Does the genus $g$ surface for $g\geq 2$ cover the surface of genus 2?

Question

I recently came across a proof that seemed to rely on the fact that $\pi_1(S_g)$ is a subgroup of $\pi_1(S_2)$ for $g\geq 2$, where $S_g$ is the surface of genus $g$. Specifically it was a proof that surface groups are fully residually free and went along the lines of 'since finitely generated subgroups of fully residually free groups are fully residually free, all surface groups for genus $g\geq 2$ are as $\pi_1(S_2)$ is.'

The easiest way I can think of proving this is constructing a covering map $S_g\to S_2$. I can do this for $g=3$ but can't quite figure out how to generalise this to higher genus. Is there a general construction that works for higher genus?

Thanks for any help.

Answer 1

Draw $S_g$ as a "flower petal" with one central donut surrounded by $g-1$ donuts on the outside. Below would be $S_6$.

The picture is terrible, but imagine it is nice and symmetric. There is action of the cyclic group on this space by rotations around the $z$-axis. The map takes small open neighborhoods to disjoint small open neighborhoods, so quotienting by the group action gives a covering map, and the quotient space is naturally isomorphic to $S_2$.

Answer 2

Tom has shown one way of visualizing it. An alternative is just twist them like $S_3\to S_2$, i.e., align the holes, make half-twists and fold over each of the $g-2$ holes in the middle.