Is morphisms in category equivalent to binary relation?

Question

Composition of relations is associative

https://ncatlab.org/nlab/show/relation#binary_relations

https://en.wikipedia.org/wiki/Binary_relation

https://en.wikipedia.org/wiki/Composition_of_relations#Properties

On the other hand, in a category

https://ncatlab.org/nlab/show/category#OneCollectionOfMorphisms

for every pair of morphisms f and g, where t(f)=s(g), a morphism g∘f, called their composite (also written gf or sometimes f;g— see diagrammatic order);

If composition of morphism (and associativity) is required to form a cateogory, are morphisms in category equivalent to binary relation?

Or, is there anything else than binary relation that is composable which can also be a morphism of cateogry?

Thanks.

Answer 1

It is not quite the case that every category can be represented by a relation, but every (small) category can be represented by a categorified relation, known as a profunctor.

Consider a category with object-set $C_0$. A reflexive transitive endo-relation $R : C_0 \times C_0 \to 2$ on $C_0$ describes whether two objects $X, Y \in C_0$ are connected by a morphism, i.e. $R(X, Y) \implies \exists f : X \to Y$. However, note that such categories are necessary thin/posetal: we may have at most one morphism between any two objects. These relations turn out to be monads in the 2-category $\mathbf{Rel}$ of relations.

To represent categories, we may instead move to profunctors, which are categorified relations (i.e. we replace the two-element set $2$ with the category of sets $\mathbf{Set}$). An endo-profunctor on a small discrete category $C_0$ (i.e. a set), which is a monad in the bicategory $\mathbf{Prof}$ of profunctors, is exactly a category with object-set $C_0$. (Monads in $\mathbf{Prof}$ are sometimes called "promonads".)

Answer 2

It is not perfectly clear to me what you are asking for.

Given a category being composable defines a relation on the class/set of morphisms, but I don’t think this is your question.

A category need not consist of sets and relations. For example you can draw a finite category like $\bullet \rightarrow \bullet$, where the indicated morphisms is the only non-identity. There is no real choice for composition here. Another example would be the category $BM$ associated to a monoid $M$, which consists of one object and an endomorphism for every monoid element. The composition is defined by the monoid multiplication. To make this work one does not really need to know, what this one object is made of, so depending on how you model it, the morphisms need not be relations.

Answer 3

I suspect that you are asking if a morphisms in a category must be a function.

The answer is no

You may regard a group as a one object category, where morphisms are the elements of a group and composition is the group multiplication. The same is true for a monoid. So morphisms in a category need not be (structure preserving) function between sets (with structures).

Anyhow one could notice that in such cases an element can actually be identified with a function: a multiplication for that element. So in some sense we fall back in the previous case.

At the same time there are cases where this is not the case: for instance take index categories, which are basically graphs with compositional arrows and identity arrows at any node. By the way they are also very useful: any commutative diagram in a category C is a functor from an index category J into C.