At least Anindya understood the questions well enough to correctly answer them!

The point of the questions was this: after we think about it a minute, the definition of feasibility relation says that

\[ \text{if } (x,y) \le (x',y') \text{ in } X^{\text{op}} \times Y, \text{ then }\Phi(x,y) \text{ implies } \Phi(x',y') .\]

But the theorem claims that this one condition is equivalent to _two_ conditions:

1. if \\(x' \le x \\) in \\(X\\), then \\( \Phi(x,y) \\) implies \\(\Phi(x',y)\\) for all \\(y \in Y\\),

2. if \\(y \le y'\\) in \\(Y\\), then \\( \Phi(x,y) \\) implies \\( \Phi(x,y')\\) for all \\(x \in X\\).

So, somehow we must be able to take a _single_ inequality in \\( X^{\text{op}} \times Y\\) and break it into _two_ inequalities, one in \\(X\\) and one in \\(Y\\).

The proof of the theorem explains how... but it might not be clear to everyone. So I wanted you to figure it out:

**Puzzle 169.** The one inequality \\((x,y) \le (x',y')\\) in \\( X^{\text{op}} \times Y\\) is equivalent to two inequalities, one in \\(X\\) and one in

\\(Y\\). What are they?

But preorders are just a special case of categories, and this fact is just a special case of a fact about categories:

**Puzzle 170.** Any morphism in a product of categories, \\(\mathcal{C} \times \mathcal{D}\\), is a composite of two morphisms, one built from a morphism in \\(\mathcal{C}\\) and one built from a morphism in \\(\mathcal{D}\\). What are they?

Both puzzles have _two correct answers!_ Why?

This stuff isn't very deep, but it's good to know when you're working with products of categories, or preorders.

The point of the questions was this: after we think about it a minute, the definition of feasibility relation says that

\[ \text{if } (x,y) \le (x',y') \text{ in } X^{\text{op}} \times Y, \text{ then }\Phi(x,y) \text{ implies } \Phi(x',y') .\]

But the theorem claims that this one condition is equivalent to _two_ conditions:

1. if \\(x' \le x \\) in \\(X\\), then \\( \Phi(x,y) \\) implies \\(\Phi(x',y)\\) for all \\(y \in Y\\),

2. if \\(y \le y'\\) in \\(Y\\), then \\( \Phi(x,y) \\) implies \\( \Phi(x,y')\\) for all \\(x \in X\\).

So, somehow we must be able to take a _single_ inequality in \\( X^{\text{op}} \times Y\\) and break it into _two_ inequalities, one in \\(X\\) and one in \\(Y\\).

The proof of the theorem explains how... but it might not be clear to everyone. So I wanted you to figure it out:

**Puzzle 169.** The one inequality \\((x,y) \le (x',y')\\) in \\( X^{\text{op}} \times Y\\) is equivalent to two inequalities, one in \\(X\\) and one in

\\(Y\\). What are they?

But preorders are just a special case of categories, and this fact is just a special case of a fact about categories:

**Puzzle 170.** Any morphism in a product of categories, \\(\mathcal{C} \times \mathcal{D}\\), is a composite of two morphisms, one built from a morphism in \\(\mathcal{C}\\) and one built from a morphism in \\(\mathcal{D}\\). What are they?

Both puzzles have _two correct answers!_ Why?

This stuff isn't very deep, but it's good to know when you're working with products of categories, or preorders.