Homeomorphisms of the Open Disk Does there exist a homeomorphism $\phi$ of the open unit disk in the plane such that $\phi$ has no fixed point but there exists $n$ such that the $n$-fold composition $\phi^n$ is the identity?
(To save some of you the bother of asking what I've done so far: nothing. Seems clear the answer is no; it's also clear to me that I simply don't have the tools to try to answer a question like this. If it helps I can say how the question arises: If the answer is no, as I suspect, then any non-trivial covering map $f:\mathbb D\to X$ is infinitely sheeted - this would say that holomorphicity is irrelevant to something that came up yesterday.)
 A: No, there does not. This might not be the easiest argument, but it follows from Harold Bell, A fixed point theorem for plane homeomorphisms, Bull. AMS, 82 (5), (1976), 778-780. Bell's much more general result is that every homeomorphism of the plane leaving some continuum $M$ invariant has a fixed point in $T(M)$, where $T(M)$ is the "filled-in hull" of $M$, i.e., the complement of the unbounded component of the complement of $M$. This result itself is an extension of the Cartwright-Littlewood theorem (which is the same result only for orientation-preserving homeomorphisms), which in itself is an extension of the Brouwer fixed point theorem. In your case, take any large closed disk $K$ contained in the unit disk such that $\phi(K) \cap K \ne \emptyset$, and let $M= \bigcup_{j=0}^{n-1} \phi^j(K)$
A: This is a special case of the Brouwer plane translation theorem. 
See the paper of John Franks entitled "A new proof of the Brouwer plane translation theorem", Ergodic Theory Dynam. Systems 12 (1992), no. 2, 217–226. 
