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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, 2015 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. $\endgroup$ Apr 24, 2015 at 2:12
  • $\begingroup$ Maybe I should add that mathematicians are often quite lax about this point, so you're in respectable company. ${}\qquad{}$ $\endgroup$ Apr 24, 2015 at 2:15

2 Answers 2

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$f : S^1 \to S^1$ 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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Here is a slightly longer argument that doesn't go through the antipodal map:

Suppose $f:S^1\rightarrow S^1$ has no fixed points. Identify $S_1$ with $\mathbb{R}/\mathbb{Z}$ and let $\pi:\mathbb{R}\rightarrow S_1$ be the quotient map. Let $\gamma:[0,1] \rightarrow S^1$ be the loop $t \mapsto f(t+\mathbb{Z})$, starting at $f(\bar{0})$. Choosing $x_0 \in \pi^{-1}(f(\bar{0}))\cap [0,1)$, we have a unique lift $\overline{\gamma}:[0,1]\rightarrow \mathbb{R}$ which starts at $x_0$. Now we can define the function \begin{align*} F:S^1 \times [0,1] &\longrightarrow S^1 \\ (t+\mathbb{Z}, \lambda) &\longmapsto \pi \left( \lambda \cdot \overline{\gamma} (t) + (1-\lambda)\cdot t \right) \end{align*} where $t$ is chosen such that $t\in [0,1)$. This is clearly continuous if $t\neq 0$. Remains to check continuity at $\bar{0}$.

By the definition of a lift, we know that $\overline{\gamma}(1) \in \{x_0+n\;|\; n\in \mathbb{Z}\}$. Claim for all $t$, we have $t < \overline{\gamma}(t) < t + 1 $. If not, suppose (wlog) the lower bound doesn't hold. But we chose $f(0)=x_0>0$, so by the intermediate value theorem we have $\overline{\gamma}(t)=t$ for some $t$ -- we have $\pi\circ \overline{\gamma}(t) = \pi(t)$ i.e. $f(\bar{t}) = \bar{t}$, giving us a fixed point of $f$ [a contradiction]. This shows $\overline{\gamma}(1) = x_0 + 1$, hence $\lim_{t\rightarrow 1} F(t+\mathbb{Z}, \lambda) = \lim_{t\rightarrow 0} F(t+\mathbb{Z},\lambda)$ as required. Thus (F) is a homotopy $f\simeq \text{id}$.

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