Proving that a map $f\colon S^1\to S^2$ is not surjective Let $f\colon S^1\to S^2$ be a differentiable map. I want to show that $f$ is not surjective. 
My idea is to use Sard's theorem as follows. By Sard, $f$ has a regular value. If this value is not in the range of $f$, we are done. So let's assume that $p\in S^1$ is a regular point for this regular value. Then $f$ has rank $1$ at $p$ and hence it should be a local immersion into $S^2$. (Therefore, $f$ is not a local surjection.)
I don't know how to proceed from here to show that $f$ is not a surjective map. Any help is appreciated. 
As a separate question, I'd like to ask if there are other (perhaps easier) ways of proving the same statement about differentiable maps $f\colon S^1\to S^2$.
 A: It is actually impossible for a regular value to be in the image of $f$.  To see this, just look back to the definition of a regular point.  A point $p\in S^1$ is a regular point of $f$ if the differential $df_p$ at $p$ is surjective.  But the tangent space of $S^1$ at $p$ is $1$-dimensional, and the tangent space of $S^2$ at $p$ is $2$-dimensional, and a linear map from a $1$-dimensional vector space to a $2$-dimensional vector space cannot be surjective.
So actually, $f$ has no regular points at all.  Thus the only way a point in $S^2$ can be a regular value is if it is not in the image of $f$ at all.  Since by Sard's theorem a regular value must exist, this means $f$ is not surjective.
A: Suppose $f$ is surjective, and consider local coordinates $\phi : \mathcal{U}\subset S^2 \rightarrow \mathbb{R}^2$ and $\psi: \mathcal{V}\subset f^{-1}(\mathcal{U}) \rightarrow \mathbb{R}$. Then the map
$$
\phi \circ f \circ \psi^{-1} : \mathbb{R} \rightarrow \mathbb{R}^2
$$
is a smooth surjection, contradicting Sard's theorem.$^{(*)}$
$^{(*)}$: The differential of $g := \phi \circ f \circ \psi^{-1}$ is of rank at most $1$, and therefore the set of critical points of $g$ is all of $\mathbb{R}$. By Sard's theorem, this implies $g(\mathbb{R}) = \mathbb{R}^2$ is of measure $0$ in $\mathbb{R}^2$, which is clearly impossible.
