Morphism composition is the property of morphism or the structure of category?

Question

In the definition of category, there is a morphism composition law. If A, B, C are objects, and if f is a morphism from A to B, g is a morphism from B to C, then there is a corresponding morphism from A to C called the composition of f and g.

I'm wondering this law should be classified as the property of the morphism of its own, or it belongs to the structure of the whole category.

Even though both are OK for further study, but in philosophy, which is better?

More explanation about my question:

In my understanding, the definition of category is composed of 5 parts.

1. The data of objects. These objects should be considered as points, and they have no property at all.

2. The data of morphisms. There is a property of a morphism: It is from which object to which object.

3. The composition law of morphisms as I said above.

4. For every object, there is a unit morphisms satisfy ... (omitted as we all know)

5. The associative law of morphisms.

Part 1 and 2 are the data of the category.

Part 4 and 5 can be considered as some constraints of the category.

But part 3 is special. It is not like the data, nor the constraint of category. What should it be?

Answer 1

As I indicated in my comment on MO, this question is well-addressed by the category-theoretic distinction between stuff, structure, and properties. A category comprises

• some stuff: the objects and morphisms; for simplicity we'll also consider the domain and codomain maps to be stuff (there's a bit of arbitrariness here). So the stuff of a category is its underlying graph.

• some structure: the identity morphisms and composition maps. Let's call a schmategory any graph equipped with "identity edges" and "composition maps", even if they're not unital or associative.

• some properties: the associativity and unit equations. So a category is a schmategory which is unital and associative.

There's a category $Cat$ of categories. Likewise, there is a category $Schmat$ of schmategories, and a category $Gph$ of graphs. There are forgetful functors $Cat \xrightarrow U Schmat \xrightarrow V Gph$. The yoga of stuff, structure, and properties tells us that $U$ forgets only properties because it is fully faithful and that $V$ forgets at most structure because it is faithful. For this reason we may say that an object of $Cat$ is an object of $Schmat$ satisfying extra properties and that an object of $Schmat$ is an object of $Gph$ equipped with extra structure.

Answer 2

I'm no expert, and I hope that Alec Rhea will say something better than this. But while we wait, here is a thought. My intuitive feeling is that the composition law is part of the data of the category. To specify a category you have to specify the composition law, along with everything else.

Consider for example categories with three objects, $A,B,C$, and the following maps. There is a unique map $f: A \to B$, a unique map $g: B \to C$, identity maps $i_A,i_B,i_C$, and two maps $h_1,h_2: A \to C$.

We get one category if we decide $g \circ f = h_1$, and a (slightly) different category if $g \circ f = h_2$.

In this simple example the two resulting categories are indistinguishable, isomorphic (just by relabeling $h_1$ and $h_2$). But with more data it might be interesting, perhaps. Maybe the objects are Hegelian zorks, and $h_1$ has the quasi-floop property while $h_2$ does not.

I guess in some settings the specification of the composition law is more or less brushed away. If the objects of the category have sets attached to them, morphisms between them are functions between those sets (maybe with some restrictions such as homomorphism property or continuity), and if a morphism is determined by that set function – in other words, if the category has a forgetful functor to the category of sets, and that forgetful functor is faithful – then there's no choice about what the composition law will be. It has to be ordinary set composition. So you can just say "morphisms are set functions [maybe with some restriction], and composition is ordinary function composition", except I suppose sometimes people "forget" to say that last part.

So, what if the category doesn't have a faithful forgetful functor? Sorry, but I can only think of examples like affine varieties over a finite field $\mathbb{F}_q$. In fact, just take the category with the single variety $\mathbb{A}^1$ and regular maps $\mathbb{A}^1 \to \mathbb{A}^1$. Any polynomial $p \in \mathbb{F}_q[X]$ corresponds to a regular map $\mathbb{A}^1 \to \mathbb{A}^1$, sending the point with coordinate $x$ to the point with coordinate $p(x)$. Usually (for various reasons) the definition of regular map includes not just the map of sets, but also the polynomial, and this is needed because $p(x)=x$ and $p(x)=x^q$ define the same map (Fermat's Little Theorem).

And now, how do you compose regular maps? No doubt the best way is to compose by composing the defining polynomials: $p_1$ composed with $p_2$ is $p_1 \circ p_2$, or $p_1(p_2(x))$. But if you are firmly committed to working over $\mathbb{F}_q$ and never extending that field, and all you care about is that set of points, then you could declare that $p_1$ composed with $p_2$ is $p_1 \circ p_2 \bmod x^q-x$. (Then you also probably should have defined regular maps as elements of $\mathbb{F}_q[x]/(x^q-x)$, but whatever.) This reduced polynomial defines the same function on the set of points. It will be bad news if you care about sheaves or base change or anything, but if you really only care about the set of points it's the same, and it might be simpler (compositions have lower degree because of the reduction modulo $x^q-x$).

So there's an example where there's a choice of how to compose maps. To say which category you want to work in, you have to say how you plan to compose maps. In that sense I think the composition law feels like part of the data of the category.