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

Question

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 $f$?

Answer 1

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.

Answer 2

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).