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?)
 A: 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 $id$.
