# Twisted Alexander polynomials detect fibered 3–manifolds

Stefan Friedl  University of Warwick, Coventry, UK<EMAIL_ADDRESS>and  Stefano Vidussi  Department of Mathematics, University of California,
Riverside, CA 92521, USA<EMAIL_ADDRESS>
###### Abstract.

A classical result in knot theory says that for a fibered knot the Alexander
polynomial is monic and that the degree equals twice the genus of the knot.
This result has been generalized by various authors to twisted Alexander
polynomials and fibered 3–manifolds. In this paper we show that the conditions
on twisted Alexander polynomials are not only necessary but also sufficient
for a 3–manifold to be fibered. By previous work of the authors this result
implies that if a manifold of the form $S^{1}\times N^{3}$ admits a symplectic
structure, then $N$ fibers over $S^{1}$. In fact we will completely determine
the symplectic cone of $S^{1}\times N$ in terms of the fibered faces of the
Thurston norm ball of $N$.

S. Friedl was supported by a CRM–ISM Fellowship and by CIRGET

S. Vidussi was partially supported by a University of California Regents’
Faculty Fellowships and by NSF grant #0906281.

##  1\. Introduction

###  1.1. Twisted Alexander polynomials and fibered 3–manifolds

Let $N$ be a compact, connected, oriented $3$–manifold with empty or toroidal
boundary. Given a nontrivial class $\phi\in
H^{1}(N;\mathbb{Z})=\mbox{Hom}(\pi_{1}(N),\mathbb{Z})$ we say that $(N,\phi)$
fibers over $S^{1}$ if there exists a fibration $f:N\to S^{1}$ such that the
induced map $f_{*}:\pi_{1}(N)\to\pi_{1}(S^{1})=\mathbb{Z}$ agrees with $\phi$.
Stated otherwise, the homotopy class in $[N,S^{1}]=H^{1}(N;\mathbb{Z})$
identified by $\phi$ can be represented by a fibration.

It is a classical result in knot theory that if a knot $K\subset S^{3}$ is
fibered, then the Alexander polynomial is monic (i.e. the top coefficient
equals $\pm 1$), and the degree of the Alexander polynomial equals twice the
genus of the knot. This result has been generalized in various directions by
several authors (e.g. [McM02, Ch03, GKM05, FK06, Ki07]) to show that twisted
Alexander polynomials give necessary conditions for $(N,\phi)$ to fiber.

To formulate this kind of result more precisely we have to introduce some
definitions. Let $N$ be a 3–manifold with empty or toroidal boundary and let
$\phi\in H^{1}(N;\mathbb{Z})$. Given $(N,\phi)$ the _Thurston norm_ of $\phi$
(cf. [Th86]) is defined as

$||\phi||_{T}=\min\\{\chi_{-}(S)\,|\,S\subset N\mbox{ properly embedded
surface dual to }\phi\\}.$

Here, given a surface $S$ with connected components $S_{1}\cup\dots\cup
S_{k}$, we define $\chi_{-}(S)=\sum_{i=1}^{k}\max\\{-\chi(S_{i}),0\\}$.

In the following we assume that $\phi\in H^{1}(N;\mathbb{Z})$ is non–trivial.
Let $\alpha:\pi_{1}(N)\to G$ be a homomorphism to a finite group. We have the
permutation representation $\pi_{1}(N)\to\mbox{Aut}(\mathbb{Z}[G])$ given by
left multiplication, which we also denote by $\alpha$. We can therefore
consider the twisted Alexander polynomial
$\Delta_{N,\phi}^{\alpha}\in\mathbb{Z}[t^{\pm 1}]$, whose definition is
detailed in Section 2.3. We denote by $\phi_{\alpha}$ the restriction of
$\phi\in H^{1}(N;\mathbb{Z})=\mbox{Hom}(\pi_{1}(N),\mathbb{Z})$ to
$\mbox{Ker}(\alpha)$. Note that $\phi_{\alpha}$ is necessarily non–trivial. We
denote by $\mbox{div}\phi_{\alpha}\in\mathbb{N}$ the divisibility of
$\phi_{\alpha}$, i.e.

$\mbox{div}\phi_{\alpha}=\max\\{n\in\mathbb{N}\,|\,\phi_{\alpha}=n\psi\mbox{
for some }\psi:\mbox{Ker}(\alpha)\to\mathbb{Z}\\}.$

We can now formulate the following theorem which appears as [FK06, Theorem 1.3
and Remark p. 938].

######  Theorem 1.1.

Let $N\neq S^{1}\times S^{2},S^{1}\times D^{2}$ be a $3$–manifold with empty
or toroidal boundary. Let $\phi\in H^{1}(N;\mathbb{Z})$ a nontrivial class. If
$(N,\phi)$ fibers over $S^{1}$, then for any homomorphism
$\alpha:\pi_{1}(N)\to G$ to a finite group the twisted Alexander polynomial
$\Delta_{N,\phi}^{\alpha}\in\mathbb{Z}[t^{\pm 1}]$ is monic and

$\mbox{deg}(\Delta_{N,\phi}^{\alpha})=|G|\,\|\phi\|_{T}+(1+b_{3}(N))\mbox{div}\phi_{\alpha}.$

It is well known that in general the constraint of monicness and degree for
the ordinary Alexander polynomial falls short from characterizing fibered
3–manifolds. The main result of this paper is to show that on the other hand
the collection of all twisted Alexander polynomials does detect fiberedness,
i.e. the converse of Theorem 1.1 holds true:

######  Theorem 1.2.

Let $N$ be a $3$–manifold with empty or toroidal boundary. Let $\phi\in
H^{1}(N;\mathbb{Z})$ a nontrivial class. If for any homomorphism
$\alpha:\pi_{1}(N)\to G$ to a finite group the twisted Alexander polynomial
$\Delta_{N,\phi}^{\alpha}\in\mathbb{Z}[t^{\pm 1}]$ is monic and

$\mbox{deg}(\Delta_{N,\phi}^{\alpha})=|G|\,\|\phi\|_{T}+(1+b_{3}(N))\mbox{div}\phi_{\alpha}$

holds, then $(N,\phi)$ fibers over $S^{1}$.

Note that alternatively it is possible to rephrase this statement in terms of
Alexander polynomials of the finite regular covers of $N$, using the fact that
$\Delta_{N,\phi}^{\alpha}=\Delta_{{\tilde{N}},p^{*}(\phi)}$ (cf. [FV08a]),
where $p:{\tilde{N}}\to N$ is the cover of $N$ determined by
$\mbox{Ker}(\alpha)$.

Note that this theorem asserts that twisted Alexander polynomials detect
whether $(N,\phi)$ fibers under the assumption that $||\phi||_{T}$ is known;
while it is known that twisted Alexander polynomials give lower bounds (cf.
[FK06, Theorem 1.1]), it is still an open question whether twisted Alexander
polynomials determine the Thurston norm.

In the case where $\phi$ has trivial Thurston norm, this result is proven in
[FV08b], using subgroup separability. Here, following a different route (see
Section 1.3 for a summary of the proof), we prove the general case.

###  1.2. Symplectic $4$–manifolds and twisted Alexander polynomials

In 1976 Thurston [Th76] showed that if a closed $3$–manifold $N$ admits a
fibration over $S^{1}$, then $S^{1}\times N$ admits a symplectic structure,
i.e. a closed, nondegenerate $2$–form $\omega$. It is natural to ask whether
the converse to this statement holds true. In its simplest form, we can state
this problem in the following way:

######  Conjecture 1.3.

Let $N$ be a closed $3$–manifold. If $S^{1}\times N$ is symplectic, then there
exists a $\phi\in H^{1}(N;\mathbb{Z})$ such that $(N,\phi)$ fibers over
$S^{1}$.

Interest in this question was motivated by Taubes’ results in the study of
Seiberg-Witten invariants of symplectic $4$–manifolds (see [Ta94, Ta95]), that
gave initial evidence to an affirmative solution of this conjecture. In the
special case where $N$ is obtained via $0$–surgery along a knot in $S^{3}$,
this question appears also in [Kr98, Question 7.11]. Over the last ten years
evidence for this conjecture was given by various authors [Kr98, CM00, Et01,
McC01, Vi03].

In [FV08a] the authors initiated a project relating Conjecture 1.3 to the
study of twisted Alexander polynomials. The outcome of that investigation is
that if $S^{1}\times N$ is symplectic, then the twisted Alexander polynomials
of $N$ behave like twisted Alexander polynomials of a fibered $3$–manifold.
More precisely, the following holds (cf. [FV08a, Theorem 4.4]):

######  Theorem 1.4.

Let $N$ be an irreducible closed $3$–manifold and $\omega$ a symplectic
structure on $S^{1}\times N$ such that $\omega$ represents an integral
cohomology class. Let $\phi\in H^{1}(N;\mathbb{Z})$ be the Künneth component
of $[\omega]\in H^{2}(S^{1}\times N;\mathbb{Z})$. Then for any homomorphism
$\alpha:\pi_{1}(N)\to G$ to a finite group the twisted Alexander polynomial
$\Delta_{N,\phi}^{\alpha}\in\mathbb{Z}[t^{\pm 1}]$ is monic and

$\mbox{deg}(\Delta_{N,\phi}^{\alpha})=|G|\,\|\phi\|_{T}+2\mbox{div}\phi_{\alpha}.$

Note that it follows from McCarthy’s work [McC01] (see also Lemma 7.1) and
Perelman’s proof of the geometrization conjecture (cf. e.g. [MT07]) that if
$S^{1}\times N$ is symplectic, then $N$ is prime, i.e. either irreducible or
$S^{1}\times S^{2}$. The proof of Theorem 1.4 relies heavily on the results of
[Kr98] and [Vi03], which in turn build on results of Taubes [Ta94, Ta95] and
Donaldson [Do96].

As the symplectic condition is open, the assumption that a symplectic manifold
admits an integral symplectic form is not restrictive. Therefore, combining
Theorem 1.2 with Theorem 1.4, we deduce that Conjecture 1.3 holds true. In
fact, in light of [FV07, Theorems 7.1 and 7.2], we have the following more
refined statement:

######  Theorem 1.5.

Let $N$ be a closed oriented 3–manifold. Then given $\Omega\in
H^{2}(S^{1}\times N;\mathbb{R})$ the following are equivalent:

  1. (1)

$\Omega$ can be represented by a symplectic structure;

  2. (2)

$\Omega$ can be represented by a symplectic structure which is
$S^{1}$–invariant;

  3. (3)

$\Omega^{2}>0$ and the Künneth component $\phi\in H^{1}(N;\mathbb{R})$ of
$\Omega$ lies in the open cone on a fibered face of the Thurston norm ball of
$N$.

Note that the theorem allows us in particular to completely determine the
symplectic cone of a manifold of the form $S^{1}\times N$ in terms of the
fibered cones of $N$.

Combined with the results of [FV07, FV08a], Theorem 1.2 shows in particular
that the collection of the Seiberg-Witten invariants of all finite covers of
$S^{1}\times N$ determines whether $S^{1}\times N$ is symplectic or not. In
particular, we have the following corollary (we refer to [Vi99, Vi03] for the
notation and the formulation in the case that $b_{1}(N)=1$).

######  Corollary 1.6.

Let $N$ be a closed 3–manifold with $b_{1}(N)>1$. Then given a spinc structure
$K\in H^{2}(S^{1}\times N;\mathbb{Z})$ there exists a symplectic structure
representing a cohomology class $\Omega\in H^{2}(S^{1}\times N;\mathbb{R})$
with canonical class $K$ if and only if the following conditions hold:

  1. (1)

$K\cdot\phi=\|\phi\|_{T}$, where $\phi\in H^{1}(N;\mathbb{R})$ is the Künneth
component of $\Omega$,

and for any regular finite cover $p:{\tilde{N}}\to N$

  1. (2)

$SW_{S^{1}\times{\tilde{N}}}(p^{*}(K))=1$,

  2. (3)

for any Seiberg–Witten basic class $\kappa\in
H^{2}(S^{1}\times{\tilde{N}};\mathbb{Z})$ we have

$|p_{*}(\kappa)\cdot\phi|\leq\mbox{deg}(p)~{}K\cdot\phi,$

(where $p_{*}$ is the transfer map) and the latter equality holds if and only
if $\kappa=\pm p^{*}K$.

(Note that, under the hypotheses of the Corollary, all basic classes of
$S^{1}\times{\tilde{N}}$ are the pull–back of elements of
$H^{2}({\tilde{N}};\mathbb{Z})$.)

Remark. A different approach to Conjecture 1.3 involves a deeper investigation
of the consequence of the symplectic condition on $S^{1}\times N$, that goes
beyond the information encoded in Theorem 1.4. A major breakthrough in this
direction has recently been obtained by Kutluhan and Taubes ([KT09]). They
show that if $N$ is a 3–manifold such that $S^{1}\times N$ is symplectic,
under some cohomological assumption on the symplectic form, then the Monopole
Floer homology of $N$ behaves like the Monopole Floer homology of a fibered
3–manifold. On the other hand it is known, due to the work of Ghiggini,
Kronheimer and Mrowka, and Ni that Monopole Floer homology detects fibered
3–manifolds ([Gh08, Ni09, KM08, Ni08]). The combination of the above results
proves in particular Conjecture 1.3 in the case that $b_{1}(N)=1$.

###  1.3. Fibered 3–manifolds and finite solvable groups: outline of the proof

In this subsection we will outline the strategy of the proof of Theorem 1.2.
It is useful to introduce the following definition.

######  Definition.

Let $N$ be a $3$–manifold with empty or toroidal boundary, and let $\phi\in
H^{1}(N;\mathbb{Z})$ be a nontrivial class. We say that $(N,\phi)$ satisfies
Condition ($*$) if for any homomorphism $\alpha:\pi_{1}(N)\to G$ to a finite
group the twisted Alexander polynomial
$\Delta_{N,\phi}^{\alpha}\in\mathbb{Z}[t^{\pm 1}]$ is monic and

$\mbox{deg}(\Delta_{N,\phi}^{\alpha})=|G|\,\|\phi\|_{T}+(1+b_{3}(N))\mbox{div}\phi_{\alpha}.$

It is well–known (see [McC01] for the closed case, and 7.1 for the general
case) that Condition ($*$) implies, using geometrization, that $N$ is prime,
so we can restrict ourself to the case where $N$ is irreducible.

Note that McMullen [McM02] showed that, when the class $\phi$ is primitive,
the condition $\Delta_{N,\phi}\neq 0$ implies that there exists a connected
Thurston norm minimizing surface $\Sigma$ dual to $\phi$. It is well–known
that to prove Theorem 1.2 it is sufficient to consider a primitive $\phi$, and
we will assume that in the following. Denote $M=N\setminus\nu\Sigma$; the
boundary of $M$ contains two copies $\Sigma^{\pm}$ of $\Sigma$ and throughout
the paper we denote the inclusion induced maps $\Sigma\to\Sigma^{\pm}\to M$ by
$\iota_{\pm}$.

By Stallings’ theorem [St62] the surface $\Sigma$ is a fiber of a fibration
$N\to S^{1}$ if and only if $\iota_{\pm}:\pi_{1}(\Sigma)\to\pi_{1}(M)$ are
isomorphisms. Hence to prove Theorem 1.2 we need to show that if $(N,\phi)$
satisfies Condition ($*$), then the monomorphisms
$\iota_{\pm}:\pi_{1}(\Sigma)\to\pi_{1}(M)$ are in fact isomorphisms. Using
purely group theoretic arguments we are not able to show directly that
Condition ($*$) implies the desired isomorphism; however, we have the
following result:

######  Proposition 1.7.

Assume that $(N,\phi)$ satisfies Condition $(*)$ and that $\phi$ is primitive.
Let $\Sigma\subset N$ be a connected Thurston norm minimizing surface dual to
$\phi$ and let $\iota$ be either of the two inclusion maps of $\Sigma$ into
$M=N\setminus\nu\Sigma$. Then $\iota:\pi_{1}(\Sigma)\to\pi_{1}(M)$ induces an
isomorphism of the prosolvable completions.

We refer to Section 2.4 for information regarding group completions.
Proposition 1.7 translates the information from Condition ($*$) into
information regarding the maps $\iota_{\pm}:\pi_{1}(\Sigma)\to\pi_{1}(M)$.
From a purely group theoretic point of view it is a difficult problem to
decide whether a homomorphism which gives rise to an isomorphism of
prosolvable completions has to be an isomorphism itself (cf. [Gr70], [BG04],
[AHKS07] and also Lemma 4.7). But in our 3–dimensional setting we can use a
recent result of Agol [Ag08] to prove the following theorem.

######  Theorem 1.8.

Let $N$ be an irreducible 3–manifold with empty or toroidal boundary. Let
$\Sigma\subset N$ be a connected Thurston norm minimizing surface. We write
$M=N\setminus\nu\Sigma$. Assume the following hold:

  1. (1)

the inclusion induced maps $\iota_{\pm}:\pi_{1}(\Sigma)\to\pi_{1}(M)$ give
rise to isomorphisms of the respective prosolvable completions, and

  2. (2)

$\pi_{1}(M)$ is residually finite solvable,

then $\iota_{\pm}:\pi_{1}(\Sigma)\to\pi_{1}(M)$ are isomorphisms, hence
$M=\Sigma\times I$.

In light of Proposition 1.7, the remaining obstacle for the proof of Theorem
1.2 is the condition in Theorem 1.8 that $\pi_{1}(M)$ has to be residually
finite solvable. It is well–known that linear groups (and hence in particular
hyperbolic 3–manifolds groups) are virtually residually $p$ for all but
finitely many primes $p$ (cf. e.g. [We73, Theorem 4.7] or [LS03, Window 7,
Proposition 9]), in particular they are residually finite solvable. Thurston
conjectured that 3–manifold groups in general are linear (cf. [Ki, Problem
3.33]), but this is still an open problem. Using the recent proof of the
geometrization conjecture (cf. e.g. [MT07]) we will prove the following
result, which will be enough for our purposes.

######  Theorem 1.9.

Let $N$ be a closed prime 3–manifold. Then for all but finitely many primes
$p$ there exists a finite cover $N^{\prime}$ of $N$ such that the fundamental
group of any component of the JSJ decomposition of $N^{\prime}$ is residually
a $p$–group.

We can now deduce Theorem 1.2 as follows: We first show in Lemmas 7.1 and 7.2
that it suffices to show Theorem 1.2 for closed prime 3–manifolds. Theorem 1.2
in that situation now follows from combining Theorems 1.7, 1.8 and 1.9 with a
more technical theorem which allows us to treat the various JSJ pieces
separately (cf. Theorem 6.4).

Added in proof: In a very recent paper ([AF10]) Matthias Aschenbrenner and the
first author showed that any 3–manifold group is virtually residually $p$.
This simplifies the proof of Theorem 1.2 as outlined in [FV10].

This paper is structured as follows. In Section 2 we recall the definition of
twisted Alexander polynomials and some basics regarding completions of groups.
In Section 3 we will prove Proposition 1.7 and in Section 4 we give the proof
of Theorem 1.8. In Section 5 we prove Theorem 1.9 and in Section 6 we provide
the proof for Theorem 6.4. Finally in Section 7 we complete the proof of
Theorem 1.2.

Conventions and notations. Throughout the paper, unless otherwise stated, we
will assume that all manifolds are oriented and connected, and all homology
and cohomology groups have integer coefficients. Furthermore all surfaces are
assumed to be properly embedded and all spaces are compact and connected,
unless it says explicitly otherwise. The derived series of a group $G$ is
defined inductively by $G^{(0)}=G$ and $G^{(n+1)}=[G^{(n)},G^{(n)}]$.

Acknowledgments. We would like to thank Ian Agol, Matthias Aschenbrenner,
Steve Boyer, Paolo Ghiggini, Taehee Kim, Marc Lackenby, Alexander Lubotzky,
Kent Orr, Saul Schleimer, Jeremy van Horn–Morris and Genevieve Walsh for many
helpful comments and conversations. We also would like to thank the referee
for suggesting several improvements to the paper and pointing out various
inaccuracies.

##  2\. Preliminaries: Twisted invariants and completions of groups

###  2.1. Twisted homology

Let $X$ be a CW–complex with base point $x_{0}$. Let $R$ be a commutative
ring, $V$ a module over $R$ and $\alpha:\pi_{1}(X,x_{0})\to\mbox{Aut}_{R}(V)$
a representation. Let $\tilde{X}$ be the universal cover of $X$. Note that
$\pi_{1}(X,x_{0})$ acts on the left on $\tilde{X}$ as group of deck
transformations. The cellular chain groups $C_{*}(\tilde{X})$ are in a natural
way right $\pi_{1}(X)$–modules, with the right action on $C_{*}(\tilde{X})$
defined via $\sigma\cdot g:=g^{-1}\sigma$, for $\sigma\in C_{*}(\tilde{X})$.
We can form by tensoring the chain complex
$C_{*}(\tilde{X})\otimes_{\mathbb{Z}[\pi_{1}(X,x_{0})]}V$, which is a complex
of $R$–modules. Now define
$H_{i}(X;V):=H_{i}(C_{*}(\tilde{X})\otimes_{\mathbb{Z}[\pi_{1}(X,x_{0})]}V)$.
The isomorphism type of the $R$–module $H_{i}(X;V)$ does not depend on the
choice of the base point, in fact it only depends on the homotopy type of $X$
and the isomorphism type of the representation.

In this paper we will also frequently consider twisted homology for a finitely
generated group $\Gamma$; its definition can be reduced to the one above by
looking at the twisted homology of the Eilenberg-Maclane space $K(\Gamma,1)$.

The most common type of presentation we consider in this paper is as follows:
Let $X$ be a topological space, $\alpha:\pi_{1}(X)\to G$ a homomorphism to a
group $G$ and $H\subset G$ a subgroup of finite index. Then we get a natural
action of $\pi_{1}(X)$ on $\mbox{Aut}_{\mathbb{Z}}(\mathbb{Z}[G/H])$ by
left–multiplication, which gives rise to the homology groups
$H_{i}(X;\mathbb{Z}[G/H])$.

We will now study the $\mathbb{Z}[\pi_{1}(X)]$–module $\mathbb{Z}[G/H]$ in
more detail. We write $C:=\alpha(\pi_{1}(X))$. Consider the set of double
cosets $C\backslash G/H$. By definition $g,g^{\prime}\in G$ represent the same
equivalence class if and only if there exist $c,c^{\prime}\in C$ and
$h,h^{\prime}\in H$ such that $cgh=c^{\prime}g^{\prime}h^{\prime}$. Note that
$g_{1},\dots,g_{k}\in G$ are a complete set of representatives of $C\backslash
G/H$ if and only if $G$ is the disjoint union of $Cg_{1}H,\dots,Cg_{k}H$. The
first part of the following lemma is an immediate consequence of [Br94,
II.5.2.], the second part follows either from Shapiro’s lemma or a
straightforward calculation.

######  Lemma 2.1.

Let $g_{1},\dots,g_{k}\in G$ be a set of representatives for the equivalence
classes $C\backslash G/H$. For $i=1,\dots,k$ write ${\tilde{C}}_{i}=C\cap
g_{i}Hg_{i}^{-1}$. We then have the following isomorphisms of left
$\mathbb{Z}[C]$–modules:

$\mathbb{Z}[G/H]\cong\bigoplus_{i=1}^{k}\mathbb{Z}[C/{\tilde{C}}_{i}].$

In particular $H_{0}(X;\mathbb{Z}[G/H])$ is a free abelian group of rank
$k=|C\backslash G/H|$.

###  2.2. Induced maps on low dimensional homology groups

In this section we will give criteria when maps between groups give rise to
isomorphisms between low dimensional twisted homology groups. We start out
with a study of the induced maps on 0–th twisted homology groups.

######  Lemma 2.2.

Let $\varphi:A\to B$ be a monomorphism of finitely generated groups. Suppose
that $B$ is a subgroup of a group $\pi$ and let $\tilde{\pi}\subset\pi$ be a
subgroup of finite index. Let $g_{1},\dots,g_{k}\in\pi$ be a set of
representatives for the equivalence classes $B\backslash\pi/\tilde{\pi}$. For
$i=1,\dots,k$ we write $\tilde{B}_{i}=B\cap g_{i}\tilde{\pi}g_{i}^{-1}$ and
$\tilde{A}_{i}=\varphi^{-1}(\tilde{B}_{i})$. Then

$\varphi_{*}:H_{0}(A;\mathbb{Z}[{\pi/\tilde{\pi}}])\to
H_{0}(B;\mathbb{Z}[{\pi/\tilde{\pi}}])$

is an epimorphism of free abelian groups and it is an isomorphism if and only
if $\varphi:A/\tilde{A}_{i}\to B/\tilde{B}_{i}$ is a bijection for any $i$.

###### Proof.

It is well–known that the induced map on 0–th twisted homology groups is
always surjective (cf. e.g. [HS97, Section 6]) and by Lemma 2.1 both groups
are free abelian groups. Now note that without loss of generality we can
assume that $A\subset B$ and that $\varphi$ is the inclusion map. It follows
from Lemma 2.1 that $H_{0}(B;\mathbb{Z}[\pi/\tilde{\pi}])$ is a free abelian
group of rank $k=|B\backslash\pi/\tilde{\pi}|$. By the same Lemma we also have

$\mathbb{Z}[\pi/\tilde{\pi}]\cong\bigoplus_{i=1}^{k}\mathbb{Z}[B/{\tilde{B}}_{i}]$

as left $\mathbb{Z}[B]$–modules and hence also as left
$\mathbb{Z}[A]$–modules. By applying Lemma 2.1 to the $\mathbb{Z}[A]$–modules
$\mathbb{Z}[B/{\tilde{B}}_{i}]$ we see that
$H_{0}(A;\mathbb{Z}[\pi/\tilde{\pi}])$ is a free abelian group of rank $k$ if
and only if $|A\backslash B/\tilde{B}_{i}|=1$ for any $i$. It is
straightforward to see that this is equivalent to $A/\tilde{A}_{i}\to
B/\tilde{B}_{i}$ being a bijection for any $i$. ∎

We will several times make use of the following corollary.

######  Corollary 2.3.

Let $\varphi:A\to B$ be a monomorphism of finitely generated groups. Let
$\beta:B\to G$ be a homomorphism to a finite group. Then

$\varphi_{*}:H_{0}(A;\mathbb{Z}[G])\to H_{0}(B;\mathbb{Z}[G])$

is an epimorphism of free abelian groups and it is an isomorphism if and only
if

$\mbox{Im}\\{A\to B\to G\\}=\mbox{Im}\\{B\to G\\}.$

###### Proof.

Let $\pi^{\prime}=B\times G$ and $\tilde{\pi}^{\prime}=B$. We can then apply
Lemma 2.2 to $A^{\prime}=A,B^{\prime}=\\{(g,\beta(g))\,|\,g\in
B\\}\subset\pi^{\prime}$ and
$\varphi^{\prime}(a)=(\varphi(a),\beta(\varphi(a)),a\in A^{\prime}$. It is
straightforward to verify that the desired equivalence of statements follows.
∎

We now turn to the question when group homomorphisms induce isomorphisms of
the 0–th and the first twisted homology groups at the same time.

######  Lemma 2.4.

Let $\varphi:A\to B$ be a monomorphism of finitely generated groups. Suppose
that $B$ is a subgroup of a group $\pi$ and let $\tilde{\pi}\subset\pi$ be a
subgroup of finite index. Let $g_{1},\dots,g_{k}\in\pi$ be a set of
representatives for the equivalence classes $B\backslash\pi/\tilde{\pi}$. For
$i=1,\dots,k$ we write $\tilde{B}_{i}=B\cap g_{i}\tilde{\pi}g_{i}^{-1}$ and
$\tilde{A}_{i}=\varphi^{-1}(\tilde{B}_{i})$. Then

$\varphi_{*}:H_{i}(A;\mathbb{Z}[{\pi/\tilde{\pi}}])\to
H_{i}(B;\mathbb{Z}[{\pi/\tilde{\pi}}])$

is an isomorphism for $i=0$ and $i=1$ if and only if the following two
conditions are satisfied:

  1. (1)

$\varphi:A/\tilde{A}_{i}\to B/\tilde{B}_{i}$ is a bijection for any $i$,

  2. (2)

$\varphi:A/[\tilde{A}_{i},\tilde{A}_{i}]\to B/[\tilde{B}_{i},\tilde{B}_{i}]$
is a bijection for any $i$.

###### Proof.

Without loss of generality we can assume that $A\subset B$ and that $\varphi$
is the inclusion map. By Lemmas 2.1 and 2.2 it suffices to show for any $i$
the following: If $A/\tilde{A}_{i}\to B/\tilde{B}_{i}$ is a bijection, then
the map $H_{1}(A;\mathbb{Z}[{B/\tilde{B}_{i}}])\to
H_{1}(B;\mathbb{Z}[{B/\tilde{B}_{i}}])$ is an isomorphism if and only if
$\varphi:A/[\tilde{A}_{i},\tilde{A}_{i}]\to B/[\tilde{B}_{i},\tilde{B}_{i}]$
is a bijection.

Using the above and using Shapiro’s Lemma we can identify

$\begin{array}[]{rcl}H_{1}(A;\mathbb{Z}[B/\tilde{B}_{i}])=H_{1}(A;\mathbb{Z}[A/\tilde{A}_{i}])&=&\tilde{A}_{i}/[\tilde{A}_{i},\tilde{A}_{i}]\hskip
28.45274pt\mbox{and}\\\
H_{1}(B;\mathbb{Z}[{B/\tilde{B}_{i}}])&=&\tilde{B}_{i}/[\tilde{B}_{i},\tilde{B}_{i}].\end{array}$

Note that $A/\tilde{A}_{i},B/\tilde{B}_{i},A/[\tilde{A}_{i},\tilde{A}_{i}]$
and $B/[\tilde{B}_{i},\tilde{B}_{i}]$ are in general not groups, but we can
view them as pointed sets. We now consider the following commutative diagram
of exact sequences of pointed sets:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{1}(A;\mathbb{Z}[{B/\tilde{B}_{i}}])\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{{A}/[\tilde{A}_{i},\tilde{A}_{i}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{{A/\tilde{A}_{i}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{1}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{1}(B;\mathbb{Z}[{B/\tilde{B}_{i}}])\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{B}/[\tilde{B}_{i},\tilde{B}_{i}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{B/\tilde{B}_{i}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1.}$

Recall that the map on the right is a bijection. It now follows from the
5–lemma for exact sequences of pointed sets that the middle map is a bijection
if and only if the left hand map is a bijection. ∎

We will several times make use of the following corollary which can be deduced
from Lemma 2.4 the same way as Corollary 2.3 is deduced from Lemma 2.2.

######  Corollary 2.5.

Let $\varphi:A\to B$ be a monomorphism of finitely generated groups, and
assume we are given a homomorphism $\beta:B\to G$ to a finite group $G$. Then

$\varphi_{*}:H_{i}(A;\mathbb{Z}[G])\to H_{i}(B;\mathbb{Z}[G]),\ \ i=0,1$

is an isomorphism if and only if the following two conditions hold:

  1. (1)

$\mbox{Im}\\{A\to B\to G\\}=\mbox{Im}\\{B\to G\\}$,

  2. (2)

$\varphi$ induces an isomorphism

$A/[\mbox{Ker}(\beta\circ\varphi),\mbox{Ker}(\beta\circ\varphi)]\to
B/[\mbox{Ker}(\beta),\mbox{Ker}(\beta)].$

Under extra conditions we can also give a criterion for a map between groups
to induce an isomorphism of second homology groups.

######  Lemma 2.6.

Let $\varphi:A\to B$ be a homomorphism between two groups such that $X=K(A,1)$
and $Y=K(B,1)$ are finite 2–complexes with vanishing Euler characteristic. Let
$\beta:B\to G$ be a homomorphism to a finite group such that

$\varphi_{*}:H_{i}(A;\mathbb{Z}[G])\to H_{i}(B;\mathbb{Z}[G]),\ \ i=0,1$

is an isomorphism, then

$\varphi_{*}:H_{2}(A;\mathbb{Z}[G])\to H_{2}(B;\mathbb{Z}[G])$

is also an isomorphism.

###### Proof.

We can and will view $X$ as a subcomplex of $Y$. It suffices to show that
$H_{2}(Y,X;\mathbb{Z}[G])=0$. Note that our assumption implies that
$H_{i}(Y,X;\mathbb{Z}[G])=0$ for $i=0,1$. Now note that
$H_{2}(Y,X;\mathbb{Z}[G])$ is a submodule of $C_{2}(Y,X;\mathbb{Z}[G])$, in
particular $H_{2}(Y,X;\mathbb{Z}[G])$ is a free $\mathbb{Z}$–module. We
therefore only have to show that $\mbox{rank}H_{2}(Y,X;\mathbb{Z}[G])=0$. Now
note that

$\begin{array}[]{rcl}\mbox{rank}H_{2}(Y,X;\mathbb{Z}[G])&=&\mbox{rank}H_{2}(Y,X;\mathbb{Z}[G])-\mbox{rank}H_{1}(Y,X;\mathbb{Z}[G])+\mbox{rank}H_{0}(Y,X;\mathbb{Z}[G])\\\
&=&|G|\chi(Y,X)\\\ &=&|G|(\chi(Y)-\chi(X))\\\ &=&0.\end{array}$

∎

We conclude this section with the following lemma.

######  Lemma 2.7.

Let $\varphi:A\to B$ be a homomorphism. Let $\hat{B}\subset\tilde{B}\subset B$
be two subgroups. Suppose that $\hat{B}\subset B$ is normal. We write
$\hat{A}:=\varphi^{-1}(\hat{B})$ and $\tilde{A}:=\varphi^{-1}(\tilde{B})$.
Assume that

$\varphi:A/\hat{A}\to B/\hat{B}\mbox{ and }\varphi:A/[\hat{A},\hat{A}]\to
B/[\hat{B},\hat{B}]$

are bijections, then

$\varphi:A/\tilde{A}\to B/\tilde{B}\mbox{ and
}\varphi:A/[\tilde{A},\tilde{A}]\to B/[\tilde{B},\tilde{B}]$

are also bijections.

###### Proof.

In the following let $n=0$ or $n=1$. Suppose that $\varphi:A/\hat{A}^{(n)}\to
B/\hat{B}^{(n)}$ is a bijection. Note that $\hat{A}^{(n)}\subset A$ and
$\hat{B}^{(n)}\subset B$ are normal, in particular $\varphi:A/\hat{A}^{(n)}\to
B/\hat{B}^{(n)}$ is in fact an isomorphism. We have to show that
$\varphi:A/\tilde{A}^{(n)}\to B/\tilde{B}^{(n)}$ is a bijection.

######  Claim.

The map $\varphi$ induces a bijection
$\tilde{A}^{(n)}/\hat{A}^{(n)}\to\tilde{B}^{(n)}/\hat{B}^{(n)}$.

We write $\overline{A}:=A/\hat{A}^{(n)}$, $\overline{B}:=B/\hat{B}^{(n)}$ and
we denote by $\overline{\varphi}:\overline{A}\to\overline{B}$ the induced map
which by assumption is an isomorphism. We denote by $\overline{H}$ the
subgroup $\tilde{B}/\hat{B}^{(n)}\subset\overline{B}$. Note that
$\overline{\varphi}$ restricts to isomorphisms
$\overline{\varphi}^{-1}(\overline{H})\to\overline{H}$ and
$\overline{\varphi}^{-1}(\overline{H}^{(n)})\to\overline{H}^{(n)}$. Since
$\overline{\varphi}^{-1}$ is an isomorphism it follows that
$\left(\overline{\varphi}^{-1}(\overline{H})\right)^{(n)}=\overline{\varphi}^{-1}\big{(}\overline{H}^{(n)}\big{)}$.
Now recall that $\overline{H}=\tilde{B}/\hat{B}^{(n)}$, hence
$\overline{H}^{(n)}=\tilde{B}^{(n)}/\hat{B}^{(n)}$. We clearly have
$\overline{\varphi}^{-1}(\overline{H})=\tilde{A}/\hat{A}^{(n)}$ and therefore
$\overline{\varphi}^{-1}(\overline{H})^{(n)}=\tilde{A}^{(n)}/\hat{A}^{(n)}$.
This shows that the isomorphism
$\overline{\varphi}:\overline{\varphi}^{-1}\big{(}\overline{H}^{(n)}\big{)}\to\overline{H}^{(n)}$
is precisely the desired isomorphism
$\tilde{A}^{(n)}/\hat{A}^{(n)}\to\tilde{B}^{(n)}/\hat{B}^{(n)}$. This
concludes the proof of the claim.

Now consider the following commutative diagram of short exact sequences of
pointed sets:

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\tilde{A}^{(n)}/\hat{A}^{(n)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{A/\hat{A}^{(n)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{A/\tilde{A}^{(n)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{1}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\tilde{B}^{(n)}/\hat{B}^{(n)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B/\hat{B}^{(n)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B/\tilde{B}^{(n)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1.}$

The middle vertical map is a bijection by assumption and we just verified that
the vertical map on the left is a bijection. It now follows from the 5–Lemma
for exact sequences of pointed sets that the vertical map on the right is also
a bijection. ∎

###  2.3. Twisted Alexander polynomials

In this section we are going to recall the definition of twisted Alexander
polynomials. These were introduced, for the case of knots, by Xiao-Song Lin in
1990 (published in [Li01]), and his definition was later generalized to
$3$–manifolds by Wada [Wa94], Kirk–Livingston [KL99] and Cha [Ch03].

Let $N$ be a compact manifold. Let $R$ be a commutative, Noetherian unique
factorization domain (in our applications $R=\mathbb{Z}$ or
$R=\mathbb{F}_{p}$, the finite field with $p$ elements) and $V$ a finite free
$R$–module Let $\alpha:\pi_{1}(N)\to\mbox{Aut}_{R}(V)$ a representation and
let $\phi\in H^{1}(N;\mathbb{Z})=\mbox{Hom}(\pi_{1}(N),\mathbb{Z})$ a
nontrivial element. We write $V\otimes_{R}R[t^{\pm 1}]=:V[t^{\pm 1}]$. Then
$\alpha$ and $\phi$ give rise to a representation
$\alpha\otimes\phi:\pi_{1}(N)\to\mbox{Aut}_{R[t^{\pm 1}]}(V[t^{\pm 1}])$ as
follows:

$\big{(}(\alpha\otimes\phi)(g)\big{)}(v\otimes p):=(\alpha(g)\cdot
v)\otimes(\phi(g)\cdot p)=(\alpha(g)\cdot v)\otimes(t^{\phi(g)}p),$

where $g\in\pi_{1}(N),v\otimes p\in V\otimes_{R}R[t^{\pm 1}]=V[t^{\pm 1}]$.

Note that $N$ is homotopy equivalent to a finite CW–complex, which, by abuse
of notation, we also denote by $N$. Then we consider
$C_{*}(\tilde{N})\otimes_{\mathbb{Z}[\pi_{1}(N)]}V[t^{\pm 1}]$ which is a
complex of finitely generated $R[t^{\pm 1}]$–modules. Since $R[t^{\pm 1}]$ is
Noetherian it follows that for any $i$ the $R[t^{\pm 1}]$–module
$H_{i}(N;V[t^{\pm 1}])$ is a finitely presented $R[t^{\pm 1}]$–module. This
means $H_{i}(N;V[t^{\pm 1}])$ has a free $R[t^{\pm 1}]$–resolution

$R[t^{\pm 1}]^{r_{i}}\xrightarrow{S_{i}}R[t^{\pm 1}]^{s_{i}}\to
H_{i}(N;V[t^{\pm 1}])\to 0.$

Without loss of generality we can assume that $r_{i}\geq s_{i}$.

######  Definition.

The _$i$ –th twisted Alexander polynomial_ of $(N,\alpha,\phi)$ is defined to
be the order of the $R[t^{\pm 1}]$–module $H_{i}(N;V[t^{\pm 1}])$, i.e. the
greatest common divisor (which exists since $R[t^{\pm 1}]$ is a UFD as well)
of the $s_{i}\times s_{i}$ minors of the $s_{i}\times r_{i}$–matrix $S_{i}$.
It is denoted by $\Delta_{N,\phi,i}^{\alpha}\in R[t^{\pm 1}]$.

Note that $\Delta_{N,\phi,i}^{\alpha}\in R[t^{\pm 1}]$ is well–defined up to a
unit in $R[t^{\pm 1}]$, i.e. up to an element of the form $rt^{i}$ where $r$
is a unit in $R$ and $i\in\mathbb{Z}$. We say that $f\in R[t^{\pm 1}]$ is
monic if its top coefficient is a unit in $R$. Given a nontrivial
$f=\sum_{i=r}^{s}a_{s}t^{i}$ with $a_{r}\neq 0,a_{s}\neq 0$ we write
$\mbox{deg}f=s-r$. For $f=0$ we write $\mbox{deg}(f)=-\infty$. Note that
$\mbox{deg}\Delta_{N,\phi,i}^{\alpha}$ is well–defined.

We now write $\pi=\pi_{1}(N)$. If we are given a homomorphism $\alpha:\pi\to
G$ to a finite group, then this gives rise to a finite dimensional
representation of $\pi$, that we will denote by
$\alpha:\pi\to\mbox{Aut}_{R}(R[G])$ as well. In the case that we have a finite
index subgroup $\tilde{\pi}\subset\pi$ we get a finite dimensional
representation $\pi\to\mbox{Aut}_{R}(R[\pi/\tilde{\pi}])$ given by
left–multiplication. When $R=\mathbb{Z}$, the resulting twisted Alexander
polynomials will be denoted by
$\Delta_{N,\phi,i}^{\pi/\tilde{\pi}}\in\mathbb{Z}[t^{\pm 1}]$, while for
$R=\mathbb{F}_{p}$ we will use the notation
$\Delta_{N,\phi,i}^{\pi/\tilde{\pi},p}\in\mathbb{F}_{p}[t^{\pm 1}]$. See
[FV08a] for the relation between these polynomials.

Finally, in the case that $\alpha:\pi\to\mbox{GL}(1,\mathbb{Z})$ is the
trivial representation we drop the $\alpha$ from the notation, and in the case
that $i=1$ we drop the subscript “$,1$” from the notation.

We summarize some of the main properties of twisted Alexander polynomials in
the following lemma. It is a consequence of [FV08a, Lemma 3.3 and 3.4] and
[FK06, Proposition 2.5].

######  Lemma 2.8.

Let $N$ be a 3–manifold with empty or toroidal boundary. Let $\phi\in
H^{1}(N;\mathbb{Z})$ nontrivial and $\tilde{\pi}\subset\pi:=\pi_{1}(N)$ a
finite index subgroup. Denote by $\phi_{\tilde{\pi}}$ the restriction of
$\phi$ to $\tilde{\pi}$, then the following hold:

  1. (1)

$\Delta_{N,\phi,0}^{\pi/\tilde{\pi}}=(1-t^{div\,{\phi}_{\tilde{\pi}}})$,

  2. (2)

if $\Delta_{N,\phi,1}^{\pi/\tilde{\pi}}\neq 0$, then
$\Delta_{N,\phi,2}^{\pi/\tilde{\pi}}=(1-t^{div\,\phi_{\tilde{\pi}}})^{b_{3}(N)}$,

  3. (3)

$\Delta_{N,\phi,i}^{\pi/\tilde{\pi}}=1$ for any $i\geq 3$.

Assume we also have a subgroup $\pi^{\prime}$ with
$\tilde{\pi}\subset\pi^{\prime}\subset\pi$. Denote the covering of $N$
corresponding to $\pi^{\prime}$ by $N^{\prime}$ and denote by $\phi^{\prime}$
the restriction of $\phi$ to $\pi^{\prime}$, then

$\Delta_{N,\phi,i}^{\pi/\tilde{\pi}}=\Delta_{N^{\prime},\phi^{\prime},i}^{\pi^{\prime}/\tilde{\pi}}$

for any $i$. Finally note that the statements of the lemma also hold for the
polynomial $\Delta_{N,\phi,i}^{\pi/\tilde{\pi},p}\in\mathbb{F}_{p}[t^{\pm
1}]$.

We also recall the following well–known result (cf. e.g. [Tu01]).

######  Lemma 2.9.

Let $N$ be a 3–manifold with empty or toroidal boundary. Let $\phi\in
H^{1}(N;\mathbb{Z})$ nontrivial and $\tilde{\pi}\subset\pi:=\pi_{1}(N)$ a
finite index subgroup. Then given $i$ the following are equivalent:

  1. (1)

$\Delta_{N,\phi,i}^{\pi/\tilde{\pi}}\neq 0$,

  2. (2)

$H_{i}(N;\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}])$ is $\mathbb{Z}[t^{\pm
1}]$–torsion,

  3. (3)

$H_{i}(N;\mathbb{Q}[\pi/\tilde{\pi}][t^{\pm 1}])$ is $\mathbb{Q}[t^{\pm
1}]$–torsion,

  4. (4)

the rank of the abelian group $H_{i}(N;\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm
1}])$ is finite.

In fact if any of the four conditions holds, then

$\mbox{deg}\Delta_{N,\phi,i}^{\pi/\tilde{\pi}}=\mbox{rank}\,H_{i}(N;\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm
1}])=\mbox{dim}\,H_{i}(N;\mathbb{Q}[\pi/\tilde{\pi}][t^{\pm 1}]).$

###  2.4. Completions of groups

Throughout the paper it is convenient to use the language of completions of
groups. Although the proof of Theorem 1.2 does not explicitly require this
terminology, group completions are the natural framework for these results. We
recall here the definitions and some basic facts, we refer to [LS03, Window 4]
and [Wi98, RZ00] for proofs and for more information.

Let $\mathcal{C}$ be a variety of groups (cf. [RZ00, p. 20] for the
definition). Examples of varieties of pertinence to this paper are given by
any one of the following:

  1. (1)

finite groups;

  2. (2)

$p$–groups for a prime $p$;

  3. (3)

the variety $\mathcal{FS}(n)$ of finite solvable groups of derived length at
most $n$;

  4. (4)

the variety $\mathcal{FS}$ of finite solvable groups.

In the following we equip a finitely generated group $A$ with its _pro–
$\mathcal{C}$ topology_, this topology is the translation invariant topology
uniquely defined by taking as a fundamental system of neighborhoods of the
identity the collection of all normal subgroups of $A$ such that the quotient
lies in $\mathcal{C}$. Note that in particular all groups in $\mathcal{C}$ are
endowed with the discrete topology.

Given a group $A$ denote by $\hat{A}_{\mathcal{C}}$ its pro–$\mathcal{C}$
completion, i.e. the inverse limit

$\hat{A}_{\mathcal{C}}=\underleftarrow{\lim}\,A\,/A_{i}$

where $A_{i}$ runs through the inverse system determined by the collection of
all normal subgroups of $A$ such that $A/A_{i}\in\mathcal{C}$. Then
$\hat{A}_{\mathcal{C}}$, which we can view as a subgroup of the direct product
of all $A/A_{i}$, inherits a natural topology. Henceforth by homomorphisms
between groups we will mean a homomorphism which is continuous with respect to
the above topologies. Using the standard convention we refer to the
pro–$\mathcal{FS}$ completion of a group as the prosolvable completion.

Note that by the assumption that $\mathcal{C}$ is a variety, the
pro–$\mathcal{C}$ completion is a covariant functor, i.e. given $\varphi:A\to
B$ we get an induced homomorphism
$\hat{\varphi}:\hat{A}_{\mathcal{C}}\to\hat{B}_{\mathcal{C}}$.

A group $A$ is called residually $\mathcal{C}$ if for any nontrivial $g\in A$
there exists a homomorphism $\alpha:A\to G$ where $G\in\mathcal{C}$ such that
$\alpha(g)\neq e$. It is easily seen that $A$ is residually $\mathcal{C}$ if
and only if the map $A\to\hat{A}_{\mathcal{C}}$ is injective. In particular,
if we are given a homomorphism $\varphi:A\to B$ between residually
${\mathcal{C}}$ groups $A,B$ such that
$\hat{\varphi}:\hat{A}_{\mathcal{C}}\to\hat{B}_{\mathcal{C}}$ is an injection,
then it follows from the following commutative diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\hat{A}_{\mathcal{C}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\hat{\varphi}}$$\textstyle{\hat{B}_{\mathcal{C}}}$

that $\varphi$ is injective as well.

The following well–known lemma gives sufficient and necessary conditions for a
homomorphism $\varphi:A\to B$ to induce an isomorphism of pro–$\mathcal{C}$
completions.

######  Lemma 2.10.

Let $\mathcal{C}$ be a variety of groups and assume that there is a
homomorphism $\varphi:A\to B$. Then the following are equivalent:

  1. (1)

$\hat{\varphi}:\hat{A}_{\mathcal{C}}\to\hat{B}_{\mathcal{C}}$ is an
isomorphism,

  2. (2)

for any $G\in\mathcal{C}$ the induced map

$\varphi^{*}:\mbox{Hom}(B,G)\to\mbox{Hom}(A,G)$

is a bijection.

We also note the following well–known lemma.

######  Lemma 2.11.

Let $\mathcal{C}$ be an extension–closed variety and let $\varphi:A\to B$ a
homomorphism of finitely generated groups which induces an isomorphism of
pro–$\mathcal{C}$ completions. Then for any homomorphism $\beta:B\to G$ to a
$\mathcal{C}$–group the restriction of $\varphi$ to
$\mbox{Ker}(\beta\circ\varphi)\to\mbox{Ker}(\beta)$ induces an isomorphism of
pro–$\mathcal{C}$ completions.

When a homomorphism $\varphi:A\to B$ of finitely generated groups induces an
isomorphism of their pro–$\mathcal{C}$ completions, then we have a relation of
the twisted homology with coefficients determined by ${\mathcal{C}}$–groups.
More precisely, we have the following.

######  Lemma 2.12.

Let $\mathcal{C}$ be a variety of groups and let $\varphi:A\to B$ be a
homomorphism of finitely generated groups which induces an isomorphism of
pro–$\mathcal{C}$ completions. Then for any homomorphism $\beta:B\to G$ to a
$\mathcal{C}$–group the map $\varphi_{*}:H_{0}(A;\mathbb{Z}[G])\to
H_{0}(B;\mathbb{Z}[G])$ is an isomorphism. Furthermore, if $\mathcal{C}$ is an
extension–closed variety containing all finite abelian groups, the map
$\varphi_{*}:H_{1}(A;\mathbb{Z}[G])\to H_{1}(B;\mathbb{Z}[G])$ is an
isomorphism.

###### Proof.

Observe that, by Corollary 2.3, the first part of the statement is equivalent
to the assertion that, for any element $\beta\in\mbox{Hom}(B,G)$,

$\mbox{Im}\\{\beta\circ\varphi:A\to G\\}=\mbox{Im}\\{\beta:B\to G\\}.$

Without loss of generality, we can reduce the proof of this isomorphism to the
case where $\beta$ is surjective. Denote
$\alpha=\beta\circ\varphi\in\mbox{Hom}(A,G)$. Assume to the contrary that
$\alpha(A)\subsetneq G$; then
$\alpha\in\mbox{Hom}(A,\alpha(A))\subset\mbox{Hom}(A,G)$ and as
$\alpha(A)\in\mathcal{C}$ there exists by hypothesis a map
$\beta^{\prime}\in\mbox{Hom}(B,\alpha(A))\subset\mbox{Hom}(B,G)$ such that
$\alpha=\beta^{\prime}\circ\varphi$. Now the two maps
$\beta,\beta^{\prime}\in\mbox{Hom}(B,G)$ (that must differ as they have
different image) induce the same map $\alpha\in\mbox{Hom}(A,G)$, contradicting
the bijectivity of $\mbox{Hom}(B,G)$ and $\mbox{Hom}(A,G)$.

We now turn to the proof of the second part of the statement. Let $\beta:B\to
G$ a homomorphism to a $\mathcal{C}$–group. Again, without loss of generality,
we can assume that $\beta:B\to G$ is surjective. Note that by the above the
homomorphism $\beta\circ\varphi:A\to G$ is surjective as well. We now write
$B^{\prime}=\mbox{Ker}(\beta)$ and $A^{\prime}=\mbox{Ker}(\beta\circ\varphi)$.
By Shapiro’s Lemma, we have the commutative diagram

$\textstyle{H_{1}(A^{\prime};\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{H_{1}(A;\mathbb{Z}[G])\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{1}(B^{\prime};\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{H_{1}(B;\mathbb{Z}[G]).}$

The claim amounts therefore to showing that the map
$\varphi_{*}:H_{1}(A^{\prime};\mathbb{Z})\to H_{1}(B^{\prime};\mathbb{Z})$ is
an isomorphism. As $A$ and $B$ are finitely generated, $A^{\prime}$ and
$B^{\prime}$ are finitely generated as well. When $\mathcal{C}$ is extension
closed, and contains all finite abelian groups, Lemma 2.11 asserts that the
map $\varphi$ induces a bijection between $\mbox{Hom}(B^{\prime},\Gamma)$ and
$\mbox{Hom}(A^{\prime},\Gamma)$ for any finite abelian group $\Gamma$; the
desired isomorphism easily follows.∎

##  3\. Monic twisted Alexander polynomials and solvable groups

The aim of this section is to prove Proposition 1.7.

###  3.1. Preliminary results

We will often make use of the following proposition (cf. [McM02, Section 4 and
Proposition 6.1]).

######  Proposition 3.1.

Let $N$ be a $3$–manifold with empty or toroidal boundary and let $\phi\in
H^{1}(N;\mathbb{Z})$ a primitive class. If $\Delta_{N,\phi}\neq 0$, then there
exists a connected Thurston norm minimizing surface $\Sigma$ dual to $\phi$.

Given a connected oriented surface $\Sigma\subset N$ we will adopt the
following conventions for the rest of the paper. We choose a neighborhood
$\Sigma\times[-1,1]\subset N$ and write $\nu\Sigma=\Sigma\times(-1,1)$. Let
$M:=N\setminus\nu\Sigma$; we will write $\Sigma^{\pm}=\Sigma\times\\{\pm
1\\}\subset\partial M$, and we will denote the inclusion induced maps
$\Sigma\to\Sigma^{\pm}\subset M$ by $\iota_{\pm}$.

We pick a base point in $M$ and endow $N$ with the same base point. Also, we
pick a base point for $\Sigma$ and endow $\Sigma^{\pm}$ with the corresponding
base points. With these choices made, we will write $A=\pi_{1}(\Sigma)$ and
$B=\pi_{1}(M)$. We also pick paths in $M$ connecting the base point of $M$
with the base points of $\Sigma^{-}$ and $\Sigma^{+}$. We now have inclusion
induced maps $\iota_{\pm}:A\to B$ for either inclusion of $\Sigma$ in $M$ and,
using the constant path, a map $\pi_{1}(M)\to\pi_{1}(N)$. Under the assumption
that $\Sigma$ is incompressible (in particular, whenever $\Sigma$ is Thurston
norm minimizing) these maps are injective. Since $M$ and $N$ have the same
base point we can view $B$ canonically as a subgroup of $\pi_{1}(N)$.

Before we state the first proposition we have to introduce a few more
definitions. Let $N$ be a $3$–manifold with empty or toroidal boundary and let
$\phi\in H^{1}(N;\mathbb{Z})$ a nontrivial class. Let $\tilde{\pi}\subset\pi$
be a finite index subgroup. We denote by $\phi_{\tilde{\pi}}$ the restriction
of $\phi\in H^{1}(N;\mathbb{Z})=\mbox{Hom}(\pi,\mathbb{Z})$ to $\tilde{\pi}$.
Note that $\phi_{\tilde{\pi}}$ is necessarily non–trivial. We say that
$\tilde{\pi}\subset\pi$ has Property (M) if the twisted Alexander polynomial
$\Delta_{N,\phi}^{\pi/\tilde{\pi}}\in\mathbb{Z}[t^{\pm 1}]$ is monic and if

$\mbox{deg}(\Delta_{N,\phi}^{\pi/\tilde{\pi}})=[\pi:\tilde{\pi}]\,\|\phi\|_{T}+(1+b_{3}(N))\mbox{div}\phi_{\tilde{\pi}}$

holds. Note that a pair $(N,\phi)$ satisfies Condition ($*$) if and only if
Property (M) is satisfied by all normal subgroups of $\pi_{1}(N)$.

The following proposition is the key tool for translating information on
twisted Alexander polynomials into information on the maps $\iota_{\pm}:A\to
B$. The proposition is well known in the classical case. In the case of normal
subgroups a proof for the ‘only if’ direction of the proposition is given by
combining [FV08a, Section 8] with [FV08b, Section 4].

######  Proposition 3.2.

Let $N$ be a 3–manifold with empty or toroidal boundary with $N\neq
S^{1}\times D^{2},N\neq S^{1}\times S^{2}$. Let $\phi\in H^{1}(N;\mathbb{Z})$
a primitive class which is dual to a connected Thurston norm minimizing
surface $\Sigma$. Let $\tilde{\pi}\subset\pi$ be a finite index subgroup. Then
$\tilde{\pi}$ has Property (M) if and only if the maps
$\iota_{\pm}:H_{i}(A;\mathbb{Z}[{\pi/\tilde{\pi}}])\to
H_{i}(B;\mathbb{Z}[{\pi/\tilde{\pi}}])$ are isomorphisms for $i=0,1$.

###### Proof.

Let $R=\mathbb{Z}$ or $R=\mathbb{F}_{p}$ with $p$ a prime. We have canonical
isomorphisms $H_{i}(\Sigma;R[\pi/\tilde{\pi}])\cong
H_{i}(A;R[\pi/\tilde{\pi}])$ and $H_{i}(M;R[\pi/\tilde{\pi}])\cong
H_{i}(B;R[\pi/\tilde{\pi}])$ for $i=0,1$. It follows from [FK06, Proposition
3.2] that splitting $N$ along $\Sigma$ gives rise to the following
Mayer–Vietoris type exact sequence

$\begin{array}[]{cccccccccccccc}&&&\dots&\to&H_{2}(N;R[\pi/\tilde{\pi}][t^{\pm
1}])\\\\[2.84544pt] \to&H_{1}(A;R[\pi/\tilde{\pi}])\otimes R[t^{\pm
1}]&\xrightarrow{t\iota_{+}-\iota_{-}}&H_{1}(B;R[\pi/\tilde{\pi}])\otimes
R[t^{\pm 1}]&\to&H_{1}(N;R[\pi/\tilde{\pi}][t^{\pm 1}])&\to&\\\\[2.84544pt]
\to&H_{0}(A;R[\pi/\tilde{\pi}])\otimes R[t^{\pm
1}]&\xrightarrow{t\iota_{+}-\iota_{-}}&H_{0}(B;R[\pi/\tilde{\pi}])\otimes
R[t^{\pm 1}]&\to&H_{0}(N;R[\pi/\tilde{\pi}][t^{\pm 1}])&\to&0.\end{array}$

which we refer to as the Mayer–Vietoris sequence of $(N,\Sigma)$ with
$R[\pi/\tilde{\pi}][t^{\pm 1}]$–coefficients. First note that by Shapiro’s
lemma the groups $H_{i}(A;R[\pi/\tilde{\pi}])$ are the $i$–th homology with
$R$–coefficients of a (possibly) disconnected surface. It follows that
$H_{i}(A;R[\pi/\tilde{\pi}])$ is a free $R$–module, in particular the
$R[t^{\pm 1}]$–modules $H_{i}(A;R[\pi/\tilde{\pi}])\otimes R[t^{\pm 1}]$ are
free $R[t^{\pm 1}]$–modules. We will several times make use of the observation
that if $H_{i}(N;R[\pi/\tilde{\pi}][t^{\pm 1}])$ is $R[t^{\pm 1}]$–torsion,
then the map $H_{i}(N;R[\pi/\tilde{\pi}][t^{\pm 1}])\to
H_{i-1}(A;R[\pi/\tilde{\pi}])\otimes R[t^{\pm 1}]$ is necessarily zero.

We first assume that $\tilde{\pi}$ has Property (M). Since
$\Delta_{N,\phi}^{\pi/\tilde{\pi}}\neq 0$ we have that the module
$H_{1}(N;\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}])\otimes_{\mathbb{Z}[t^{\pm
1}]}\mathbb{Q}(t)$ is trivial. Note that by Lemma 2.9 we have that
$H_{0}(N;\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}])$ is also $\mathbb{Z}[t^{\pm
1}]$–torsion. We now consider the Mayer–Vietoris sequence of $(N,\Sigma)$ with
$\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}]$–coefficients. Tensoring the exact
sequence with $\mathbb{Q}(t)$ we see that

$\begin{array}[]{rcl}\mbox{rank}_{\mathbb{Z}}(H_{0}(A;\mathbb{Z}[\pi/\tilde{\pi}]))&=&\mbox{rank}_{\mathbb{Q}(t)}(H_{0}(A;\mathbb{Z}[\pi/\tilde{\pi}])\otimes_{\mathbb{Z}}\mathbb{Q}(t))\\\
&=&\mbox{rank}_{\mathbb{Q}(t)}(H_{0}(B;\mathbb{Z}[\pi/\tilde{\pi}])\otimes_{\mathbb{Z}}\mathbb{Q}(t))=\mbox{rank}_{\mathbb{Z}}(H_{0}(B;\mathbb{Z}[\pi/\tilde{\pi}])).\end{array}$

Using this observation and using Lemma 2.2 we see that the maps
$\iota_{\pm}:H_{0}(A;\mathbb{Z}[\pi/\tilde{\pi}])\to
H_{0}(B;\mathbb{Z}[\pi/\tilde{\pi}])$ are epimorphism between free abelian
groups of the same rank. Hence the maps are in fact isomorphisms.

In order to prove that the maps
$\iota_{\pm}:H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])\to
H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])$ are isomorphisms we first consider the
following claim.

######  Claim.

$H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])$ and
$H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])$ are free abelian groups of the same
rank.

Let $p$ be a prime. We consider the Mayer–Vietoris sequence of $(N,\Sigma)$
with $\mathbb{F}_{p}[\pi/\tilde{\pi}][t^{\pm 1}]$–coefficients. Denote by
$\Delta_{N,\phi}^{{\pi/\tilde{\pi},p}}\in\mathbb{F}_{p}[t^{\pm 1}]$ the
twisted Alexander polynomial with coefficients in $\mathbb{F}_{p}$. It follows
from $\Delta^{\pi/\tilde{\pi}}_{N,\phi}$ monic and from [FV08a, Proposition
6.1] that $\Delta_{N,\phi}^{{\pi/\tilde{\pi},p}}\neq 0\in\mathbb{F}_{p}[t^{\pm
1}]$. Furthermore by Lemma 2.8 we have that
$\Delta_{N,\phi,2}^{{\pi/\tilde{\pi},p}}\neq 0\in\mathbb{F}_{p}[t^{\pm 1}]$.
In particular $H_{i}(N;\mathbb{F}_{p}[\pi/\tilde{\pi}][t^{\pm 1}])$ is
$\mathbb{F}_{p}[t^{\pm 1}]$–torsion for $i=1,2$. It follows from the fact that
$H_{i}(A;\mathbb{F}_{p}[\pi/\tilde{\pi}])\otimes_{\mathbb{F}_{p}}\mathbb{F}_{p}[t^{\pm
1}]$ is a free $\mathbb{F}_{p}[t^{\pm 1}]$–module and the above observation
that $H_{i}(N;\mathbb{F}_{p}[\pi/\tilde{\pi}][t^{\pm 1}])$ is
$\mathbb{F}_{p}[t^{\pm 1}]$–torsion for $i=1,2$ that the Mayer–Vietoris
sequence gives rise to the following short exact sequence

$0\to H_{1}(A;\mathbb{F}_{p}[\pi/\tilde{\pi}])\otimes\mathbb{F}_{p}[t^{\pm
1}]\xrightarrow{t\iota_{+}-\iota_{-}}H_{1}(B;\mathbb{F}_{p}[\pi/\tilde{\pi}])\otimes\mathbb{F}_{p}[t^{\pm
1}]\to H_{1}(N;\mathbb{F}_{p}[\pi/\tilde{\pi}][t^{\pm 1}])\to 0.$

Tensoring with $\mathbb{F}_{p}(t)$ we see that in particular
$H_{1}(A;\mathbb{F}_{p}[\pi/\tilde{\pi}])\cong
H_{1}(B;\mathbb{F}_{p}[\pi/\tilde{\pi}])$ as $\mathbb{F}_{p}$–vector spaces.
The homology group $H_{0}(A;\mathbb{Z}[\pi/\tilde{\pi}])$ is
$\mathbb{Z}$–torsion free. It follows from the universal coefficient theorem
applied to the complex of $\mathbb{Z}$–modules
$C_{*}(\tilde{\Sigma})\otimes_{\mathbb{Z}[A]}{\mathbb{Z}[\pi/\tilde{\pi}]}$
that

$H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])\otimes_{\mathbb{Z}}\mathbb{F}_{p}\cong
H_{1}(A;\mathbb{F}_{p}[\pi/\tilde{\pi}])$

for every prime $p$. The same statement holds for $B$. Combining our results
we see that
$H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])\otimes_{\mathbb{Z}}\mathbb{F}_{p}$ and
$H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])\otimes_{\mathbb{Z}}\mathbb{F}_{p}$ are
isomorphic for any prime $p$. Since $H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])$ is
free abelian it follows that $H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])\cong
H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])$. This completes the proof of the claim.

In the following we equip the free $\mathbb{Z}$–modules
$H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])$ and
$H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])$ with a choice of basis. We now study
the Mayer–Vietoris sequence for $(N,\Sigma)$ with
$\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}]$–coefficients. Using an argument
similar to the above we see that it gives rise to the following exact sequence

$H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])\otimes\mathbb{Z}[t^{\pm
1}]\xrightarrow{t\iota_{+}-\iota_{-}}H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])\otimes\mathbb{Z}[t^{\pm
1}]\to H_{1}(N;\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}])\to 0.$

Since $H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])$ and
$H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])$ are free abelian groups of the same
rank it follows that the above exact sequence is a resolution of
$H_{1}(N;\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}])$ by free $\mathbb{Z}[t^{\pm
1}]$–modules and that
$\Delta^{\pi/\tilde{\pi}}_{N,\phi}=\det(t\iota_{+}-\iota_{-})$. Recall that
Property (M) states in particular that

(1)
$\mbox{deg}\Delta_{N,\phi}^{{\pi/\tilde{\pi}}}=|\pi/\tilde{\pi}|\,\|\phi\|_{T}+(1+b_{3}(N))\mbox{div}\phi_{\tilde{\pi}}.$

Recall that we assumed that $N\neq S^{1}\times D^{2}$ and $N\neq S^{1}\times
S^{2}$, in particular $\chi(\Sigma)\leq 0$ and therefore
$-\chi(\Sigma)=||\phi||_{T}$. Writing
$b_{i}=\mbox{rank}_{\mathbb{Z}}(H_{i}(\Sigma;\mathbb{Z}[\pi/\tilde{\pi}]))=\mbox{rank}_{\mathbb{Z}}(H_{i}(A;\mathbb{Z}[\pi/\tilde{\pi}]))$
a standard Euler characteristic argument now shows that

$-b_{0}+b_{1}-b_{2}=-|\pi/\tilde{\pi}|\chi(\Sigma)=|\pi/\tilde{\pi}|\cdot||\phi||_{T}.$

By [FK06, Lemma 2.2] we have
$b_{i}=\mbox{deg}\Delta^{{\pi/\tilde{\pi}}}_{N,\phi,i}$ for $i=0$ and $i=2$.
We also have
$\mbox{deg}\Delta^{{\pi/\tilde{\pi}}}_{N,{\phi},0}=\mbox{div}\phi_{\pi/\tilde{\pi}}$
and
$\mbox{deg}\Delta^{{\pi/\tilde{\pi}}}_{N,{\phi},2}=b_{3}(N)\mbox{div}\phi_{\tilde{\pi}}$
by Lemma 2.8. Combining these facts with (1) we conclude that
$\mbox{deg}\Delta^{{\pi/\tilde{\pi}}}_{N,\phi}=b_{1}$. So we now have
$\mbox{deg}(\det(t\iota_{+}-\iota_{-}))=b_{1}$. Since $\iota_{+}$ and
$\iota_{-}$ are $b_{1}\times b_{1}$ matrices over $\mathbb{Z}$ it now follows
that $\det(\iota_{+})$ equals the top coefficient of
$\Delta^{\pi/\tilde{\pi}}_{N,\phi}$, which by Property (M) equals $\pm 1$. By
the symmetry of twisted Alexander polynomials we have that the bottom
coefficient of $\Delta^{\pi/\tilde{\pi}}_{N,\phi}$ also equals $\pm 1$, we
deduce that $\det(\iota_{-})=\pm 1$. This shows that
$\iota_{+},\iota_{-}:H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])\to
H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])$ are isomorphisms. We thus showed that if
$\tilde{\pi}$ has Property (M), then the maps
$\iota:H_{i}(A;\mathbb{Z}[{\pi/\tilde{\pi}}])\to
H_{i}(B;\mathbb{Z}[{\pi/\tilde{\pi}}])$ are isomorphisms for $i=0,1$.

Now assume that we are given a finite index subgroup $\tilde{\pi}\subset\pi$
such that the maps $\iota_{\pm}:H_{i}(A;\mathbb{Z}[{\pi/\tilde{\pi}}])\to
H_{i}(B;\mathbb{Z}[{\pi/\tilde{\pi}}])$ are isomorphisms for $i=0,1$. It
follows from the assumption that
$\iota_{\pm}:H_{0}(A;\mathbb{Z}[{\pi/\tilde{\pi}}])\to
H_{0}(B;\mathbb{Z}[{\pi/\tilde{\pi}}])$ are isomorphisms that the map

$H_{0}(A;\mathbb{Z}[\pi/\tilde{\pi}])\otimes\mathbb{Z}[t^{\pm
1}]\xrightarrow{t\iota_{+}-\iota_{-}}H_{0}(B;\mathbb{Z}[\pi/\tilde{\pi}])$

is injective. In particular the Mayer–Vietoris sequence of $(N,\Sigma)$ with
$\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}]$–coefficients gives rise to the
following exact sequence

$H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])\otimes\mathbb{Z}[t^{\pm
1}]\xrightarrow{t\iota_{+}-\iota_{-}}H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])\otimes\mathbb{Z}[t^{\pm
1}]\to H_{1}(N;\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}])\to 0.$

As above $H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])$ is a free abelian group and by
our assumption $H_{1}(B;\mathbb{Z}[\pi/\tilde{\pi}])\cong
H_{1}(A;\mathbb{Z}[\pi/\tilde{\pi}])$ is also free abelian. In particular the
above exact sequence defines a presentation for
$H_{1}(N;\mathbb{Z}[\pi/\tilde{\pi}][t^{\pm 1}])$ and we deduce that

$\Delta_{N,\phi}^{\pi/\tilde{\pi}}=\det(t\iota_{+}-\iota_{-}).$

Since $\iota_{-}$ and $\iota_{+}$ are isomorphisms it follows that
$\Delta_{N,\phi}^{\pi/\tilde{\pi}}$ is monic of degree $b_{1}$. An argument
similar to the above now shows that

$\mbox{deg}\Delta_{N,\phi}^{{\pi/\tilde{\pi}}}=|\pi/\tilde{\pi}|\,\|\phi\|_{T}+(1+b_{3}(N))\mbox{div}\phi_{\tilde{\pi}}.$

∎

###  3.2. Finite solvable quotients

Given a solvable group $S$ we denote by $\ell(S)$ its derived length, i.e. the
length of the shortest decomposition into abelian groups. Put differently,
$\ell(S)$ is the minimal number such that $S^{(\ell(S))}=\\{e\\}$. Note that
$\ell(S)=0$ if and only if $S=\\{e\\}$.

For sake of comprehension, we briefly recall the notation. We are considering
a $3$–manifold $N$ with empty or toroidal boundary, and we fix a primitive
class $\phi\in H^{1}(N;\mathbb{Z})$. We denote by $\Sigma$ a connected
Thurston norm minimizing surface dual to $\phi$, and write $A=\pi_{1}(\Sigma)$
and $B=\pi_{1}(M)$ (with $M=N\setminus\nu\Sigma$) and we denote the two
inclusion induced maps $A\to B$ with $\iota_{\pm}$. We also write
$\pi=\pi_{1}(N)$. Note that $\pi=\langle
B,t|\iota_{-}(A)=t\iota_{+}(A)t^{-1}\rangle$.

Given $n\in\mathbb{N}\cup\\{0\\}$ we denote by $\mathcal{S}(n)$ the statement
that for any finite solvable group $S$ with $\ell(S)\leq n$ we have that for
$\iota=\iota_{-},\iota_{+}$ the map

$\iota^{*}:\mbox{Hom}(B,S)\to\mbox{Hom}(A,S)$

is a bijection. This is equivalent by Lemma 2.10 to assert that $\iota:A\to B$
induces an isomorphism of pro–$\mathcal{FS}(n)$ completions. Recall that by
Corollary 2.3 and Lemma 2.12 statement $\mathcal{S}(n)$ implies then that for
any homomorphism $\beta:B\to S$ to a finite solvable group $S$ with
$\ell(S)\leq n$ we have $\mbox{Im}\\{\beta\circ\iota:A\to B\to
S\\}=\mbox{Im}\\{\beta:B\to S\\}$.

Our goal is to show that $\mathcal{S}(n)$ holds for all $n$. We will show this
by induction on $n$. For the induction argument we use the following auxiliary
statement: Given $n\in\mathbb{N}\cup\\{0\\}$ we denote by $\mathcal{H}(n)$ the
statement that for any homomorphism $\beta:B\to S$ where $S$ is finite
solvable with $\ell(S)\leq n$ we have that for $\iota=\iota_{-},\iota_{+}$ the
homomorphism

$\iota_{*}:H_{1}(A;\mathbb{Z}[S])\to H_{1}(B;\mathbb{Z}[S])$

is an isomorphism.

In the next two sections we will prove the following two propositions:

######  Proposition 3.3.

If $\mathcal{H}(n)$ and $\mathcal{S}(n)$ hold, then $\mathcal{S}(n+1)$ holds
as well.

######  Proposition 3.4.

Assume that $(N,\phi)$ satisfies Condition $(*)$. If $\mathcal{S}(n)$ holds,
then $\mathcal{H}(n)$ holds as well.

We can now prove the following corollary, which amounts to Proposition 1.7.

######  Corollary 3.5.

Assume that $(N,\phi)$ satisfies Condition $(*)$ and that $\phi$ is primitive.
Let $\Sigma\subset N$ be a connected Thurston norm minimizing surface dual to
$\phi$ and let $\iota:A\to B$ be one of the two injections. Then for any
finite solvable group $G$ the map

$\mbox{Hom}(B,G)\xrightarrow{\iota^{*}}\mbox{Hom}(A,G)$

is a bijection, i.e. $\iota:A\to B$ induces an isomorphism of prosolvable
completions.

###### Proof.

The condition $\mathcal{S}(0)$ holds by fiat. It follows from Proposition 3.2
applied to the trivial group that if $(N,\phi)$ satisfies Condition ($*$),
then $\iota_{\pm}:H_{1}(A;\mathbb{Z})\to H_{1}(B;\mathbb{Z})$ are
isomorphisms, i.e. $\mathcal{H}(0)$ holds. The combination of Propositions 3.3
and 3.4 then shows that $\mathcal{H}(n)$ and $\mathcal{S}(n)$ hold for all
$n$. The corollary is now immediate. ∎

###  3.3. Proof of Proposition 3.3

In this section we will prove Proposition 3.3. Let $\iota=\iota_{-}$ or
$\iota=\iota_{+}$. Since $\mathcal{S}(n)$ holds we only have to consider the
case of $G$ a finite solvable group with $\ell(G)=n+1$. By definition $G$ fits
into a short exact sequence

$1\to I\to G\to S\to 1,$

where $I=G^{(n)}$ is finite abelian and $S=G/G^{(n)}$ finite solvable with
$\ell(S)=n$.

We will construct a map $\Phi:\mbox{Hom}(A,G)\to\mbox{Hom}(B,G)$ which is an
inverse to $\iota^{*}:\mbox{Hom}(B,G)\to\mbox{Hom}(A,G)$. Let $\alpha:A\to G$
be a homomorphism. Without loss of generality we can assume that $\alpha$ is
an epimorphism. Denote $A\xrightarrow{\alpha}G\to S$ by $\alpha^{\prime}$ and
denote the map $A\to A/\mbox{Ker}(\alpha^{\prime})^{(1)}$ by $\rho$. Note that
$\alpha$ sends $\mbox{Ker}(\alpha^{\prime})$ to the abelian group $I$, hence
$\alpha$ vanishes on $\mbox{Ker}(\alpha^{\prime})^{(1)}$. This shows that
$\alpha$ factors through $\rho$, in particular $\alpha=\psi\circ\rho$ for some
$\psi:A/\mbox{Ker}(\alpha^{\prime})^{(1)}\to G$.

Recall that $\ell(S)=n$, therefore by $\mathcal{S}(n)$ we have that
$\alpha^{\prime}:A\to S$ equals $\iota^{*}(\beta^{\prime})$ for some
$\beta^{\prime}:B\to S$. By Lemma 2.12, $\mathcal{S}(n)$ guarantees that
$i_{*}:H_{0}(A;\mathbb{Z}[S])\to H_{0}(B;\mathbb{Z}[S])$ is an isomorphism; on
the other hand $\mathcal{H}(n)$ asserts that $i_{*}:H_{1}(A;\mathbb{Z}[S])\to
H_{1}(B;\mathbb{Z}[S])$ is an isomorphism as well. By Corollary 2.5 this
implies that $\iota$ induces an isomorphism

$\iota:A/\mbox{Ker}(\alpha^{\prime})^{(1)}\xrightarrow{\cong}B/\mbox{Ker}(\beta^{\prime})^{(1)}.$

The various homomorphisms can be summarized in the following commutative
diagram:

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{\iota}$$\scriptstyle{\alpha^{\prime}}$$\scriptstyle{\rho}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta^{\prime}}$$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{=}$$\textstyle{S}$$\textstyle{A/\mbox{Ker}(\alpha^{\prime})^{(1)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\scriptstyle{\iota}$$\scriptstyle{\psi}$$\scriptstyle{\alpha^{\prime}}$$\textstyle{B/\mbox{Ker}(\beta^{\prime})^{(1)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta^{\prime}}$$\textstyle{G.}$

Now we define $\Phi(\alpha)\in\mbox{Hom}(B,G)$ to be the homomorphism

$B\to
B/\mbox{Ker}(\beta^{\prime})^{(1)}\xrightarrow{\iota^{-1}}A/\mbox{Ker}(\alpha^{\prime})^{(1)}\xrightarrow{\psi}G.$

It is now straightforward to check that $\Phi$ and $\iota^{*}$ are inverses to
each other.

###  3.4. Proof of Proposition 3.4

In this section we will prove Proposition 3.4. So let $\beta:B\to S$ be a
homomorphism to a finite solvable group $S$ with $\ell(S)\leq n$. If $\beta$
extends to $\pi_{1}(N)$, $\mathcal{H}(n)$ will follow immediately from
Proposition 3.2. In general $\beta$ though will not extend; however using
$\mathcal{S}(n)$ we will construct a homomorphism $\pi=\langle
B,t|\iota_{-}(A)=t\iota_{+}(A)t^{-1}\rangle\to G$ to a finite group $G$ ‘which
contains $\beta:B\to S$’ to get the required isomorphism.

We first need some notation. Given groups $C$ and $H$ we define

$C(H)=\bigcap\limits_{\gamma\in Hom(C,H)}\mbox{Ker}(\gamma).$

We summarize a few properties of $C(H)\subset C$ in the following lemma.

######  Lemma 3.6.

Let $C$ be a finitely generated group. Then the subgroup $C(H)\subset C$ has
the following properties:

  1. (1)

$C(H)\subset C$ is normal and characteristic.

  2. (2)

If $H$ is finite and solvable, then $C/C(H)$ is finite and solvable with
$\ell(C/C(H))\leq\ell(H)$.

###### Proof.

Statement (1) is immediate. To prove the rest, consider the injection

$C/C(H)=C/\bigcap\limits_{\gamma\in
Hom(C,H)}\mbox{Ker}(\gamma)\to\prod\limits_{\gamma\in
Hom(C,H)}C/\mbox{Ker}(\gamma).$

If $H$ is finite, then $\mbox{Hom}(C,H)$ is a finite set (since $C$ is
finitely generated), hence $C/C(H)$ is finite. If $H$ is furthermore solvable,
then for any $\gamma\in\mbox{Hom}(C,H)$ the groups $C/\mbox{Ker}(\gamma)$ are
solvable, hence $C/C(H)$ is solvable as well. Moreover for any
$\gamma\in\mbox{Hom}(C,H)$ we have $\ell(C/\mbox{Ker}(\gamma))\leq\ell(H)$. We
therefore get

$\ell(C/C(H))\leq\max_{\gamma\in
Hom(C,H)}\ell(C/\mbox{Ker}(\gamma))\leq\ell(H).$

∎

We will also need the following group homomorphism extension lemma.

######  Lemma 3.7.

Assume that $\mathcal{S}(n)$ holds and that $S$ is a finite solvable group
with $\ell(S)\leq n$. Let $\beta:B\to S$ be a homomorphism.

Then there exists a $k\in\mathbb{N}$, a semidirect product
$\mathbb{Z}/k\ltimes B/B(S)$ and a homomorphism

$\pi=\langle B,t|\iota_{-}(A)=t\iota_{+}(A)t^{-1}\rangle\to\mathbb{Z}/k\ltimes
B/B(S)$

which extends $B\to B/B(S)$, i.e. we have the following commutative diagram:

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B/B(S)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{Z}/{k}\ltimes
B/B(S)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{Z}/{k}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi.\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

###### Proof.

Assume that $\mathcal{S}(n)$ holds and that $S$ is a finite solvable group
with $\ell(S)\leq n$. Let $\beta:B\to S$ be a homomorphism. We denote the
projection map $B\to B/B(S)$ by $\rho$.

######  Claim.

There exists an automorphism $\gamma:B/B(S)\to B/B(S)$ such that
$\rho(\iota_{+}(a))=\gamma(\rho(\iota_{-}(a)))$ for all $a\in A$.

Let $\iota=\iota_{-}$ or $\iota=\iota_{+}$. By Lemma 3.6 we know that $B/B(S)$
is finite solvable with $\ell(B/B(S))\leq n$. It follows from $\mathcal{S}(n)$
that

$\iota_{*}:A/\mbox{Ker}\\{A\xrightarrow{\iota}B\xrightarrow{\rho}B/B(S)\\}\to
B/B(S)$

is an isomorphism. On the other hand it is also a straightforward consequence
of $\mathcal{S}(n)$ that

$\mbox{Ker}\\{A\xrightarrow{\iota}B\xrightarrow{\rho}B/B(S)\\}=A(S).$

Combining these two observations we see that $\iota$ gives rise to an
isomorphism $\iota_{*}:A/A(S)\to B/B(S)$. We now take
$\gamma:=\iota_{+*}\circ(\iota_{-*})^{-1}$. This concludes the proof of the
claim.

We now write $H=B/B(S)$. It is now straightforward to verify that

$\begin{array}[]{rcl}\pi=\langle
B,t|\iota_{-}(A)=t\iota_{+}(A)t^{-1}\rangle&\to&\mathbb{Z}\ltimes H=\langle
H,t|H=t\gamma(H)t^{-1}\rangle\\\ b&\mapsto&\rho(b),\,\,\,b\in B,\\\
t&\mapsto&t\end{array}$

defines a homomorphism. Since $H=B/B(S)$ is a finite group it follows that the
automorphism $\gamma$ has finite order $k$, in particular the projection map
$\mathbb{Z}\ltimes B/B(S)\to\mathbb{Z}/k\ltimes B/B(S)$ is a homomorphism.
Clearly the resulting homomorphism $\pi\to\mathbb{Z}/k\ltimes B/B(S)$ has all
the required properties. ∎

We are in position now to prove Proposition 3.4.

###### Proof of Proposition 3.4.

Recall that we assume that $(N,\phi)$ satisfies Condition $(*)$ and that
$\mathcal{S}(n)$ holds. We have to show that $\mathcal{H}(n)$ holds as well.
So let $\beta:B\to S$ be a homomorphism to a finite solvable group $S$ with
$\ell(S)\leq n$. We have to show that for $\iota=\iota_{-},\iota_{+}$ the
homomorphism

$\iota_{*}:H_{1}(A;\mathbb{Z}[S])\to H_{1}(B;\mathbb{Z}[S])$

is an isomorphism. Without loss of generality we can assume that $\beta$ is
surjective. Recall that $\mathcal{S}(n)$ implies that $\beta\circ\iota:A\to S$
is surjective as well.

We now apply Lemma 3.7 to find a homomorphism

$\pi=\langle B,t|\iota_{-}(A)=t\iota_{+}(A)t^{-1}\rangle\to\mathbb{Z}/k\ltimes
B/B(S)$

which extends $B\to B/B(S)$. Note that

(2) $\begin{array}[]{rcl}\mbox{Ker}\\{\gamma:B\to\pi\to\mathbb{Z}/k\ltimes
B/B(S)\\}&=&\mbox{Ker}\\{B\to B/B(S)\\}\\\ \mbox{Ker}\\{\gamma\circ\iota:A\to
B\to\pi\to\mathbb{Z}/k\ltimes B/B(S)\\}&=&\mbox{Ker}\\{\iota:A\to B\to
B/B(S)\\}.\end{array}$

We let

$\begin{array}[]{rclcl}\hat{B}&=&\mbox{Ker}\\{B\to B/B(S)\\},\\\
\tilde{B}&=&\mbox{Ker}(\beta).\end{array}$

Clearly $\hat{B}\subset\tilde{B}$ by the definition of $B/B(S)$. We also write
$\hat{A}=\iota^{-1}(\hat{B})$ and $\tilde{A}=\iota^{-1}(\tilde{B})$. We now
consider the epimorphism $\pi_{1}(N)=\pi\to\mathbb{Z}/k\ltimes B/B(S)$. By
Condition ($*$), Equation (2), Proposition 3.2 and Corollaries 2.5 and 2.5 it
follows that the maps

$\iota:A/\hat{A}\to B/\hat{B}\mbox{ and }\iota:A/[\hat{A},\hat{A}]\to
B/[\hat{B},\hat{B}]$

are isomorphisms. It now follows from Lemma 2.7 and Corollary 2.5 that the
maps

$\iota:A/\tilde{A}\to B/\tilde{B}\mbox{ and }\iota:A/[\tilde{A},\tilde{A}]\to
B/[\tilde{B},\tilde{B}]$

are also isomorphisms. ∎

##  4\. A product criterion

In this section we will apply a theorem of Agol to prove a criterion for a
manifold to be a product, which complements Proposition 1.7.

In order to state our result, we first recall the definition of a sutured
manifold (cf. [Ga83, Definition 2.6] or [CC03, p. 364]). A _sutured manifold_
$(M,\gamma)$ is a compact oriented 3–manifold $M$ together with a set
$\gamma\subset\partial M$ of pairwise disjoint annuli $A(\gamma)$ and tori
$T(\gamma)$. Furthermore, the structure of a sutured manifold consists of the
following choices of orientations:

  1. (1)

For each $A\in A(\gamma)$ a choice of a simple closed, oriented curve in $A$
(called _suture_) such that $A$ is the tubular neighborhood of the curve, and

  2. (2)

the choice of an orientation for each component of $\partial M\setminus
A(\gamma)$.

The orientations must be compatible, i.e. the orientation of the components of
$\partial M\setminus A(\gamma)$ must be coherent with the orientations of the
sutures.

Given a sutured manifold $(M,\gamma)$ we define $R_{+}(\gamma)$ as the
components of $\overline{\partial M\setminus\gamma}$ where the orientation
agrees with the orientation induced by $M$ on $\partial M$, and
$R_{-}(\gamma)$ as the components of $\overline{\partial M\setminus\gamma}$
where the two orientations disagree. We define also
$R(\gamma)=R_{+}(\gamma)\cup R_{-}(\gamma)$.

A sutured manifold $(M,\gamma)$ is called _taut_ if $M$ is irreducible and if
each component of $R(\gamma)$ is incompressible and Thurston norm–minimizing
in $H_{2}(M,\gamma;\mathbb{Z})$ (we refer to [Sc89] for information regarding
the Thurston norm on sutured manifolds).

An example of a taut sutured manifold is given by taking an oriented surface
$\Sigma$ and considering $\Sigma\times I$ with sutures given by the annuli
$\partial\Sigma\times I$. The sutures are oriented by the orientation of
$\partial\Sigma$. We can pick orientations such that
$R_{-}(\gamma)=\Sigma\times 0$ and $R_{+}(\gamma)=\Sigma\times 1$. If a
sutured manifold $(M,\gamma)$ is diffeomorphic (as a sutured manifold) to such
a product then we say that $(M,\gamma)$ is a _product sutured manifold_.

Another example of a taut sutured manifold comes from considering an oriented
incompressible Thurston norm minimizing surface $\Sigma\subset N$ in an
irreducible 3–manifold with empty or toroidal boundary. We let
$(M,\gamma)=(N\setminus\nu\Sigma,\partial N\cap(N\setminus\nu\Sigma))$. With
appropriate orientations $(M,\gamma)$ is a taut sutured manifold such that
$R_{-}(\gamma)=\Sigma^{-}$ and $R_{+}(\gamma)=\Sigma^{+}$.

The following theorem immediately implies Theorem 1.8.

######  Theorem 4.1.

Assume we have a taut sutured manifold $(M,\gamma)$ which has the following
properties:

  1. (1)

$R_{\pm}(\gamma)$ consist of one component $\Sigma^{\pm}$ each, and the
inclusion induced maps $\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)$ give rise to
isomorphisms of the respective prosolvable completions,

  2. (2)

$\pi_{1}(M)$ is residually finite solvable,

then $(M,\gamma)$ is a product sutured manifold.

The key ingredient to the proof of Theorem 4.1 is a result of Agol’s [Ag08]
which we recall in Section 4.1. We will then provide the proof for Theorem 4.1
in Sections 4.2 and 4.3.

######  Remark.

(1) It is an immediate consequence of ‘peripheral subgroup separability’
[LN91] that the theorem holds under the assumption that the inclusion induced
maps $\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)$ give rise to isomorphisms of the
respective _profinite_ completions. It is not clear how the approach of [LN91]
can be adapted to prove Theorem 4.1.
(2) It is also interesting to compare Theorem 4.1 with a result of
Grothendieck. In [Gr70, Section 3.1] Grothendieck proves that if $\varphi:A\to
B$ is a homomorphism between finitely presented, residually finite groups
which induces an isomorphism of the profinite completions, and if $A$ is
arithmetic (e.g. a surface group), then $\varphi$ is an isomorphism. It is an
interesting question whether Theorem 4.1 can be proved using purely group
theoretic arguments. We refer to [AHKS07] for more information regarding this
question.

###  4.1. Agol’s theorem

Before we can state Agol’s result we have to introduce more definitions. A
group $G$ is called _residually finite $\mathbb{Q}$–solvable_ or _RFRS_ if
there exists a filtration of groups $G=G_{0}\supset G_{1}\supset G_{2}\dots$
such that the following hold:

  1. (1)

$\cap_{i}G_{i}=\\{1\\}$,

  2. (2)

$G_{i}$ is a normal, finite index subgroup of $G$ for any $i$,

  3. (3)

for any $i$ the map $G_{i}\to G_{i}/G_{i+1}$ factors through $G_{i}\to
H_{1}(G_{i};\mathbb{Z})/\mbox{torsion}$.

Note that RFRS groups are in particular residually finite solvable, but the
RFRS condition is considerably stronger than being residually finite solvable.
The notion of an RFRS group was introduced by Agol [Ag08], we refer to Agol’s
paper for more information on RFRS groups.

Given a sutured manifold $(M,\gamma)$ the double $DM_{\gamma}$ is defined to
be the double of $M$ along $R(\gamma)$, i.e. $DM_{\gamma}=M\cup_{R(\gamma)}M$.
Note that the annuli $A(\gamma)$ give rise to toroidal boundary components of
$DM_{\gamma}$. We denote by $r:DM_{\gamma}\to M$ the retraction map given by
‘folding’ the two copies of $M$ along $R(\gamma)$.

We are now in a position to state Agol’s result. The theorem as stated here is
clearly implicit in the proof of [Ag08, Theorem 6.1].

######  Theorem 4.2.

Let $(M,\gamma)$ be a connected, taut sutured manifold such that $\pi_{1}(M)$
satisfies property RFRS. Write $W=DM_{\gamma}$. Then there exists an
epimorphism $\alpha:\pi_{1}(M)\to S$ to a finite solvable group, such that in
the covering $p:\widetilde{W}\to W$ corresponding to $\alpha\circ
r_{*}:\pi_{1}(W)\to S$ the pull back of the class $[R_{-}(\gamma)]\in
H_{2}(W,\partial W;\mathbb{Z})$ lies on the closure of the cone over a fibered
face of $\widetilde{W}$.

Note that $[R_{+}(\gamma)]=\pm[R_{-}(\gamma)]$ in $H_{2}(W,\partial
W;\mathbb{Z})$, i.e. $[R_{-}(\gamma)]$ is a fibered class if and only if
$[R_{+}(\gamma)]$ is a fibered class. In case that $\widetilde{W}$ has
vanishing Thurston norm, then we adopt the usual convention that by the
fibered face we actually mean
$H^{1}(\widetilde{W},\mathbb{R})\setminus\\{0\\}$.

###  4.2. Proof of Theorem 4.1

From now on assume we have a taut sutured manifold $(M,\gamma)$ with the
following properties:

  1. (1)

$R_{\pm}(\gamma)$ consist of one component $\Sigma^{\pm}$ each and the
inclusion induced maps $\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)$ give rise to
isomorphisms of the respective prosolvable completions.

  2. (2)

$\pi_{1}(M)$ is residually finite solvable.

Since Theorem 4.1 is obvious in the case $M=S^{2}\times[0,1]$ we will
henceforth assume that $M\neq S^{2}\times[0,1]$.

Our main tool in proving Theorem 4.1 is Theorem 4.2. In order to apply it we
need the following claim.

######  Claim.

The group $\pi_{1}(M)$ is RFRS.

###### Proof.

By assumption the group $\pi_{1}(M)$ is residually finite solvable. This means
that we can find a sequence $\pi_{1}(M)=B_{0}\supset B_{1}\supset B_{2}\dots$
with the following properties:

  1. (1)

$\cap_{i}B_{i}=\\{1\\}$,

  2. (2)

$B_{i}$ is a normal, finite index subgroup of $\pi_{1}(M)$ for any $i$,

  3. (3)

for any $i$ the map $B_{i}\to B_{i}/B_{i+1}$ factors through $B_{i}\to
H_{1}(B_{i};\mathbb{Z})$.

It remains to show that $B_{i}\to B_{i}/B_{i+1}$ factors through
$H_{1}(B_{i};\mathbb{Z})/\mbox{torsion}$. In fact we claim that
$H_{1}(B_{i};\mathbb{Z})$ is torsion–free. Indeed, first note that by
Shapiro’s lemma $H_{1}(B_{i};\mathbb{Z})\cong
H_{1}(B;\mathbb{Z}[B/B_{i}])\cong H_{1}(M;\mathbb{Z}[B/B_{i}])$. Furthermore,
by Lemma 2.12 we have

$H_{1}(\Sigma^{-};\mathbb{Z}[B/B_{i}])\xrightarrow{\cong}H_{1}(M;\mathbb{Z}[B/B_{i}]),$

but the first group is clearly torsion–free as it is the homology of a finite
cover of a surface. ∎

In the following we write $W=DM_{\gamma}$. By the above claim we can apply
Theorem 4.2 which says that there exists an epimorphism $\alpha:\pi_{1}(M)\to
S$ to a finite solvable group, such that in the covering $p:\widetilde{W}\to
W$ corresponding to $\alpha\circ r_{*}:\pi_{1}(W)\to S$ the pull back of the
class $[R_{-}(\gamma)]=[\Sigma^{-}]\in H_{2}(W,\partial W;\mathbb{Z})$ lies on
the closure of the cone over of a fibered face of $\widetilde{W}$.

Note that we can view $\widetilde{W}$ as the double of the cover
$(\widetilde{M},\tilde{\gamma})$ of $(M,\gamma)$ induced by
$\alpha:\pi_{1}(M)\to S$. We summarize the main properties of
$\tilde{\Sigma}^{\pm}$ and $\widetilde{W}$ in the following lemma.

######  Lemma 4.3.

  1. (1)

$\tilde{\Sigma}^{\pm}:=p^{-1}(\Sigma^{\pm})$ are connected surfaces,

  2. (2)

the inclusion induced maps
$\pi_{1}(\tilde{\Sigma}^{\pm})\to\pi_{1}(\widetilde{M})$ give rise to
isomorphisms of prosolvable completions,

  3. (3)

if $\tilde{\Sigma}^{-}$ is the fiber of a fibration
$\widetilde{W}=D\widetilde{M}_{\tilde{\gamma}}\to S^{1}$, then $\widetilde{M}$
is a product over $\tilde{\Sigma}^{-}$,

  4. (4)

$M$ is a product over $\Sigma^{-}$ if and only if $\widetilde{M}$ is a product
over $\tilde{\Sigma}^{-}$.

###### Proof.

First note that it follows from Lemma 2.12 and Corollary 2.3 and the
assumption that $\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)$ give rise to isomorphisms
of the respective prosolvable completions that
$\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)\to S$ is surjective, i.e. the preimages
$\tilde{\Sigma}^{\pm}:=p^{-1}(\Sigma^{\pm})$ are connected. The second claim
follows from Lemma 2.11 since the maps
$\pi_{1}({\Sigma}^{\pm})\to\pi_{1}({M})$ give rise to isomorphisms of their
prosolvable completions.

For the third claim consider the following commutative diagram

$\textstyle{\pi_{1}(\tilde{\Sigma}^{-})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(\widetilde{W}\setminus\nu\tilde{\Sigma}^{-})}$$\textstyle{\pi_{1}(\widetilde{M}).\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

If $\tilde{\Sigma}^{-}$ is the fiber of a fibration
$D\widetilde{M}_{\tilde{\gamma}}\to S^{1}$, then the top map in the above
commutative diagram is an isomorphism. We can think of
$\widetilde{W}\setminus\nu\tilde{\Sigma}^{-}$ as
$\widetilde{M}\cup_{\tilde{\Sigma}^{+}}\widetilde{M}$. It is now clear that
the lower two maps are injective. But then the lower left map also has to be
an isomorphism, i.e. $\widetilde{M}$ is a product over $\tilde{\Sigma}^{-}$.
The last claim is well–known, it is for example a consequence of [He76,
Theorem 10.5]. ∎

Using the above lemma it is now clear that the following lemma implies Theorem
4.1.

######  Lemma 4.4.

Let $(M,\gamma)$ be a taut sutured manifold such that $R_{\pm}(\gamma)$
consist of one component $\Sigma^{\pm}$ each. Assume the following hold:

  1. (A)

The inclusion induced maps $\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)$ give rise to
isomorphisms of the respective prosolvable completions.

  2. (B)

The class in $H^{1}(DM_{\gamma};\mathbb{Z})$ represented by the surface
$\Sigma^{-}$ lies on the closure of the cone over a fibered face of
$DM_{\gamma}$.

Then $\Sigma^{-}$ is the fiber of a fibration $DM_{\gamma}\to S^{1}$.

In the following we write $W=DM_{\gamma}$. Note that we have a canonical
involution $\tau$ on $W$ with fix point set $R(\gamma)$. From now on we think
of $W=DM_{\gamma}$ as $M\cup_{R(\gamma)}\tau(M)$.

Our main tool in the proof of Lemma 4.4 will be the interplay between the
Thurston norm and McMullen’s Alexander norm [McM02]. Recall that given a
3–manifold $V$ with $b_{1}(V)\geq 2$ the Alexander norm
$||-||_{A}:H^{1}(V;\mathbb{R})\to\mathbb{R}_{\geq 0}$ has the following
properties:

  1. (a)

The Alexander norm ball is dual to the Newton polyhedron defined by the
symmetrized Alexander polynomial
$\Delta_{V}\in\mathbb{Z}[H_{1}(V;\mathbb{Z})/\mbox{torsion}]$.

  2. (b)

The Alexander norm ball is a (possibly noncompact) polyhedron with finitely
many faces.

  3. (c)

For any $\phi\in H^{1}(V;\mathbb{R})$ we have $||\phi||_{A}\leq||\phi||_{T}$,
and equality holds for fibered classes.

  4. (d)

Let $C\subset H^{1}(V;\mathbb{R})$ be a fibered cone, i.e. the cone on a
fibered face of the Thurston norm ball, then $C$ is contained in the cone on
the interior of a top–dimensional face of the Alexander norm ball.

  5. (e)

Let $C_{1},C_{2}\subset H^{1}(V;\mathbb{R})$ be fibered cones which are
contained in the same cone on the interior of a top–dimensional face of the
Alexander norm ball, then $C_{1}=C_{2}$.

Our assumption that the induced maps $\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)$ give
rise to isomorphisms of the respective prosolvable completions implies that
$W=DM_{\gamma}$ ‘looks algebraically’ the same as $\Sigma^{-}\times S^{1}$.
More precisely, we have the following lemma which we will prove in Section
4.3.

######  Lemma 4.5.

Let $(M,\gamma)$ be a taut sutured manifold with the property that
$R_{\pm}(\gamma)$ consist of one component $\Sigma^{\pm}$ each. Assume that
(A) holds. Then the following hold:

  1. (1)

There exists an isomorphism

$f:\mathbb{R}\oplus H_{1}(\Sigma^{-},\partial\Sigma^{-};\mathbb{R})\to
H_{2}(W,\partial W;\mathbb{R})$

such that $f(1,0)=[\Sigma^{-}]$ and such that $\tau(f(r,h))=f(r,-h)$.

  2. (2)

The class $\phi=PD(\Sigma^{-})\in H^{1}(W;\mathbb{Z})$ lies in the cone $D$ on
the interior of a top–dimensional face of the Alexander norm ball.

Note that (1) implies in particular that $b_{1}(W)\geq 2$. Assuming this lemma
we are now in a position to prove Lemma 4.4.

###### Proof of Lemma 4.4.

Let $(M,\gamma)$ be a taut sutured manifold with the property that
$R_{\pm}(\gamma)$ consist of one component $\Sigma^{\pm}$ each. Assume that
(A) and (B) hold.

By Lemma 4.5 there exists a cone $D\subset H^{1}(W;\mathbb{R})$ on the
interior of a top–dimensional face of the Alexander norm ball which contains
$\phi=PD([\Sigma^{-}])$. We denote the map

$\mathbb{R}\oplus
H_{1}(\Sigma^{-},\partial\Sigma^{-};\mathbb{R})\xrightarrow{f}H_{2}(W,\partial
W;\mathbb{R})\xrightarrow{PD}H^{1}(W;\mathbb{R})$

by $\Phi$.

By (B) we can find $h\in H_{1}(\Sigma^{-},\partial\Sigma^{-};\mathbb{R})$ such
that $\Phi(1,h)$ and $\Phi(1,-h)$ lie in $D$ and such that $\Phi(1,h)$ lies in
the cone $C$ on a fibered face $F$ of the Thurston norm ball. Note that the
$\tau_{*}:H^{1}(W;\mathbb{R})\to H^{1}(W;\mathbb{R})$ sends fibered classes to
fibered classes and preserves the Thurston norm. In particular
$\tau(\Phi(1,h))=\Phi(1,-h)$ is fibered as well and it lies in $\tau(C)$ which
is the cone on the fibered face $\tau(F)$ of the Thurston norm ball. Recall
that $\tau(\Phi(1,h))=\Phi(1,-h)$ lies in $D$, it follows from Property (d) of
the Alexander norm that $\tau(C)\subset D$. We then use (e) to conclude that
$C=\tau(C)$. In particular $\Phi(1,h)$ and $\Phi(1,-h)$ lie in $C$. Since $C$
is convex it follows that $\phi=\Phi(1,0)\in C$ i.e. $\phi$ is a fiber class.
∎

###  4.3. Proof of Lemma 4.5

By Lemma 2.12 the following lemma is just the first statement of Lemma 4.5.

######  Lemma 4.6.

Let $(M,\gamma)$ be a taut sutured manifold with the property that
$R_{\pm}(\gamma)$ consist of one component $\Sigma^{\pm}$ each. Assume that
$\iota_{\pm}:H_{1}(\Sigma^{\pm};\mathbb{R})\to H_{1}(M;\mathbb{R})$ are
isomorphisms. Then there exists an isomorphism

$f:\mathbb{R}\oplus H_{1}(\Sigma^{-},\partial\Sigma^{-};\mathbb{R})\to
H_{2}(W,\partial W;\mathbb{R})$

such that $f(1,0)=[\Sigma^{-}]$ and such that $\tau(f(b,c))=f(b,-c)$.

###### Proof.

We start out with the following two claims.

######  Claim.

$M$ has no toroidal sutures.

###### Proof.

Denote the toroidal sutures by $T_{1},\dots,T_{n}$. Recall that for any
compact 3–manifold $X$ we have $b_{1}(\partial X)\leq 2b_{1}(X)$. In our case
it is easy to see that we have $b_{1}(\partial
M)=b_{1}(\Sigma^{-})+b_{1}(\Sigma^{+})+\sum_{i=1}^{n}b_{1}(T_{i})=2b_{1}(\Sigma)+2n$.
On the other hand, since $H_{1}(\Sigma^{\pm};\mathbb{R})\to
H_{1}(M;\mathbb{R})$ are isomorphisms we have $b_{1}(M)=b_{1}(\Sigma)$. It now
follows from $b_{1}(\partial M)\leq 2b_{1}(M)$ that $n=0$. ∎

######  Claim.

The inclusion induced maps
$H_{1}(\Sigma^{\pm},\partial\Sigma^{\pm};\mathbb{R})\to
H_{1}(M,A(\gamma);\mathbb{R})$ are isomorphisms.

Now consider the following commutative diagram:

$\textstyle{H_{1}(\partial\Sigma^{-};\mathbb{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{1}(\Sigma^{-};\mathbb{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{H_{1}(\Sigma^{-},\partial\Sigma^{-};\mathbb{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{0}(\partial\Sigma^{-};\mathbb{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{0}(\Sigma^{-};\mathbb{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{H_{1}(A(\gamma);\mathbb{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{1}(M;\mathbb{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{1}(M,A(\gamma);\mathbb{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{0}(A(\gamma);\mathbb{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{0}(M;\mathbb{R}).}$

Note that by the compatibility condition in the definition of sutured
manifolds we have that for each component $A$ of $A(\gamma)$ the subset
$\partial A\cap\Sigma^{-}=A\cap\partial\Sigma^{-}\subset\partial A$ consists
of exactly one boundary component of $A$. This implies that the maps
$H_{i}(\partial\Sigma^{-}\cap A;\mathbb{R})\to H_{i}(A;\mathbb{R})$ are
isomorphisms. The claim now follows immediately from the above commutative
diagram and from the assumption that $H_{1}(\Sigma^{\pm};\mathbb{R})\to
H_{1}(M;\mathbb{R})$ are isomorphisms.

We now define

$g:H_{1}(\Sigma^{-},\partial\Sigma^{-};\mathbb{R})\to H_{2}(W,\partial
W;\mathbb{R})$

as follows: given an element $c\in
H_{1}(\Sigma^{-},\partial\Sigma^{-};\mathbb{R})$ represent it by a chain
$c^{-}$, since the maps
$H_{1}(\Sigma^{\pm},\partial\Sigma^{\pm};\mathbb{R})\to
H_{1}(M,A(\gamma);\mathbb{R})$ are isomorphisms we can find a chain $c^{+}$ in
$\Sigma^{+}$ such that $[c^{-}]=[c^{+}]\in H_{2}(M,A(\gamma);\mathbb{R})$. Now
let $d$ be a $2$–chain in $M$ such that $\partial d=c^{-}\cup-c^{+}$. Then
define $g(c)$ to be the element in $H_{2}(W,\partial W;\mathbb{R})$
represented by the closed $2$–chain $d\cup-\tau(d)$. It is easy to verify that
$g$ is a well–defined homomorphism. Note that $\partial
W=A(\gamma)\cup\tau(A(\gamma))$ since $W$ has no toroidal sutures. It is now
straightforward to check, using a Mayer–Vietoris sequence, that the map

$\begin{array}[]{rclcl}f:\mathbb{R}&\oplus&H_{1}(\Sigma^{-},\partial\Sigma^{-};\mathbb{R})&\to&H_{2}(W,\partial
W;\mathbb{R})\\\ (\,b&,&c\,)&\mapsto&b[\Sigma^{-}]+g(c)\end{array}$

is an isomorphism. Clearly $f(1,0)=[\Sigma^{-}]$. It is also easy to verify
that $\tau(f(b,c))=f(b,-c)$. This shows that $f$ has all the required
properties. ∎

The second statement of Lemma 4.5 is more intricate. We start with the
following lemma which in light of [Gr70], [BG04] and [AHKS07] has perhaps some
independent interest.

######  Lemma 4.7.

Let $\varphi:A\to B$ be a homomorphism of finitely generated metabelian groups
which induces an isomorphism of prosolvable completions. Then $\varphi$ is
also an isomorphism.

###### Proof.

We first show that $A\to B$ is an injection. We consider the following
commutative diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\widehat{A}_{\mathcal{FS}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\widehat{B}_{\mathcal{FS}}.}$

The vertical maps are injections since metabelian groups are residually finite
(cf. [Ha59]). The bottom map is an isomorphism by assumption. It now follows
that the top map is an injection.

Now suppose that the homomorphism $A\to B$ is not surjective. We identify
$H_{1}(A;\mathbb{Z})=H_{1}(A;\mathbb{Z})\xrightarrow{\cong}H_{1}(B;\mathbb{Z})=H_{1}(B;\mathbb{Z})$
via $\varphi$ and refer to the group as $H$. Let $g^{\prime}\in
B\setminus\varphi(A)$. We can pick an $a\in A$ such that $\varphi(a)$ and
$g^{\prime}$ represent the same element in $H$. Let
$g=\varphi(a)^{-1}g^{\prime}$. Then $g$ represents the trivial element in $H$
but $g$ is also an element in $B\setminus\varphi(A)$.

We will show that there exists a homomorphism $\alpha:B\to G$ to a finite
metabelian group such that $\alpha$ separates $g$ from $\varphi(A)$, i.e. such
that $\alpha(g)\not\in\alpha(\varphi(A))$. This then immediately contradicts,
via Lemma 2.10, our assumption that $\varphi:A\to B$ induces an isomorphism of
prosolvable completions. Our construction of finding such $\alpha$ builds on
some ideas of the proof of [LN91, Theorem 1].

We write ${B_{1}}=B_{2}={B}$. We denote the inclusion maps $A\to B_{i}=B$ by
$\varphi_{i}$. We let $C=B_{1}*_{A}B_{2}$. It is straightforward to see that
the homomorphisms $B_{i}\to C$ give rise to an isomorphism
$H_{1}(B_{i};\mathbb{Z})=H\to H_{1}(C;\mathbb{Z})$. Denote by

$s:B_{1}*_{A}B_{2}\to B_{1}*_{A}B_{2}$

the switching map, i.e. the map induced by $s(b)=b\in B_{2}$ for $b\in B_{1}$
and $s(b)=b\in B_{1}$ for $b\in B_{2}$. Note that $s$ acts as the identity on
$A\subset C$. Also note that $s$ descends to a map $C/C^{(2)}\to C/C^{(2)}$
which we also denote by $s$. We now view $g$ as an element in $B_{1}$ and
hence as an element in $C$. Note that the fact that $g$ represents the trivial
element in $H$ implies that $g$ represents an element in $C^{(1)}/C^{(2)}$. We
will first show that $s(g)\neq g\in C/C^{(2)}$. Consider the following
commutative diagram of exact sequences

$\begin{array}[]{ccccccccccccccc}H_{1}(A;\mathbb{Z}[H])&\to&H_{1}(B_{1};\mathbb{Z}[H])\oplus
H_{1}(B_{2};\mathbb{Z}[H])&\to&H_{1}(C;\mathbb{Z}[H])&\to&0\\\\[5.69054pt]
\,\,\,\downarrow\cong&&\,\,\,\downarrow\cong&&\,\,\downarrow\cong&&\downarrow&&\\\\[5.69054pt]
A^{(1)}&\to&B_{1}^{(1)}\times B_{2}^{(1)}&\to&C^{(1)}/C^{(2)}&\to&0.\\\
h&\mapsto&(\varphi_{1}(h),\varphi_{2}(h)^{-1})&\end{array}$

Since $g\in B_{1}^{(1)}\setminus\varphi(A^{(1)})$ it follows that $(g,g^{-1})$
does not lie in the image of $A^{(1)}$ in $B_{1}^{(1)}\times B_{2}^{(1)}$. It
therefore follows from the above diagram that $gs(g)^{-1}\neq e\in
C^{(1)}/C^{(2)}$.

Note that $C/C^{(2)}$ is metabelian, and hence by [Ha59] residually finite. We
can therefore find an epimorphism $\alpha:C/C^{(2)}\to G$ onto a finite group
$G$ (which is necessarily metabelian) such that $\alpha(gs(g)^{-1})\neq e$.
Now consider $\beta:C/C^{(2)}\to G\times G$ given by
$\beta(h)=(\alpha(h),\alpha(s(h)))$. Then clearly
$\beta(g)\not\in\beta(A)\subset\\{(g,g)\,|\,g\in G\\}$. The restriction of
$\beta$ to $B=B_{1}$ now clearly separates $g$ from $A$. ∎

######  Corollary 4.8.

Let $\varphi:A\to B$ be a homomorphism of finitely generated groups which
induces an isomorphism of prosolvable completions. Then the induced map
$A/A^{(2)}\to B/B^{(2)}$ is an isomorphism.

###### Proof.

It follows immediately from Lemma 2.10 that $\varphi$ induces an isomorphism
of the prosolvable completions of the metabelian groups $A/A^{(2)}$ and
$B/B^{(2)}$. It now follows from Lemma 4.7 that the induced map $A/A^{(2)}\to
B/B^{(2)}$ is an isomorphism. ∎

We now turn to the proof of the second claim of Lemma 4.5. For the remainder
of this section let $(M,\gamma)$ be a taut sutured manifold with the property
that $R_{\pm}(\gamma)$ consist of one component $\Sigma^{\pm}$ each. We assume
that (A) holds, i.e. the inclusion induced maps
$\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)$ give rise to isomorphisms of the
respective prosolvable completions. We have to show that the class
$\phi=PD(\Sigma^{-})\in H^{1}(W;\mathbb{Z})$ lies in the cone on the interior
of a top–dimensional face of the Alexander norm ball.

For the remainder of the section we pick a base point $x^{-}\in\Sigma^{-}$ and
a base point $x^{+}\in\Sigma^{+}$. We endow $W,M$ and $\tau(M)$ with the base
point $x^{-}$. Furthermore we pick a path $\gamma$ in $M$ connecting
$x^{-}\in\Sigma^{-}$ to $x^{+}\in\Sigma^{+}$.

Our goal is to understand the Alexander norm ball of $W$. In order to do this
we first have to study $H=H_{1}(W;\mathbb{Z})$. Let $t$ denote the element in
$H$ represented by the closed path $\gamma\cup-\tau(\gamma)$. It follows from
a straightforward Mayer–Vietoris sequence argument that we have an isomorphism

$\begin{array}[]{rclcl}f:H_{1}(\Sigma^{-};\mathbb{Z})&\oplus&\langle
t\rangle&\to&H_{1}(W;\mathbb{Z})\\\
(\,b&,&t^{k}\,)&\mapsto&\iota(b)+kt.\end{array}$

In particular $H$ is torsion–free. We write $F=H_{1}(\Sigma^{-};\mathbb{Z})$.
We use $f$ to identify $H$ with $F\times\langle t\rangle$ and to identify
$\mathbb{Z}[H]$ with $\mathbb{Z}[F][t^{\pm 1}]$.

We now consider the Alexander module $H_{1}(W;\mathbb{Z}[H])$. Recall that
$H_{1}(W;\mathbb{Z}[H])$ is the homology of the covering of $W$ corresponding
to $\pi_{1}(W)\to H_{1}(W;\mathbb{Z})=H$ together with the
$\mathbb{Z}[H]$–module structure given by deck transformations.

In the following claim we compare $W$ with $\Sigma\times S^{1}$. We also write
$F=H_{1}(\Sigma;\mathbb{Z})$ and we can identify $H_{1}(\Sigma\times
S^{1};\mathbb{Z})$ with $H=F\times\langle t\rangle$. In particular we identify
$H_{1}(\Sigma\times S^{1};\mathbb{Z})$ with $H_{1}(W;\mathbb{Z})$. With these
identifications we can now state the following lemma.

######  Lemma 4.9.

The $\mathbb{Z}[H]$–module $H_{1}(W;\mathbb{Z}[H])$ is isomorphic to the
$\mathbb{Z}[H]$–module $H_{1}(\Sigma\times S^{1};\mathbb{Z}[H])$.

###### Proof.

In the following we identify $\Sigma$ with $\Sigma^{-}\subset W$. We denote by
$X$ the result of gluing $M$ and $\tau(M)$ along $\Sigma=\Sigma^{-}$. Note
that we have two canonical maps $r:\Sigma^{+}\to M\to X$ and
$s:\Sigma^{+}\to\tau(M)\to X$. We furthermore denote the canonical inclusion
maps $\Sigma\to M,\Sigma\to\tau(M)$ and $\Sigma=\Sigma^{-}\to X$ by $i$.
Throughout this proof we denote by $i,r,s$ the induced maps on solvable
quotients as well.

_Claim A._ The map $i:\pi_{1}(\Sigma)\to\pi_{1}(X)$ gives rise to an
isomorphism
$\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}\to\pi_{1}(X)/\pi_{1}(X)^{(2)}$.

In the following let $M^{\prime}$ be either $M$ or $\tau(M)$. Recall that we
assume that $\pi_{1}(\Sigma)\to\pi_{1}(M^{\prime})$ gives rise to isomorphisms
of the prosolvable completions. It now follows from Corollary 4.8 that
$\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}\to\pi_{1}(M^{\prime}/\pi_{1}(M^{\prime})^{(2)}$
is an isomorphism. Now let
$g:\pi_{1}(X)=\pi_{1}(M\cup_{\Sigma}\tau(M))\to\pi_{1}(M))$ be the ‘folding
map’. Note that

$\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}\xrightarrow{i}\pi_{1}(X)/\pi_{1}(X)^{(2)}\xrightarrow{g}\pi_{1}(M)/\pi_{1}(M)^{(2)}\xleftarrow{\cong}\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}$

is the identity map. In particular
$\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}\xrightarrow{i}\pi_{1}(X)/\pi_{1}(X)^{(2)}$
is injective. On the other hand it follows from the van Kampen theorem that

$\pi_{1}(X)=\pi_{1}(M)*_{\pi_{1}(\Sigma)}\pi_{1}(M^{\prime}),$

in particular $\pi_{1}(X)/\pi_{1}(X)^{(2)}$ is generated by the images of
$\pi_{1}(M)$ and $\pi_{1}(\tau(M))$ in $\pi_{1}(X)/\pi_{1}(X)^{(2)}$. But it
follows immediately from the following commutative diagram

$\textstyle{\pi_{1}(\Sigma)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$------------$\textstyle{\pi_{1}(M^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{\pi_{1}(M^{\prime})/\pi_{1}(M^{\prime})^{(2)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(X)/\pi_{1}(X)^{(2)}}$

that image of $\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}$ in
$\pi_{1}(X)/\pi_{1}(X)^{(2)}$ also generates the group. This concludes the
proof of the claim A.

_Claim B._ For any $g\in\pi_{1}(\Sigma^{+})/\pi_{1}(\Sigma^{+})$ we have

$r(g)=s(g)\in\pi_{1}(X)/\pi_{1}(X)^{(2)}$

Denote by $\tau:X=M\cup_{\Sigma}\tau(M)\to X=M\cup_{\Sigma}\tau(M)$ the map
given by switching the two copies of $M$. Clearly $r(g)=\tau_{*}(s(g))$. But
$\tau_{*}$ acts trivially on image of $\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}$
in $\pi_{1}(X)/\pi_{1}(X)^{(2)}$. By the above claim this means that
$\tau_{*}$ acts trivially on $\pi_{1}(X)/\pi_{1}(X)^{(2)}$. This concludes the
proof of the claim.

We now view $W$ as the result of gluing the two copies of $\Sigma^{+}$ in
$\partial X$ by the identity map. First note that by the van Kampen theorem we
have

$\pi_{1}(W)=\langle
t,\pi_{1}(X)\,|\,ts(g)t^{-1}=r(g),g\in\pi_{1}(\Sigma^{+})\rangle.$

Note that by Claim B the obvious assignments give rise to a well–defined map

$\pi_{1}(W)=\langle
t,\pi_{1}(X)\,|\,ts(g)t^{-1}=r(g),g\in\pi_{1}(\Sigma^{+})\rangle\to\langle
t\rangle\times\pi_{1}(X)/\pi_{1}(X)^{(2)}.$

Since $\pi_{1}(X)/\pi_{1}(X)^{(2)}$ is metabelian this map descends to a map

$\Phi:\langle
t,\pi_{1}(X)\,|\,ts(g)t^{-1}=r(g),g\in\pi_{1}(\Sigma^{+})\rangle/(\dots)^{(2)}\to\langle
t\rangle\times\pi_{1}(X)/\pi_{1}(X)^{(2)}.$

_Claim C._ The map $\Phi:\pi_{1}(W)/\pi_{1}(W)^{(2)}\to\langle
t\rangle\times\pi_{1}(X)/\pi_{1}(X)^{(2)}$ is an isomorphism.

We denote by $\Psi$ the following map:

$\begin{array}[]{rcl}\langle
t\rangle\times\pi_{1}(X)/\pi_{1}(X)^{(2)}&\to&\langle
t,\pi_{1}(X)/\pi_{1}(X)^{(2)}\,|\,ts(g)t^{-1}=r(g),g\in\pi_{1}(\Sigma^{+})\rangle\\\\[2.84526pt]
&=&\langle
t,\pi_{1}(X)\,|\,ts(g)t^{-1}=r(g),g\in\pi_{1}(\Sigma^{+}),\pi_{1}(X)^{(2)}\rangle\\\\[2.84526pt]
&\to&\langle
t,\pi_{1}(X)\,|\,ts(g)t^{-1}=r(g),g\in\pi_{1}(\Sigma^{+})\rangle/(\dots)^{(2)}.\end{array}$

Clearly $\Psi$ is surjective and we have $\Phi\circ\Psi=\mbox{id}$. It follows
that $\Phi$ is an isomorphism. This concludes the proof of the claim.

Finally note that we have a canonical isomorphism

$\pi_{1}(\Sigma\times S^{1})/\pi_{1}(\Sigma\times S^{1})^{(2)}=\langle
t\rangle\times\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}.$

It now follows from the above discussion that we have an isomorphism

$\begin{array}[]{rcl}\pi_{1}(W)/\pi_{1}(W)^{(2)}&\xrightarrow{\Phi}&\langle
t\rangle\times\pi_{1}(X)/\pi_{1}(X)^{(2)}\\\ &\cong&\langle
t\rangle\times\pi_{1}(\Sigma)/\pi_{1}(\Sigma)^{(2)}\\\ &=&\pi_{1}(\Sigma\times
S^{1})/\pi_{1}(\Sigma\times S^{1})^{(2)}\end{array}$

which we again denote by $\Phi$. Note that under the abelianization the map
$\Phi$ descends to the above identification $H_{1}(\Sigma\times
S^{1};\mathbb{Z})=H=H_{1}(W;\mathbb{Z})$. We now get the following commutative
diagram of exact sequences

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{1}(W;\mathbb{Z}[H])\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(W)/\pi_{1}(W)^{(2)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Phi}$$\textstyle{H:=H_{1}(W;\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{=}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{1}(\Sigma\times
S^{1};\mathbb{Z}[H])\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(\Sigma\times
S^{1})/\pi_{1}(\Sigma\times
S^{1})^{(2)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{1}(\Sigma\times
S^{1};\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

The lemma is now immediate. ∎

We are now ready to prove the second statement of Lemma 4.5. Note that the
isomorphism of Alexander modules implies that the Alexander polynomials
$\Delta_{W}$ and $\Delta_{\Sigma\times S^{1}}$ agree in $\mathbb{Z}[H]$. It is
well–known that $\Delta_{\Sigma\times
S^{1}}=(t-1)^{||\phi||_{T}}\in\mathbb{Z}[H]=\mathbb{Z}[F][t^{\pm 1}]$. Recall
that we are interested in $\phi=PD([\Sigma])$, and that $\phi$ as an element
in $\mbox{Hom}(H,\mathbb{Z})=\mbox{Hom}(F\times\langle t\rangle),\mathbb{Z})$
is given by $\phi(t)=1,\phi|_{F}=0$. It is now obvious from
$\Delta_{W}=\Delta_{\Sigma\times S^{1}}=(t-1)^{||\phi||_{T}}$ that $\phi$ lies
in the interior of a top–dimensional face of the Alexander norm ball of $W$.
This concludes the proof of the second statement of Lemma 4.5 modulo the proof
of the claim.

##  5\. Residual properties of 3–manifold groups

Proposition 1.7 and Theorem 4.1 are almost enough to deduce Theorem 1.2, but
we still have to deal with the assumption in Theorem 4.1 that $\pi_{1}(W)$ has
to be residually finite solvable.

Using well–known arguments (see Section 7 for details) one can easily see that
Proposition 1.7 and Theorem 4.1 imply Theorem 1.2 for 3–manifolds $N$ which
have virtually residually finite solvable fundamental groups. Here we say that
a group $\pi$ has virtually a property if a finite index subgroup of $\pi$ has
this property. It seems reasonable to conjecture that all 3–manifold groups
are virtually residually finite solvable. For example linear groups (and hence
fundamental groups of hyperbolic 3–manifolds and Seifert fibered spaces) are
virtually residually finite solvable and (virtually) fibered 3–manifold groups
are easily seen to be (virtually) residually finite solvable.

It is not known though whether all 3–manifold groups are linear. In the case
of 3–manifolds with non–trivial JSJ decomposition we therefore use a slightly
different route to deduce Theorem 1.2 from Proposition 1.7 and Theorem 4.1. In
Lemmas 7.1 and 7.2 we first show that it suffices in the proof of Theorem 1.2
to consider closed prime 3–manifolds. In this section we will show that given
a closed prime 3–manifold $N$, there exists a finite cover $N^{\prime}$ of $N$
such that all pieces of the JSJ decomposition of $N^{\prime}$ have residually
finite solvable fundamental groups (Theorem 5.1). Finally in Section 6 we will
prove a result which allows us in the proof of Theorem 1.2 to work with each
JSJ piece separately (Theorem 6.4).

###  5.1. Statement of the theorem

We first recall some definitions. Let $p$ be a prime. A $p$–group is a group
such that the order of the group is a power of $p$. Note that any $p$–group is
in particular finite solvable. A group $\pi$ is called _residually $p$_ if for
any nontrivial $g\in\pi$ there exists a homomorphism $\alpha:\pi\to P$ to a
$p$–group such that $\alpha(g)\neq e$. A residually $p$ group is evidently
also residually finite solvable.

For the reader’s convenience we recall the statement of Theorem 1.9 which we
will prove in this section.

######  Theorem 5.1.

Let $N$ be a closed irreducible 3–manifold. Then for all but finitely many
primes $p$ there exists a finite cover $N^{\prime}$ of $N$ such that the
fundamental group of any JSJ component of $N^{\prime}$ is residually $p$.

######  Remark.

  1. (1)

Note that this theorem relies on the geometrization results of Thurston and
Perelman.

  2. (2)

A slight modification of our proof shows that the statement of the theorem
also holds for irreducible 3–manifolds with toroidal boundary.

###  5.2. Proof of Theorem 5.1

The proof of the theorem combines in a straightforward way ideas from the
proof that finitely generated subgroups of $\mbox{GL}(n,\mathbb{C})$ are
virtually residually $p$ for all but finitely many primes $p$ (cf. e.g. [We73,
Theorem 4.7] or [LS03, Window 7, Proposition 9]) with ideas from the proof
that 3–manifold groups are residually finite (cf. [He87]). Since all technical
results can be found in either [We73] or [He87], and in order to save space,
we only give an outline of the proof by referring heavily to [We73] and
[He87].

In the following recall that given a positive integer $n$ there exists a
unique characteristic subgroup of
$\mathbb{Z}\oplus\mathbb{Z}=\pi_{1}(\text{torus})$ of index $n^{2}$, namely
$n(\mathbb{Z}\oplus\mathbb{Z})$.

######  Definition 5.2.

Let $N$ be a $3$-manifold which is either closed or has toroidal boundary.
Given a prime $p$ we say that a subgroup $\Gamma\subset\pi_{1}(N)$ has
Property $(p)$ if it satisfies the following two conditions:

  1. (1)

$\Gamma$ is residually $p$, and

  2. (2)

for any torus $T\subset\partial N$ the group $\Gamma\cap\pi_{1}(T)$ is the
unique characteristic subgroup of $\pi_{1}(T)$ of index $p^{2}$.

######  Proposition 5.3.

Let $N$ be a compact orientable $3$-manifold with empty or toroidal boundary
such that the interior has a complete hyperbolic structure of finite volume.
Then for all but finitely many primes $p$ there exists a finite index subgroup
of $\pi_{1}(N)$ which has Property $(p)$.

###### Proof.

This theorem is essentially a straightforward combination of [He87, Lemma 4.1]
with the proof that finitely generated linear groups are virtually residually
$p$. We will use throughout the notation of the proof of [He87, Lemma 4.1].
First we pick a finitely generated subring $A\subset\mathbb{C}$ as in [He87,
Proof of Lemma 4.1]. In particular we can assume that
$\pi_{1}(N)\subset\mbox{SL}(2,A)$ where $A\subset\mathbb{C}$. We pick a prime
$p$ and a maximal ideal $\mathfrak{m}\subset A$ as in [He87, p. 391]. We then
have in particular that $\mbox{char}(A/\mathfrak{m})=p$. For $i\geq 1$ we now
let
$\Gamma_{i}=\mbox{Ker}\\{\pi_{1}(N)\to\mbox{SL}(2,A/\mathfrak{m}^{i})\times
H/p^{i}H\\}$ where $H=H_{1}(N;\mathbb{Z})/\mbox{torsion}$. We claim that
$\Gamma_{1}\subset\pi_{1}(N)$ is a finite index subgroup which has Property
$(p)$. Clearly $\Gamma_{1}$ is of finite index in $\pi_{1}(N)$ and by [He87,
p. 391] the subgroup $\Gamma_{1}$ also satisfies condition (2). The proof that
finitely generated linear groups are virtually residually $p$ (cf. [We73,
Theorem 4.7] or [LS03, Window 7, Proposition 9]) then shows immediately that
all the groups $\Gamma_{1}/\Gamma_{i},i\geq 1$ are $p$–groups and that
$\cap_{i=1}^{\infty}\Gamma_{i}=\\{1\\}$. In particular $\Gamma_{1}$ is
residually $p$. ∎

######  Proposition 5.4.

Let $N$ be a Seifert fibered space. Then for all but finitely many primes $p$,
there exists a finite index subgroup of $\pi_{1}(N)$ which has Property $(p)$.

###### Proof.

If $N$ is a closed Seifert fibered space, then it is well–known that
$\pi_{1}(N)$ is linear, and the proposition immediately follows from the fact
that linear groups are virtually residually $p$ for almost all primes $p$.

Now consider the case that $N$ has boundary. It is well–known (cf. for example
[Ha01, Lemma 6] and see also [He87, p. 391]) that there exists a finite cover
$q:N^{\prime}\to N$ with the following two properties:

  1. (1)

$N^{\prime}=S^{1}\times F$ for some surface $F$,

  2. (2)

for any torus $T\subset\partial N$ the group
$\pi_{1}(N^{\prime})\cap\pi_{1}(T)$ is the unique characteristic subgroup of
$\pi_{1}(T)$ of index $p^{2}$.

We now write $\Gamma:=\pi_{1}(N^{\prime})\subset\pi_{1}(N)$. The group
$\Gamma$ is residually $p$ since free groups are residually $p$. It now
follows from (2) that $\Gamma$ has the required properties. ∎

We are now in a position to prove Theorem 5.1.

###### Proof of Theorem 5.1.

Let $N$ be a closed irreducible 3–manifold. Let $N_{1},\dots,N_{r}$ be the JSJ
components. For all but finitely many primes $p$ we can by Propositions 5.3
and 5.4 find finite index subgroups $\Gamma_{i}\subset\pi_{1}(N_{i})$ for
$i=1,\dots,r$ which have Property $(p)$. We denote by $N_{i}^{\prime}$ the
cover of $N_{i}$ corresponding to $\Gamma_{i}$.

By the second condition of Property $(p)$ the intersections of the subgroups
$\Gamma_{i},i=1,\dots,r$ with the fundamental group of any torus of the JSJ
decomposition coincide. We can therefore appeal to [He87, Theorem 2.2] to find
a finite cover $N^{\prime}$ of $N$ such that any component in the JSJ
decomposition of $N^{\prime}$ is homeomorphic to some
$N_{i}^{\prime},i\in\\{1,\dots,r\\}$. Recall that
$\pi_{1}(N_{i}^{\prime})=\Gamma_{i}$ is residually $p$ for any $i$, hence the
cover $N^{\prime}$ of $N$ has the desired properties. ∎

##  6\. The JSJ decomposition and prosolvable completions

Let $N$ be a closed 3–manifold and let $\phi\in H^{1}(N;\mathbb{Z})$ primitive
with $||\phi||_{T}>0$. If $(N,\phi)$ fibers, and if $\Sigma\subset N$ is a
surface dual to $\phi$ which is the fiber of the fibration, then it is
well–known (cf. e.g. [EN85]) that the JSJ tori of $N$ cut the product
$N\setminus\nu\Sigma\cong\Sigma\times[0,1]$ into smaller products.

If $(N,\phi)$ satisfies Condition ($*$), and if $\Sigma\subset N$ is a
connected Thurston norm minimizing surface dual to $\phi$, then we will see in
Lemma 6.3 and Theorem 6.4 that the JSJ tori of $N$ cut the manifold
$N\setminus\nu\Sigma$ into smaller pieces which look like products ‘on the
level of prosolvable completions’. This result will play an important role in
the proof of Theorem 1.2 as it allows us to work with each JSJ piece
separately.

###  6.1. The statement of the theorem

Throughout this section let $N$ be a closed irreducible 3–manifold.
Furthermore let $\phi\in H^{1}(N;\mathbb{Z})$ be a primitive class which is
dual to a _connected_ Thurston norm minimizing surface. (Recall that by
Proposition 3.1 this is in particular the case if $(N,\phi)$ satisfies
Condition $(*)$.) Finally we assume that $||\phi||_{T}>0$.

We now fix once and for all embedded tori $T_{1},\dots,T_{r}\subset N$ which
give the JSJ decomposition of $N$. (Recall that the $T_{1},\dots,T_{r}$ are
unique up to reordering and isotopy.)

We will make several times use of the following well–known observations:

######  Lemma 6.1.

Let $\Sigma\subset N$ be an incompressible surface in general position with
the JSJ torus $T_{i},i\in 1,\dots,r$. Let $c$ be a component of $\Sigma\cap
T_{i}$. Then $c$ represents a non–trivial element in $\pi_{1}(T_{i})$ if and
only if $c$ represents a non–trivial element in $\pi_{1}(\Sigma)$.

######  Lemma 6.2.

There exists an embedded Thurston norm minimizing surface $\Sigma\subset N$
dual to $\phi$ with the following three properties:

  1. (1)

$\Sigma$ is connected,

  2. (2)

the tori $T_{i},i=1,\dots,r$ and the surface $\Sigma$ are in general position,

  3. (3)

any component of $\Sigma\cap T_{i},i=1,\dots,r$ represents a nontrivial
element in $\pi_{1}(T_{i})$.

Now, among all surfaces dual to $\phi$ satisfying the properties of the lemma
we pick a surface $\Sigma$ which minimizes the number
$\sum_{i=1}^{r}b_{0}(\Sigma\cap T_{i})$.

Given $\Sigma$ we can and will fix a tubular neighborhood
$\Sigma\times[-1,1]\subset N$ such that the tori $T_{i},i=1,\dots,r$ and the
surface $\Sigma\times t$ are in general position for any $t\in[-1,1]$. We from
now on write $M=N\setminus\Sigma\times(-1,1)$ and
$\Sigma^{\pm}=\Sigma\times\pm 1$.

We denote the components of $N$ cut along $T_{1},\dots,T_{r}$ by
$N_{1},\dots,N_{s}$. Let $\\{A_{1},\dots,A_{m}\\}$ be the set of components of
the intersection of the tori $T_{1}\cup\dots\cup T_{r}$ with $M$. Note that
the surfaces $A_{i}\subset M,i=1,\dots,m$ are properly embedded since we
assumed that the tori $T_{i}$ and the surfaces $\Sigma^{\pm}=\Sigma\times\pm
1$ are in general position. We also let $\\{M_{1},\dots,M_{n}\\}$ be the set
of components of the intersection of $N_{i}$ with $M$ for $i=1,\dots,s$. Put
differently, $M_{1},\dots,M_{n}$ are the components of $M$ cut along
$A_{1},\dots,A_{m}$. For $i=1,\dots,n$ we furthermore write
$\Sigma_{i}^{\pm}=M_{i}\cap\Sigma^{\pm}$.

Let $i\in\\{1,\dots,m\\}$. If the surface $A_{i}$ is an annulus, then we say
that $A_{i}$ _connects $\Sigma^{-}$ and $\Sigma^{+}$_ if one boundary
component of $A_{i}$ lies on $\Sigma^{-}$ and the other boundary component
lies on $\Sigma^{+}$. The following lemma will be proved in Section 6.2

######  Lemma 6.3.

Assume that $(N,\phi)$ satisfies Condition $(*)$, then for $i=1,\dots,m$ the
surface $A_{i}$ is an annulus which connects $\Sigma^{-}$ and $\Sigma^{+}$.

We can now formulate the main theorem of this section. The proof will be given
in Sections 6.2, 6.3 and 6.4.

######  Theorem 6.4.

Assume that for $i=1,\dots,m$ the surface $A_{i}$ is an annulus which connects
$\Sigma^{-}$ and $\Sigma^{+}$. Furthermore assume that the inclusion induced
maps $\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)$ give rise to an isomorphism of
prosolvable completions. Then for $i=1,\dots,n$ the following hold:

  1. (1)

The surfaces $\Sigma_{i}^{\pm}$ are connected.

  2. (2)

Given $j\in\\{1,\dots,n\\}$ with $M_{i}\subset N_{j}$ the inclusion induced
map $\pi_{1}(M_{i})\to\pi_{1}(N_{j})$ is injective.

  3. (3)

The inclusion induced maps $\pi_{1}(\Sigma_{i}^{\pm})\to\pi_{1}(M_{i})$ give
rise to isomorphisms of the respective prosolvable completions.

We would like to remind the reader that at the beginning of the section we
made the assumption that $||\phi||_{T}>0$.

###  6.2. Proof of Lemma 6.3

We first recall the following theorem from an earlier paper (cf. [FV08b,
Theorem 5.2]).

######  Theorem 6.5.

Let $Y$ be a closed irreducible 3–manifold. Let $\psi\in H^{1}(Y;\mathbb{Z})$
a primitive class. Assume that for any epimorphism $\alpha:\pi_{1}(Y)\to G$
onto a finite group $G$ the twisted Alexander polynomial
$\Delta_{Y,\psi}^{\alpha}\in\mathbb{Z}[t^{\pm 1}]$ is nonzero. Let $T\subset
Y$ be an incompressible embedded torus. Then either $\psi|_{T}\in
H^{1}(T;\mathbb{Z})$ is nonzero, or $(Y,\psi)$ fibers over $S^{1}$ with fiber
$T$.

With this theorem we are now able to prove Lemma 6.3. We use the notation from
the previous section. So assume that $(N,\phi)$ is a pair which satisfies
Condition $(*)$. In particular we have that $\Delta_{N,\phi}^{\alpha}\neq 0$
for any epimorphism $\alpha:\pi_{1}(N)\to G$ onto a finite group $G$. We can
therefore apply Theorem 6.5 to the tori $T_{1},\dots,T_{r}\subset N$ to
conclude that either $(N,\phi)$ fibers over $S^{1}$ with toroidal fiber, or
$\phi|_{T_{i}}\in H^{1}(T_{i};\mathbb{Z})$ is nonzero for $i=1,\dots,r$.
Recall that we assumed that $||\phi||_{T}>0$, we therefore only have to deal
with the latter case. From $\phi|_{T_{i}}\in H^{1}(T_{i};\mathbb{Z})$ nonzero
we obtain that $\Sigma$ (which is dual to $\phi$) necessarily intersects
$T_{i}$ in at least one curve which is homologically essential on $T_{i}$. In
fact by our assumption on $\Sigma$ and $T_{1},\dots,T_{r}$ any intersection
curve $\Sigma\cap T_{i}\subset T_{i}$ is essential, in particular the
components of $T_{i}$ cut along $\Sigma$ are indeed annuli.

In order to prove Lemma 6.3 it now remains to show that each of the annuli
$A_{i}$ connects $\Sigma^{-}$ and $\Sigma^{+}$. So assume there exists an
$i\in\\{1,\dots,m\\}$ such that the annulus $A_{i}$ does not connect
$\Sigma^{-}$ and $\Sigma^{+}$. Without loss of generality we can assume that
$\Sigma^{+}\cap A_{i}=\emptyset$. We equip $A_{i}$ with an orientation. Denote
the two oriented components of $\partial A_{i}$ by $c$ and $-d$. By our
assumption $c$ and $d$ lie in $\Sigma^{-}$, and they cobound the annulus
$A_{i}\subset M$.

Now recall that by Proposition 3.2 our assumption that $(N,\phi)$ satisfies
Condition $(*)$ implies in particular that $H_{1}(\Sigma^{-};\mathbb{Z})\to
H_{1}(M;\mathbb{Z})$ is an isomorphism. Note that $c,d$ are homologous in $M$
via the annulus $A:=A_{i}$, and since $H_{1}(\Sigma^{-};\mathbb{Z})\to
H_{1}(M;\mathbb{Z})$ is an isomorphism we deduce that $c$ and $d$ are
homologous in $\Sigma^{-}$ as well. Since $\Sigma^{-}$ is closed we can now
find two subsurfaces $\Sigma_{1},\Sigma_{2}\subset\Sigma^{-}$ such that
$\partial\Sigma_{1}=-c\cup d$, and such that (with the orientations induced
from $\Sigma^{-}$) the following hold: $\Sigma_{1}\cup\Sigma_{2}=\Sigma$,
$\partial\Sigma_{2}=c\cup-d$ and $\Sigma_{1}\cap\Sigma_{2}=c\cup d$. Note that
possibly one of $\Sigma_{1}$ or $\Sigma_{2}$ is disconnected.

######  Claim.

The surfaces $\overline{\Sigma}_{1}=\Sigma_{1}\cup A$ and
$\overline{\Sigma}_{2}=\Sigma_{2}\cup-A$ are closed, orientable and connected.
Furthermore, there exists a $j\in\\{1,2\\}$ such that
$\mbox{genus}(\overline{\Sigma}_{j})=\mbox{genus}(\Sigma)$ and such that
$\overline{\Sigma}_{j}$ is homologous to $\Sigma$ in $N$.

###### Proof.

It is clear that $\overline{\Sigma}_{1}$ and $\overline{\Sigma}_{2}$ are
closed, orientable and connected. We give $\overline{\Sigma}_{k},k=1,2$ the
orientation which restricts to the orientation of $\Sigma_{k}$. We therefore
only have to show the second claim.

Recall that Condition $(*)$ implies that the inclusion induced maps
$H_{j}(\Sigma^{-};\mathbb{Z})\to H_{j}(M;\mathbb{Z}),j=0,1$ are isomorphisms.
It follows from Lemma 2.6 that we also have an isomorphism
$H_{2}(\Sigma^{-};\mathbb{Z})\to H_{2}(M;\mathbb{Z})$, in particular
$H_{2}(M;\mathbb{Z})$ is generated by $[\Sigma^{-}]$. Now note that
$\overline{\Sigma}_{1}$ and $\overline{\Sigma}_{2}$ represent elements in
$H_{2}(M;\mathbb{Z})$. We can write
$[\overline{\Sigma}_{k}]=l_{k}[\Sigma^{-}],k=1,2$ for some
$l_{k}\in\mathbb{Z}$. Note that
$[\overline{\Sigma}_{1}]+[\overline{\Sigma}_{2}]=[\Sigma^{-}]$, i.e.
$l_{1}+l_{2}=1$.

Now let $k\in\\{1,\dots,r\\}$ such that $A_{i}\subset T_{k}$. Recall that we
assume that any component of $\Sigma\cap T_{k}$ represents a nontrivial
element in $\pi_{1}(T_{k})$. By Lemma 6.1 any component of $\Sigma\cap T_{k}$
therefore also represents a nontrivial element in $\pi_{1}(\Sigma)$. In
particular $c$ and $d$ do not bound disks on $\Sigma$, which in turn implies
that $\chi(\Sigma_{k})\leq 0,k=1,2$. It follows that

(3) $-\chi(\overline{\Sigma}_{k})=-\chi((\Sigma^{-}\setminus\Sigma_{3-k})\cup
A)=-\chi(\Sigma)+\chi(\Sigma_{3-k})\leq-\chi(\Sigma),\quad k=1,2.$

On the other hand, by the linearity of the Thurston norm and the genus
minimality of $\Sigma$ we have

(4) $-\chi(\overline{\Sigma}_{k})\geq-|l_{k}|\chi(\Sigma),\quad k=1,2.$

Now recall our assumption that $\chi(\Sigma)=||\phi||_{T}>0$. It follows that
$l_{1}+l_{2}=1$ and the inequalities (3) and (4) can only be satisfied if
there exists a $j$ with $l_{j}=1$ and
$\chi(\overline{\Sigma}_{j})=\chi(\Sigma)$. (Note that necessarily $l_{3-j}=0$
and $\overline{\Sigma}_{3-j}$ is a torus.) ∎

Note that there exists a proper isotopy of $A\subset M$ to an annulus
$A^{\prime}\subset M$ such that $\partial A^{\prime}$ lies entirely in
$\Sigma_{j}$ and such that $A^{\prime}$ is disjoint from all the other
$A_{j},j=1,\dots,r$. We then let $\Sigma_{j}^{\prime}\subset\Sigma_{j}$ be the
subsurface of $\Sigma_{j}$ such that $\partial\Sigma_{j}^{\prime}=\partial
A^{\prime}$. Clearly $\Sigma^{\prime}:=\Sigma_{j}^{\prime}\cup-A^{\prime}$ is
isotopic to $\Sigma_{j}\cup-A$, in particular by the claim $\Sigma^{\prime}$
is a closed connected surface homologous to $\Sigma$ in $N$ with
$\mbox{genus}(\Sigma^{\prime})=\mbox{genus}(\Sigma)$ which satisfies all the
properties listed in Lemma 6.2. On the other hand we evidently have
$b_{0}(\Sigma^{\prime}\cap T_{j})\leq b_{0}(\Sigma)-2$. Since we did not
create any new intersections we in fact have
$\sum_{i=1}^{r}b_{0}(\Sigma^{\prime}\cap T_{i})<\sum_{i=1}^{r}b_{0}(\Sigma\cap
T_{i})$. But this contradicts the minimality of
$\sum_{i=1}^{r}b_{0}(\Sigma\cap T_{i})$ in our choice of the surface $\Sigma$.
We therefore showed that the assumption that $A_{i}$ does not connect
$\Sigma^{-}$ and $\Sigma^{+}$ leads to a contradiction. This concludes the
proof of Lemma 6.3.

###  6.3. Preliminaries on the components $M_{1},\dots,M_{n}$

We continue with the notation from the previous sections. Using Lemma 6.3 we
can now prove the following lemma, which in particular implies the first
statement of Theorem 6.4.

######  Lemma 6.6.

Assume that the inclusion induced maps $\pi_{1}(\Sigma^{\pm})\to\pi_{1}(M)$
give rise to isomorphisms of the prosolvable completions. Let
$i\in\\{1,\dots,n\\}$. Then the following hold:

  1. (1)

The surfaces $\Sigma_{i}^{\pm}$ are connected.

  2. (2)

For any homomorphism $\alpha:\pi_{1}(M)\to S$ to a finite solvable group the
inclusion maps induce isomorphisms

$H_{j}(\Sigma_{i}^{\pm};\mathbb{Z}[S])\to H_{j}(M_{i};\mathbb{Z}[S])$

for $j=0,1$.

###### Proof.

We first consider statement (1). Recall that $M_{1},\dots,M_{n}$ are the
components of $M$ split along $A_{1},\dots,A_{m}$. We therefore get the
following commutative diagram of Mayer–Vietoris sequences

$\textstyle{\dots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bigoplus\limits_{k=1}^{m}H_{j}(A_{k}\cap\Sigma^{\pm};\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bigoplus\limits_{l=1}^{n}H_{j}(M_{l}\cap\Sigma^{\pm};\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{j}(\Sigma^{\pm};\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\dots}$$\textstyle{\dots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bigoplus\limits_{k=1}^{m}H_{j}(A_{k};\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bigoplus\limits_{l=1}^{n}H_{j}(M_{l};\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{j}(M;\mathbb{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\dots.}$

Note that the vertical homomorphisms on the left are isomorphisms since by
Lemma 6.3 for any $i=1,\dots,m$ the $A_{i}$ is an annulus which connects
$\Sigma^{-}$ and $\Sigma^{+}$, i.e. $A_{i}$ is a product on
$A_{i}\cap\Sigma^{\pm}$. Also note that the vertical homomorphisms on the
right are isomorphisms for $j=0,1$ by Proposition 3.2 and for $j=2$ by Lemma
2.6. We can now appeal to the 5–lemma to deduce that the middle homomorphisms
are isomorphisms as well. But for any $j$ the middle homomorphism is a direct
sum of homomorphisms, it therefore follows in particular that the maps
$H_{j}(\Sigma_{i}^{\pm};\mathbb{Z})\to H_{j}(M_{i};\mathbb{Z}),j=0,1$ are
isomorphisms for any $i\in\\{1,\dots,n\\}$. In particular
$b_{0}(\Sigma_{i}^{\pm})=b_{0}(M_{i})=1$, i.e. the surfaces $\Sigma_{i}^{\pm}$
are connected.

We now prove statement (2). Let $\alpha:\pi_{1}(M)\to S$ be a homomorphism to
a finite solvable group. Recall that by Lemmas 2.12 and 2.6 we have that the
inclusion induced maps $H_{j}(\Sigma^{\pm};\mathbb{Z}[S])\to
H_{j}(M;\mathbb{Z}[S])$ are isomorphisms for $j=0,1,2$. It now follows from
the commutative diagram of Mayer–Vietoris sequences as above, but with
$\mathbb{Z}[S]$–coefficients (cf. [FK06] for details) that

$H_{j}(\Sigma_{i}^{\pm};\mathbb{Z}[S])\to H_{j}(M_{i};\mathbb{Z}[S])$

is an isomorphism for any $i\in\\{1,\dots,n\\}$ and $j=0,1$.

∎

The following lemma in particular implies the second statement of Theorem 6.4.

######  Lemma 6.7.

For any pair $(i,j)$ such that $M_{i}\subset N_{j}$ we have a commutative
diagram of injective maps as follows:

$\textstyle{\pi_{1}(\Sigma_{i}^{\pm})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(M_{i})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(N_{j})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(\Sigma^{\pm})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(M)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(N).}$

###### Proof.

First note that since $\Sigma$ is incompressible we know that the two bottom
maps are injective. Furthermore recall that $N_{j}$ is a JSJ component of $N$,
i.e. a component of the result of cutting $N$ along incompressible tori, hence
$\pi_{1}(N_{j})\to\pi_{1}(N)$ is injective.

######  Claim.

For any $k\in\\{1,\dots,n\\}$ the maps
$\pi_{1}(\Sigma^{\pm}_{k})\to\pi_{1}(\Sigma)$ are injective.

Let $c$ be a component of $\Sigma\cap T_{l}$ for some $l\in\\{1,\dots,r\\}$.
Recall that by our choice of tori $T_{1},\dots,T_{r}$ the curve $c$ represents
a nontrivial element in $\pi_{1}(T_{l})$. By Lemma 6.1 the curve $c$ also
represents a nontrivial element in $\pi_{1}(\Sigma)$. In particular none of
the components of $\Sigma^{\pm}\setminus\Sigma_{k}^{\pm}$ are disks and
therefore the maps $\pi_{1}(\Sigma_{k}^{\pm})\to\pi_{1}(\Sigma^{\pm})$ are
injective. This concludes the proof of the claim.

Now let $K=\\{k\in\\{1,\dots,n\\}\,|\,M_{k}\subset N_{j}\\}$. It follows from
the claim and the above commutative diagram that for any $k\in K$ the
inclusion induced map $\pi_{1}(\Sigma_{k})\to\pi_{1}(N_{j})$ is injective,
i.e. for any $k\in K$ the surface $\Sigma_{k}\subset N_{j}$ is incompressible.
Since $M_{i}$ is a component of cutting $N_{j}$ along the incompressible
surfaces $\Sigma_{k}^{-}\subset N_{j},k\in K$ we have that
$\pi_{1}(M_{i})\to\pi_{1}(N_{j})$ is injective.

By commutativity of the above diagram we now obtain that all other maps are
injective as well. ∎

###  6.4. The conclusion of the proof of Theorem 6.4