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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.

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  • $\begingroup$ Would you please clarify your notation? Is $\Sigma^1_3$ as surface with $1$ boundary component? $\endgroup$ – Cheerful Parsnip Dec 6 '13 at 20:39
  • $\begingroup$ 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. $\endgroup$ – Cheerful Parsnip Dec 6 '13 at 20:40
  • $\begingroup$ @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. $\endgroup$ – D Wiggles Dec 6 '13 at 20:55
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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 $\geq 7$.

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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.

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