Preorder Category is Category

Theorem
Let $\left({S, \precsim}\right)$ be a preordered set.

Let $\mathbf S$ be its associated preorder category.

Then $\mathbf S$ is a category.

Proof
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.

Note that for objects $a, b$ of $\mathbf S$, there is at most one morphism $a \to b$, by definition of $\mathbf S$.

Suppose that $a \to b$ and $b \to c$ are morphisms of $\mathbf S$.

Then we have $a \precsim b$ and $b \precsim c$, and as $\precsim$ is a preordering:


 * $a \precsim c$

which is to say that there is a unique morphism $a \to c$.

By uniqueness, this morphism is the composite of $a \to b$ and $b \to c$.

Also, for any $a \in S$, there is a morphism $a \to a$ by virtue of:


 * $a \precsim a$

since $\precsim$ is a preordering.

The uniqueness of the morphisms mentioned above implies directly that $a \to a$ classifies as the identity morphism $1_a$ for $a$.

The uniqueness of morphisms also immediately implies associativity.

Hence $\mathbf S$ is a metacategory.

Since both $S$ and $\precsim$ are sets, $\mathbf S$ is also a category.