Check or falsify: there is a $C^\infty$ surjective map $S^1 \to S^1 \times S^1$. Check or falsify: There is a $C^\infty$-map  $S^1 \to S^1 \times S^1$ that is surjective.
Comments: I think it's not true, because $\dim S^1 = 1$ and $\dim S^1 \times S^1 = 2$ but I do not know how to justify.
 A: I assume you mean to check whether or not there is a smooth map $S^1\to S^1\times S^1$.
In this case, the answer is no.  Suppose $f:S^1\to S^1\times S^1$ is surjective.  By Sard's theorem, there is a regular value $y\in S^1\times S^1$ of $f$.  Since $f$ is surjective, $f(x)=y$ for some $x\in S^1$.  But then $df_x$ is surjective, so that 
$$1=\dim T_xS^1=\dim\ker df_x+\dim T_{y}(S^1\times S^1)\geq\dim T_{y}(S^1\times S^1)=2,$$
a contradiction.
Note that this argument applies in greater generality:  If $N^n$ and $M^m$ are smooth manifolds with $m<n$, then any smooth map $f:M\to N$ is not surjective.
A: Aweygan showws there is no smooth surjective map, while Danny uses cardinality to show there is a surjection $S^1\rightarrow S^1\times S^1$.  Since cardinality ignores things like the topology, the kinds of surjections that Danny is talking about are typically horribly discontinuous.
These leaves open an intermediate question:  is there a continuous surjective map from $S^1$ onto $S^1\times S^1$?  The answer is that there are continuous surjective maps.  Here is one way of building them.
Start with a space-filling curve, that is, a surjective continuous map $[0,1]:I\rightarrow [0,1]\times [0,1]$.  The topological space $[0,1]\times [0,1]$ has a natural surjective continous map to $S^1\times S^1$, given by identifying the boundary in the usual way to get a torus.  Composing gives a continuous surjection $f:[0,1]\rightarrow S^1\times S^1$.  We will imagine the domain of $f$ as the northern hemisphere of $S^1$.
Now, choose any continuous path $\gamma:[0,1]\rightarrow S^1\times S^1$ which connects $f(0)$ to $f(1)$.  We will think of the domain of $\gamma$ as the southern hemisphere of $S^1$.
Then, define $F:S^1\rightarrow S^1\times S^1$ by $$F(x) = \begin{cases} f(x) & x \text{ is in northern hemisphere}\\ \gamma(x) & x \text{ is in southern hemisphere}\end{cases}.$$  Because $f$ and $\gamma$ are continuous, and $f(0) = \gamma(0)$ and $f(1) = \gamma(1)$, $F$ is continuous.  Because $f$ is surjective, so is $F$.
A: No. There is such a map unless we impose restriction on that map.
The reason is simple. $S^1$ and $S^1\times S^1$, as sets, have the same cardinality as $\mathbb{R}$. Hence there is a bijection from $S^1$ onto $S^1\times S^1$.
