Ok, here I'll give it a go.

> (a) Consider the category where the objects are natural numbers and where there is a unique morphism from m to n if m divides n. Given two numbers, for example 42 and 27, what is their product? What is the name of this binary operation?

Given two objects A, B in a category, a product of A and B consists of another object, call it \$$A \times B\$$, along with two morphisms \$$\pi_1: A \times B \rightarrow A\$$, and \$$\pi_2: A \times B \rightarrow B\$$, which satisfies a certain universal property in the category.

But before talking about the universal property, let's instantiate this first part of the definition to the category at hand.

Here, object A = m, and object B = n.

We have to now posit what the object \$$A \times B\$$ is, and provide the two morphisms \$$\pi_1: A \times B \rightarrow A\$$ and \$$\pi_2: A \times B \rightarrow B\$$.

@FabricioOlivetti posited that \$$A \times B\$$ is the greatest common divisor of m and n, gcd(m,n). Indeed it is, as we are now proving.

Claim: \$$A \times B = m \times n = gcd(m,n)\$$

(Note, don't be thrown off by the odd appearance of the above equation. Here the symbol \$$\times\$$ is being used in a generic categorical sense, which in the case is being used to describe an operation that is not multiplication.)

Now we must produce \$$\pi_1: m \times n \rightarrow m\$$ and \$$\pi_2: m \times n \rightarrow n\$$.

But because this category is "thin," which means that there is at most one arrow between any two objects, there could only be one possible choice for \$$pi_1\$$ and \$$\pi_2\$$. Let's start with \$$\pi_1: gcd(m,n) \rightarrow m\$$. Does any such arrow exist going from gcd(m,n) to m? Well yes, because gcd(m,n) divides m! So that is what \$$\pi_1\$$ is. Similarly, \$$\pi_2\$$ is the unique arrow going from gcd(m,n) to n.