# Why must commuting maps (of an interval) without common fixed points have at least 11 fixed points for the composition?

I've been looking at the examples of commuting functions on a closed interval which have no common fixed points. These were discovered in 1967 by William M Boyce and J Philip Huneke.

Earlier work by Glen Baxter showed that each of the functions for an example must induce a permutation on the fixed points of the composition which preserves the "crossing type" of the fixed point (assuming that the fixed point set is finite). Boyce studied these "Baxter permutations", which have now found other applications in combinatorics. Boyce stated without proof that the number of fixed points of the composition must be at least 11. He indicates that this can be shown by considering all the possible Baxter permutations for smaller numbers of fixed points, and proving that each one can't occur for an example. A later paper of his in the Houston Journal gives a proof for the case of 7 fixed points. I can't seem to generalize this to other cases. Maybe I'm just missing something simple.

In particular, I can't handle the case of 5 "crossing" fixed points. (A "crossing" fixed point is one where the graph of the function changes from one side of the diagonal to the other. The number of such points must be odd.) This reduces to the following statement, which I'm unable to prove:

Claim: There do not exist continuous functions $f$ and $g$ mapping the closed interval $[0,6]$ to itself which satisfy the following:

1. $f\circ g=g\circ f$, i.e., $f$ and $g$ commute.
2. $fg(n)=n$ for $n\in\{1,2,3,4,5\}$
3. $fg(x)\ge x$ if $x\in[0,1]\cup[2,3]\cup[4,5]$
$fg(x)\le x$ if $x\in[1,2]\cup[3,4]\cup[5,6]$
4. The restriction of $f$ to $\{1,2,3,4,5\}$ is the permutation whose cycle structure is $(135)(24)$, and the restriction of $g$ to this set is the inverse permutation $(153)(24)$

The Boyce/Huneke examples actually have 13 fixed points for the composition. It is an open question as to whether or not an example with 11 exists. Boyce states, again without proof, that there are only three possible permutations for the case of 11 fixed points, but that it seems to be difficult to get an approximating sequence of functions to converge uniformly for examples based on those permutations.

There is an excellent survey paper by Eric McDowell available for free from Auburn's Topology Proceedings site that includes references to just about every paper related to this question.

I should also note that the term "Baxter permutation" is used here in a slightly different way than is common in combinatorics. It turns out that the permutation of the "up-crossing" fixed points is determined by the permutation of the "down-crossing" points. The corresponding permutation of the down-crossing points (1,3, and 5 in the example above) was originally called a "reduced" Baxter permutation, but is now what is usually meant by the term.