Circle to circle homotopic to the constant map?

Question

How to prove that a continuous function, homotopic to the constant map $f:S^1\to S^1$ (a) has a fixed point and that (b) has a point $x$, such that $f$ maps $x$ to its antipodal point $-x$?

Answer 1

For the first part, let $i\colon S^1\to D^2$ be the inclusion of the circle into the unit disk and, since $f$ is null-homotopic, let $\tilde{f}\colon D^2\to S^1$ be an extension of $f$ to the whole disk (which exists). Since $f$ has no fixed points, and the image of $\tilde{f}$ lies within $S^1$, what can we say about $i\circ \tilde{f}\colon D^2\to D^2$ and what theorem about maps on disks does this contradict?

For the second part, just prove that the composition of a nullhomotopic map with the map which rotates the circle by $\pi$ is also nullhomotopic (hint: rotation is homotopic to the identity and if $f\simeq f'$ and $g\simeq g'$ then $f\circ g\simeq f'\circ g'$), and then use part a.

Answer 2

Lemma: Show that if $A$ is a retract of $B^2$, then every continuous map $f : A \to A$ has a fixed point.

Proof: Suppose that $A$ is a retract of $B^2$, then by definition there exists a continuous map $r : B^2 \to A$ such that $r(a) = a$ for all $a \in A$. Let $f : A \to A$ be an arbitrary continuous map. Define $g : B^2 \to B^2$ by $g = j \circ f \circ r$ where $j : A \to B^2$ is the inclusion map. By the Brouwer fixed-point theorem for the disk, there exists $x \in B^2$ such that $g(x) = x$. But notice that $g(x) = j(f(r(x))) = f(r(x)) = x$ which means that $x \in A$ since $x \in \operatorname{Im}(f) \subseteq A$. Since $r$ is a retraction of $B^2$ onto $A$, $r(x) = x$ and hence $f(r(x)) = f(x) = x$. Conclude that $f$ has a fixed point.

Theorem: Show that if $h : S^1 \to S^1$ is nulhomotopic, then $h$ has a fixed point and $h$ maps some point $x$ to its antipode $-x$.

Proof: Since $h : S^1 \to S^1$ is nulhomotopic there exists a continuous extension $k : B^2 \to S^1$ of $h$ into $B^2$. Define $g : B^2 \to B^2$ by $g = j \circ k$ where $j : S^1 \to B^2$ is the inclusion map. $g$ is continuous, so by the fixed point theorem, there exists a fixed point $x \in B^2$ such that $g(x) = x$. But notice that $x = g(x) = j(k(x)) = k(x) \in S^1$ so $x \in S^1$ and hence $k(x) = h(x) = x$ and thus $h$ has a fixed point.

Define $\alpha : S^1 \to S^1$ by $\alpha(x) = -x$. By hypothesis, $h$ is nulhomotopic, so there exists $c \in S^1$ such that $h$ is homotopic to $e_c$. In particular, there exists a homotopy $F : S^1 \times I \to S^1$ such that $F(s, 0) = h(s)$ and $F(s, 1) = e_c(s) = c$. Since $\alpha$ is continuous, then $\alpha \circ F$ is a homotopy between $\alpha \circ h$ and $\alpha \circ e_c = e_{-c}$. Hence $\alpha \circ h$ is nulhomotopic. By previous discussion, there exists a fixed point $x$ such that $\alpha(h(x)) = -h(x) = x$. Multiply both sides by -1 and we get $h(x) = -x$. Conclude that $h$ maps some point $x$ to its antipode $-x$.

Answer 3

Munkres (Topology(2nd ed)) has the following theorem (Theorem 55.5)

Given a non-vanishing vector field $\tilde{f}:\mathbb{D}^2 \rightarrow \mathbb{R}^2\setminus \{0\}$, there are points on $S^1$ where it respectively points directly inwards and outwards, i.e, $\exists s_1 \in S^1, \ \exists s_2 \in S^1,$ $\tilde{f}(s_1) = t_1s_1, \ \tilde{f}(s_2) = - t_2s_2; \ t_1, t_2 > 0$.

The statement in question is actually equivalent to this.

$(\Rightarrow)$ if $f:S^1 \rightarrow S^1$ is nullhomotopic, $f$ can be extended to a continuous map $\tilde{f}:\mathbb{D}^2 \rightarrow S^1 \subset \mathbb{R}^2 \setminus \{0\}$.
So, $\exists s_1 \in S^1, \ \exists s_2 \in S^1,$ $f(s_1) = t_1s_1, \ f(s_2) = - t_2s_2; \ t_1, t_2 > 0$.
But, $f(S^1) \subset S^1$ $\Longrightarrow$ $(f(s_1) = t_1 s_1 \in S^1 \Rightarrow t_1 =1)\wedge(f(s_2) = -t_2 s_2 \in S^1 \Rightarrow t_2 =1)$.
$\Longrightarrow$ $f(s_1) = s_1$; $f(s_2) = - s_2$. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \blacksquare$

$(\Leftarrow)$ Given $\tilde{f}:\mathbb{D}^2 \rightarrow \mathbb{R}^2\setminus \{0\}$, we describe $\tilde{g}:\mathbb{D}^2 \rightarrow S^1$ as $\tilde{g} = \frac{\tilde{f}}{\parallel \tilde{f} \parallel}$ .
Then, $\tilde{g}|_{S^1}:S^1 \rightarrow S^1$ is nullhomotopic. $\Longrightarrow$ $\exists s_1,s_2 \in S^1, \ \ \ \tilde{g}(s_1)=s_1; \ \tilde{g}(s_2) = s_2$.
$\Longrightarrow$ $\tilde{f}(s_1)= \parallel \tilde{f}(s_1) \parallel s_1, \ \parallel\tilde{f}(s_1)\parallel > 0; \ \ \ \ \ \tilde{f}(s_2) = - \parallel \tilde{f}(s_2) \parallel s_2, \ \parallel \tilde{f}(s_2) \parallel>0$. $\ \ \ \ \blacksquare$

Munkres actually goes on to use the aforesaid theorem to prove Brouwer's fixed point theorem, which I believe all other proofs have used.

Answer 4

for a different, very short proof: work in $\mathbb C$ and use winding numbers.

with $\gamma(t)=\exp\big(2\pi i \cdot t\big)$ for $t\in[0,1]$ you have
$n\big(f\circ\gamma, 0\big)=0$ since $f\circ \gamma$ is null-homotopic and with the inversion map $h:z\mapsto z^{-1}$, consider the curve $\sigma$ given by $\sigma(t) := \big(h\circ \gamma(t)\big) \cdot \big(f\circ\gamma(t)\big)$
$\implies n\big(\sigma,0\big)=n\big(h\circ\gamma, 0\big)+n\big(f\circ\gamma, 0\big)=-1+0=-1\neq 0$
$\implies$the image of $\sigma$ is all of the unit circle, and in particular $1=\sigma(t')$ for some $t'\in [0,1]$
i.e. $\frac{1}{\gamma(t')}\cdot f\big(\gamma(t')\big)=1\implies f\big(\gamma(t')\big)=\gamma(t')$, so $f$ has a fixed point.