Check or falsify: there is a $C^\infty$ surjective map $S^1 \to S^1 \times S^1$.

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.

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 $F$.

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 $S^1\times S^1$.