# Circle to circle homotopic to the constant map?

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 $-x$?

• What are some of the approaches you've tried already? – Dan Rust Nov 23 '14 at 20:44
• Note that $0\notin S^1=\{\,z\in\mathbb C:|z|=1\,\}$. But parts of the problem statement seem to be unclear (so please check them again carefully): There are many $f\colon S^1\to S^1$ that are homotopic to the constant map - and $x\mapsto -x$ is not among them – Hagen von Eitzen Nov 23 '14 at 20:45
• Well, indeed. Then how can I show that $f$ has a constant point? A constant point must satisfy $f(x)=x$. Let me try to re-write question. You are right, I ve not written it down well. – Marion Nov 23 '14 at 20:48
• A typical trick s to assume that $f(x)\ne x$ for all $x$ and then consider $x\mapsto \frac{f(x)-x}{|f(x)-x|}$ – Hagen von Eitzen Nov 23 '14 at 20:49
• I believe the right phrasing would be "maps some $x$ to it's antipodal point" – Arthur Nov 23 '14 at 21:40

## 3 Answers

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.

• To get the extension do you apply some generalisation to $\mathbb R^2$ of Tietze's extension theorem? – Rudy the Reindeer Nov 23 '14 at 23:29
• @RudytheReindeer Usually the first thing you prove in an alegbraic topology class is that a null-homotopic map $S^1\to X$ can be extended to a map $D^2\to X$. This is also a sufficient condition for the map to be null-homotopic. Basically, just clue one end of the mapping cylinder, given by the homotopy one the 'constant end'. This 'mapping cone' can then be mapped to the disk by flattening it in the obvious way. – Dan Rust Nov 23 '14 at 23:46
• Ah, great, I understand. I think I can write the proof. Thank you very much for your comment! – Rudy the Reindeer Nov 24 '14 at 1:13

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 $-x$.

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.