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

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.

• I’m not sure what you mean by a linear map in this context. Can you find a simple map of $S^1$, the unit circle, onto itself that has no fixed point? If you can do that, you can easily modify it to get a map of the torus onto itself that has no fixed point. Oct 13, 2014 at 17:17
• A rotation. ${}{}{}$ Oct 13, 2014 at 17:17
• so the map of the circle is $f: S^1 \to S^1$ by $f(x,y)=(\frac{x}{2}, \frac{y}{2})$ will do? And I need to rotate this circle around the origin. What map do I need for rotation? Antipodal map ? Oct 13, 2014 at 17:33
• Your $f$ is not a map of $S^1$ to itself: it maps $S^1$ to the circle of radius $\frac12$ centred at the origin, not to $S^1$. Oct 13, 2014 at 17:53
• @DianeVanderwaif Doesn't work either. This maps $S^1$ to a circle of radius $2$. Instead of trial and error, try to use the hint give to you above by copper.hat. Also, you're right that a retraction from the solid torus onto its boundary doesn't exist. However, I think you need a stronger justification that saying that the torus has a hole. Oct 13, 2014 at 18:07

Define $f:S^1 \to S^1$ by $f(x,y)=(\cos(x+y+1),\sin(x+y+1))$
Rotation $g: R^k \to R^k$, $g(z)= \cos(z)+\sin(z)$
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.