am trying to now understand Category theory. I usually understand most of mathematics through set theory, like I understand well the set theory definition of functions, relations, sets and that helps me a lot to understand objects like groups. Now I am trying to understand what a category is but I do not understand what a morphisms is since I have been told they are not relations. Can morphisms be described with the language of set theory? can morphisms be formalized as sets or classes? If it can it would be very helpful to see a set theory definition of a morphism. Formal definitions are best but they do not need to be completely formal. I have seen that morphisms are like arrows between elements in a category and that they are elements of a set that contains all morphisms between two elements of a category. But I still would like to see if they can be understood through set theory since I have heard axiomatic set theory can be used as a foundation for category theory. Thanks in advance!

Answer 1

Category theory is extremely general, and (IMHO) this generality makes it difficult to understand.

One way to visualize what a category is by generalizing the intuitive notion of a graph.

Categories have vertices (objects) and directed edges (morphisms OR EQUIVALENTLY arrows).

Each object has a distinguished identity morphism. Equivalently, given any morphism, you can tell whether it is an identity morphism or not ... just inherently.

Categories are also equipped with a notion of composition. For any two morphisms where the destination of the second feeds into the source of the first, you have a new morphism. Composition is required to be associative. Composition is also required to treat identity morphisms as identity elements with respect to composition.

Most instances of categories are structure-preserving maps between members of some kind of class (e.g. groups, rings, topological spaces). The underlying objects form a proper class rather than a set, and the arrows are organized into homsets. You normally don't talk about all the arrows at once, so you can avoid questions about whether they form a proper class or a set when you gather them together (or whether two arrows from different homsets are the same or different). In this setting, the objects are frequently "sets with some additional structure" and the arrows are "maps", but in a category theoretic setting you seldom "peer inside" arrows or objects and just treat them as opaque.

---

Here is the theory of a single category, expressed as a first-order theory with a single sort, the sort of morphisms. This axiomatization is similar to presentations of set theory that you may have seen before (in that it is a first-order theory).

Typically, a category is presented as having both objects and morphisms, but you can "forget" the distinction and identify an object $A$ with its identity morphism $\mathrm{id}_A$.

Let $R(\cdot, \cdot, \cdot)$ be a 3-place predicate denoting reverse composition.

$R(a, b, c)$ holds if and only if $b \circ a = c$. More visually, this says that in the picture $ \text{Obj} \stackrel{a}{\to} \text{Obj} \stackrel{b}{\to} \text{Obj}$, $c$ is the morphism that you get by following the morphism $a$ and then the morphism $b$.

I will first axiomatize the notion of a semicategory (or semigroupoid).

Composition is associative.

$$ (\exists z \mathop. R(a, b, z) \land R(z, c, u)) \iff (\exists w\mathop. R(b, c, w) \to R(a, w, u)) $$

Composition is a partial function

$$ R(a, b, c) \land R(a, b, d) \implies c = d $$

And that's it. That's a semicategory.

In order to be a category, we need to assert the existence of identity morphisms that are identities with respect to composition.

Let $I(\cdot)$ be a one-place predicate that identifies identity morphisms.

Every identity is a left identity.

$$ I(a) \land (\exists c \mathop. R(a, b, c)) \implies R(a, b, b) $$

Every identity is a right identity.

$$ I(b) \land (\exists c \mathop. R(a, b, c)) \implies R(a, b, a) $$

There's a unique identity that post-composes with every morphism.

$$ \exists! a \mathop. I(a) \land (\exists c \mathop. R(a, b, c)) $$

There's a unique identity that pre-composes with every morphism.

$$ \exists! b \mathop. I(b) \land (\exists c \mathop. R(a, b, c)) $$