Chain-complete and least element iff every order-preserving map has least fixed point

Question

Let $P$ be a poset. I want to show the following are equivalent.

1. $P$ is chain-complete and it has a least element.
2. For every order-preserving map $f:P\to P$, the set $P_f$ of fixed points of $f$ has a least element and is chain complete (in the inherited ordering).

$1.\implies 2.$ I struggle here because given a chain $C\subseteq P_f$ the supremum $x$ (with respect to $P$ need not be a fixed point itself. I can only show $x\leq f(x)$, now I need to find a fixed point above $x$. And how do I find the least fixed point?

$2.\implies 1.$ Taking $f$ as the identity we get the least element of $P$. However, how can I show $P$ is chain-complete? Is this even true?

(note that I don't regard $\emptyset$ as a chain, so the least element property really is an extra condition)

Answer 1

$2. \Longrightarrow 1.$ You're almost there. You are given that $P_f$ is chain complete, so since $f = \mathrm{id}_P$ we have $P_f = P$ (since $\mathrm{id}_P$ fixes the whole poset) is chain complete.

$1. \Longrightarrow 2.$ The statement about the least fixed point is one of the versions of the Tarski-Knaster theorem. For the proof see, for example, theorem 1.3 here.

To show that $P_f$ is chain complete let $C \subseteq P_f$ be a chain with $\bigvee C = x$. You can show that $x \leq f(x)$. Let $X = \{y \in P \mid y \geq x\}$. For any $y \in X$ we have $x \leq y \Rightarrow x \leq f(x) \leq f(y)$ so $f(y) \in X$ implying $f\restriction{X} \colon X \to X$. Since $X$ is also chain complete, the map $f\restriction{X}$ has the least fixed point $x_* \in X \Rightarrow x \leq x_*$, meaning that $x_*$ is an upper bound for $C$. But since it is the least fixed point in $X$ it is the least upper bound for $C$.