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!

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    $\begingroup$ Do you understand what the elements of a group are? If so, then congratulations, you understand what the morphisms of a category are. There is no difference besides that the axioms they satisfy are a bit different. $\endgroup$ – Eric Wofsey Jul 22 at 3:45
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    $\begingroup$ There is nothing "special" about categories that requires a different understanding of what they "really are". They are just an abstract algebraic structure like any other kind you've seen. $\endgroup$ – Eric Wofsey Jul 22 at 3:46
  • $\begingroup$ The only part I dont completely understand about categories are morphisms. I understand the part where a category consists of a class of objects, but not the elements of the set formed by every two of such objects aka morphisms and I was wondering if morphisms could be defined using set theory so that classes or sets are the only primitive notions. Thank you for the response! $\endgroup$ – lklk Jul 22 at 4:35
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    $\begingroup$ The specific set theoretic definition of what a morphism is is going to vary from category to category, The way what a "point" is varies from topological space to topological space, or what a "group element" is varies from group to group, or "element of a poset" varies from poset to poset, or... $\endgroup$ – Malice Vidrine Jul 22 at 6:18
  • $\begingroup$ In the category of all sets, the morphisms from $A$ to $B$ are all functions $A\to B$. In categories of sets with additional structure, the morphisms are structure-preserving functions: homomorphisms, continuous functions, and so on. But morphisms don't have to be functions: any preorder is a category, with exactly one morphism ("arrow") $A\to B$ iff $A\le B$. We don't really care just what that arrow is as a set, as long as it's unique. Similarly, it's just a curiosity that, say, $\pi$ is "really" a set of rationals, which in turn are eq classes of pairs of signed integers, etc. $\endgroup$ – BrianO Jul 25 at 20:19

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.

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

  • $\begingroup$ The axiomatization was helpful, so are you sayin morphism are any objects that make the axioms true? I also was wondering if morphisms could be expressed as sets or classes, so where sets/classes are the primitive terms and morphisms are expressed in terms of sets/classes. I was thinking that a morphism could be expressed as some sort of tuple such that if A is a morphism between objects b and c then A would be some sort of tuple that includes b and c and maybe some other information. Thank you for the help! $\endgroup$ – lklk Jul 22 at 4:29
  • $\begingroup$ The axiomatization I gave you is an axiomatization for a category as a whole, not individual morphisms. Expressing objects and morphisms as sets or classes is not commonly done. They are normally thought of as opaque. You can encode the identity relation $I$ and the composition relation $R$ as sets and classes if that helps, but trying to encode individual objects and morphisms that way, I think, imposes too many restrictions on the category. $\endgroup$ – Gregory Nisbet Jul 22 at 5:04

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