# No Smooth Onto Map from Circle to Torus

My professor was lecturing today and he made this statement which I was unable to verify. (I worded it nicer)

There is no map which is both smooth and onto from $S^1$ to $S^1$$\times$ $S^1$.

When he said this he included the original "one can clearly see" statement which is usually when the most people seem to be confused in his class. Thus, I was hesitant to ask why. So, please no one lecture me on the fact that I should have asked anyway. If it is really that obvious, I would rather look stupid here than in class :)

-
I am increasingly becoming of the opinion that the phrase "clearly ____ is true" should be forever removed from mathematicians' vocabularies. –  Keaton Mar 13 at 5:20

This follows from Sard's theorem, which implies that the image of a smooth map $S^1 \to S^1 \times S^1$ has Lebesgue measure zero. More generally, Sard's theorem implies that there are no smooth onto maps from a smooth manifold of dimension $n$ to a smooth manifold of dimension $m > n$.

The obvious statement is that there are no diffeomorphisms between such manifolds, which just follows by computing differentials (the differential of a diffeomorphism is an isomorphism). The differential of a surjective smooth map is not surjective in general (consider the surjection $\mathbb{R} \ni x \mapsto x^3 \in \mathbb{R}$) so the obvious argument doesn't work (or at least it doesn't obviously work).

Note also that this statement is false if "smooth" is replaced by "continuous" due to the existence of space-filling curves.

-
Ah good point. I figured it fell out easily from some theorem. I just couldn't think of what theorem. Remembering Sard's then does make the statement obvious. Thanks –  Leo Spencer Mar 13 at 5:49
A smooth map $S^1 \rightarrow S^1 \times S^1$ is a differentiable curve; by differentiability it must have finite arc length. But a curve of finite length can't cover the 2-dimensional torus.
Why not? ${}{}$ –  Qiaochu Yuan Mar 13 at 7:09