Degree and maps between closed orientable surfaces

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?

• Yes, this makes sense. Thank you. – 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.
• 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 $\alpha=t^{\ast}\tilde{\alpha}$. – Luc Aug 17 '13 at 22:52