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Suppose $f:S^1\rightarrow S^1$ is a map not homotopic to the identity map. Show there exists $x,y\in S^1$ such that $f(x)=x$ and $f(y)=-y$? (If there are no fixed points, then $f$ is homotopic to the identity map? and if no such $y$ exists then $f$ is homotopic to the identity? How can I show this?)

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    $\begingroup$ Hint: Consider the homotopy $F(x,t) = \frac{f(x) + \mu x}{|f(x)+\mu x|}$ for $\mu\in \{1,-1\}$. $\endgroup$ Commented Apr 11, 2019 at 20:54
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    $\begingroup$ @JasonDeVito Where is $t$ on the right hand side? $\endgroup$
    – Paul Frost
    Commented Apr 11, 2019 at 22:20
  • $\begingroup$ Related -math.stackexchange.com/q/1249307/588038 $\endgroup$
    – cqfd
    Commented Apr 12, 2019 at 2:27
  • $\begingroup$ @Paul: Oops! I meant $tf(x) + (1-t)\mu x$ in both the numerator and denominator. $\endgroup$ Commented Apr 12, 2019 at 14:19

1 Answer 1

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Let us prove the following slightly more general result:

Suppose $f:S^1\rightarrow S^1$ is a map not homotopic to the identity map. Then for each $a \in S^1$ there exists $z \in S^1$ such that $f(z) = az$.

You consider the special cases $a =1, -1$.

Proof by contradiction:

Assume there exists $a \in S^1$ such that $f(z) \ne az$ for all $z \in S^1$. Let us first observe that the line segment $s(x,y) = \{ (1-t)x + ty \mid t \in [0,1] \} \subset \mathbb C$ connecting two points $x, y \in S^1$ contains $0$ if and only if $y = -x$. This implies that $$H : S^1 \times I \to S^1, H(z,t) = \dfrac{(1-t)f(z) +t(-az)}{\lvert (1-t)f(z) +t(-az) \rvert} .$$ is well-defined because $-az \ne -f(z)$ for all $z$. This shows that $f$ is homotopic to the map $g : S^1\to S^1, g(z) = -az$. Write $-a = e^{i\alpha}$ with $\alpha \in [0,2\pi)$ and define $$G : S^1 \times I \to S^1, G(z,t) = e^{i\alpha t}z .$$ This is a homotopy from the identity to $g$.

We have shown that $f \simeq id$ which is a contradiction.

Alternative approach:

Consider the map $f^* : S^1 \to S^1, f^*(z) = \frac{f(z)}{z}$. Clearly $f^*$ is surjective if and only for each $a \in S^1$ there exists $z \in S^1$ such that $f(z) = az$.

Assume that $f^*$ is not surjective. Then there exists $a \in S^1$ such that $f^*(S^1) \subset S^1 \setminus \{ a \}$. The latter space is homeomorphic to an open interval, hence contractible and we conclude that $f^*$ is homotopic to a constant map. Since all constant maps into a path connected space are homotopic, we find a homotopy $H^* : S^1 \times I \to S^1$ such that $H^*(z,0) = f^*(z)$ and $H^*(z,1) = 1$. Then $H(z,t) = zH^*(z,t)$ is a homotopy from $f$ to $id$.

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    $\begingroup$ Can you please tell me the meaning of $az$? Is it dot product? $\endgroup$
    – cqfd
    Commented Apr 12, 2019 at 14:57
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    $\begingroup$ It is the product of complex numbers ($S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1 \}$). Similarly $\frac{f(z)}{z}$ is the quotient of complex numbers. $\endgroup$
    – Paul Frost
    Commented Apr 12, 2019 at 15:17

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