11
$\begingroup$

How to show that every continuous function $f:S^1\to S^1$ without fixed points is homotopic to the identity? (without using homology nor the concept of degree).

$\endgroup$
  • 2
    $\begingroup$ You might want to draw a graph, identifying $S^1$ with $[0,1)$. Since there are no fixed points, the graph does not meet the diagonal. Can you see an approach from there? $\endgroup$ – Mike F Apr 24 '15 at 2:06
  • $\begingroup$ I'd have phrased this with "every" instead of "any". The way the word "any" is used in English is rather odd: "How to show that there is any function that blah blah blah, then$\ldots$?" In that case "any" means in effect "some". But "How to show that any function blah blah blah$\ldots$?" could be construed differently. "Could", not "must". "Any" unambiguously implies "every" in some contexts, but then a phrase gets copied into a slightly differently constructed sentence and becomes ambiguous. "Every" suffers no such problem. $\endgroup$ – Michael Hardy Apr 24 '15 at 2:12
  • $\begingroup$ Maybe I should add that mathematicians are often quite lax about this point, so you're in respectable company. ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 24 '15 at 2:15
17
$\begingroup$

$f : S^1 \to S^1$ be a map with no fixed points. Consider the projection of the straightline homotopy $$H(s, t) = \frac{(1-t)f(s) - ts}{\left \lVert(1-t)f(s)-ts\right \rVert}$$ between $f$ and the antipodal map $-\text{id}$, which is well-defined since $f(x) \neq x$ for all $x$. Compose this with the homotopy $$H(s, t) = e^{i\pi (1-t)} s$$ which rotates $-\text{id}$ to the identity map $\text{id}$. Thus, by transitivity, $f \sim \text{id}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.