**Puzzle 76.** I agree with Marius that \\(f\\) must be injective. In fact I think this is a necessary and sufficient condition! But before the proof, a small example.

In John's example where \\(X\\) is a set of groceries and \\(f\\) maps groceries to their cost, $$\text{grocery 1} \leq \text{grocery 2}$$ iff the cost of grocery 1 is less than or equal to the cost of grocery 2. If \\(f\\) is not injective then there exist two different groceries (let's say apples and oranges) with the same cost (let's say \$1). Since in the world of cost \$1 \\(\leq\\) \$1 by reflexivity, in the world of groceries we have apples \\(\leq\\) oranges and oranges \\(\leq \\) apples. This means that apples and oranges are equivalent (which makes sense because they are equivalent in terms of cost). But of course apples \\(\neq \\) oranges. So the groceries with the relation induced by \\(f\\) does not form a poset!

This argument works in general to show that if \\(f\\) induces a poset relation \\( (X, \leq_X) \\), then \\(f\\) is injective.

*Proof by contradiction:* Suppose that \\(f\\) is not injective. Then there exists \\(x,x' \in X\\) such that \\(f(x) = f(x')\\) where \\(x \neq x'\\). By reflexivity \\[ f(x) \leq_Y f(x') \text{ and } f(x') \leq_Y f(x).\\] By definition of \\( \leq_X \\), this means that $$x \leq_X x' \text{ and } x' \leq_X x.$$ Since \\( x \neq x' \\), this means that \\( (X, \leq_X) \\) is not a poset.

The converse is also true: If \\(f\\) is injective, then it induces a poset relation \\( (X, \leq_X) \\).

*Proof:* Suppose that $$x \leq_X x' \text{ and } x' \leq_X x.$$ Then by the definition of \\(\leq_X\\) \\[f(x) \leq_Y f(x') \text{ and } f(x') \leq_Y f(x).\\] Since \\( (Y, \leq_Y) \\) is a poset, this implies that \\(f(x) = f(x')\\). And since \\(f\\) is injective, this means that \\(x = x'\\).