Least fixed point of restricted function Let $P$ be a poset with the property that every order-preserving map $f:P\to P$ has a least fixed point $\mu(f)$.
Now for any $p\in P$, the poset $\downarrow(p)=\{x\in P|x\leq p\}$ must also have this property.
Let $f,g:P\to P$ be two order-preserving maps such that $f(x)\leq g(x)$. I'm supposed to show that $\mu(f)\leq\mu(g)$. However in the following proof I don't use $f(x)\leq g(x)$, so there must be a mistake.
"Proof" for $\mu(f)\leq\mu(g)$: The restriction of $f$ to $\downarrow(\mu(g))$ has a least fixed point $p$. But then $p=\mu(f)$ and $p\leq\mu(g)$ so we are done.
Is the conclusion $p=\mu(f)$ wrong?? It feels like it should be wrong but $\mu(f)$ is the least fixed point so $\mu(f)\leq p$ so $\mu(f)\in \downarrow(\mu(g))$ so $\mu(f) = p$. Yes? No?

You are using $f \leq g$ when considering the restriction of $f$ to $\downarrow(\mu(g)) = Q$. To apply the property to $Q$, you need to check that the restriction $f\restriction{Q}$ maps $Q$ to itself (i.e., $f\restriction{Q} \colon Q \to Q$). In other words, you need to check that $\forall x \in Q$ you have $f(x) \in Q$. 
Let $x \in \downarrow(\mu(g))$ (and hence $x \leq \mu(g)$). Then since $f \leq g$ and $g$ is order-preserving we have $$f(x) \leq g(x) \leq g(\mu(g)) = \mu(g),$$
that is $f(x) \in \downarrow(\mu(g))$.
