If $f : S^1 \to S^1$ is a homeomorphism, what is the degree of it?

I was told the answer is 1 or -1, but I can't prove it. Can anyone help me?

  • $\begingroup$ You may want to look at this thread for some inspiration. $\endgroup$ – t.b. Oct 28 '11 at 18:10
  • 7
    $\begingroup$ What definition of degree are you using? $\endgroup$ – Cheerful Parsnip Oct 28 '11 at 18:25

One definition of degree of a map from $S^1$ to $S^1$ is to look at the image of $1\in \mathbb Z\cong \pi_1(S^1)$. Hence a map of degree $n$ induces multiplication by $n$ on $\pi_1(S^1)\cong \mathbb Z$. Now a homeomorphism $h$ has an inverse $g$ so that $gh=hg=id_{S^1}$. Applying $\pi_1$, $g_*h_*=h_*g_*=id_{\mathbb Z}$. So if $h_*(1)=\deg(h)$, this implies that $\deg(h)$ evenly divides $1$. So it is $\pm 1$.

The other notion of degree I know is to perturb $f$ to a smooth map, take a small neighborhood of a generic point $p$ and count the preimages of $p$ with sign determined by whether $f$ locally flips or preserves orientation. In this case, since a homeomorphism is injective, every point will have at most one preimage. I guess we need to check that a homeomorphism can be perturbed to a diffeomorphism for this argument to work.

Edit: I think I now know what the definition of degree is that the OP is considering. Namely, consider $f\colon I\to S^1$ with $f(0)=f(1)$ as representing a map from $S^1$ to $S^1$. Now lift $f$ to the universal cover to get $\tilde{f}\colon I\to \mathbb R$, where the covering projection $p\colon\mathbb R\to S^2$ is given by $p(t)=e^{2\pi i t}$. Then define $\deg(f)=\tilde{f}(1)-\tilde{f}(0)$. Now suppose $|\deg(f)|>1$. Then the image of $\tilde{f}$ contains one other lift, say $\tilde{f}(a)$, of $f(0)$ between $\tilde{f}(0)$ and $\tilde{f}(1)$. So $f=p\circ \tilde{f}$ is not injective since $f(a)=f(0)$. On the other hand, if $\deg(f)=0$, then either $\tilde{f}$, and hence $f$ is constant, or $\tilde{f}$ is not injective. But then $f=p\circ\tilde{f}$ is also not injective. So the only possibility for a homeomorphism is $\deg=\pm 1$.

  • $\begingroup$ I have not seen the sign $\pi_1(S^1)$, does it mean the lifting of $f$? $\endgroup$ – hxhxhx88 Oct 29 '11 at 0:24
  • $\begingroup$ Evidently this isn't the definition of degree you were using then! It means the fundamental group of $S^1$. $\endgroup$ – Cheerful Parsnip Oct 29 '11 at 0:26
  • $\begingroup$ @hxhxhx88: I've edited to try to address your definition better. Hopefully I succeeded. $\endgroup$ – Cheerful Parsnip Oct 30 '11 at 17:58

Hint: Use that $S^1$ is the 1-point compactification of the real line.

So, we get a homeorphism, $g:\mathbb{R}\rightarrow \mathbb{R}$.

Find a homotopy between such a $g$ and either $g_0(x)=x$ or $g_1(x)=-x$, such that it can be extended to the 1-point compactification.

(Hint: $g$ must be either strictly increasing or strictly decreasing, and, obviously, unbounded.)

Now show that the result of extending $g_0$ and $g_1$ to the 1-point compactification is either of degree $-1$ or degree $1$.

Advantage of this proof is that it directly finds a homotopy between $f$ and a particular map of degree $1$ or $-1$, without even knowing what "degree" means. Basically, we can say if $f$ is a homeomorphism, then $f$ is homotopic with one of two maps. We don't even know if those two maps are homotopic or not.

  • $\begingroup$ Nice. $\hspace{1em}$ $\endgroup$ – Cheerful Parsnip Oct 28 '11 at 19:45
  • $\begingroup$ I'm sorry I'm a new to topology and I have no idea about 'order'....I will look at it~ $\endgroup$ – hxhxhx88 Oct 29 '11 at 0:26
  • $\begingroup$ I meant degree, not order, @user797225 $\endgroup$ – Thomas Andrews Oct 29 '11 at 0:30

It would help if you told us which definition of degree are you using, as Jim Conant asked you.

Anyway, one of the properties of your degree must be being multiplicative with respect to composition. That is:

$$ \mathrm{deg}\ (g\circ f) = \mathrm{deg}\ g \cdot \mathrm{deg}\ f $$


$$ \mathrm{deg}\ \mathrm{id} = 1 \ . $$

So, if $f: S^1 \longrightarrow S^1 $ is a homeomorphism, you have an inverse $f^{-1}$. Applying the degree to the equality

$$ f \circ f^{-1} = \mathrm{id} $$

you get

$$ \mathrm{deg}\ f \cdot \mathrm{deg}\ f^{-1} = 1 \ . $$

That is, $\mathrm{deg}\ f$ must be an integer number which has an inverse (namely, $\mathrm{deg}\ f^{-1} $). But the only integer numbers with an inverse are $\pm 1$.

  • $\begingroup$ I'm reading a book named $Introduction\ to\ Topology\ Pure \ and\ Applied$, and the degree I refer to is defined by circle function in Chapter 9 in that book. $\endgroup$ – hxhxhx88 Oct 29 '11 at 4:57

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