Smooth surjective map from torus to sphere I’m not sure if this is a dumb question but I can’t seem to get my way around it for several days now.
Is there a surjective smooth function from the 2-torus to the 2-sphere?
I’d be greatly happy if an example can be provided with some explanation.
On my own, I attempted to use Sard’s Theorem but I can’t seem to know how to draw a conclusion from there.
Thanks a lot in advance.
PS: This is my first question here and I’m sorry if the question fails some obvious expectations. I searched the site first but obtained the reverse question ( i.e., there being no such map from the sphere to the torus).
 A: The general result here is

Proposition: Suppose $M^n$ and $N^n$ are two connected closed manifolds of the same dimension.  Then there is a smooth surjective map $f:M\rightarrow N$.

Here's one way of proving it.
Lemma 1.  Let $B(0,1)\subseteq \mathbb{R}^n$ be the closed ball of radius $1$ centered at $0\in \mathbb{R}^n$.  Then there is a radially symmetric smooth surjective map $g:B(0,1)\rightarrow B(0,1)$ with the property that $g$ maps the points of distance $\geq 3/4$ from $0$ to $0$.
Proof:  Let $\psi$ be a smooth bump function which is supported on $[1/4, 3/4]$.  The function $x\psi(x)$ is continuous, so achieves a maximum value $K$ on $[1/4,3/4]$.
Thinking of $B(0,1)$ in polar coordinates $(r,\omega)$ with $\omega\in S^{n-1}$, define $g(r,\omega) = ( \psi(r)r/K, \omega)$.  Then $g$ is identically $0$ in a neighborhood of $0$, so is smooth at $0$ (which is always a concern when defining things in polar coordinates).  In addition, $K$ is chosen so that $\psi(r)r/K\leq 1$ for any $r$ (so the image really does lie in $B(0,1)$, and that for some specific $r$, $\psi(r)r/K = 1$, so the image really is all of $B(0,1)$.  $\square$
Lemma 2:  There is a smooth surjective map $h:M^n\rightarrow B(0,1)$.
Proof:  Given $p\in M$, let $U\subseteq M$ be an open neighborhood of $p$.  By shrinking $U$ if necessary, we may assume that the closure of $U$ is diffeomorphic to the closed ball $B(0,1)$.  Let $h_1:\overline{U}\rightarrow B(0,1)$ be such a diffeomorphism.  Then the function $g\circ h_1$ (with $g$ from Lemma 1) is smooth on $U$ and is identically equal to $0\in B(0,1)$ for points near $\partial U$.  In particular, we can extend $g\circ h_1$ to a map $h:M\rightarrow B(0,1)$ by defining $h(x) = \begin{cases}g(h_1(x)) & x\in U\\ 0 & x\notin U\end{cases}$ $.\square$
Lemma 3:  There is a smooth surjective map $j:B(0,1)\rightarrow N$.
Proof:  Picking a background Riemannian metric on $N$, because $N$ is closed and connected, we can rescale the metric so that the diameter of $N$ is strictly smaller than $1$.  Further, because $N$ is closed, this metric is automatically complete, so by the Hopf-Rinow theorem, given any two points, there is a minimizing geodesic connecting them.
So, the exponential map $\exp_n:T_n N\rightarrow N$ is surjective when restricted to a ball $B$ of radius $1$ in $T_n N$.  Of course, $T_n N$ is isometric (as an inner product space), there is a diffeomorphism $j_1:B(0,1)\rightarrow B$.  Then $j :=\exp\circ j_1$ is the desired map.  $\square$
Now, to prove the Proposition, just use $f = j\circ h.$  Since a composition of smooth surjective maps is smooth and surjective, we are done.
