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Let $f:S^1\to S^1$ be continous. Show that if $f$ is a homotopy to the Constant function, there exists a point $x\in S^1$ such that the length between $x$ and $f(x)$ is $\frac{\pi}{6}$.

Hint: You can lift a Path from $S^1$ to $\mathbb{R}$.

I have no Idea how to even start solving this. Any intuition would be very helpful.

SlyxBrd
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  • take $f$ be rotation by $\pi/12$ clockwise then it is continuous and homotopic to identity and there is no point $x$ such that that $d(x,f(x))< \pi/6$ both in $\mathbb{R}^2$ and $S^1$ so the question is wrong. – MasM Oct 17 '18 at 10:15
  • Or easier: take $f(x)=x$. Obviously distance from $x$ to $f(x)$ is always $0$ regardless of used metric. – freakish Oct 17 '18 at 10:34
  • @SlyxBrd Indeed, the statement is wrong. You have counterexamples in two comments above. Perhaps it should've been "homotopic to a constant function" instead of "homotopic to the identity"? I think this is true under this assumption. – freakish Oct 17 '18 at 10:37
  • You are correct, I am very sorry for misleading you, english isn't my native language. Thank you for your help! – SlyxBrd Oct 17 '18 at 10:43
  • Can you be more accurate ? – Marco Lecci Oct 17 '18 at 11:03
  • You should edit the title. And what do you mean by "can create an arc of any length"? – Paul Frost Oct 18 '18 at 16:02

1 Answers1

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Throughout I assume that by "length" you mean the Euclidean distance. This is important because on the unit sphere the maximal Euclidean distance is achieved on a pair of antipodal points. And also because that maximal distance is $2$ which is greater than $\pi/6$.

So first of all we need the following

Lemma. If $f:S^1\to S^1$ is a null-homotopic continuous map then $f$ has a fixed point, i.e. there exists $x\in S^1$ such that $f(x)=x$.

The proof can be found here.

So let $p_1$ be a fixed point of $f$.

Now let $\Lambda:S^1\to S^1$ be the antipodal map $\Lambda(x)=-x$. It follows that $\Lambda\circ f$ is also null homotopic and thus by our Lemma it has a fixed point as well. I.e. there exists $p_2\in S^1$ such that $f(p_2)=-p_2$.

Now consider the function

$$D:S^1\to\mathbb{R}$$ $$D(x)=d(x,f(x))$$

where $d$ is the Euclidean distance. It follows that

$$D(p_1)=0$$ $$D(p_2)=2$$

and since $D$ is continous and $S^1$ is connected then the image of $D$ is connected as well. In this case it is equal to $[0,2]$. In particular for any $r\in[0,2]$ there is $x\in S^1$ such that $d(x,f(x))=r$.

freakish
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