Smooth map from $S^1$ to $S^1$ with degree zero Is there any example of a smooth map $f:S^1\to S^1$ that has degree zero that is not the constant map? 
Either the map would have no regular values or every regular value would have an even number of pre-images with cancelling degrees, but I'm having a hard time seeing what the regular values should be. 
 A: Here's a specific map which is actually surjective.  Considering $S^1$ as $\{(x, y): x^2 + y^2 = 1\}$ in the Euclidean plane, take the projection $\pi_x: \mathbb{R}^2 \to \mathbb{R}, (x, y) \mapsto x$.  Define $f: \mathbb{R} \to \mathbb{R}^2, t \mapsto (\cos \pi t, \sin \pi t)$.  Finally define $g$ as the composition restricted to $S^1$: $f \circ (\pi_x | S^1): S^1 \to S^1$.  $\pi_x$ and $f$ are both smooth, so $g$ is too.  $\pi_x$ is nullhomotopic, so $g$ is nullhomotopic and therefore has degree $0$.  And it's clear that (roughly speaking) $g$ goes through all angles in the range $[-\pi, \pi]$ and is therefore onto.
Edit: fixed scaling mistake, clarified domain and range.
A: Yes, identify $S^1$ with $[-\pi,\pi]$ and consider the map:
$$f(t) = \cos(t):S^1\rightarrow S^1$$
It is obviously smooth and not surjective, and thus has degree zero.
A: Degree is a homotopy invariant, which is to say that if $f$ and $g$ are homotopic maps from $S^1$ to $S^1$ then they have the same degree. Then any map $f:S^1\to S^1$ which is null-homotopic (homotopic to a constant map) will have degree zero, even if it itself is not constant. An example of such a map is given by flattening $S^1$ down onto the unit interval (say, by projecting it onto the $x$-axis, thinking of $S^1$ as embedded in $\mathbb{R}^2$) and then identifying this interval with a hemisphere of $S^1$. Contracting this hemisphere to a point gives the desired null-homotopy.
