Homotopy on the unit circle

Question

I am trying understand why the identity function on the unit circle $X=\{(x,y): x^2+y^2=1\}$ is not homotopic to $f: X \to X$ where $f(z)=(1,0)$ for all $z\in X$.

Answer 1

A contractible space is simply connected. If the identity map on the circle was homotopic to the constant map at some point, then it would be contractible. This can't be true since we know that the fundamental group of the circle is isomorphic to $\mathbb{Z}$.

Answer 2

(Other users have given answers using the fact that $\pi_1(S^1)=\Bbb Z$. Depending on how you prove this fact, the argument may be circular, since one of the proofs of the fundamental group uses the fact (in a more general form) that the loop cannot be null-homotoped. But there are other proofs for $\pi_1(S^1)=\Bbb Z$, for example by a van Kampen theorem for the fundamental groupoid, so I don't say their answers are not valid. My answer makes use of the essential ingredient of that proof which uses the universal cover $\Bbb R$ of $S^1$.)

The map $p:\Bbb R→S^1, x\mapsto e^{2\pi ix}$ is a covering map, and this implies that given a commutative square $$\begin{matrix} X & \xrightarrow{f} & \Bbb R\\ ↓ & \nearrow & ↓ \\ X×I & \xrightarrow{H} & S^1 \end{matrix}$$ where the left-hand map is $x\mapsto(x,0)$, there is a unique map $\tilde H:X\times I\to\Bbb R$ making the triangles commute. So $p\tilde H=H$ and this is often expressed by saying $\tilde H$ lifts $H$. In other words: Given a homotopy $H$ in $S^1$, and a map $f$ in $\Bbb R$ lifting $H_0$, there is a unique homotopy $\tilde H$ in $\Bbb R$ lifting $H$ and starting with the given $f$.

If you take $X=\{*\}$, and $f(*)=\tilde x_0$, then $h:I\to S^1$ is simply a path starting at $p(\tilde x_0)$ and the property implies that there is a unique path starting at $\tilde x_0$ and lifting $h$.
It is easy to see that a path homotopy in $S^1$ is lifted by a path homotopy in $\Bbb R$, given an $\tilde x_0$ lifting the starting point of the path(s) in $S^1$.

Now if $a(t)=e^{2\pi i t}$ were path homotopic to $(1,0)=x_0$, then its lift $\tilde a(t)=t$ at $\tilde x_0=0$ had to be path homotopic to a constant path, which is clearly impossible.

Answer 3

This is the standard reasoning why $S^{1}$ can not retract to a point. It doesn't use path arguments except we take for granted the fundamental groups of the given spaces. It relies on the properties of the induced homomorphisms, which work for homology too, as well as for higher homotopy groups.

We will use the facts that $\pi_{1}(S^{1})=\mathbb{Z}$, $(f\circ g)_{*}=f_{*}\circ g_{*}$ and $(\mathrm{id_{X}})_{*}=\mathrm{id}_{\pi_{1}(X)}$, and that homotopic maps induce the same homomorphism between fundamental groups. Here $f_{*}$ denotes the induced map between fundamental groups given by a function $f$.

Assume towards contradiction that your constant function $f:S^{1}\to S^{1}$ would be homotopic to the identity map. Note that $f=i\circ g$ where $g:S^{1}\to\{(0,1)\}$ is the constant map and $i:\{(1,0)\}\to S^{1}$ the inclusion map. Now you get $$\mathrm{id}_{\pi_{1}(S^{1})}=(\mathrm{id}_{S^{1}})_{*}=f_{*}=(i\circ g)_{*}=i_{*}\circ g_{*}.$$ Since $g_{*}$ has a left inverse, this means precisely that $g_{*}$ is injective. But $g_{*}:\pi_{1}(S^{1})\to \pi_{1}(\{(0,1)\})$ where $\pi_{1}(S^{1})=\mathbb{Z}$ and $\pi_{1}(\{(0,1)\})=\{0\}$. So we have shown that there is an injection from $\mathbb{Z}$ to $\{0\}$, which is a contradiction as such functions can not exist. We thus conclude that your map $f$ can not be homotopic to the identity function.

Additional exercise for you: Take $S^{1}\subseteq \mathbb{R}P^{2}$ as the equator of $S^{2}$ which gets identified to itself when defining $\mathbb{R}P^{2}$. Show that there is no continuous function $r:\mathbb{R}P^{2}\to S^{1}$ so that $r\circ i=id_{S^{1}}$, where $i:S^{1}\to\mathbb{R}P^{2}$ is the inclusion map. Hint: use similar reasoning as above.