Let $M_g, M_h$ be closed orientable surfaces of genus $g,h$ respectively.

If $g>h$, we know there exists a map $M_g \rightarrow M_h$ of degree 1: just think of $M_g=M_h\#M_{g-h}$ and consider the map $M_g=M_h\#M_{g-h} \rightarrow M_h$ that pinches $M_{g-h}$ to a point, which can be easily seen to have degree 1.

How about the case $g<h$? I have read that any map $f:M_g \rightarrow M_h$ with $g<h$ must have zero degree (and hence be homotopic to a constant map).

One way to see this would be to show that $f$ is non-surjective. How can we do this? The induced map in homology $f_{\ast}:H_1(M_g)\simeq \mathbb{Z}^{2g} \rightarrow H_1(M_h) \simeq \mathbb{Z}^{2h}$ is clearly non-surjective, but how about $f$? Is this the right way to proceed?

  • $\begingroup$ Yes, this makes sense. Thank you. $\endgroup$ – Luc Aug 17 '13 at 22:42

Are you familiar with cohomology and the cup product structure for surfaces? If so, show there exists an $\alpha\in H^1(M_h)\setminus\{0\}$ with $f^∗\alpha=0$. Then, show there is a $\beta\in H^1(M_h)$ with $\alpha\smile\beta\neq 0$. Poincare duality and the naturality of the cup product gets you the rest of the way.

  • 1
    $\begingroup$ An argument for the existence of $\alpha$ could be: think of $M_h=M_g\#M_{h-g}$ and compose $f:M_g\rightarrow M_h$ with $t:M_h\simeq M_g\#M_{h-g} \rightarrow M_g \wedge M_{h-g}$ that collapses the $\mathbb{S}^1$ separating both. Then any class $\tilde{\alpha}\in H^1(M_g \wedge M_{h-g})$ coming from $H^1(M_{h-g})$ pulls back to 0 under the composition $t\circ f$, so we may take $\alpha=t^{\ast}\tilde{\alpha}$. $\endgroup$ – Luc Aug 17 '13 at 22:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.