Surface groups are residually finite

Question

I am looking at the short paper by Hempel "Residual finiteness of surface groups". Let $F$ be a compact orientable surface and let $f : S^1 \to F$ be a map that represents a nontrivial element of the fundamental group of $F$. Further assume that $f$ is an embedding. We want to construct a covering of $F$ by a compact surface $\tilde{F}$ such that the map $f$ can not be lifted to this covering.

First suppose that $f$ is a parameterization for one of the "standard generators" - then Hempel claims that we can construct a two-sheeted covering of $F$ that $f$ does not lift to. What is this construction??

Alternatively, suppose that the image of $f$ is homologically trivial (i.e. the image of $f$ is a product of commutators in $\pi_1(F)$. Then Hempel claims that we can construct a six-sheeted cover of $F$ corresponding to the kernel of a suitable map from $\pi_1(F)$ to the fundamental group of a genus 3 surface. What is this map and what is this construction?

This question was asked before and some parts of it were answered but I am confused about some of the unanswered parts:

John Hempel's proof of residual finiteness of surface groups

Answer 1

Correction: Hempel used $\Sigma_3$ to denote symmetric group of three letters not the surface of genus three. Its order is $6$, therefore any surjective map will have a kernel of index $6$.

Now to answer to your question, let us consider the possibilities of the components of a given curve $x$. The complement of the curve $x$ can be either connected (non-separating) or disconnected (separating). If the complement is connected then by the classification of surfaces, any two such curves are homeomorphic, hence start with a standard generator.

Case-1: Non-separating.

Recall that the algebraic intersection number $\hat{i}(x,y)$ is defined as the sum of the indices of the intersection points of $x$ and $y$, where an intersection point is of index $+1$ if orientation agrees with the surface, $-1$ otherwise. Also $\hat{i}(x,y)$ only depends on the homology classes of $x$ and $y$.

Now given any non-separating curve $x$ you can always find $y$ such that $\hat{i}(x,y)=1$.

Define the homomorphism $\hat{i}(\,\,,y)$ from $\pi_1(F)$ to $\mathbb{Z}_2$. The kernel is the required subgrop ad the cover ($x$ does not belong to the kernel).

Case-2: Separating.

Let $x=\prod_{i=1}^n[x_i,y_i]$. Define the homomorphism $\phi:\pi_1{F}\rightarrow\Sigma_3$ as:

1) $\phi(x_1)=(1,2),\,\, \phi(y_1)=(1,3).$

2) $\phi(x_2)=(1,2), \,\, \phi(y_2)=(2,3).$

3) $\phi(x_i)=\phi(y_i)=e$ for all $i\neq 1,2.$

Check that the product of the commutators is $e$, therefore this is a homomorphism and $x$ is not in the kernel.

Answer 2

Perhaps a partial answer will already help you:

Whether or not lifting a map to a covering is possible is determined by the image on fundamental groups. The map lifts if and only if the image on $\pi_1$ is contained in the characteristic subgroup of the covering (which is the subgroup that $\pi_1$ of the covering maps to under the projection). So if we have a map $f:Y\to X$ and a covering $p:E\to X$ $f$ lifts to $E$ precisely if $f_*\pi_1(Y,y_0)\subset p_*\pi_1(E,e_0)$ (where $p(e_0)=x_0$, $x_0$ the chosen base point of $X$).

Then, the claims by Hempel reduce to the claims that

• There exists an index 2 subgroup of $\pi_1(F)$ which does not contain the subgroup generated by one of the standard generators.

• There exists and index 6 subgroup which is the kernel of a map from $\pi_1(F)$ to $\pi_1$ of a genus three surface.