My ultimate goal is to show that if $\Sigma_g$ and $\Sigma_h$ are compact, orientable surfaces of genus $g$ and $h$ respectively, and $g<h$, then any map $f: \Sigma_g \to \Sigma_h$ must have degree $0$. There is an answer here that uses the ring structure of cohomology, but I was hoping there might be some more elementary proof.
So I started by thinking about the simplest case: A map from the sphere to the torus:
The OP of the post linked above suggests showing that in general, if we can show that any (continuous) map $f: \Sigma_g \to \Sigma_h$ must be non-surjective, the result would follow.
This naturally leads one to consider what maps there are from the sphere $S^2$ to the torus $T$. The only non-trivial one I can think of is a projection $S^2 \to S^1$ composed with an embedding $S^1 \hookrightarrow T$, which is clearly not surjective. But are all non-trivial maps $S^2 \to T$ of this form? Can we demonstrate this?
So: Does anyone know of a (relatively) elementary way of showing that there cannot exist a surjective map from $S^2$ to $T$ (other than saying that it is clear), and more generally, from $\Sigma_g$ to $\Sigma_h$ when $g<h$?
Addendum: On a second thought, the projection $S^2 \to S^1$ is evidently not continuous - my mistake. So I add to the above: Is there any non-trivial map from $S^2$ to $T$?