TITLE: Periodic point-free maps and free ultrafilters.
QUESTION [6 upvotes]: Let $X$ be a set and $u$ be a free ultrafilter on $X$. We can consider a topology on $X$ by declaring every element of $u \cup \{\emptyset \}$ to be open. 
El'kin's original motivation for looking at this topology is that $X$ is a space without isolated points which can't be split into disjoint dense subspaces. Another quirky feature: if $X$ is say $\mathbb{R}$ then translations are not continuous. This last fact makes me wonder: are there any periodic-point free continuous self maps at all on $X$ with this topology? In other words:

Is there a free ultrafilter $u$ on a set $X$ and a periodic point-free onto map $f: X \to X$ such that: $f^{-1}(A) \in u$, for every $A \in u$? Is there at least one such fixed-point free map?

REPLY [8 votes]: Although the question has already been thoroughly answered, it might be worthwhile to point out that the result can be separated into the combinatorial "meat", which doesn't involve ultrafilters, and a small corollary where the meat is fed to the ultrafilter problem.  The meat is the following theorem, which is, if I remember correctly, explicit in the early references; it is essentially proved (though not stated) in Joel Hamkins's answer here.  Given any set $X$ and any function $f:X\to X$, there is a partition of $X$ into four disjoint (possibly empty) sets $X=A_0\sqcup A_1\sqcup A_2\sqcup A_3$ such that $A_0$ is the set of fixed points of $f$ and each of the other three $A_i$'s is disjoint from its image under $f$.  Once one has this result, one immediately sees that any ultrafilter on $X$ must contain one of the four $A_i$'s and if that $A_i$ isn't $A_0$ then the ultrafilter can't be $f$-invariant as in the question.