Can one tell what objects or morphisms represent in a category by just looking at the category structure (topology)?

Question

Does the topology(structure)(just objects and morphisms between objects) in a category define and indicate what the objects and morphisms are representing? Because if not then the examples in this answer (examples of morphisms that are not function) would only make sense if we augment the structure of the category with the description about what those objects and morphisms represent explicitly, otherwise one seems to be able to use the same category with the same structure for representing sets and functions between them.

Answer 1

Just as the 'same' group can arise in multiple ways, the 'same' category can arise in multiple ways. The question you linked (indirectly) gives an example of this. We can view $\mathbb{N}$ as a poset (with it natural order) and so it is a category. Here the objects are natural numbers, and the arrows just given by the relation $\leq$. We get the same category if we take as objects sets of the form $\{1, 2, \ldots, n\}$ for all $n \in \mathbb{N}$ and as arrows only the inclusion functions.

It seems to me you are also interested in this question about arrows that are not functions of sets. So far, it may seem like we can always represent our category in some way where the objects are sets and the arrows are just some kind of functions of sets (as I just did in my example). There is a way to make this precise: concrete categories (Wikipedia, nLab).

A category $\mathcal{C}$ is called a concrete category if it admits a faithful functor $U: \mathcal{C} \to \mathbf{Set}$. We call this functor $U$ the forgetful functor.

Recall that a faithful functor is a functor that is injective on arrows. So a forgetful functor is essentially a way of making precise how we can think of our objects in $\mathcal{C}$ as sets and of our arrows as functions of sets.

The question then becomes: is every category a concrete category? The answer to this is no. The most famous example is the category $\mathbf{hTop}$ where the objects are topological spaces and the arrows are homotopy classes of continuous functions. This is not a concrete category (note, it is far from trivial to see this, but you might convince yourself by just trying).

Note that every small category is a concrete category. This is the answer by Dan Rust. What actually happens here is the following. Given a small category $\mathcal{C}$ we have the Yoneda embedding $Y: \mathcal{C} \to \mathbf{Set}^{\mathcal{C}^\text{op}}$, which is (full and) faithful. Then we also have a faithful functor $F: \mathbf{Set}^{\mathcal{C}^\text{op}} \to \mathbf{Set}$ that sends a presheaf $X$ (an object in $\mathbf{Set}^{\mathcal{C}^\text{op}}$) to $\coprod_{C \in \mathcal{C}} X(C)$. So we obtain a faithful functor $U: \mathcal{C} \to \mathbf{Set}$ by setting $U = FY$.

Edit: by the way, the way you use topology here is not the way it is usually used. You explained what you mean afterwards, but there are notions of a topology on a category (e.g. Grothendieck topology) that are really something different than what you mean here.

Answer 2

I think the following should work:

Let $\mathcal{C} = (Obj, Mor)$ be a small category and let's define a new category $\mathcal{C}' = (Obj', Mor')$ by defining the objects as $$Obj' = \{O' = Hom(-,O) \mid O \in Obj\}$$ and for every morphism $f \in Hom(O_1,O_2)$, define a function $f' \colon O_1' \to O_2'$ by $f'(g) = f \circ g$ where $g \in Hom(-,O_1)$. Then define $$Mor' = \{f' \mid f \in Mor\}.$$

This is now a concrete category as the objects are sets, and the morphisms are functions, and it is clear that $\mathcal{C}$ and $\mathcal{C}'$ are isomorphic categories.

Ok, requiring that $\mathcal{C}$ is small is a pretty heavy restriction, but I don't see a way around that.