Thanks Marius, Tobias, and John for the responses! I had a lot of fun working on these problems.

Marius, I really like the example of getting a discount instead of a bag charge! Your comment about opposite categories also made me think that given a function \\(f: X \to Y\\) we can define a relation on \\(X\\) in an opposite way by \\[x \leq_X x' \iff f(x) \geq_Y f(x').\\]

I also wanted to check my thinking about Puzzle 77 again.

I showed that the property \\(f(x) \otimes_Y f(x') = f(x \otimes_X x') \\) is sufficient for making \\( (X, \leq_X, \otimes_X, 1_x) \\) a monoidal pre-order. But the examples of a bag cost and coupon discount that Marius and I suggested, show that this is not a necessary condition, since in both of those cases we only have \\(f(x) \otimes_Y f(x') \leq f(x \otimes_X x') \\) and \\(f(x) \otimes_Y f(x') \geq f(x \otimes_X x') \\) respectively. So as of yet, we don't have a nice necessary**and** sufficient condition on \\(f\\) for making \\( (X, \leq_X, \otimes_X, 1_x) \\) a monoidal pre-order. Is that correct?

Marius, I really like the example of getting a discount instead of a bag charge! Your comment about opposite categories also made me think that given a function \\(f: X \to Y\\) we can define a relation on \\(X\\) in an opposite way by \\[x \leq_X x' \iff f(x) \geq_Y f(x').\\]

I also wanted to check my thinking about Puzzle 77 again.

I showed that the property \\(f(x) \otimes_Y f(x') = f(x \otimes_X x') \\) is sufficient for making \\( (X, \leq_X, \otimes_X, 1_x) \\) a monoidal pre-order. But the examples of a bag cost and coupon discount that Marius and I suggested, show that this is not a necessary condition, since in both of those cases we only have \\(f(x) \otimes_Y f(x') \leq f(x \otimes_X x') \\) and \\(f(x) \otimes_Y f(x') \geq f(x \otimes_X x') \\) respectively. So as of yet, we don't have a nice necessary