Find a map of the solid torus into itself having no fixed point. Where does the proof of the Brouwer theorem fail.

Question

Find a map of the solid torus into itself having no fixed point. Where does the proof of the Brouwer theorem fail.

I know that the proof is fail because the torus has a hole, so we can't construct the retraction to its boundary. I think the only fixed point is at the origin, but the torus doesn't cover the origin, so any linear map will do. But I'm not so sure. Any hint will be appreciated.

Answer 1

Define $f:S^1 \to S^1$ by $f(x,y)=(\cos(x+y+1),\sin(x+y+1))$

Rotation $g: R^k \to R^k$, $g(z)= \cos(z)+\sin(z)$

Answer 2

The Brouwer fixed-point theorem requires convexity of the domain being mapped to itself. A solid torus, considered as a subset of $\mathbb{R}^3$, is not convex, which is why a rotation of angle other than $2\pi$ (fixing the axis in $\mathbb{R}^3$ perpendicular to the plane of the core circle of the torus) does not have a fixed point.