**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'\$$.