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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.

Diane Vanderwaif
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    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. – Brian M. Scott Oct 13 '14 at 17:17
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    A rotation. ${}{}{}$ – copper.hat Oct 13 '14 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 ? – Diane Vanderwaif Oct 13 '14 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 $S^1$. – Brian M. Scott Oct 13 '14 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. – Ayman Hourieh Oct 13 '14 at 18:07
  • About the retraction I can say that torus is a compact manifold with boundary, and any compact manifold with boundary has no retraction from itself onto the boundary. I tried to understand copper hint too. But I understand the torus is a circle that is rotated by $360^o$. So I need a map of the circle first, then the map for rotation. No ? – Diane Vanderwaif Oct 13 '14 at 18:16

2 Answers2

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

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

Neal
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XiaoXiao Zhen
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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.

Neal
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  • what phrase in the statement of the theorem in p.65 in Guillemin and Pollack said that "The Brouwer fixed-point theorem requires convexity of the domain being mapped to itself" ? ................ is it because the closed unit ball is convex? – Intuition Nov 25 '18 at 19:39
  • So what does it mean to have a fixed point by drawing? – Intuition Nov 25 '18 at 19:59