What are two continuous maps from $S^1$ to $S^1$ which are not homotopic?

Question

This is an exam question I encountered while studying for my exam for our topology course:

Give two continuous maps from $S^1$ to $S^1$ which are not homotopic. (Of course, provide a proof as well.)

The only continuous maps from $S^1$ to $S^1$ I can think of are rotations, and I thought rotations on a circle can be continuously morphed into one another.

Answer 1

The identity is not homotopic to a constant map; otherwise, $S^1$ would be contractible, which would imply $\pi_1(S^1)=0$.

Answer 2

Recall that for any map $f : S^1 \to S^1$ that we have the induced map $f_\ast : H_1(S^1) \to H_1(S^1)$ which is a map between two infinite cyclic groups. Any such map is given by multiplication by $n \in \Bbb{Z}$, so we define the degree of $f$ to this integer $n$.

If you think of $S^1$ as lying in $\Bbb{R}^2$ then what about $f : S^1 \to S^1$ that is reflection about the $x$ - axis and $g : S^1 \to S^1$ that is the identity? $f$ cannot be homotopic to $g$ because $\deg f = - 1$ while $\deg g= 1$.

Added for OP: Here is why the degree of the reflection map is $-1$. We construct $S^1$ as a simplicial structure as shown in the picture below.

$\hspace{4.5 cm}$

The orientations are according to the arrows. Then our simplicial chain complex is just

$$0 \to \Bbb{Z}\{c,d\} \stackrel{\partial}{\to} \Bbb{Z}\{a,b\} \to 0$$

where the boundary map $\partial$ is just given by $\partial(c) = b - a$ and $\partial(d) = a - b$. Then $\partial(c+d) = 0$ and we easily see that $\ker \partial = \langle c +d\rangle $. It will now follow that

$$H_1(S^1)\cong \ker \partial/0 \cong \Bbb{Z}\{c+d\} \cong \Bbb{Z}.$$

From this it is clear that the degree of the reflection map is $-1$, since it sends the generator of the homology to its negative ($c\mapsto -d$ while $d \mapsto -c$).

Answer 3

If you think of $S^1$ as the unit circle in $\mathbb C$, each homotopy class of loops in $S^1$ with base point $1$ is represented by the map $f_n(z) = z^n$ where $n$ can be any integer, and it's not too hard to show that $f_m$ and $f_n$ are in the same homotopy class if and only if $m=n$.

Answer 4

$F_1:S^1\to S^1 $

$F_1(s)=(\cos(2\pi s),\sin(2\pi s))$ and

$F_2:S^1\to S^1$

$F_2(s)=(\cos(4\pi s),\sin(4\pi s))$ where $s$ goes from $0$ to $1$, are not homotopic.