Is there a self-homeomorphism of the 2-sphere with exactly 3 fixed points? I don't believe so, but I'm not sure how to prove it.
The Lefschetz-Hopf theorem says in this case that the sum of the fixed point indices is 0 or 2 (since our map is a self-homeomorphism). My initial thought was to use this theorem and a parity argument; showing that the index of a fixed point of a homeomorphism must be $\pm 1$, but this won't work for trivial reasons (there are self-homeomorphisms of the 2-sphere with exactly 1 fixed point).
 A: By the compactness of $S^2$, a smooth vector field that vanishes at precisely three points will determine a smooth 1-parameter family of diffeomorphisms with three fixed points. In cylindrical coordinates $(\phi,z)$ on $S^2$, we can accomplish this by summing a pair of vectors fields --- one in the $\partial \phi$ direction and a perpendicular one in the $\partial z$ direction --- whose sets of zeroes intersect in three points.
We can also construct a suitable homeomorphism directly: Consider cylindrical coordinates $(\phi,z)$ on $S^2$ and the map $(\phi,z)\mapsto (\phi+\sin(\phi/2),z^3)$. Away from the poles $z=\pm1$ (which are clearly fixed), a fixed point $(\phi,z)$ must satisfy $\sin(\phi/2)=0$ and $z^3=z$, hence $\phi \equiv 0 \operatorname{mod} 2\pi$ and $z=0$. It follows that the only fixed points are $(0,0)$ and the poles $z=\pm1$.
Remark. We can easily tweak the above homeomorphism to produce a diffeomorphism, but this does require a bit more work (e.g. using a periodic bump function for the coordinate $\phi$).
A: Project the sphere (minus its poles) to the infinite cylinder $x^2+y^2=1$, then map $(x,y,z)\mapsto (x,y,z+x-1)$, then project back to the sphere
