**Puzzle 103.** Figure out the best way to take any category \\(\mathcal{C}\\) and "squash it down", getting a preorder whose elements are the objects of \\(\mathcal{C}\\).

Matthew wrote:

> I would define \\(\leq_{\mathcal{C}}\\) with:

> \[ X \leq_{\mathcal{C}} Y \equiv \mathrm{Hom}(X,Y) \neq \varnothing \]

> where \\(\mathrm{Hom}(X,Y)\\) is the [set of morphisms](https://en.wikipedia.org/wiki/Morphism#Hom-set) between \\(X\\) and \\(Y\\).

Right! Here are two equivalent ways to think about this trick:

1. Given a category, create a preorder by decreeing \\(x \le y\\) iff there's a morphism from \\(x\\) to \\(y\\).

2. Given a category, identify all morphisms from \\(x\\) to \\(y\\) - that is, decree them all to be equal - so that there's at most one morphism from \\(x\\) to \\(y\\).

These are equivalent because we can think of a preorder as a category with at most one morphism from any object \\(x\\) to any object \\(y\\): we write \\(x \le y\\) iff such a morphism exists.

Matthew wrote:

> I would define \\(\leq_{\mathcal{C}}\\) with:

> \[ X \leq_{\mathcal{C}} Y \equiv \mathrm{Hom}(X,Y) \neq \varnothing \]

> where \\(\mathrm{Hom}(X,Y)\\) is the [set of morphisms](https://en.wikipedia.org/wiki/Morphism#Hom-set) between \\(X\\) and \\(Y\\).

Right! Here are two equivalent ways to think about this trick:

1. Given a category, create a preorder by decreeing \\(x \le y\\) iff there's a morphism from \\(x\\) to \\(y\\).

2. Given a category, identify all morphisms from \\(x\\) to \\(y\\) - that is, decree them all to be equal - so that there's at most one morphism from \\(x\\) to \\(y\\).

These are equivalent because we can think of a preorder as a category with at most one morphism from any object \\(x\\) to any object \\(y\\): we write \\(x \le y\\) iff such a morphism exists.