This is not homework, but a sample test question. The question is:
Show that there is no surjective smooth function $$S^1 \to S^1 \times S^1 \times S^1.$$
Now I can see that, for example $$S^1 \to S^1$$ is infinitely differentiable with the identity map from the 1-D circle in 2-D space to itself.
Now I am wondering, due to the nature of the test material, if somehow I am supposed to either apply Gaussian curvature or the Gauss-Bonett formula. I am also familiar a little bit with the fundamental forms.
While it looks like there should not be a surjective function (not even considering smooth), as all the points on $S^1$ could themselves just be mapped to the $S^1$ of the $S^1 \times S^1 \times S^1$. What is this surface called (I am not even sure what it looks like).
Thanks, Brian.
If you know a bit about Sard's theorem, it is straightforward. Assume that there exists a smooth map $f:S^1\rightarrow S^1\times S^1\times S^1=\mathbb T^3$.
On one side, Sard's theorem tells you that it must exist a regular value of $f$: that is a point $y\in \mathbb T^3$ such that $\forall x\in f^{-1}(\{y\})$, the derivative $d_xf:T_xS^1\rightarrow T_{f(x)}\mathbb T^3$ is surjective.
On the other side, if $f$ is surjective then for any $y\in \mathbb T^3$, the set $f^{-1}(y)$ is not empty and for every $x\in f^{-1}(y)$, the rank of $d_xf$ is at most $1$ : a contradiction.
About geometric arguments, most of the stuff you want to use are about surfaces but the question seems to be far from two dimensional things (by the way which surface are you talking about at the end?).