degree of smooth maps from 2-sphere to 2-torus Why any smooth map from the 2-sphere to the 2-torus has zero degree?
Can we show that there is no surjective smooth map from 2-sphere to 2-torus?
There can't be a surjective map from the two sphere to the torus. There exists a surjective map from $f$ from $T^2$ to $S^2$ obtained by collapsing 1-cells in the canonical CW structure of $T^2$ see:  Is it possible to obtain a sphere from a quotient of a torus?

If you have a surjective map $g:S^2\rightarrow T^2$, then $f\circ g:S^2\rightarrow S^2$ is surjective. This is impossible since $\pi_2(T^2)=1$, $f$ is homotopic to a constant and so is $f\circ g$, so its degree is zero. the answer here

https://mathoverflow.net/questions/20712/do-surjective-degree-zero-maps-exist

implies that $f\circ g$ must miss a point
Here's a proof in the spirit of Schoen &amp; Yau's 2017 proof of the positive-mass theorem in general dimensions. Schoen &amp; Yau's proof crucially uses (and shows) the following fact:


  If $M$ is a compact oriented smooth $n$-manifold (without boundary) and there is
  a smooth map $F: M \rightarrow T^n$ to the $n$-torus with non-zero degree,
  then $M$ cannot possibly carry a metric with strictly positive scalar
  curvature. (In particular, $M$ cannot be a sphere.)


In contrast to the higher-dimensional case, for $n=2$ the proof is easy and answers OP's claim right away:

Suppose $M$ carried a positive scalar curvature metric. Then every connected component of $M$ has total curvature $&gt; 0$ and is thus a topological 2-sphere by Gauß-Bonnet's theorem and the classification of compact 2-surfaces without boundary. 

Now consider the form $dx^1 \wedge dx^2$ on $T^2$ where $x^1, x^2$ are canonical coordinates on $T^2 = S^1 × S^1$, normalized such that $\int_{T^2} dx^1 \wedge dx^2 = 1$. Consider the closed 1-forms $\theta^i = F^*(dx^i)$ on $M$ and compute:

$$
  \int_M \theta^1 \wedge \theta^2 = \int_M F^*(dx^1 \wedge dx^2) = \operatorname{deg}(F) \int_{T^2} dx^1 \wedge dx^2 = \operatorname{deg}(F) \neq 0
$$

But this means that $\theta^1 \wedge \theta^2$ has non-zero integral on some connected component $S^2$ of $M$.

We will show that the closed 1-forms $\theta^i$ restricted to $S^2$ lie in linearly independent (in particular non-zero) cohomology classes $[\theta^1], [\theta^2]$ of $S^2$. But then $H^1(S^2)$ is at least 2-dimensional, a contradiction since $H^1(S^2) = 0$.

So suppose their cohomology classes were linearly dependent, i.e. $\lambda [\theta^1] = [\theta^2]$ for some $\lambda \in \mathbb{R}\setminus\{0\}$ (without loss of generality). Then $\lambda \theta^1 - \theta^2 = df$ for some function $f$ on $S^2$. Then:

$$
  \theta^1 \wedge \theta^2 = \theta^1 \wedge (\lambda^{-1} \theta^1 - df) = 0 - \lambda^{-1} \theta^1 \wedge df = d(\lambda^{-1} f \theta^1)
$$

So the 2-form $\theta^1 \wedge \theta^2$ would be exact on $S^2$, so in particular it would have zero integral on $S^2$ by Stokes' theorem – a contradiction.