# lifting a product of commutators of standard generators on 2-manifolds

I have a problem with understand the proof http://www.ams.org/journals/proc/1972-032-01/S0002-9939-1972-0295352-2/S0002-9939-1972-0295352-2.pdf

I don't understand this part: "(...) we can easily construct $p: \tilde{F} \rightarrow F$ to be a six sheeted covering corresponding to the kernel of an appropriate map $\pi_1(F) \rightarrow \Sigma_3$ (such covering that: if $f$ represents a loop without sef-intersections and which is a product of commutators of "standard generators" then $f$ does not lift to a loop in $\tilde{F}$)."

$\hspace{1.5cm}$ 1. why we know that such covering exist?

$\hspace{1.5cm}$ 2. why this covering is six sheeted?

$\hspace{1.5cm}$ 3. why kernel doesn't contain $f$?

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## migrated from mathoverflow.netApr 25 at 22:05

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I think that talking about "products of commutators" is not useful (probably the fashion of the time was more algebraic). Cut along your null-homologous curve, so you get two pieces each with one boundary component, and restrict your attention to one of these (say, $S_1$). You want to find a nontrivial covering of $S_1$ with a single boundary component (which will be of length a multiple of the original boundary). You can then verify Hempel's claim by some educational cutting and pasting.

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Justin Malestein and I give a very explicit construction of such a cover in the proof of Lemma 2.1 of our paper

Malestein, Justin, Putman, Andrew; On the self-intersections of curves deep in the lower central series of a surface group. Geom. Dedicata 149 (2010), 73–84.

which is available here. We actually construct an $8$-fold cover instead of a $6$-fold cover because it was important for us that the deck group be nilpotent (in this case, a $2$-group). This was needed because our paper (among other things) was generalizing Hempel's argument to show that surface groups were residually nilpotent. But the $8$-fold cover we construct is good enough to make Hempel's argument go through (and in any case once you see what is happening you'll have no problem finding the $6$-fold cover if that is what you really want).

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thank you for your reply, but I have some new questions to your proof: 1. why always there is a surjection $\phi': H_1(\Sigma; \mathbb{Z}) \rightarrow \mathbb{Z}/8\mathbb{Z}$ such that $\phi'(\phi(x))\neq 0$ (for example when we have $\alpha_1^7 \beta_1$ it belongs to $H_1(\Sigma; \mathbb{Z})$ and I think it goes to 0)? Could you explain how look the surjection $\phi'$ –  Filip Parker Apr 26 at 18:35
2.I could believe that $\pi_1(\Sigma,*)= \langle \alpha_1,\beta_1,...,\alpha_g,\beta_g,x_1,...,x_g \mid [\alpha_1,\beta_1] \cdots [\alpha_g,\beta_g] = x_1 \cdots x_b \rangle$ but I have little experience in algebraic topology so I would like to see a proof of this fact. Could you tell me where I can read about it or could you explain this. –  Filip Parker Apr 26 at 18:35
3.I think I found a small mistakes at the end of "case 1". In one place there is : "$x \notin \{\alpha_1,\beta_1,\alpha_g,\beta_g\}$" and later there is "$s \notin \{\alpha_1,\beta_1,\alpha_g,\beta_g\}$" - I guess in the first place there should be $s$ instead of $x$, but in the second " $s \notin \{\alpha_1,\beta_1\}$" am I right? –  Filip Parker Apr 26 at 18:36
@FilipParker : 1. The surjection depends on $x$; for example, if you are dealing with $\alpha_1^7 \beta_1$ then you choose the homomorphism that takes $[\alpha_i]$ to $0$ for all $i$, that takes $[\beta_i]$ to $0$ for all $i \neq 1$, and that takes $[\beta_1]$ to $1$. 2. There are 2 points here. The first is that this is a presentation for $\pi_1$ of a punctured surface, which is discussed in any book that deals with surfaces and the fundamental group (e.g. Massey's book on the fundamental group). (continued) –  Andy Putman Apr 26 at 22:53
(continued) The second is that the surface is of this form. This can be derived from the classification of surfaces. I recommend perusing Section 1.3 of Farb-Margalit's "Primer on mapping class groups" for the details. 3. You are correct that these are typos; thanks for pointing them out! –  Andy Putman Apr 26 at 22:53
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