>**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 \$$