Can the identity $ab=\gcd(a,b)\text{lcm}(a,b)$ be recovered from this category?

Question

Define the category $\mathcal{C}$ as follows. The objects are defined as $\text{Obj}(\mathcal{C})=\mathbb{Z}^+$, and a lone morphism $a\to b$ exists if and only if $a\mid b$. Otherwise $\text{Hom}(a,b)=\varnothing$. It can be shown that $\gcd(a,b)$ is a product in $\mathcal{C}$ and $\text{lcm}(a,b)$ is a coproduct in $\mathcal{C}$. My question is:

Can we recover the identity $$ab=\gcd(a,b)\text{lcm}(a,b)$$ from the properties of this category?

My initial thought was to denote a morphism $a\to ak$ by simply $k$ (or a triple $(a,k,ak)$ if we are concerned about disjointness), and then of course composition would just be multiplication: $k:a\to b,\ k':b\to c$, then $kk':a\to c$. Then, by the universal properties of products and coproducts, the diagram: $$ \require{AMScd} \begin{CD} \gcd(a,b)=d @>{a/d}>> a\\ @V{b/d}VV @VV{\ell/a}V \\ b @>>{\ell/b}> \ell=\text{lcm}(a,b) \end{CD} $$ commutes. But of course composition only yields $$\frac{a}{d}\frac{\ell}{a}=\frac{b}{d}\frac{\ell}{b}.$$ I'm also curious about other identities this category might "know about".

Edit: I'm thinking we could just show that $$\frac{ab}{\ell}$$ satisfies the universal property for products (or dually, that $ab/d$ is a coproduct) which would imply that $ab/\ell$ is isomorphic to $d$. By definition this implies that $d$ and $ab/\ell$ divide each other, and hence must be equal (hopefully that's correct). Of course, I am open to any other methods involving categorical machinery that I'm not familiar with (limits, pullbacks, etc.) and any other identities that $\mathcal{C}$ may be aware of.

Answer 1

The problem here is that $\mathcal C$ does not know what "multiplication" is. The entity $ab$ is entirely meaningless to it.

It is similarly meaningless to say that $\dfrac{ab}\ell$ is the GCD, because as far as $\mathcal C$ is concerned, this object is not connected to $a$, $b$ or $\ell$.

The only thing that $\mathcal C$ can hope to express things about is number-theoretic statements that are solely expressible in terms of divisibility (such as the $\gcd$ and $\rm lcm$).

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However, a more interesting situation arises when we consider $\mathcal C$ as a (strictly) monoidal category with multiplication as its "tensor" product $\otimes$.

All of a sudden, we find ourselves in a situation where a meaningful "product" exists. We can even "divide", because given $a \mid c$, we can recover $b := \dfrac c a$ as the coproduct over those $d$ such that $a \otimes d \mid c$. Naturally, then, $a \otimes b = c$.

In standard categorical notation, we can then write the desired identity as: $$a \otimes b = (a + b) \otimes (a \times b)$$

Answer 2

In fact, every partially ordered set can be viewed as a category in the way you describe; your example is obtained by applying this construction to the poset $(\mathbb{Z}_{\geq 1},\mid)$. This process always turns meets into products and joins into coproducts. So your observations are essentially the fact that $\mathrm{gcd}$ is the meet in $(\mathbb{Z}_{\geq 1},\mid)$ and $\mathrm{lcm}$ is the join.

As far as I can tell, there's no truly sensible way to recover the multiplicative structure of $\mathbb{Z}_{\geq 1}$ from the poset $(\mathbb{Z}_{\geq 1},\mid)$ alone. The problem is that there are certain distinguished chains in $(\mathbb{Z}_{\geq 1},\mid),$ namely those of the form $\{n^k \mid k \in \mathbb{Z}_{\geq 0}\}.$ Speaking vaguely: if you know these, there's a sensible way to recover the multiplicative structure. But the poset $(\mathbb{Z}_{\geq 1},\mid)$ alone doesn't "know" these.

Note that, if we identify each element of $\mathbb{Z}_{\geq 1}$ with its multiset of prime factors, then the identity you're interested in becomes:

$$A+B = (A \cup B) + (A \cap B)$$

(in multiset notation), which makes it really obvious why its true!