# Problem related to fixed point on $S^1$

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?)

• Hint: Consider the homotopy $F(x,t) = \frac{f(x) + \mu x}{|f(x)+\mu x|}$ for $\mu\in \{1,-1\}$. – Jason DeVito Apr 11 '19 at 20:54
• @JasonDeVito Where is $t$ on the right hand side? – Paul Frost Apr 11 '19 at 22:20
• – Shivering Soldier Apr 12 '19 at 2:27
• @Paul: Oops! I meant $tf(x) + (1-t)\mu x$ in both the numerator and denominator. – Jason DeVito Apr 12 '19 at 14:19

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 $$a =1, -1$$.

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 $$id$$.

• Can you please tell me the meaning of $az$? Is it dot product? – Shivering Soldier Apr 12 '19 at 14:57
• 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. – Paul Frost Apr 12 '19 at 15:17