>**Puzzle 102.** What is the category with the fewest morphisms that is not a preorder?

A monoid \\(\mathcal{M}\\) with two morphisms, the identity \\(id_\mathcal{M}\\) and a non-identity morphism \\( f \\), such that \\(f \circ f = f \\). \\(\qquad \square \\)

>**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}\\).

Let \\(\mathcal{C}\\) be a category, then we can make a preorder, \\(\mathcal{C}_\leq \\), via the following construction:

The objects of \\( \mathcal{C}_\leq \\) are the same as \\( \mathcal{C} \\),

\\[
Ob( \mathcal{C}_\leq ) = Ob( \mathcal{C} )
\\]

the preordering comes from there existing at least one morphism between objects (ignore how many there are if there are more than one),

\\[
\begin{align}
x \leq_\mathcal{C} y = [\exists f \mid f\in \mathcal{C}(x,y)] \\\\
= \min(1,|\mathcal{C}(x,y)|)
\end{align}
\\]

this gives reflexivity from identities,
\\[
\begin{align}
x \leq_\mathcal{C} x = [\exists id_x \mid id_x \in \mathcal{C}(x,x)] \\\\
= \min(1,|\mathcal{C}(x,x)|)
\end{align}
\\]

and transitivity from composition,
\\[
\begin{align}
x \leq_\mathcal{C} y \wedge y \leq_\mathcal{C} z \implies x \leq_\mathcal{C} z \\\\
= [\exists f \mid f \in \mathcal{C}(x,y)] \wedge [\exists g \mid g \in \mathcal{C}(y,z)] \implies [\exists h \mid h = g \circ f \wedge h \in \mathcal{C}(x,z)] \\\\
= \min(1,|\mathcal{C}(x,y)|) \cdot \min(1,|\mathcal{C}(y,z)|) = \min(1,|\mathcal{C}(x,z)|)
\end{align} \\]

\\(\qquad \square \\)