# Map from sphere to torus

My ultimate goal is to show that if $$\Sigma_g$$ and $$\Sigma_h$$ are compact, orientable surfaces of genus $$g$$ and $$h$$ respectively, and $$g, then any map $$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?

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 $$T$$?

There are many surjective map $$S^2 \rightarrow T$$, clearly there is a surjective map $$[0,1] \times [0,1] \rightarrow T$$ and let $$S^2 \rightarrow [0,1] \times [0,1]$$ be any surjective map. For example let $$S^2 \rightarrow D^2$$ be the projection onto the first two coordinates (which is surjective), then since $$D^2 \approx [0,1] \times [0,1]$$ we are done. There are also many surjective maps $$\Sigma_g \rightarrow \Sigma_h$$ in general but I will leave this without a proof.
We can prove that any map $$S^2 \rightarrow T$$ is null-homotopic by showing that if $$[-,-]$$ denotes homotopy classes then $$[S^2,T] = [S^2, S^1 \times S^1] = [S^2, S^1] \times [S^2, S^1]$$ and then it is standard using covering space theory to prove that every map $$f:S^2 \rightarrow S^1$$ has to have a lift along $$e^{2\pi i t}:\mathbb R \rightarrow S^1$$ which means that $$f$$ has to be null-homotopic since $$\mathbb R$$ is contractible. Put all the pieces together to get that $$[S^2,T]$$ only has one element. I.e every map $$S^2 \rightarrow T$$ is nullhomotopic.
• No they won't. $T$ is a quotient space of $[0,1] \times [0,1]$ and the map $[0,1] \times [0,1] \rightarrow T$ is just the quotient map. And the map $S^2 \rightarrow D^2$ is just the restriction of the projection map $\mathbb R ^2 \rightarrow \mathbb R ^2$ onto $S^2 \subset \mathbb R ^3$ – Noel Lundström May 6 '20 at 15:47
• You are completely right. My mistake was that I confused the continuity of $f$ with the continuity of its (potential) inverse. Many thanks. – Heinrich Wagner May 6 '20 at 16:29