Maps between surface groups

Question

Let $S_g$ denote the fundamental group of a genus $g$ closed orientable surface. Are all maps $S_g \to S_h$ where $g<h$ necessarily trivial?

Answer 1

No, see my comment for a nontrivial construction. But in a more positive direction, every map $S_g \to S_h$, $g < h$, factors (up to homotopy) through the $1$-skeleton of $S_h$.

1) Consider the image of $\pi_1(S_g)$ in $\pi_1(S_h)$. This subgroup corresponds to the fundamental group of some covering space of $\pi_1(S_h)$. Since finite degree covering spaces of $S_h$ have higher genus than $S_h$, so it suffices to see that no quotient of $\pi_1(S_g)$ can be a surface group of higher genus. To see this, note that $\pi_1(S_g)$ is $2g$-generated, and cannot be generated by fewer elements (look at the abelianization), and taking quotients only decreases the minimal number of generators necessary. Thus the image of $\pi_1(S_g)$ cannot have finite index in $\pi_1(S_h)$. So the map $S_g \to S_h$ factors through some noncompact covering space $S$ of $S_h$.

Pick the cell structure of $S$ so that the map $S \to S_h$ is cellular. Now (nontrivial fact): any noncompact surface deformation retracts to a 1-dimensional subcomplex. Call that $C \subset S$. Thus the map $S_g \to S_h$ factors through $C$, and thus has image inside the 1-skeleton of $S_h$, as desired.