Does every continuous map $f$ from $D^2$ minus $k$ disjoint open disks to itself have a fixed point?

Let $A_k$ denote $D^2$ with $k\geq 0$ disjoint open disks removed. For $k=0$, the answer is positive by Brouwer's fixed point theorem. For $1\leq k\neq 2$, it's not difficult to see that the answer is negative: by arranging the disks in a symmetrical way, I can apply certain rotations to obtain self maps with no fixed point, e.g. if $k=4$, I can set one disk at the center, and position the other three in an equilateral triangle around the center; then a 120 degree rotation is a map from $A_4$ to itself with no fixed point. This sort of arrangement can always be done for $k\neq 2$. Note that 120 degree rotation of $A_4$ can be extended to a continuous map of $D^2$ to itself.

The tricky case is $k=2$, which is where I'm stuck. My thinking here is to consider any possible extension

$$\bar f:D^2\rightarrow D^2\ |\ \bar f|_{A_2}=f.$$

Since $\bar f$ is continuous, it must map each of the two missing disks to either itself or to the other. In the case where $\bar f$ maps one disk to the other, then no fixed points lie in either of the disks (since they are disjoint), hence, by Brouwer, there must be a fixed point of $\bar f$ lying in $A_2$, which will be a fixed point of $f$. In the case where $\bar f$ maps the two disks to themselves, then by applying Brouwer to the restriction of $\bar f$ to each disk, we have that $\bar f$ has at least two fixed points in $D^2$.

If I had holomorphicity to work with, then this would be enough, since a holomorphic map from $D^2$ to itself with two fixed points must be the identity. Am I on the right track, or is there some arrangement of the missing disks and a transformation that I'm just not seeing?

• You have to take a look at the Lefschetz Fixed Point Theorem. – Peter Saveliev Apr 25 '18 at 20:01

A disk with two holes is homeomorphic to a "pair of pants": a $2$-sphere with three discs removed. In a symmetric version where the centres of the discs lie on the equator and form the vertices of an equilateral triangle, and where the discs have equal radius, you see that a reflection about the plane going through the three centres of the discs, followed by an obvious $120$ rotation, will have no fixed point.
The south hemisphere is mapped to the north hemisphere, so the only possible fixed points are on the equator. But the equator is rotated by $120$ degrees, so there are no fixed points there.