How is a preorder a thin category?
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.
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)