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).

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    $\begingroup$ 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

$f : S^1 \to S^1$ 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, $f \sim \text{id}$.

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