How is a preorder a thin category?

Question

I'm reading Category Theory for Programmers. In the chapter Categories Great and Small, it says:

A preorder is a category where there is at most one morphism going from any object a to any object b. Another name for such a category is “thin.” A preorder is a thin category.

My question is, if I have a <= b, a <= c, b <= d, c <= d like below:

doesn't that mean I have two morphisms going from object a to object d (i.e. a -> b -> d and a -> c -> d)?

What am I misunderstanding here?

Answer 1

No, those two routes from $a$ to $d$ are different, but when you compose them, you get equal morphisms $a\to d$. For instance, you can consider the preorder on the natural numbers that says $n\geq m$ when $n$ is divisible by $m$. Then $6\geq 3,6\geq 2,3\geq 1,$ and $2\geq 1$. This gives us two routes by which to show that $6\geq 1$, but that doesn't mean there are two different morphisms $6\to 1$ in this category.

More broadly, this is no different than the observation that $31=0+31=7+24$, or that smoking causes lung cancer, which causes death, and that smoking causes heart disease, which causes death, so there are (at least) two different reasons why smoking causes death; none of this stops you from defining a preorder in which $x\leq y$ means "$x$ causes $y$ (in at least one way.)"

Closer to your example, your picture could describe at least two different categories: $C_1$, the category freely generated by that picture, is the "non-commutative" square in which there are two distinct morphisms $a\to d$, while $C_2$, the commutative square, is given by $C_2$ together with the relation that both paths $a\to d$ are equal. $C_2$ is a preorder, but $C_1$ is not. These notions of generators and relations are very much the same as you may have encountered in group theory, or maybe not, if you're coming from programming.

Answer 2

You did not mention how we turn a preorder (a set $S$ with a reflexive transitive relation ${\le}\subseteq S\times S$) into a category $\mathcal C$. Once you make this explicit, the thin-ness gets clear:

• The class of objects is the set $S$
• The class of morphisms is the relation $\le$, where $a\le b$ (i.e., the pair $(a,b)\in S\times S$) is a morphism from $a$ to $b$.

Apparently, as we deal with a relation, there either is or is not a morphism (ond only one) from one object to another, i.e., $\mathcal C$ is certainly thin. (That it is a category to begin with follows from the preorder properties: reflexivity gives us identities, transitivity gives us composition)