The p point such that map from $S^{1}$ to $S^{1}$ satisfies $f(p) = f(-p)$ Let $f: S^{1} \rightarrow S^{1}$ be a continuous map homotopic to a constant. Show that:
1) there exists a point $p \in S^{1}$ such that $f(p) = f(-p)$.
2) the same, while $\deg(f) \in 2 \mathbb{Z}$ 
3) what if $\deg(f) \in 2\mathbb{Z}+1$?
My idea for a) was to use Brouwer's theorem ($f$ is homotopic to a constant, so it enduces a map $\bar{f}: B^{2} \rightarrow S^{1}$, and therefore it has a fixed point...), but i don't really know how to continue this.
 A: Consider the covering projection $exp : \mathbb R \to S^1, exp(t) = e^{2 \pi it}$. Any map $f  : X \to S^1$ which is homotopic to a constant map $c : X \to S^1$ has a lift $f' : X \to \mathbb  R$. To see this, let $H : X \times I \to S^1$ be a homotopy such that $H_0 = H(-,0) = c$ and $H_1 = H(-,1) = f$. Since $H_0$ has a lift $c' : S^1 \to \mathbb R$, there exists a lift $H'$ of $H$ (homotopy lifting property of covering projections). Then $f' = H'_1$ is a lift of $f$. 


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*Let $f' : S^1 \to \mathbb R$ be a lift of $f$. Define
$$g : S^1 \to \mathbb R, g(z) = f'(z) - f'(-z) .$$
$g(S^1)$ is a connected subset of $\mathbb R$, thus an interval. Therefore if $z_1, z_2 \in S^1$, then $\frac{g(z_1) + g(z_2)}{2} \in g(S^1)$. We have $g(-z) = -g(z)$ for all $z$. Hence $0 = \frac{g(z) + g(-z)}{2} \in g(S^1)$. This shows that $f'(p) - f'(-p) = 0$ for some $p$ and we conclude $f(p) = f(-p)$.

*The degree of a map $f : S^1 \to S^1$ can be computed as follows. Let $q : I  = [0,1] \to S^1, q(t) = exp(t)$. Then $f q : I \to S^1$ is homotopic to a constant map and hence has a lift $l : I \to  \mathbb R$. We then have $\deg(f) = l(1) - l(0)$. This number is in $\mathbb Z$ and does not depend on the choice of $l$. The map $\mu_n(z) = z^n$ has degree $n$ because $\mu_n q$ has $l_n(t) = nt$ as a lift. Moreover, given maps $f_1, f_2 : S^1 \to S^1$, we can define their product
$$f_1 \cdot f_2 : S^1 \to S^1, (f_1 \cdot f_2)(z) = f_1(z) \cdot f_2(z) .$$
If $l_i$ is a lift of $f_i q$, then clearly $l_1 + l_2$ is a lift of $f_1 \cdot f_2$ which implies $\deg(f_1 \cdot f_2) = \deg(f_1) + \deg(f_2)$.  Now let $\deg(f) = n$. Then the map $h = \mu_{-n} \cdot f$ has degree $0$ and by 1. we know that there exists $p$ such that $p^{-n}f(p) = h(p) = h(-p) = (-p)^{-n}f(-p) = (-1)^{-n}p^{-n}f(-p)$ which is equivalent to $f(p) = (-1)^nf(-p)$. If $n$ is even, we have $f(p) = f(-p)$, and if $n$ is odd, we have $f(p) = -f(-p)$.

*If $n$ is odd, then $\mu_n(z) = z^n \ne - z^n = (-z)^n = \mu_n(-z)$ for all $z$. Thus in general there does not exist $p$ such that $f(p) = f(-p)$ if $\deg(f)$ is odd.
