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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.

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4 Answers 4

up vote 18 down vote accepted

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

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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.

$\hspace{4.5 cm}$enter image description here

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 $d \mapsto -c$).

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Hm I understand that $\deg g = 1$, but I don't really understand why $\deg f = -1$. Could you please explain why this is the case? Also, why are maps of different degrees never homotopic? –  Max Muller Jun 5 '13 at 11:08
    
o.k., Thank you! Unfortunately my knowledge of algebraic topology is not very deep, so I don't really understand anything related to homology (yet). Thank you for your answers though. –  Max Muller Jun 5 '13 at 11:20
    
@MaxMuller You can define degree of a map $\varphi:S^1\to S^1$ by looking at the behavior of the composite $\pi_1(S^1,x)\xrightarrow{\varphi_*}\pi_1(S^1,f(x))\to\pi_1(S^1,x)$ on the generator $g$ of $\pi_1(S^1,x)$. I.e. $g\mapsto(\deg \varphi)g$. Though this definition actually matches what BenjaLim added, this might help you see why $\deg \varphi\ne\deg\psi$ implies they are not homotopic. –  Karl Kronenfeld Jun 5 '13 at 11:35
    
@user1 It matches my definition because I believe in this case the Hurewicz homomorphism is actually an isomorphism. –  user38268 Jun 5 '13 at 11:50
    
Exactly, you go from one to the other by composing the homomorphism and its inverse on either side. –  Karl Kronenfeld Jun 5 '13 at 11:53

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 $m=n$.

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That's a nice general case, but "it's not too hard" doesn't suggest how you show they're not homotopic. And you could just use $f_n(z) = nz$, right? –  alexis Jun 6 '13 at 8:54

$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.

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