# Is there a Covering Map $\Sigma_3^1\to \Sigma_2^1$

Let $S_{g,n}^b$ be a genus $g$ surface with $b$ boundary components and $n$ punctures. I'm having some trouble with these past qualifying exam questions:

Is there a covering map $p\colon \Sigma_3^1\to \Sigma_2^1$?
Is there a covering map $p\colon \Sigma_{3,1}\to\Sigma_{2,1}?$

I believe that the answer is no in both cases. Both $\Sigma_{3,1}$ and $Sigma_3^1$ deformation retracts onto a wedge of $6$ circles, while $\Sigma_{2,1}$ and $\Sigma_2^1$ deformation retract onto a wedge of $4$ circles. Any covering $p\colon\Sigma_3^1\to\Sigma_2^1$ would then induce an injection $p_*\colon F_6\to F_4$ whose image has finite index, because the target is compact. I know that we can inject $F_6\hookrightarrow F_4$, but I don't think the image can have finite index. I don't know how to prove this, though.

I haven't made much progress on the first question. Any help would be greatly appreciated.

• Would you please clarify your notation? Is $\Sigma^1_3$ as surface with $1$ boundary component? Dec 6, 2013 at 20:39
• If so, then the fact that $F_6$ would have to inject with finite index would come from compactness, which applies to the first question and not to the second. Dec 6, 2013 at 20:40
• @GrumpyParsnip $\Sigma_3^1$ is a genus 3 surface with 1 boundary component. You are correct about the finite index coming from compactness. I will edit the statement to correct this. Dec 6, 2013 at 20:55

This is a slightly modified version of Grumpy Parsnip's answer to my first question. If $p\colon X\to Y$ is a $d$-sheeted covering map, then $\chi(Y)=d\chi(X)$. This can easily be seen by triangulating $X$ (by sufficiently small triangles) and lifting this triangulation to a triangulation of $Y$. We can calculate $\chi(\Sigma_3^1)=2-2\cdot 3-1=-5$ and $\chi(\Sigma_2^1)=2-2\cdot 2-1=-3$. Since $3\nmid 5$, we cannot have a covering map $p\colon\Sigma_3^1\to\Sigma_2^1$.
The second question wasn't actually on the qual, but seemed like a natural thing to ask, though apparently it is more difficult. According to someone much smarter than me, there do exist covering maps $\Sigma_{3,1}\to\Sigma_{2,1}$. The Euler characteristic argument implies that any such map must be infinite sheeted. He didn't give me any indication of how to prove it though.
Your argument for surfaces with boundary is fine. To see that $F_6$ does not embed in $F_4$ with finite index, it suffices to show that no finite sheeted cover of $\vee _4 S^1$ has fundamental group of ranks $6$. Note that a $k$-sheeted cover will have $k$ vertices and $4k$ edges. The Euler characteristic is then $k-4k$ and the rank of the first homology $b_1$ satisfies $b_0-b_1=1-b_1=k-4k$. Thus $b_1=3k+1$. In particular, if $k\geq 2$, the rank of the fundamental group is $\geq 7$.