What is the degree of a homeomorphism from $S^1$ to $S^1$? 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?

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 $\deg=\pm 1$.
