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I have the following problem:

Show that if $f:S^1\to S^1$ is a continuous map, and $f$ is homotopic to a constant, then $\exists p\in S^1 : f(p)=p$.

My approach is to show that if for all $p, \ $ $f(p)\neq p$, then $f$ is homotopic to $\mathrm {id}_{S^1}$. To prove this I thought of using the parametrization of the segment $pf(p)$, and projecting outwards to $S^1$. However this creates problems if the points are antipodes. I suppose then, that this assumption is too strong, and I should either prove that $f$ is homotopic to $z^n$ (by contradiction) for non zero $n$, or bring in some theorem such as Borsak-Ulam, which applies here because the map is non-surjective, and can be thought of as a map into $\mathbb R$, but I can't think of how this would help. I would appreciate some help.

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An alternative argument would be to show that if $f$ has no fixed point, then $f$ is homotopic to the antipodal map $h(x)=-x$.

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  • $\begingroup$ I was able to show this, thanks. But then shouldn't it also be homotopic to the identity? How come it's easy to show homotopy to the antipodal map, but not so for the identity? $\endgroup$ – GPerez May 27 '14 at 13:16
  • $\begingroup$ Because the antipodal argument works in every dimension, and only in odd dimensions are the identity and antipodal maps homotopic (why?). $\endgroup$ – Ted Shifrin May 27 '14 at 13:17
  • $\begingroup$ Hehe, no idea actually. In 1D, the tool I have is winding number, without it I'm lost (I'm just beginning algebraic topology). $\endgroup$ – GPerez May 27 '14 at 13:33
  • $\begingroup$ You can write down a nowhere-vanishing vector field on $S^n$ if and only if $n$ is odd (the lack of one when $n$ is even is often called the Hairy Ball Theorem), and a (smooth) homotopy between the identity and the antipodal map is equivalent to the existence of such a vector field. A much harder result is that two maps from a compact, oriented $n$-manifold to $S^n$ are homotopic if and only if they have the same degree. ... More for you to learn down the line :) $\endgroup$ – Ted Shifrin May 27 '14 at 14:09
  • $\begingroup$ One further comment: You can, by replacing $f$ with $-f$, change the condition of no fixed point to the condition of never equalling the antipodal point. In the latter case you then get a homotopy to the identity. $\endgroup$ – Ted Shifrin May 27 '14 at 14:10
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There is a proposition that you will want to use here which is the following:

Proposition. A map $g\colon S^1\to X$ is null-homotopic if and only if there exists a map $\tilde{g}\colon D^2\to X$ such that the restriction satisfies $\tilde{g}|_{S^1}=g$. That is, $g$ can be extended to a map on the disk.

Now, if $f\colon S^1\to S^1$ is null-homotopic, let us suppose that it does not have a fixed point. Let $i\colon S^1\to D^2$ be the inclusion of the circle into the disk $D^2$. Let $\tilde{f}\colon D^2\to S^1$ be the extension of $f$ that must exist by the above proposition. Clearly if $f$ has no fixed points, then $$i\circ \tilde{f}\colon D^2\to S^1 \to D^2$$ also has no fixed points but this contradicts Brouwer's fixed-point theorem.

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  • $\begingroup$ Thank you, this works. Although the proposition seems non-trivial, and I'm doing exercises as if they were test questions. Since we haven't seen this in class, I would have to prove it myself; do you know if the proof is simple? $\endgroup$ – GPerez May 27 '14 at 13:06
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    $\begingroup$ The proof is very simple actually! If there is a homotopy $H\colon S^1\times [0,1]\to X$ with $H(x,1)=g(x)$ and $H(x,0)=c(x)$ then is we let $q\colon S^1\times [0,1]\to D^2$ be the quotient map where we shrink the circle $S^1\times\{0\}$ to a point (this 'cone' and the disk are obviously homeomorphic), then by the universal property of the quotient map, $H$ factors through the quotient because it is constant on $S^1\times \{0\}$ so we have $S^1\times[0,1]\stackrel{q}{\to}D^2\stackrel{h}{\to} X$ is the same map as $H$ and the map $h$ satisfies the properties we need for $\tilde{g}$. $\endgroup$ – Dan Rust May 27 '14 at 13:21
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    $\begingroup$ because $h$ acts on the boundary of $D^2$ the same way that the restriction $H|_{S^1\times\{1\}}=g$ does. The other direction is just a consequence of the fact that $D^2$ is contractible. This question goes into more detail math.stackexchange.com/questions/47397/… $\endgroup$ – Dan Rust May 27 '14 at 13:23

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