Confusion about an answer - why does this element have to be unique? I'm reading this answer, and I'm stuck at "... the unique element that some elements of the chain map it to". Why does this element have to be unique? For example, consider the chain of functions $\mathbb N\to \mathbb N$ under the order described in the answer the first few  elements of which are described as follows:
$f_1$ has fixed points $0,1,2,3$ (and only them)
$f_2$ has fixed points $0,1,3$ (and only them)
$f_3$ has fixed point $1$ (and only them)
The condition $f_2\leq f_3$ means $\{1\}\subseteq \{0,1,3\}$ and the restriction of $f_2$ to $\mathbb N\setminus \{0,1,3\}$ equals the restriction of $f_3$ to the same set. In particular, it might be the case that $f_3(0)=15$ whereas $f_2(0)=0$. If I call the function that is being constructed in the answer $L$, what should be $L(0)$? Should it be $f_3(0)$ or should it be $f_2(0)$, or something else?
Also, just to make sure I understand the other part of the definition of $L$ correctly, the fixed points of $L$ in my example should just contain $1$ (since it's the only point where "all elements of the chain have fixed points")?
 A: The function defined in the answer is the least upper bound of the chain. If your chain is $f_2\leq f_3$, then the least upper bound is just $f_3$.
The expression "the unique element that some elements of the chain map it to" is not very well formulated, it should say "the unique element that all functions in the chain map it to for big enough indices".
More precisely, let $C$ is a chain of bijections $A\to A$ for this partial order. We can define $X=\bigcap_{f\in C}Fix(f)$. Then I claim the following: for any $a\in A$, all functions $f$ that are far enough in the chain take the same value at $a$; let us write $L(a)\in A$ for this value. We may call this value the "eventual value" of the chain at $a$ (because all functions in the chain will "eventually" take this value). Let us write this value $L(a)$.
Proof of the claim: Let $a\in A$. If $a\in X$, then $a$ is a fixed point of all $f\in C$, so of course they all take the same value at $a$, namely $L(a):=a$. Now if $a\not\in X$, that means by definition of $X$ that there exists some $f\in C$ such that $a$ is not a fixed point of $f$. But then for any $g\geq f$, $a$ is also not a fixed point of $g$, and $f(a)=g(a)$, by definition of the order. So starting from (at worst) $f$, all functions in the chain indeed take the same value $L(a):=f(a)$ at $a$.
This defines our function $L:A\to A$ which is the least upper bound of $C$. Its fixed points are exactly the elements of $X$. To be clear, for a fixed function $f\in C$, and some $a\in A$, there are 3 possibilities:

*

*$a\in X$: then $a$ is a fixed point of $L$ and $f$, and so $f(a)=L(a)=a$;

*$a\not\in X$ but $a$ is a fixed point of $f$: then $f(a)=a$, but $L(a)\neq a$, as $a$ is not an eventual fixed point of the chain, so $f(a)\neq L(a)$;

*$a$ is not a fixed point of $f$: then $f(a)=L(a)$, as $f$ already takes the eventual value of the chain at $a$.

