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 $T$?


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.

  • $\begingroup$ But wouldn't all these maps be discontinuous? I should clarify that by map I mean continuous function $\endgroup$ – Heinrich Wagner May 6 '20 at 15:11
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    $\begingroup$ 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 $S^2 \subset \mathbb R ^3$ $\endgroup$ – Noel Lundström May 6 '20 at 15:47
  • $\begingroup$ You are completely right. My mistake was that I confused the continuity of $f$ with the continuity of its (potential) inverse. Many thanks. $\endgroup$ – Heinrich Wagner May 6 '20 at 16:29

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