# Free representations of outer automorphism groups of free products via
characteristic abelian coverings

Alexis Marchand

###### Abstract

Given a free product $G$, we investigate the existence of faithful free
representations of the outer automorphism group $\operatorname{Out}(G)$, or in
other words of embeddings of $\operatorname{Out}(G)$ into
$\operatorname{Out}\left(F_{m}\right)$ for some $m$. We base our work on a
paper of Bridson and Vogtmann in which they construct embeddings of
$\operatorname{Out}\left(F_{n}\right)$ into
$\operatorname{Out}\left(F_{m}\right)$ for some values of $n$ and $m$ by
interpreting $\operatorname{Out}\left(F_{n}\right)$ as the group of homotopy
equivalences of a graph $X$ of genus $n$, and by lifting homotopy equivalences
of $X$ to a characteristic abelian cover of genus $m$. In order to adapt their
method to a free product $G$, we view $G$ as the fundamental group of a graph
of groups, and we explain how to obtain notions of homotopy and coverings for
graphs of groups by considering the action on a Bass-Serre tree. Using these
notions, as well as a presentation of $\operatorname{Out}(G)$ due to Fuchs-
Rabinovich, we show for instance that $\operatorname{Out}(G)$ has a faithful
free representation when $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with
$F_{d}$ free of rank $d$ and $G_{i}$ finite abelian of order coprime to $n-1$.

##  1 Introduction

####

Given a group $G$, denote by $\operatorname{Aut}(G)$ the group of
automorphisms of $G$. Write $\operatorname{Inn}(G)$ for the normal subgroup of
$\operatorname{Aut}(G)$ consisting of conjugations by elements of $G$ and
consider the quotient

$\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn}(G).$

The group $\operatorname{Out}(G)$ is called the _outer automorphism group_ of
$G$.

The study of $\operatorname{Out}\left(F_{n}\right)$, where $F_{n}$ is the free
group of rank $n$, gained momentum following the work of Culler and Vogtmann
[7] who introduced the concept of outer space, an analogue for graphs of the
Teichmüller space of a surface, and on which
$\operatorname{Out}\left(F_{n}\right)$ acts. It seems natural to try to
generalise by studying $\operatorname{Out}(G)$ when $G$ is a free product. Let
us mention in this direction the construction of an outer space associated
with a free product by Guirardel and Levitt [14].

A possible strategy for studying $\operatorname{Out}(G)$ is to try to embed it
into a better-understood group, for instance
$\operatorname{Out}\left(F_{n}\right)$. Following the terminology coined by
Kielak [15], and by analogy with linear representations, we will call such a
morphism from a group $\Gamma$ to $\operatorname{Out}\left(F_{n}\right)$ a
_free representation_ of $\Gamma$.

We are led to hope that the existence of faithful free representations of a
group could lead to extend properties of
$\operatorname{Out}\left(F_{n}\right)$ to this group. For instance, we say
that a group $\Gamma$ satisfies the _Tits alternative_ relative to a class
$\mathcal{C}$ of groups if every subgroup of $\Gamma$ is either virtually in
$\mathcal{C}$ or contains a non-abelian free subgroup. Bestvina, Feighn and
Handel [2, 3] proved that $\operatorname{Out}\left(F_{n}\right)$, and
therefore any group admitting a faithful free representation, satisfies the
Tits alternative relative to abelian groups. This is parallel to Tits’
original result [20], according to which any linear group (i.e. any group
admitting a faithful linear representation) satisfies the Tits alternative
relative to solvable groups.

Furthermore, Collins [6] showed that, if $G$ is a free product of a finite
number of finite groups, then there is a free finite index subgroup $F$ of $G$
such that $\operatorname{Aut}(G)/F$ embeds into $\operatorname{Out}(F)$. This
enables Collins to use cohomological results of Culler and Vogtmann [7] on
$\operatorname{Out}\left(F_{n}\right)$ to deduce that $\operatorname{Out}(G)$
is virtually torsion free and of finite virtual cohomological dimension.

Motivated by the known structural results on
$\operatorname{Out}\left(F_{n}\right)$, we therefore set ourselves the task of
constructing faithful free representations of the group
$\operatorname{Out}(G)$, where $G$ is a free product.

####

Towards this goal, we take inspiration from work of Bridson and Vogtmann [4]
in which they explore embeddings
$\operatorname{Out}\left(F_{n}\right)\hookrightarrow\operatorname{Out}\left(F_{m}\right)$
for different values of $n$ and $m$. Their method consists in viewing $F_{n}$
as the fundamental group of a graph $X$ of genus $n$ and in analysing certain
characteristic abelian coverings $p:\hat{X}\rightarrow X$ with
$\pi_{1}\hat{X}\cong F_{m}$.

In this context, write $N=\pi_{1}\hat{X}\leqslant\pi_{1}X=F_{n}$. Consider the
Galois group $\operatorname{Deck}(p)$ of the covering, the group
$\operatorname{HE}(X)$ of homotopy equivalences of $X$ up to homotopy, and the
group $\operatorname{FHE}(p)$ of fibre-preserving homotopy equivalences of
$\hat{X}$ up to fibrewise homotopy (see 3.5 for a precise definition). Bridson
and Vogtmann [4, Prop. 1] prove that there is an equivalence of exact
sequences:

$1$$F_{n}/N$$\operatorname{Aut}\left(F_{n}\right)/N$$\operatorname{Out}\left(F_{n}\right)$$1$pt$1$pt$\operatorname{Deck}(p)$pt$\operatorname{FHE}(p)$pt$\operatorname{HE}\left(X\right)$pt$1$$\cong$$\cong$$\cong$

From this observation, it follows that we can construct an embedding
$\operatorname{Out}\left(F_{n}\right)\hookrightarrow\operatorname{Aut}\left(F_{n}\right)/N$
by splitting the projection
$\operatorname{FHE}(p)\twoheadrightarrow\operatorname{HE}\left(X\right)$, or
in more topological terms, by lifting homotopy equivalences of $X$ to the
covering space $\hat{X}$. But since $N$ is characteristic in $F_{n}$, there is
a restriction homomorphism
$\operatorname{Aut}\left(F_{n}\right)/N\rightarrow\operatorname{Out}(N)$,
which is injective in most cases, and therefore we can sometimes obtain an
embedding
$\operatorname{Out}\left(F_{n}\right)\hookrightarrow\operatorname{Out}(N)$ by
composition. This allows Bridson and Vogtmann [4, Cor. A] to show that, if
$m=r^{n}(n-1)+1$, with $r$ coprime to $n-1$, then there is an embedding
$\operatorname{Out}\left(F_{n}\right)\hookrightarrow\operatorname{Out}\left(F_{m}\right)$.

####

We aim to apply this method when the free group $F_{n}$ is replaced by a free
product $G$. In this case, $G$ becomes the fundamental group of a graph of
groups $\mathbb{X}$. Our first step will consist in giving a meaning to the
notions of homotopy equivalences and covering maps for graphs of groups.
Following an idea of Guirardel and Levitt [14], we will change viewpoints and
focus on the action of $G$ on a universal cover $\tilde{X}$ of $\mathbb{X}$
(in the sense of graphs of groups) rather than on the structure of
$\mathbb{X}$ itself. Our fundamental object is therefore a tree, $\tilde{X}$,
endowed with an action of the group $G$. This object will be called a _$G$
-tree_ or a _tree with $G$-action_. We explore this notion in Section 3,
leading us to an equivalence of exact sequences generalising the case of
graphs.

######  Theorem A.

Let $p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$ be a
characteristic covering of trees with actions. If $(A,G)$ is Grushko,
$Z_{G}\left(p_{*}N\right)=1$, and $\left(A^{\prime},N\right)$ is non-
contractible, then there is an equivalence of exact sequences:

$1$$G/N$$\operatorname{Aut}(G)/N$$\operatorname{Out}(G)$$1$pt$1$pt$\operatorname{Deck}(p)$pt$\operatorname{FHE}(p)$pt$\operatorname{HE}\left(A,G\right)$pt$1$$\cong$$\cong$$\cong$

####

We will then be able to use the method of Bridson and Vogtmann to construct
embeddings $\operatorname{Out}(G)\hookrightarrow\operatorname{Out}(N)$ for
some choices of a free product $G$ and a subgroup $N$ corresponding to a
characteristic abelian cover of a $G$-tree. This is the content of Section 4.
The main results will be the following.

######  Theorems B and C.

Let $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with $n\geqslant 2$, $F_{d}$
free of rank $d$, $G_{d+1},\dots,G_{n}$ abelian and not isomorphic to
$\mathbb{Z}$.

  1. (B)

If $d=0$, choose an integer $r_{i}\in\mathbb{Z}$ coprime to $n-1$ for each
$i\in\left\\{1,\dots,n\right\\}$, in such a way that $r_{i}=r_{j}$ as soon as
$G_{i}\cong G_{j}$, and set $N=G^{\prime}G_{1}^{r_{1}}\cdots G_{n}^{r_{n}}$.

  2. (C)

In general, choose an integer $r\in\mathbb{Z}$ coprime to $n-1$ and set
$N=G^{\prime}G^{r}$.

In both cases, there is an embedding
$\operatorname{Out}(G)\hookrightarrow\operatorname{Out}(N)$.

We are particularly interested in the cases where $N$ is free of finite rank,
because they give us faithful free representations of $\operatorname{Out}(G)$.
For example, we have the following two corollaries.

######  Corollary C1.

Let $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with $n\geqslant 2$, $F_{d}$
free of rank $d$, $G_{d+1},\dots,G_{n}$ finite abelian. We assume that $n-1$
is coprime to the order $\left|G_{i}\right|$ of each factor $G_{i}$. Then
there is a free subgroup $F$ of finite rank and of finite index in $G$ such
that there is an embedding

$\operatorname{Out}(G)\hookrightarrow\operatorname{Out}(F).$

In particular, $\operatorname{Out}(G)$ has a faithful free representation.

######  Corollary C2.

Let $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with $F_{d}$ free of rank $d$,
$G_{d+1},\dots,G_{n}$ finite abelian. Then there is an integer $k\geqslant 0$
such that $\operatorname{Out}\left(F_{k}\ast G\right)$ has a faithful free
representation.

It is already known that $\operatorname{Out}(G)$ satisfies the Tits
alternative relative to abelian groups whenever $G$ is a free product with the
assumptions of Corollary C1. In fact, to prove that a group $\Gamma$ satisfies
the Tits alternative relative to some class $\mathcal{C}$ of groups, it
suffices to construct a _virtual embedding_ of $\Gamma$ into another group
$\Delta$ satisfying the Tits alternative relative to $\mathcal{C}$, i.e. an
embedding of a finite index subgroup of $\Gamma$ into $\Delta$. In particular,
if $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$ with $G_{i}$ finite (not
necessarily abelian), then we can use a result of Carette [5, Theorem A] to
deduce that the projection
$p:\operatorname{Aut}(G)\twoheadrightarrow\operatorname{Out}(G)$ splits
virtually, i.e. there is a finite index subgroup $H$ of
$\operatorname{Out}(G)$ and an embedding
$i:H\hookrightarrow\operatorname{Aut}(G)$ such that $p\circ
i=\operatorname{id}_{H}$. This implies in particular that
$\operatorname{Out}(G)$ virtually embeds into $\operatorname{Aut}(G)$ and that
its image intersects $\operatorname{Inn}(G)$ trivially. Now, by Lemma 4.7 of
this text, if we pick a free characteristic finite index subgroup $F$ of $G$,
then the restriction map
$\operatorname{Aut}(G)\rightarrow\operatorname{Aut}(F)$ is injective, from
which it follows by composition that $\operatorname{Out}(G)$ virtually embeds
into $\operatorname{Out}(F)$. Since $\operatorname{Out}\left(F_{n}\right)$
satisfies the Tits alternative relative to abelian groups by the Bestvina-
Feighn-Handel Theorem [2, 3], so does $\operatorname{Out}(G)$. However, our
results give an alternative proof: under the assumptions of Corollary C1,
$\operatorname{Out}(G)$ has a faithful free representation, so it inherits the
Tits alternative from $\operatorname{Out}\left(F_{n}\right)$.

####

Using the above results, we go on to examine the special case of
$\operatorname{Out}\left(W_{n}\right)$ at the end of Section 4, where $W_{n}$
is the _universal Coxeter group_ of rank $n$, i.e. the free product of $n$
copies of $\mathbb{Z}/2$. If $n$ is even, then Corollary C1 applies and
$\operatorname{Out}\left(W_{n}\right)$ has a faithful free representation.
Looking at the construction more closely, we show that there is an embedding
$\operatorname{Out}\left(W_{n}\right)\hookrightarrow\operatorname{Out}\left(F_{m}\right)$
with $m=2^{n-1}(n-2)+1$. Computing an explicit value of $m$ such that
$\operatorname{Out}(G)$ embeds into $\operatorname{Out}\left(F_{m}\right)$ can
actually be done with the same method for all free products $G$ satisfying the
assumptions of Corollary C1, but we limit ourselves to the special case of
$W_{n}$ where the resulting formula is simple enough. We also use the example
of $\operatorname{Out}\left(W_{n}\right)$ to check that our method of
construction of free representations is not exhaustive: for instance, when
$n=3$, $\operatorname{Out}\left(W_{3}\right)$ admits a faithful free
representation not given by Corollary C1. We expect $\operatorname{Out}(G)$ to
have faithful free representations for many other free products $G$, as our
method is probably a very special trick that we can use to construct free
representations in specific cases.

#### Acknowledgements.

This work is based on the author’s Master’s thesis, which was written in
Institut Fourier, Grenoble. The author wishes to thank François Dahmani for
suggesting this topic, supervising his thesis and helping with the preparation
of this paper. The author is thankful to the anonymous referee for pointing
out work of Carette and its application to the Tits alternative for
$\operatorname{Out}(G)$ in a more general setting than that of this paper.
Financial support from an ENS Lyon studentship is gratefully acknowledged.

##  2 Notations and preliminaries

#### Graphs and trees.

We work with the definition of graphs used by Serre in [19]: a _graph_ is a
quadruple $X=\left(V(X),\vec{E}(X),o_{X},\iota_{X}\right)$, where $V(X)$ is
the set of _vertices_ , $\vec{E}(X)$ is the set of _arcs_ (or oriented edges),
$o_{X}:\vec{E}(X)\rightarrow V(X)$ maps an arc to its origin, and
$\iota_{X}:\vec{E}(X)\rightarrow\vec{E}(X)$ is a fixed-point-free involution,
mapping an arc to its inverse (we will write $\bar{e}=\iota_{X}(e)$). An
_edge_ of $X$ is a pair $\left\\{e,\bar{e}\right\\}$, for $e\in\vec{E}(X)$. A
_path_ in $X$ is an alternated sequence of vertices and arcs

$c=\left(v_{0},e_{1},v_{1},e_{2},v_{2},\dots,v_{k-1},e_{k},v_{k}\right),$

with $k\geqslant 0$,
$v_{i}=o_{X}\left(e_{i}\right)=o_{X}\left(\bar{e}_{i-1}\right)$. We set
$o_{X}(c)=v_{0}$; there is an inverse path given by

$\bar{c}=\iota_{X}(c)=\left(v_{k},\bar{e}_{k},v_{k-1},\dots,v_{2},\bar{e}_{2},v_{1},\bar{e}_{1},v_{0}\right).$

We denote by $P(X)$ the set of paths of $X$. We will sometimes write $v\in X$
rather than $v\in V(X)$.

If $X,Y$ are two graphs, a _morphism_ $f:X\rightarrow Y$ is the data of maps
$f_{V}:V(X)\rightarrow V(Y)$ and $f_{\vec{E}}:\vec{E}(X)\rightarrow P(Y)$ such
that $o_{Y}\circ f_{\vec{E}}=f_{V}\circ o_{X}$ and $\iota_{Y}\circ
f_{\vec{E}}=f_{\vec{E}}\circ\iota_{X}$. Note that we allow morphisms to send
edges to paths, including paths of length zero (which correspond to vertices).

The graph $X$ is said to be _connected_ if for all $u,v\in V(X)$, there exists
$c\in P(X)$ such that $o_{X}(c)=u$ and $o_{X}\left(\bar{c}\right)=v$.

A path $c=\left(v_{0},e_{1},v_{1},\dots,v_{k-1},e_{k},v_{k}\right)$ is said to
be _reduced_ if $e_{i}\neq\bar{e}_{i-1}$ for all $i$. A _loop_ is a path $c$
such that $o_{X}(c)=o_{X}\left(\bar{c}\right)$. A _tree_ is a connected graph
having no reduced loop.

Observe that a reduced morphism of trees (i.e. such that the image of every
arc is a reduced path) is determined by the image of vertices. Hence, we will
often define morphisms of trees without specifying the images of arcs, with
the implicit assumption that the morphism is chosen to be reduced.

#### Conjugation action and outer automorphisms.

Given a group $G$ and an element $\gamma\in G$, there is an automorphism
$\operatorname{ad}(\gamma)\in\operatorname{Aut}(G)$ defined by
$\operatorname{ad}(\gamma):g\mapsto\gamma g\gamma^{-1}$. The map
$\gamma\mapsto\operatorname{ad}(\gamma)$ defines a group homomorphism
$\operatorname{ad}:G\rightarrow\operatorname{Aut}(G)$. This homomorphism is
injective if $G$ has trivial centre. The group of _inner automorphisms_ of $G$
is the normal subgroup
$\operatorname{Inn}(G)=\operatorname{ad}(G)\trianglelefteq\operatorname{Aut}(G)$
and the group of _outer automorphisms_ is the quotient
$\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn}(G)$.

A subgroup $N$ of $G$ is _characteristic_ if it is preserved by all
automorphisms of $G$.

When we write an arrow $G\rightarrow\operatorname{Aut}(G)$ without further
precision, this will always refer to the morphism $\operatorname{ad}$.

#### Grushko’s Theorem.

The fundamental theorem for the sequel is a rigidity result for free products.
This result if often combined with the existence of free product
decompositions and stated as follows.

######  Theorem (Grushko).

If $G$ is a finitely generated group, then there are freely indecomposable
groups $G_{1},\dots,G_{m}$ not isomorphic to $\mathbb{Z}$, and a free group
$F_{r}$ of rank $r\geqslant 0$ such that

$G=G_{1}\ast\cdots\ast G_{m}\ast F_{r}.$

If we have another decomposition $G=H_{1}\ast\cdots\ast H_{n}\ast F_{s}$ with
$H_{1},\dots,H_{n}$ freely indecomposable not isomorphic to $\mathbb{Z}$ and
$F_{s}$ free of rank $s$, then $m=n$, $r=s$, and there is a permutation
$\sigma\in\mathfrak{S}_{m}$ and elements $\gamma_{1},\dots,\gamma_{m}\in G$
such that $H_{i}=\gamma_{i}G_{\sigma(i)}\gamma_{i}^{-1}$ for all $i$.

In fact, we will only need the uniqueness of the decomposition. It seems that
Kurosh should be credited with the uniqueness result, which is named
_Isomorphiesatz_ in [16]. It can be derived as a corollary of Kurosh’s
_Untergruppensatz_ (see [16]), at least in the case where $G$ has no free
factor. The original argument is combinatorial, but the _Untergruppensatz_ can
be proved simply with the help of Bass-Serre Theory (see Serre’s book [19, §
I.5.5]).

Grushko’s contribution [13] was to prove that the rank is additive for the
operation of free product. The existence of the decomposition follows easily.

##  3 Groups acting on trees, homotopy and coverings

###  3.1 Groups acting on trees

####

The aim of what follows is to set up notions of morphisms, homotopy and
coverings for graphs of groups. A definition of morphisms and covering maps
was given by Bass [1], but his definition will not be sufficient for our
purpose, because we will need to consider morphisms of graphs sending arcs to
paths, rather than only to arcs. Following an idea of Guirardel and Levitt
[14], we work at the level of trees to give appropriate definitions.

####

Given a graph $X$ with basepoint $x_{0}$, we have a free action of the
fundamental group $\pi_{1}\left(X,x_{0}\right)$ on the universal cover
$\tilde{X}$, which is a tree. Moreover, any morphism of graphs
$f:\left(X,x_{0}\right)\rightarrow\left(Y,y_{0}\right)$ induces a group
homomorphism
$f_{*}:\pi_{1}\left(X,x_{0}\right)\rightarrow\pi_{1}\left(Y,y_{0}\right)$ as
well as a morphism of trees
$\tilde{f}:\left(\tilde{X},\tilde{x}_{0}\right)\rightarrow\left(\tilde{Y},\tilde{y}_{0}\right)$,
which is uniquely defined after choosing lifts $\tilde{x}_{0},\tilde{y}_{0}$
of $x_{0},y_{0}$. Finally, $\tilde{f}$ is $\pi_{1}$-equivariant, meaning that

$\tilde{f}\left(\gamma\cdot\tilde{x}\right)=\left(f_{*}\gamma\right)\cdot\tilde{f}\left(\tilde{x}\right)$

for all $\tilde{x}\in\tilde{X}$ and $\gamma\in\pi_{1}\left(X,x_{0}\right)$.
Conversely, given two trees endowed with free actions of groups, any
equivariant morphism induces a morphism between the quotient graphs.

####

Generalising notions of morphisms to graphs of groups amounts to removing the
freeness assumption for the action of the fundamental group on the universal
cover.

######  Definition 3.1.

A _tree with action_ is a couple $(A,G)$, where $A$ is a tree and $G$ is a
group acting on $A$ without edge inversion. We also say that $A$ is a _$G$
-tree_. The group $G$ is the _fundamental group_ of $(A,G)$ and the tree $A$
is its _universal cover_.

A _morphism_ of trees with actions $\varphi:(A,G)\rightarrow(B,H)$ is a group
homomorphism $\varphi_{*}:G\rightarrow H$ together with a morphism of trees
$\varphi:A\rightarrow B$ that is equivariant, meaning that

$\varphi(g\cdot c)=\left(\varphi_{*}g\right)\cdot\varphi(c)$

for any path $c\in P(A)$ and any element $g\in G$.

If $A$ is a $G$-tree, then the quotient $G\backslash A$ has a natural
structure of graph of groups, where the class $[v]\in V\left(G\backslash
A\right)$ of the vertex $v\in V(A)$ is labelled by the isomorphism class of
the stabiliser $\operatorname{Stab}_{G}(v)$ and the class
$[e]\in\vec{E}\left(G\backslash A\right)$ is labelled by
$\operatorname{Stab}_{G}(e)$. If the action of $G$ on $A$ is free, then the
stabilisers are trivial and the quotient is in fact a graph. In any case,
Bass-Serre Theory tells us how to reconstruct $A$ (as well as $G$ and its
action) as the universal cover of the graph of groups $G\backslash A$ (see
Serre’s monograph [19]). Moreover, the notion of morphism of graphs of groups
introduced by Bass [1] is strictly more restrictive than ours: it amounts to
considering equivariant morphisms of trees, where morphisms send arcs to arcs,
rather than to paths (see [1, Prop. 2.4]).

####

In order to define good notions of homotopy for trees with actions, we first
need to characterise homotopy in graphs in combinatorial terms. Let
$f_{0},f_{1}:\left(X,x_{0}\right)\rightarrow\left(Y,y_{0}\right)$ be two
morphisms of graphs. If $f_{0}$ and $f_{1}$ are homotopic relative to $x_{0}$,
then the induced morphisms
${f_{0}}_{*},{f_{1}}_{*}:\pi_{1}\left(X,x_{0}\right)\rightarrow\pi_{1}\left(Y,y_{0}\right)$
are equal. If $f_{0}$ and $f_{1}$ are freely homotopic, then there is a
homotopy $f_{\bullet}$ between them, and if $c$ is the loop $t\mapsto
f_{t}\left(x_{0}\right)$ based at $y_{0}$ in $Y$, then we have
${f_{0}}_{*}=\operatorname{ad}(c)\circ{f_{1}}_{*}$ (where we recall that
$\operatorname{ad}(c):\gamma\mapsto c\gamma c^{-1}$). In fact this
characterises homotopy in graphs. This leads to the following definition.

######  Definition 3.2.

Let $\varphi_{0},\varphi_{1}:(A,G)\rightarrow(B,H)$ be two morphisms of trees
with actions.

  * •

We say that $\varphi_{0}$ and $\varphi_{1}$ are _homotopic_ , and we write
$\varphi_{0}\sim\varphi_{1}$, if
${\varphi_{0}}_{*}={\varphi_{1}}_{*}:G\rightarrow H$.

  * •

We say that $\varphi_{0}$ and $\varphi_{1}$ are _freely homotopic_ , and we
write
$\varphi_{0}\stackrel{{\scriptstyle\textnormal{free}}}{{\sim}}\varphi_{1}$, if
there is $h\in H$ such that
${\varphi_{0}}_{*}=\operatorname{ad}(h)\circ{\varphi_{1}}_{*}$.

###  3.2 Homotopy equivalence and reduction

######  Definition 3.3.

A morphism of trees with actions $\varphi:(A,G)\rightarrow(B,H)$ is a
_homotopy equivalence_ if there exists $\psi:(B,H)\rightarrow(A,G)$ such that
$\varphi\psi\sim\operatorname{id}_{(B,H)}$ and
$\psi\varphi\sim\operatorname{id}_{(A,G)}$. The morphism $\varphi$ is a _free
homotopy equivalence_ if there is $\psi$ such that
$\varphi\psi\stackrel{{\scriptstyle\textnormal{free}}}{{\sim}}\operatorname{id}_{(B,H)}$
and
$\psi\varphi\stackrel{{\scriptstyle\textnormal{free}}}{{\sim}}\operatorname{id}_{(A,G)}$.

We denote by $\operatorname{he}_{\bullet}(A,G)$ and $\operatorname{he}(A,G)$
the sets of homotopy equivalences and free homotopy equivalences
$(A,G)\rightarrow(A,G)$. We consider in addition the groups

$\operatorname{HE}_{\bullet}(A,G)=\operatorname{he}_{\bullet}(A,G)/\sim\quad\textrm{and}\quad\operatorname{HE}(A,G)=\operatorname{he}(A,G)/\stackrel{{\scriptstyle\textnormal{free}}}{{\sim}}.$

####

Let us examine a simple example to illustrate the notion of homotopy
equivalence of trees with actions. Figure 1 shows two $G$-trees, with
$G=\mathbb{Z}/2\ast\mathbb{Z}/2\ast\mathbb{Z}/4$, where we denote by $a,b,c$
the respective generators of the three factors
$\mathbb{Z}/2,\mathbb{Z}/2,\mathbb{Z}/4$. To depict a $G$-tree $A$, we draw
the tree $A$, on which we show the action of the group $G$ (or of its
generators only), and we give the quotient graph of groups $G\backslash A$,
which summarises the data of $(A,G)$. The vertices with trivial stabiliser are
drawn in black and the other ones are white; the edges in a given orbit are
drawn with the same colour. On Figure 1, there is a morphism from the $G$-tree
on the left to that on the right corresponding to the contraction of all green
edges (those corresponding to the generator $a$), with the identity morphism
on $G$; the induced morphism of graphs of groups consists in contracting the
unique green edge. In fact, this morphism of $G$-trees is a homotopy
equivalence, as will show the proof of Lemma 3.4 which generalises this
example.

$A_{1}$$G\backslash A_{1}$$c$$b$$a$$\mathbb{Z}/4$$\mathbb{Z}/2$$\mathbb{Z}/2$

$A_{2}$$G\backslash A_{2}$$c$$b$$a$$\mathbb{Z}/2$$\mathbb{Z}/4$$\mathbb{Z}/2$

Figure 1: Two homotopy equivalent $G$-trees $A_{1}$ and $A_{2}$, where
$G=\mathbb{Z}/2\ast\mathbb{Z}/2\ast\mathbb{Z}/4$

Lemma 3.4 is a reduction result for trees with actions: (ii) is a restatement
of the fact that, in a graph, we can contract a spanning tree to get a
homotopy equivalence with a bouquet of $n$ circles, and (iii) is an analogue
for trees with non-free actions, generalising the example of Figure 1.

######  Lemma 3.4.

Let $A$ be a $G$-tree with trivial edge stabilisers.

  1. (i)

If $A_{0}$ is a $G$-invariant subtree of $A$, then there is a homotopy
equivalence $(A,G)\sim\left(A_{0},G\right)$.

  2. (ii)

If the action of $G$ on $A$ is free, then there is a $G$-tree $B$, where $G$
acts simply transitively on vertices of $B$, and such that $(A,G)\sim(B,G)$.

  3. (iii)

If the action of $G$ on $A$ is not free, then there is a $G$-tree $B$, where
all vertices have non-trivial stabiliser, and such that $(A,G)\sim(B,G)$.

###### Proof.

  1. (i)

We define a morphism of trees $r:A\rightarrow A_{0}$, where $r(v)$ is the
unique vertex of $A_{0}$ minimising the distance to $v\in A$. Since $A_{0}$ is
$G$-invariant, $r:(A,G)\rightarrow\left(A_{0},G\right)$ is a morphism with
$r_{*}=\operatorname{id}_{G}$, and the inclusion
$\left(A_{0},G\right)\rightarrow\left(A,G\right)$ is a homotopy inverse of
$r$.

  2. (ii)

If the action of $G$ on $A$ is free, then we have a graph $X=G\backslash A$
whose fundamental group is $G$ and whose universal cover is $A$. We write $n$
for the genus of $X$, $B_{n}$ for the bouquet of $n$ circles (i.e. the unique
graph with one vertex and $n$ edges) and $\tilde{B}_{n}$ for its universal
cover. We then have a homotopy equivalence $X\rightarrow B_{n}$, which lifts
to a morphism of trees $A\rightarrow\tilde{B}_{n}$, which can be seen as a
homotopy equivalence of trees with actions because $G\cong\pi_{1}B_{n}$.
Hence, $(A,G)$ is homotopy equivalent to
$\left(\tilde{B}_{n},\pi_{1}B_{n}\right)$, which is simply transitive on
vertices.

  3. (iii)

Let $T$ be the set of vertices in $A$ with trivial stabiliser. Note that $T$
is stable under $G$ and denote by $\bar{T}=G\backslash T$ the set of orbits of
elements of $T$. For $Gv\in\bar{T}$, we pick an orbit $Gp(v)$, with $p(v)\in
A\smallsetminus T$ minimising the distance to $v$, and we write $c_{v}$ for
the unique path from $v$ to $p(v)$ in $A$ and $\bar{c}_{Gv}$ for its orbit
under $G$. The inner vertices of the path $c_{v}$ have to be in $T$. We assume
in addition that the choices $p(v)$ have been made in such a way that, if
$w\in c_{v}$, then $p(w)=p(v)$.

Let $B$ be the tree obtained from $A$ by contracting all the orbits
$\bar{c}_{Gv}$. Since the contracted set is $G$-stable, there is an induced
action of $G$ on $B$ such that the contraction morphism $q:A\rightarrow B$ is
$G$-equivariant, i.e. $q:(A,G)\rightarrow(B,G)$ is a morphism of trees with
actions, with $q_{*}=\operatorname{id}_{G}$.

In addition, we can define a morphism of trees $j:B\rightarrow A$ by
$qj(v)=v$, which gives a morphism of trees with actions
$j:(B,G)\rightarrow(A,G)$ with $j_{*}=\operatorname{id}_{G}$. Hence, $q,j$ are
inverse homotopy equivalences, and all vertices of $(B,G)$ have non-trivial
stabiliser.∎

####

For our purpose, we will need extra assumptions on $G$-trees, to ensure that
the chosen action of $G$ bears enough information on the free product
structure of $G$.

######  Definition 3.5.

A $G$-tree $A$ is said to be _Grushko_ if the following conditions are
satisfied:

  * •

$A$ is cocompact, i.e. the quotient graph $G\backslash A$ is finite,

  * •

Edge stabilisers are trivial,

  * •

Vertex stabilisers are freely indecomposable and not isomorphic to
$\mathbb{Z}$.

Observe that, given a finitely generated group $G$, a $G$-tree is Grushko if
and only if it is a Bass-Serre tree of a Grushko decomposition of $G$. With
this assumption, we get the following result, which is a first step towards a
correspondence between automorphisms of $G$ and homotopy equivalences of
$G$-trees.

######  Proposition 3.6.

Let $A$ be a Grushko $G$-tree.

  1. (i)

For all $\alpha\in\operatorname{Aut}(G)$, there is a morphism
$\varphi:(A,G)\rightarrow(A,G)$ such that $\varphi_{*}=\alpha$.

  2. (ii)

Let $\varphi:(A,G)\rightarrow(A,G)$. The following assertions are equivalent:

    1. (a)

$\varphi\in\operatorname{he}_{\bullet}(A,G)$.

    2. (b)

$\varphi\in\operatorname{he}(A,G)$.

    3. (c)

$\varphi_{*}\in\operatorname{Aut}(G)$.

###### Proof.

  1. (i)

According to Lemma 3.4, we can assume to be in one of the following two cases:

     * •

The action $G$ on $A$ is simply transitive. We then pick a vertex $v\in A$, so
that we have a bijection between $G$ and vertices of $A$ given by $g\mapsto
gv$. We then define $\varphi:A\rightarrow A$ by $\varphi(gv)=(\alpha g)v$,
which gives a morphism $\varphi:(A,G)\rightarrow(A,G)$ with
$\varphi_{*}=\alpha$.

     * •

All vertices of $A$ have non-trivial stabiliser. We then consider the graph of
groups $\mathbb{X}=G\backslash A$ (which is finite with trivial edge
stabilisers), which we equip with a spanning tree $T$ such that $A=\tilde{X}$
is the corresponding universal cover, and
$G=\pi_{1}\left(\mathbb{X},T\right)$. We can write

$G=G_{1}\ast\cdots\ast G_{k}\ast F_{r},$

with $G_{i}$ freely indecomposable not isomorphic to $\mathbb{Z}$, and we can
choose a section $s:V\left(\mathbb{X}\right)\rightarrow V(A)$ such that, if
$v_{1},\dots,v_{k}$ are the vertices of $\mathbb{X}$, then

$\operatorname{Stab}_{G}\left(s\left(v_{i}\right)\right)=G_{i}\leqslant G.$

The vertices of $A$ are the orbits of the $s\left(v_{i}\right)$ for
$i\in\left\\{1,\dots,k\right\\}$.

According to Grushko’s Theorem, there is a permutation
$\sigma\in\mathfrak{S}_{k}$, and an element $\gamma_{i}\in G$ for each $i$,
such that $\alpha\left(G_{i}\right)=\gamma_{i}G_{\sigma(i)}\gamma_{i}^{-1}$.
We define $\varphi:A\rightarrow A$ by

$\varphi\left(g\cdot s\left(v_{i}\right)\right)=\left(\alpha
g\right)\gamma_{i}\cdot s\left(v_{\sigma(i)}\right).$

We get a morphism $\varphi:(A,G)\rightarrow(A,G)$ with $\varphi_{*}=\alpha$.

  2. (ii)

It is clear that
$\ref{caract_he_1}\Rightarrow\ref{caract_he_2}\Rightarrow\ref{caract_he_3}$.
Conversely, assume that $\varphi_{*}\in\operatorname{Aut}(G)$. By (i), there
is $\psi:(A,G)\rightarrow(A,G)$ such that
$\psi_{*}=\left(\varphi_{*}\right)^{-1}$, so $\psi$ is a homotopy inverse of
$\varphi$.∎

###  3.3 The Birman exact sequence

####

Let $A$ be a $G$-tree. Given a homotopy equivalence
$\varphi\in\operatorname{he}_{\bullet}(A,G)$, there is an induced automorphism
$\varphi_{*}$ of $G$. Moreover, two homotopy equivalences $\varphi_{0}$ and
$\varphi_{1}$ are homotopic if and only if
${\varphi_{0}}_{*}={\varphi_{1}}_{*}$. Hence, we get a monomorphism
$\operatorname{HE}_{\bullet}(A,G)\hookrightarrow\operatorname{Aut}(G)$ given
by $\varphi\mapsto\varphi_{*}$. Likewise, there is a monomorphism
$\operatorname{HE}(A,G)\hookrightarrow\operatorname{Out}(G)$ making the
following diagram commute:

$\operatorname{HE}_{\bullet}(A,G)$$\operatorname{HE}(A,G)$pt$\operatorname{Aut}(G)$pt$\operatorname{Out}(G)$$\hookrightarrow$$\hookrightarrow$

With the assumptions of Proposition 3.6, the morphism
$\operatorname{HE}_{\bullet}(A,G)\rightarrow\operatorname{Aut}(G)$ must be
surjective, and since $\operatorname{Aut}(G)\rightarrow\operatorname{Out}(G)$
also is, $\operatorname{HE}(A,G)\rightarrow\operatorname{Out}(G)$ is
surjective, so the vertical arrows are isomorphisms.

The two horizontal arrows of the above diagram are quotient maps, we now
determine their kernel. Suppose that the centre $Z(G)$ is trivial (this is the
case for instance if $G$ is a free product of at least two factors). In this
case, the conjugation action of $G$ on itself induces an embedding
$G\hookrightarrow\operatorname{Aut}(G)$ which is the kernel of
$\operatorname{Aut}(G)\twoheadrightarrow\operatorname{Out}(G)$.

In fact, we have the following result.

######  Proposition 3.7 (Birman exact sequence).

Let $A$ be a Grushko $G$-tree, with $Z(G)=1$. Then there is a morphism
$\delta:G\rightarrow\operatorname{HE}_{\bullet}(A,G)$ such that the following
diagram commutes and has exact rows:

$1$$G$$\operatorname{HE}_{\bullet}(A,G)$$\operatorname{HE}(A,G)$$1$pt$1$pt$G$pt$\operatorname{Aut}(G)$pt$\operatorname{Out}(G)$pt$1$$=$$\cong$$\cong$$\delta$

All the arrows except for $\delta$ are the natural morphisms between the given
objects.

###### Proof.

After the above discussion, there remains only to construct $\delta$.

For $g\in G$, we define a homotopy equivalence
$\varphi_{g}\in\operatorname{he}_{\bullet}(A,G)$ by

$\displaystyle\varphi_{g}:v\in A$ $\displaystyle\longmapsto g\cdot v\in A,$
$\displaystyle{\varphi_{g}}_{*}:h\in G$ $\displaystyle\longmapsto ghg^{-1}\in
G,$

and we denote by $\delta(g)\in\operatorname{HE}_{\bullet}(A,G)$ the homotopy
class of $\varphi_{g}$. Since ${\varphi_{g}}_{*}$ acts on $G$ by conjugation,
this map $\delta$ makes the above diagram commutes. We know that the second
row is exact, and so is the first one. ∎

Seeing $\operatorname{HE}(A,G)$ as the _mapping class group_ of $(A,G)$, the
upper row is the analogue of the Birman exact sequence, and the morphism
$\delta$ corresponds to what some authors call the _point-pushing map_ (see
for instance [8, § 4.2]).

###  3.4 Normal coverings and Galois group

####

We now introduce a notion of coverings for trees with actions. In the setting
of graphs, if $p:\left(Y,y_{0}\right)\rightarrow\left(X,x_{0}\right)$ is a
(pointed) covering, then $Y$ and $X$ have the same universal cover
$\tilde{X}$, and the induced morphism
$p_{*}:\pi_{1}\left(Y,y_{0}\right)\rightarrow\pi_{1}\left(X,x_{0}\right)$ is
injective. Conversely, any subgroup $N\leqslant\pi_{1}\left(X,x_{0}\right)$
induces a covering $N\backslash\tilde{X}\rightarrow X$.

######  Definition 3.8.

A morphism of trees with actions
$p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$ is a _covering_ if
$p:A^{\prime}\rightarrow A$ is an isomorphism of trees and $p_{*}:N\rightarrow
G$ is a monomorphism of groups. The covering is said to be _normal_
(respectively _characteristic_) if $p_{*}N$ is a normal (respectively
characteristic) subgroup of $G$.

Consider a normal covering $p:\left(A^{\prime},N\right)\rightarrow(A,G)$. We
first try to understand the analogue of the Galois group for trees with
actions. Given $g\in G$, we set

$\displaystyle\theta(g):v\in A^{\prime}$ $\displaystyle\longmapsto
p^{-1}\left(g\cdot p(v)\right)\in A^{\prime}.$

Since $p_{*}N$ is normal in $G$, we can define a morphism
$\theta(g)_{*}:N\rightarrow N$ by making $g$ act by conjugation on $p_{*}N$:

$p_{*}\theta(g)_{*}n=g\left(p_{*}n\right)g^{-1}.$

We obtain a morphism of trees with actions
$\theta(g):\left(A^{\prime},N\right)\rightarrow\left(A^{\prime},N\right)$.

The map $\theta$ gives a group homomorphism
$G\rightarrow\operatorname{Aut}\left(A^{\prime},N\right)$. In the case of
graphs, this corresponds to the action of the fundamental group of the base
space on the covering space. The _Galois group_ (or _deck transformation
group_) of $p$ is defined by

$\operatorname{Deck}(p)=\theta(G)/\theta\left(p_{*}N\right).$

If we denote by $\bar{\theta}$ the composition
$G\xrightarrow{\theta}\theta(G)\rightarrow\operatorname{Deck}(p)$, we thus get
an exact sequence

$1\rightarrow
N\xrightarrow{p_{*}}G\xrightarrow{\bar{\theta}}\operatorname{Deck}(p)\rightarrow
1.$

####

Take for example $G=\mathbb{Z}/2\ast\mathbb{Z}/2$ the infinite dihedral group
(whose generators are denoted by $a,b$), and let
$N=\left<(ab)^{k}\right>\trianglelefteq G$. Let $A$ be a Bass-Serre tree of
$G$. Figure 2 depicts the covering
$p:\left(A,N\right)\rightarrow\left(A,G\right)$, where
$p=\operatorname{id}_{A}$, and $p_{*}:N\rightarrow G$ is the inclusion (for
$k=4$). The tree $A$ is depicted at the top, together with the action of the
generators; the quotient graph of groups $G\backslash A$ is in the lower right
corner, and $N\backslash A$ is in the lower left corner. Note that, in the
present case, $N$ is an infinite cyclic group acting by translation of length
$2k$ on $A$. In particular, the $N$-action on $A$ is free, so the graph of
groups $N\backslash A$ has trivial vertex stabilisers: it is in fact an usual
graph.

The Galois group is given by $\operatorname{Deck}(p)\cong G/N$; here, it is
the dihedral group $D_{2k}$. Its action on $\left(A,N\right)$ can be
interpreted as an action on the quotient graph $N\backslash A$: the generators
$a$ and $b$ act by reflection along two diagonals, as shown in the picture.

$A$$N\backslash A$$G\backslash A$$a$$b$$b$$a$$\mathbb{Z}/2$$\mathbb{Z}/2$
Figure 2: A covering $\left(A,N\right)\rightarrow\left(A,G\right)$ of trees
with actions

From now on, trees with actions will be represented by drawing only the
quotient graph of groups. With this convention, Figure 2 can be interpreted as
representing a tower of coverings
$\left(A,1\right)\rightarrow\left(A,N\right)\rightarrow\left(A,G\right)$
forming a commutative diagram.

###  3.5 Fibre-preserving homotopy equivalences

####

We now focus on the group $\operatorname{FHE}(p)$ of fibre-preserving homotopy
equivalences and on its relations with the other groups of interest. For a
covering of graphs $p:Y\rightarrow X$, a morphism $h:Y\rightarrow Y$ is fibre-
preserving if for all $y_{1},y_{2}\in Y$ such that
$p\left(y_{1}\right)=p\left(y_{2}\right)$, we have
$ph\left(y_{1}\right)=ph\left(y_{2}\right)$. Observe in addition that, for a
normal covering, $p\left(y_{1}\right)=p\left(y_{2}\right)$ if and only $y_{1}$
are $y_{2}$ are in the same $\pi_{1}X$-orbit.

######  Definition 3.9.

Let $p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$ be a normal
covering of trees with actions. A morphism
$\psi:\left(A^{\prime},N\right)\rightarrow\left(A^{\prime},N\right)$ is
_fibre-preserving_ is for any path $c\in P\left(A\right)$ and for any element
$g\in G$, there exists $g^{\prime}\in G$ such that

$p\psi p^{-1}\left(g\cdot c\right)=g^{\prime}\cdot p\psi p^{-1}(c).$

We denote by $\operatorname{fhe}(p)$ the set of homotopy equivalences
$\psi\in\operatorname{he}\left(A^{\prime},N\right)$ that are fibre-preserving.

####

Let $\psi\in\operatorname{fhe}(p)$. We aim to project $\psi$ to a morphism
$\varphi:(A,G)\rightarrow(A,G)$ making the following square commute:

$\left(A^{\prime},N\right)$$\left(A^{\prime},N\right)$pt$\left(A,G\right)$pt$\left(A,G\right)$$\psi$$p$$p$$\varphi$

The following lemma will allow us to construct $\varphi$.

######  Definition 3.10.

A $G$-tree $A$ is said to be _contractible_ if it is homotopy equivalent to a
point with a trivial group action.

######  Lemma 3.11.

Let $p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$ be a normal
covering of trees with actions, with $\left(A^{\prime},N\right)$ non-
contractible and $\left(A,G\right)$ with trivial edge stabilisers. If
$\psi\in\operatorname{fhe}(p)$, then for all $g\in G$, there is a unique
$\hat{g}\in G$ such that

$p\psi p^{-1}\left(g\cdot c\right)=\hat{g}\cdot p\psi p^{-1}(c)$

for any path $c\in P\left(A\right)$.

###### Proof.

Since $\left(A^{\prime},N\right)$ is not contractible and $\psi$ is a homotopy
equivalence, its image is not a point. In particular, there is an arc
$e\in\vec{E}\left(A^{\prime}\right)$ such that $\psi(e)$ is a non-trivial
path. But $\left(A,G\right)$ has trivial edge stabilisers, so
$\operatorname{Stab}_{G}\left(p\psi(e)\right)=1$.

Let $g\in G$. Since $\psi$ is fibre-preserving, there exists $\hat{g}\in G$
such that

$p\psi p^{-1}\left(g\cdot p(e)\right)=\hat{g}\cdot p\psi(e).$

For $v\in A^{\prime}$, there is a unique reduced path $c$ in $A^{\prime}$
starting at $v$ and ending with $e$ or $\bar{e}$. Since $\psi$ is fibre-
preserving, there exists $\hat{g}_{0}\in G$ such that

$p\psi p^{-1}\left(g\cdot p(c)\right)=\hat{g}_{0}\cdot p\psi(c).$ ($\ast$)

This implies in particular that

$\hat{g}_{0}\cdot p\psi\left(e\right)=p\psi p^{-1}\left(g\cdot
p(e)\right)=\hat{g}\cdot p\psi(e),$

so $\hat{g}^{-1}\hat{g}_{0}\in\operatorname{Stab}_{G}\left(p\psi(e)\right)=1$,
which means that $\hat{g}_{0}=\hat{g}$. Moreover, $\left(\ast\right)$ implies
that

$p\psi p^{-1}\left(g\cdot p(v)\right)=\hat{g}_{0}\cdot p\psi(v)=\hat{g}\cdot
p\psi(v).$

Since this identity is true for any vertex, it is also true for any path, and
the isomorphism $p:A^{\prime}\rightarrow A$ allows one to transfer it to $A$.
∎

######  Definition 3.12.

Let $p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$ be a normal
covering of trees with actions, with $\left(A^{\prime},N\right)$ non-
contractible and $\left(A,G\right)$ with trivial edge stabilisers. Let
$\psi\in\operatorname{fhe}(p)$. With the notations of Lemma 3.11, the
_projection_ of $\psi$ along $p$ is the morphism
$\varphi:(A,G)\rightarrow(A,G)$ given by

$\displaystyle\varphi:v\in A$ $\displaystyle\longmapsto p\psi p^{-1}(v)\in A,$
$\displaystyle\varphi_{*}:g\in G$ $\displaystyle\longmapsto\hat{g}\in G.$

In the covering of Figure 2, let us consider the morphism
$\psi:\left(A,N\right)\rightarrow\left(A,N\right)$ corresponding to the
rotation of the quotient graph $N\backslash A$ sending a vertex to its
neighbour in the clockwise direction. Formally, $\psi:A\rightarrow A$ is a
translation of length $1$ and $\psi_{*}=\operatorname{id}_{N}$. Then $\psi$ is
an element of $\operatorname{fhe}(p)$ and its projection $\varphi$ along $p$
is given by $\varphi_{*}:a\mapsto b$ and $b\mapsto bab$; it can be understood
as the morphism of the graph of groups $G\backslash A$ that permutes the two
vertices.

####

In order to define a good notion of homotopy in $\operatorname{fhe}(p)$, let
us come back to graphs once more. Given a covering of graphs $p:Y\rightarrow
X$ and two fibre-preserving homotopy equivalences
$h_{0},h_{1}\in\operatorname{fhe}(p)$, a fibrewise homotopy from $h_{0}$ to
$h_{1}$ is a free homotopy $h_{\bullet}$ such that $h_{t}$ is fibre-preserving
for all $t\in[0,1]$. But a fibre-preserving map can be seen as a map admitting
a projection to $X$; hence a fibrewise homotopy is a homotopy that can be
projected to $X$, which leads to the following definition.

######  Definition 3.13.

Let $p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$ be a normal
covering of trees with actions, with $\left(A^{\prime},N\right)$ non-
contractible and $\left(A,G\right)$ with trivial edge stabilisers. Two fibre-
preserving homotopy equivalences $\psi_{0},\psi_{1}\in\operatorname{fhe}(p)$
are _fibrewise homotopic_ , which we write
$\psi_{0}\stackrel{{\scriptstyle\textnormal{fibres}}}{{\sim}}\psi_{1}$, if the
respective projections $\varphi_{0},\varphi_{1}$ of $\psi_{0},\psi_{1}$
satisfy
$\left(\varphi_{0}\right)_{*}=\operatorname{ad}\left(p_{*}n\right)\circ\left(\varphi_{1}\right)_{*}$
for some $n\in N$. We consider the group

$\operatorname{FHE}(p)=\operatorname{fhe}(p)/\stackrel{{\scriptstyle\textnormal{fibres}}}{{\sim}},$

and we denote by
$\mu:\operatorname{FHE}(p)\rightarrow\operatorname{HE}\left(A,G\right)$ the
morphism induced by
$\psi\in\operatorname{fhe}(p)\mapsto\varphi\in\operatorname{he}(A,G)$, where
$\varphi$ is the projection of $\psi$.

###  3.6 Contractibility of $G$-trees

####

To define the group $\operatorname{FHE}(p)$ of fibre-preserving homotopy
equivalences of a normal covering
$p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$, we made use of the
non-contractibility of $\left(A^{\prime},N\right)$. Before going any further,
let us focus on this condition and see how it can be restated.

######  Lemma 3.14.

Let $A$ be a $G$-tree. The following assertions are equivalent:

  1. (i)

$(A,G)$ is contractible.

  2. (ii)

There is a vertex $v\in A$ such that $\operatorname{Stab}_{G}(v)=G$.

###### Proof.

The implication
$\ref{lmm_caract_contractile_2}\Rightarrow\ref{lmm_caract_contractile_1}$ is a
consequence of Lemma 3.4.(i) because (ii) means that $\\{v\\}$ is
$G$-invariant. Conversely, suppose that there is a homotopy equivalence
$\varphi:\left(\bullet,H\right)\rightarrow\left(A,G\right)$. If
$v=\varphi\left(\bullet\right)\in A$, then

$g\cdot
v=\left(\varphi_{*}\varphi_{*}^{-1}g\right)\cdot\varphi\left(\bullet\right)=\varphi\left(\left(\varphi_{*}^{-1}g\right)\cdot\bullet\right)=\varphi\left(\bullet\right)=v$

for all $g\in G$, so $\operatorname{Stab}_{G}(v)=G$.∎

######  Corollary 3.15.

If $p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$ is a covering of
trees with actions, then $\left(A^{\prime},N\right)$ is contractible if and
only if there is $v\in A$ such that
$p_{*}N\leqslant\operatorname{Stab}_{G}(v)$.∎

###  3.7 Characteristic coverings

####

For a characteristic covering, the following results highlight the relations
between the various protagonists introduced above, generalising the case of
graphs.

######  Proposition 3.16.

Let $p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$ be a
characteristic covering of trees with actions. If $(A,G)$ is Grushko, the
centraliser $Z_{G}\left(p_{*}N\right)$ is trivial, and
$\left(A^{\prime},N\right)$ is non-contractible, then we have a commutative
diagram with exact rows and columns:

$1$$G$$\operatorname{HE}_{\bullet}(A,G)$$\operatorname{HE}(A,G)$$1$pt$1$pt$\operatorname{Deck}(p)$pt$\operatorname{FHE}(p)$pt$\operatorname{HE}(A,G)$pt$1$pt$N$pt$N$pt$1$pt$1$pt$1$pt$1$$=$$=\joinrel=\joinrel=$$\delta$$p_{*}$$\bar{\theta}$$\lambda$$\eta$$\mu$

###### Proof.

The first column is obtained by construction of the Galois group and the first
row is the Birman exact sequence (see Proposition 3.7).

_Construction of $\lambda$._

Let $\varphi\in\operatorname{he}_{\bullet}(A,G)$. We set

$\psi:v\in A^{\prime}\longmapsto p^{-1}\varphi p(v)\in A^{\prime}.$

Since $p_{*}N$ is characteristic in $G$, we can define in addition
$\psi_{*}:N\rightarrow N$ by making $\varphi_{*}$ act by restriction on
$p_{*}N$, i.e. $p_{*}\psi_{*}=\varphi_{*}p_{*}$. We check that
$\psi\in\operatorname{fhe}(p)$, and that the fibrewise homotopy class of
$\psi$ only depends on the homotopy class of $\varphi$. Hence,
$\varphi\mapsto\psi$ defines a group homomorphism
$\lambda:\operatorname{HE}_{\bullet}(A,G)\rightarrow\operatorname{FHE}(p)$.
Moreover, $\varphi$ is the projection of $\psi$, so the right-hand square
commutes.

_Construction of $\eta$._

The morphism $\theta:G\rightarrow\operatorname{Aut}\left(A^{\prime},N\right)$
used to construct the Galois group (see 3.4) has its image contained in
$\operatorname{fhe}(p)$, so we can consider the inclusion homomorphism
$\eta_{0}:\theta(G)\rightarrow\operatorname{FHE}(p)$. Since
$\theta\left(p_{*}N\right)\subseteq\operatorname{Ker}\eta_{0}$, there is an
induced morphism $\eta:\operatorname{Deck}(p)\rightarrow\operatorname{FHE}(p)$
such that $\eta\bar{\theta}=\eta_{0}\theta$. Moreover, the left-hand square
commutes because
$\eta\bar{\theta}(g)_{*}=\eta_{0}\theta(g)_{*}=\lambda\delta(g)_{*}$.

_Exactness of the second row._

The morphism $\mu$ is surjective because
$\operatorname{HE}_{\bullet}(A,G)\rightarrow\operatorname{HE}(A,G)$ is
surjective. The morphism $\eta$ is injective: if
$\theta(g)\stackrel{{\scriptstyle\textnormal{fibres}}}{{\sim}}\operatorname{id}_{\left(A^{\prime},N\right)}$,
then there is $\nu\in N$ such that $\theta(g)_{*}=\operatorname{ad}(\nu)$.
This implies that
$\operatorname{ad}(g)_{\left|p_{*}N\right.}=\operatorname{ad}\left(p_{*}\nu\right)_{\left|p_{*}N\right.}$,
hence $g^{-1}\left(p_{*}\nu\right)\in Z_{G}\left(p_{*}N\right)=1$, so $g\in
p_{*}N$ and $\bar{\theta}(g)=1$. Moreover, $\mu\eta\bar{\theta}=1$, and
$\bar{\theta}$ is an epimorphism, so $\mu\eta=1$, i.e.
$\operatorname{Im}\eta\subseteq\operatorname{Ker}\mu$. Conversely, if
$\psi\in\operatorname{Ker}\mu$, then the projection $\varphi$ of $\psi$ is
freely homotopic to $\operatorname{id}_{\left(A,G\right)}$, so there is $g\in
G$ such that $\varphi_{*}=\operatorname{ad}(g)$, which implies that
$\psi\stackrel{{\scriptstyle\textnormal{fibres}}}{{\sim}}\theta(g)$, so
$\psi\in\operatorname{Im}\eta$.

We obtain the exactness of the second column by chasing the diagram. ∎

######  Theorem A.

Let $p:\left(A^{\prime},N\right)\rightarrow\left(A,G\right)$ be a
characteristic covering of trees with actions. If $(A,G)$ is Grushko,
$Z_{G}\left(p_{*}N\right)=1$, and $\left(A^{\prime},N\right)$ is non-
contractible, then there is an equivalence of exact sequences:

$1$$G/N$$\operatorname{Aut}(G)/N$$\operatorname{Out}(G)$$1$pt$1$pt$\operatorname{Deck}(p)$pt$\operatorname{FHE}(p)$pt$\operatorname{HE}\left(A,G\right)$pt$1$$\cong$$\cong$$\cong$

###### Proof.

We obtain the result by combining Proposition 3.16 with the Birman exact
sequence (Proposition 3.7). ∎

######  Remarks.

  1. (i)

If we restrict our attention to trees with free actions, then the theory
developed above is the theory of homotopy and coverings of graphs. Therefore,
Theorem A is a generalisation of the result used by Bridson and Vogtmann for
graphs [4, Prop. 1].

  2. (ii)

Another possible definition of fibrewise homotopy would be the following:
$\psi_{0},\psi_{1}\in\operatorname{fhe}(p)$ are fibrewise homotopic if they
are freely homotopic (in $\left(A^{\prime},N\right)$) and their respective
projections $\varphi_{0},\varphi_{1}$ are freely homotopic (in $(A,G)$). In
general, this definition is less restrictive than that given above: in the
example of Figure 2 (which does not satisfy $Z_{G}\left(p_{*}N\right)=1$), if
$\psi\in\operatorname{fhe}(p)$ is the rotation of angle $\pi/k$ of the graph
$N\backslash A$, then we can check that $\psi$ is fibrewise homotopic to
$\psi^{3}$ for the second definition but not for the first one. However, the
proofs of Proposition 3.16 and Theorem A remain valid for both definitions of
fibrewise homotopy. In particular, the second definition yields the same group
$\operatorname{FHE}(p)$ under the assumptions of Theorem A.

##  4 Free representations of $\operatorname{Out}(G)$

###  4.1 Characteristic abelian covers associated with a free product

####

Our goal is now to apply Bridson and Vogtmann’s method, using the geometric
intuition given by the above study of group actions on trees, to construct
embeddings of $\operatorname{Out}(G)$, where $G$ is a free product. We first
limit ourselves to the case where $G$ is a free product of groups that are
freely indecomposable and not isomorphic to $\mathbb{Z}$. Moreover, we will
need to assume that the factors are abelian. In 4.5 and 4.6, we will see how
to deal with the case where some of the factors are isomorphic to $\mathbb{Z}$
(i.e. where $G$ has a free factor).

The characteristic cover $\hat{X}$ of the graph $X$ chosen by Bridson and
Vogtmann in [4] to construct embeddings of
$\operatorname{Out}\left(F_{n}\right)$ is an abelian cover, i.e. a normal
cover whose Galois group is abelian. A good reason to make this choice is the
simplicity of the geometry of $\hat{X}$: it is a toric grid. If we replace the
graph $X$ by a $G$-tree $A$, we will see that abelian coverings of $(A,G)$
have a geometry that is close to that of a torus, and this analogy will guide
us towards the construction of embeddings of $\operatorname{Out}(G)$.

The first step is to choose an abelian cover of $(A,G)$ that is indeed
characteristic.

######  Lemma 4.1.

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product of groups that are freely
indecomposable and not isomorphic to $\mathbb{Z}$. Choose integers
$r_{1},\dots,r_{n}\in\mathbb{Z}$ such that $r_{i}=r_{j}$ as soon as
$G_{i}\cong G_{j}$. Then the subgroup $N=G^{\prime}G_{1}^{r_{1}}\cdots
G_{n}^{r_{n}}$ is characteristic in $G$.

###### Proof.

Let $\varphi\in\operatorname{Aut}(G)$. By Grushko’s Theorem, there is a
permutation $\sigma\in\mathfrak{S}_{n}$ and elements
$\gamma_{1},\dots,\gamma_{n}\in G$ such that

$\varphi\left(G_{i}\right)=\gamma_{i}G_{\sigma(i)}\gamma_{i}^{-1}.$

Note that $r_{i}=r_{\sigma(i)}$ for all $i$ because $G_{i}\cong
G_{\sigma(i)}$. We have

$\varphi\left(G^{\prime}\right)\subseteq G^{\prime}\subseteq N$

because $G^{\prime}$ is characteristic in $G$. Moreover,

$\varphi\left(G_{i}^{r_{i}}\right)=\gamma_{i}G_{\sigma(i)}^{r_{i}}\gamma_{i}^{-1}\subseteq\left[\gamma_{i},G_{\sigma(i)}^{r_{i}}\right]G_{\sigma(i)}^{r_{i}}\subseteq
G^{\prime}G_{\sigma(i)}^{r_{\sigma(i)}}\subseteq N.$

Hence,
$\varphi(N)\subseteq\varphi\left(G^{\prime}\right)\varphi\left(G_{1}^{r_{1}}\right)\cdots\varphi\left(G_{n}^{r_{n}}\right)\subseteq
N$. ∎

####

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product and
$N=G^{\prime}G_{1}^{r_{1}}\cdots G_{n}^{r_{n}}$ as in Lemma 4.1. We have an
action of $G$ on a Bass-Serre tree $A$, and a characteristic covering of trees
with actions

$p:\left(A,N\right)\rightarrow\left(A,G\right),$

where $p:A\rightarrow A$ is the identity and $p_{*}:N\rightarrow G$ is the
inclusion. We aim to understand the geometry of $\left(A,N\right)$.

The case where the factors $G_{i}$ are abelian is particularly interesting
because of the following lemma.

######  Lemma 4.2.

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product of abelian groups. If $A$
is a Bass-Serre tree of the free product, then the commutator subgroup
$G^{\prime}$ acts freely on $A$. In particular, $G^{\prime}$ is free.

###### Proof.

If $v$ is a vertex of $A$, then its stabiliser under $G$ is given by

$\operatorname{Stab}_{G}(v)=\gamma G_{i}\gamma^{-1},$

with $\gamma\in G$ and $i\in\left\\{1,\dots,n\right\\}$. Since the
abelianisation of $G$ is the direct product $G_{1}\times\cdots\times G_{n}$,
the stabiliser of $v$ under $G^{\prime}$ is given by

$\operatorname{Stab}_{G^{\prime}}(v)=\operatorname{Stab}_{G}(v)\cap
G^{\prime}=\gamma G_{i}\gamma^{-1}\cap\operatorname{Ker}\left(G\rightarrow
G_{1}\times\cdots\times G_{n}\right)=1.$

It follows that $G^{\prime}$ acts freely on $A$, so $G^{\prime}$ is free. ∎

####

Hence, if the factors are abelian, then the maximal abelian cover
$\left(A,G^{\prime}\right)$ is a tree with free action and corresponds to a
graph. For the sake of simplicity, we may assume that the chosen
characteristic subgroup $N=G^{\prime}G_{1}^{r_{1}}\cdots G_{n}^{r_{n}}$ is
actually equal to $G^{\prime}$ by taking $G_{i}$ finite abelian and
$r_{i}=\left|G_{i}\right|$.

$G_{1}^{\prime}\backslash A_{1}$$G_{1}\backslash
A_{1}$$v_{0}$$\mathbb{Z}/3$$\mathbb{Z}/2$ (a)
$G_{1}=\mathbb{Z}/2\ast\mathbb{Z}/3$

$G_{2}^{\prime}\backslash A_{2}$$G_{2}\backslash
A_{2}$$v_{0}$$\mathbb{Z}/4$$\mathbb{Z}/3$$\mathbb{Z}/2$ (b)
$G_{2}=\mathbb{Z}/2\ast\mathbb{Z}/3\ast\mathbb{Z}/4$

Figure 3: Examples of maximal abelian coverings associated with free products

For the tree $A$, we can take the universal cover (in the sense of graphs of
groups) of a star-shaped tree of groups $\mathbb{X}$ with $n$ vertices
labelled by $G_{1},\dots,G_{n}$, all linked to a common unlabelled vertex (see
Figure 3). To represent the graph $G^{\prime}\backslash A$, we first focus on
the fibre of the unlabelled vertex $v_{0}$ of $\mathbb{X}=G\backslash A$ (in
black on Figure 3). The group $\operatorname{Deck}(p)\cong G/G^{\prime}\cong
G_{1}\times\cdots\times G_{n}$ acts simply transitively on this fibre, so the
latter is a toric grid. Since $A$ is the universal cover (in the sense of
graphs) of $G^{\prime}\backslash A$, we understand the local geometry of
$G^{\prime}\backslash A$: each vertex of the fibre of $v_{0}$ is the endpoint
of one edge for each factor $G_{i}$. The other endpoint of such an edge is a
vertex of the fibre of the vertex labelled by $G_{i}$, which has degree
$\left|G_{i}\right|$. We know in addition that $\operatorname{Deck}(p)\cong
G_{1}\times\cdots\times G_{n}$ acts on the vertices of the fibre of $v_{0}$ in
a toric way, and $G_{i}\subseteq\operatorname{Deck}(p)$ stabilises the whole
fibre of the vertex of $\mathbb{X}$ labelled by $G_{i}$. This allows one to
obtain a drawing of $G^{\prime}\backslash A$ (which we see as a representation
of $\left(A,G^{\prime}\right)$) as in Figure 3.

We will keep in mind this image of characteristic abelian covers
$\left(A,N\right)$, remembering that in the general case where the factors
$G_{i}$ are neither abelian nor finite, the action of $N$ on $A$ may not be
free, so the graph of groups $N\backslash A$ may have labelled vertices.

###  4.2 Generators and relations of $\operatorname{Out}(G)$

####

To split the projection
$\operatorname{FHE}(p)\twoheadrightarrow\operatorname{HE}(A,G)$, we will use a
presentation of $\operatorname{HE}(A,G)\cong\operatorname{Out}(G)$, as in
Bridson and Vogtmann’s paper [4]. We use the work of Fuchs-Rabinovich [9], who
gives a presentation of $\operatorname{Aut}(G)$ when $G$ is a free product. It
will then suffice to add relations corresponding to inner automorphisms to get
a presentation of $\operatorname{Out}(G)$.

If $G$ has no free factor, Fuchs-Rabinovich [9] defines three kinds of
generators for $\operatorname{Aut}(G)$.

######  Notation 4.3.

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product of groups that are freely
indecomposable and not isomorphic to $\mathbb{Z}$.

Factor automorphisms:

For $i\in\left\\{1,\dots,n\right\\}$, and
$\bar{\varphi}_{i}\in\operatorname{Aut}\left(G_{i}\right)$, we set

$\displaystyle\varphi_{i}:g_{i}\in G_{i}$
$\displaystyle\mapsto\bar{\varphi}_{i}\left(g_{i}\right),$ $\displaystyle
g_{j}\in G_{j}$ $\displaystyle\mapsto g_{j}\quad\textrm{if $j\neq i$}.$

We write
$\Phi_{i}=\left\\{\varphi_{i},\>\bar{\varphi}_{i}\in\operatorname{Aut}\left(G_{i}\right)\right\\}$.
Elements of $\Phi_{i}$ are called _automorphisms of the factor $G_{i}$_ and we
write $\Phi=\bigcup_{i}\Phi_{i}$.

Permutation automorphisms:

Given $i,j\in\left\\{1,\dots,n\right\\}$ such that $G_{i}\cong G_{j}$, we pick
an isomorphism $\bar{\omega}_{ij}:G_{i}\rightarrow G_{j}$. We assume that
choices have been made in such a way that
$\bar{\omega}_{ii}=\operatorname{id}_{G_{i}}$ and
$\bar{\omega}_{jk}\bar{\omega}_{ij}=\omega_{ik}$. If $i\neq j$, we consider

$\displaystyle\omega_{ij}:g_{i}\in G_{i}$
$\displaystyle\mapsto\bar{\omega}_{ij}\left(g_{i}\right),$ $\displaystyle
g_{j}\in G_{j}$ $\displaystyle\mapsto\bar{\omega}_{ji}\left(g_{j}\right),$
$\displaystyle g_{k}\in G_{k}$ $\displaystyle\mapsto g_{k}\quad\textrm{if
$k\neq i,j$}.$

Note that $\omega_{ij}^{2}=\operatorname{id}_{G}$. We set
$\Omega=\left\\{\omega_{ij},\>i\neq j,\>G_{i}\cong G_{j}\right\\}$. Elements
of $\Omega$ are called _permutation automorphisms_.

Dehn twists:

Given $i\neq j\in\left\\{1,\dots,n\right\\}$ and $\gamma_{i}\in G_{i}$, we
define

$\displaystyle\alpha_{ij}^{\left(\gamma_{i}\right)}:g_{j}\in G_{j}$
$\displaystyle\mapsto\gamma_{i}^{-1}g_{j}\gamma_{i},$ $\displaystyle g_{k}\in
G_{k}$ $\displaystyle\mapsto g_{k}\quad\textrm{if $k\neq j$}.$

We set $A_{ij}=\left\\{\alpha_{ij}^{\left(\gamma_{i}\right)},\>\gamma_{i}\in
G_{i}\right\\}$ and $A=\bigcup_{i\neq j}A_{ij}$. Elements of $A$ are called
_Dehn twists_.

A system of relations for $\operatorname{Aut}(G)$ is given by the following
theorem. Observe that we consider the groups
$\operatorname{Aut}\left(G_{i}\right)\cong\Phi_{i}$ and $G_{i}\cong A_{ij}$ as
given.

######  Theorem 4.4 (Fuchs-Rabinovich [9]).

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product of groups that are freely
indecomposable and not isomorphic to $\mathbb{Z}$. Then
$\operatorname{Aut}(G)$ is generated by factor automorphisms, permutation
automorphisms and Dehn twists. Moreover, the following relations give a
presentation of $\operatorname{Aut}(G)$:

  1. (1)

$\varphi_{i}\varphi_{j}=\varphi_{j}\varphi_{i}$ for $\varphi_{i}\in\Phi_{i}$,
$\varphi_{j}\in\Phi_{j}$, $i\neq j$.

  2. (2)

$\varphi_{i}\alpha_{jk}^{\left(\gamma_{j}\right)}=\alpha_{jk}^{\left(\gamma_{j}\right)}\varphi_{i}$
for $\varphi_{i}\in\Phi_{i}$, $\gamma_{j}\in G_{j}$, $i\neq j$.

  3. (3)

$\varphi_{i}\alpha_{ik}^{\left(\gamma_{i}\right)}=\alpha_{ik}^{\left(\varphi_{i}\gamma_{i}\right)}\varphi_{i}$
for $\varphi_{i}\in\Phi_{i}$, $\gamma_{i}\in G_{i}$.

  4. (4)

$\alpha_{ij}^{\left(\gamma_{i}\right)}\alpha_{k\ell}^{\left(\gamma_{k}\right)}=\alpha_{k\ell}^{\left(\gamma_{k}\right)}\alpha_{ij}^{\left(\gamma_{i}\right)}$
for $\gamma_{i}\in G_{i}$, $\gamma_{k}\in G_{k}$, $k\neq j$ and $\ell\neq
i,j$.

  5. (5)

$\alpha_{ki}^{\left(\gamma_{k}\right)}\alpha_{ij}^{\left(\gamma_{i}\right)}\left(\alpha_{ki}^{\left(\gamma_{k}\right)}\right)^{-1}=\left(\alpha_{kj}^{\left(\gamma_{k}\right)}\right)^{-1}\alpha_{ij}^{\left(\gamma_{i}\right)}\alpha_{kj}^{\left(\gamma_{k}\right)}$
for $\gamma_{i}\in G_{i}$, $\gamma_{k}\in G_{k}$.111A misprint in [9] was
corrected in [12, § 4].

  6. (6)

$\omega_{ij}\varphi_{i}=\varphi_{j}\omega_{ij}$ if $\varphi_{i}\in\Phi_{i}$
and $\varphi_{j}\in\Phi_{j}$ are related by
$\left(\omega_{ij}\varphi_{i}\right)_{\left|G_{i}\right.}=\left(\varphi_{j}\omega_{ij}\right)_{\left|G_{i}\right.}$.

  7. (7)

$\omega_{ij}\varphi_{k}=\varphi_{k}\omega_{ij}$ for $k\neq i,j$.

  8. (8)

$\omega_{ij}\alpha_{ij}^{\left(\gamma_{i}\right)}=\alpha_{ji}^{\left(\omega_{ij}\gamma_{i}\right)}\omega_{ij}$
for $\gamma_{i}\in G_{i}$.

  9. (9)

$\omega_{ij}\alpha_{ik}^{\left(\gamma_{i}\right)}=\alpha_{jk}^{\left(\omega_{ij}\gamma_{i}\right)}\omega_{ij}$
for $\gamma_{i}\in G_{i}$.

  10. (10)

$\omega_{ij}\alpha_{ki}^{\left(\gamma_{k}\right)}=\alpha_{kj}^{\left(\gamma_{k}\right)}\omega_{ij}$
for $\gamma_{k}\in G_{k}$.

  11. (11)

$\omega_{ij}\alpha_{k\ell}^{\left(\gamma_{k}\right)}=\alpha_{k\ell}^{\left(\gamma_{k}\right)}\omega_{ij}$
for $\gamma_{k}\in G_{k}$, $k,\ell\neq i,j$.

####

To obtain a presentation of $\operatorname{Out}(G)$, it suffices to add
relations whose normal closure is the subgroup $\operatorname{Inn}(G)$ of
inner automorphisms. To write the presentation, we identify generators of
$\operatorname{Aut}(G)$ with their image in $\operatorname{Out}(G)$.

######  Corollary 4.5.

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product of groups that are freely
indecomposable and not isomorphic to $\mathbb{Z}$. Then
$\operatorname{Out}(G)$ is generated by factor automorphisms, permutation
automorphisms and Dehn twists. Moreover, relations
$\ref{thrm_pres_autg_r1}-\ref{thrm_pres_autg_r11}$ of Theorem 4.4, together
with relation (0) below, give a presentation of $\operatorname{Out}(G)$:

  1. (0)

$\varphi_{i}\prod_{j\neq i}\alpha_{ij}^{\left(\gamma_{i}\right)}=1$ for
$\gamma_{i}\in G_{i}$, with $\varphi_{i}\in\Phi_{i}$ given by
${\varphi_{i}}_{\left|G_{i}\right.}=\operatorname{ad}\left(\gamma_{i}^{-1}\right)$.∎

###  4.3 Lifting homotopy equivalences

####

Now that we have a presentation of
$\operatorname{HE}(A,G)\cong\operatorname{Out}(G)$, we are ready to lift
homotopy equivalences of the tree with action $(A,G)$ to a characteristic
abelian cover $(A,N)$. In fact, we will work in the algebraic setting, i.e. we
will split the epimorphism
$\operatorname{Aut}(G)/N\twoheadrightarrow\operatorname{Out}(G)$, but we will
follow the geometric intuition given by the correspondence with coverings of
trees with actions. The proof could also be written in the language of trees
with actions, but it would be heavier in the present case.

Let $S$ denote the generating set of $\operatorname{Out}(G)$ introduced above.
Observe that each generator $s\in S$ was defined by specifying a
representative, which we will write $s_{0}$, of $s$ in
$\operatorname{Aut}(G)$. We write $R\subseteq F(S)$ for the set of relations
of $\operatorname{Aut}(G)$ and $R_{1}$ for the extra relations of
$\operatorname{Out}(G)$. We thus have $\operatorname{Out}(G)=\left<S\mid R\cup
R_{1}\right>$. Given a word $w$ in the free group $F(S)$, we denote by
$w_{0}\in F\left(S_{0}\right)$ the corresponding word obtained after
substituting $s\mapsto s_{0}$ (with $S_{0}=\left\\{s_{0},\>s\in S\right\\}$).

We aim to define a splitting of the projection
$\operatorname{Aut}(G)/N\twoheadrightarrow\operatorname{Out}(G)$. We need to
associate an element $\hat{s}\in\operatorname{Aut}(G)$ with each generator
$s\in S$ in such a way that

$\hat{s}^{-1}s_{0}\in\operatorname{Inn}(G),$

and that for all $r\in R\cup R_{1}$, the lift $\hat{r}$ of the relation $r$ to
$\operatorname{Aut}(G)$ satisfies

$\hat{r}\in\operatorname{Inn}(N).$

If we take $\hat{s}=s_{0}$ for all $s$, the relations $r\in R$ which hold in
$\operatorname{Aut}(G)$ will automatically satisfy $\hat{r}=r_{0}=1$, but the
relations of $R_{1}$ will not lift trivially.

We thus need to correct the lifts $\hat{s}$ of the different generators $s\in
S$ in order to satisfy all the relations of $\operatorname{Out}(G)$. When $G$
is a free group, Bridson and Vogtmann [4] correct the lifts of the generators
corresponding to transvections by a translation along an axis; in other words,
they make a component of the Galois group (which splits as a direct product)
act on the lifts. Given the action of the Galois group for trees with actions
(see 3.4), this amounts to composing the lifts of some generators by a
conjugation. This idea leads to the following proof.

######  Proposition 4.6.

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product of abelian groups that are
not isomorphic to $\mathbb{Z}$. For each $i\in\left\\{1,\dots,n\right\\}$, we
pick an integer $r_{i}\in\mathbb{Z}$ coprime to $n-1$, so that $r_{i}=r_{j}$
as soon as $G_{i}\cong G_{j}$. We consider the characteristic subgroup
$N=G^{\prime}G_{1}^{r_{1}}\cdots G_{n}^{r_{n}}$ of $G$.

Then the projection
$\operatorname{Aut}(G)/N\twoheadrightarrow\operatorname{Out}(G)$ splits.

###### Proof.

For all $i\in\left\\{1,\dots,n\right\\}$, since $r_{i}$ is coprime to $n-1$,
we may pick integers $u_{i},t_{i}\in\mathbb{Z}$ such that
$u_{i}(n-1)+t_{i}r_{i}=1$. We then lift the generators of
$\operatorname{Out}(G)$ as follows:

  * •

For a factor automorphism $\varphi_{i}\in\Phi_{i}$, we take the standard lift:
$\hat{\varphi}_{i}=\left(\varphi_{i}\right)_{0}$.

  * •

For a permutation automorphism $\omega_{ij}\in\Omega$, we take the standard
lift: $\hat{\omega}_{ij}=\left(\omega_{ij}\right)_{0}$.

  * •

For a Dehn twist $\alpha_{ij}^{\left(\gamma_{i}\right)}\in A_{ij}$, we correct
the standard lift by

$\hat{\alpha}_{ij}^{\left(\gamma_{i}\right)}=\operatorname{ad}\left(\gamma_{i}^{u_{i}}\right)\circ\left(\alpha_{ij}^{\left(\gamma_{i}\right)}\right)_{0}.$

We have to check that the images of $\hat{\varphi}_{i}$, $\hat{\omega}_{ij}$
and $\hat{\varphi}_{ij}^{\left(\gamma_{i}\right)}$ in
$\operatorname{Aut}(G)/N$ satisfy the relations
$\ref{cons_pres_outg_12}-\ref{thrm_pres_autg_r11}$ of $\operatorname{Out}(G)$.

Since the lifts of factor automorphisms and permutation automorphisms are not
corrected, the relations involving these types of automorphisms only are
automatically satisfied. This is the case of relations (1), (6) and (7). There
remains four kinds of relations:

$\ref{thrm_pres_autg_r2}-\ref{thrm_pres_autg_r3}$

The relations between factor automorphisms and Dehn twists. We check by
computation that they are satisfied in $\operatorname{Aut}(G)$, and therefore
in $\operatorname{Aut}(G)/N$.

$\ref{thrm_pres_autg_r8}-\ref{thrm_pres_autg_r11}$

The relations between permutation automorphisms and Dehn twists. We check by
computation that they are satisfied in $\operatorname{Aut}(G)$, and therefore
in $\operatorname{Aut}(G)/N$.

$\ref{thrm_pres_autg_r4}-\ref{thrm_pres_autg_r5}$

The relations between Dehn twists. We check that

$\displaystyle\hat{\alpha}_{ij}^{\left(\gamma_{i}\right)}\hat{\alpha}_{k\ell}^{\left(\gamma_{k}\right)}$
$\displaystyle=\operatorname{ad}\left(\left[\gamma_{i}^{u_{i}},\gamma_{k}^{u_{k}}\right]\right)\circ\hat{\alpha}_{k\ell}^{\left(\gamma_{k}\right)}\hat{\alpha}_{ij}^{\left(\gamma_{i}\right)},$
$\displaystyle\hat{\alpha}_{ki}^{\left(\gamma_{k}\right)}\hat{\alpha}_{ij}^{\left(\gamma_{i}\right)}\left(\hat{\alpha}_{ki}^{\left(\gamma_{k}\right)}\right)^{-1}$
$\displaystyle=\operatorname{ad}\left(\left[\gamma_{k}^{u_{k}-1},\gamma_{i}^{u_{i}}\right]\left[\gamma_{i}^{u_{i}},\gamma_{k}^{-u_{k}}\right]\right)\circ\left(\hat{\alpha}_{kj}^{\left(\gamma_{k}\right)}\right)^{-1}\hat{\alpha}_{ij}^{\left(\gamma_{i}\right)}\hat{\alpha}_{kj}^{\left(\gamma_{k}\right)},$

so these relations are satisfied in $\operatorname{Aut}(G)/N$ (but not in
$\operatorname{Aut}(G)$).

(0)

The relation of $\operatorname{Out}(G)$. If $\varphi_{i}\in\Phi_{i}$ is given
by
${\varphi_{i}}_{\left|G_{i}\right.}=\operatorname{ad}\left(\gamma_{i}^{-1}\right)$,
we see that

$\hat{\varphi}_{i}\prod_{j\neq
i}\hat{\alpha}_{ij}^{\left(\gamma_{i}\right)}=\operatorname{ad}\left(\gamma_{i}^{u_{i}(n-1)-1}\right)\circ\hat{\varphi}_{i}=\operatorname{ad}\left(\gamma_{i}^{t_{i}r_{i}}\right)\circ\hat{\varphi}_{i}.$

Since $G_{i}$ is abelian, $\hat{\varphi}_{i}=\operatorname{id}_{G}$, and since
$G_{i}^{r_{i}}\subseteq N$, we have $\gamma_{i}^{t_{i}r_{i}}\in N$, so the
last relation is satisfied in $\operatorname{Aut}(G)/N$.

Hence, the map defined on the generators of $\operatorname{Out}(G)$ by
$s\mapsto\hat{s}$ induces a group homomorphism
$\operatorname{Out}(G)\rightarrow\operatorname{Aut}(G)/N$, which is a
splitting of the projection since each generator $s$ is the image of $\hat{s}$
in $\operatorname{Out}(G)$. ∎

###  4.4 Embedding $\operatorname{Out}(G)$ into $\operatorname{Out}(N)$

####

Under certain assumptions, we have constructed a monomorphism
$\operatorname{Out}(G)\hookrightarrow\operatorname{Aut}(G)/N$ by splitting the
projection $\operatorname{Aut}(G)/N\twoheadrightarrow\operatorname{Out}(G)$.
Since $N$ is a characteristic subgroup of $G$, we have in addition a
restriction homomorphism
$\operatorname{Aut}(G)\rightarrow\operatorname{Aut}(N)$ given by
$\psi\mapsto\psi_{\left|N\right.}$, which induces a morphism
$\operatorname{Aut}(G)/N\rightarrow\operatorname{Out}(N)$. The following lemma
generalises a result of Collins [6] on the injectivity of the restriction
homomorphism. We adapt an argument given by Bridson and Vogtmann [4, Lem. 1]
in the case where $G$ is a free group.

######  Lemma 4.7.

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product of non-trivial groups,
with $n\geqslant 2$. We assume that $G\not\cong\mathbb{Z}/2\ast\mathbb{Z}/2$.
If $N$ is a characteristic finite index subgroup of $G$, then the restriction
homomorphism $\operatorname{Aut}(G)\rightarrow\operatorname{Aut}(N)$ is
injective.

###### Proof.

Let $\psi\in\operatorname{Aut}(G)$ such that
$\psi_{\left|N\right.}=\operatorname{id}_{N}$. We write $k=\left[G:N\right]$.
Hence, for all $x\in G$, since $x^{k}\in N$, we have

$\psi(x)^{k}=\psi\left(x^{k}\right)=x^{k}.$

If $x$ does not belong to a conjugate of one of the factors $G_{i}$, then
considering the normal forms of $x$ and $\psi(x)$ shows that $\psi(x)=x$.

Let us now observe that, for $g_{i}\in G_{i}\smallsetminus\\{1\\}$ and
$g_{j}\in G_{j}\smallsetminus\\{1\\}$, with $i\neq j$, we have

$\psi\left(g_{i}\right)\psi\left(g_{j}\right)^{-1}=\psi\left(g_{i}g_{j}^{-1}\right)=g_{i}g_{j}^{-1},$

hence $g_{i}^{-1}\psi\left(g_{i}\right)=g_{j}^{-1}\psi\left(g_{j}\right)$.
Hence, there must be an element $\gamma\in G$ such that for all
$x\in\bigcup_{i}G_{i}\smallsetminus\\{1\\}$,

$\psi(x)=x\gamma.$

For $g_{i}\in G_{i}\smallsetminus\\{1\\}$ and $g_{j}\in
G_{j}\smallsetminus\\{1\\}$, with $i\neq j$, we have in addition

$g_{j}\gamma
g_{i}\gamma=\psi\left(g_{j}\right)\psi\left(g_{i}\right)=\psi\left(g_{j}g_{i}\right)=g_{j}g_{i},$

so that $g_{i}=\gamma g_{i}\gamma$.

If $\gamma\neq 1$, we write $\gamma$ in normal form as

$\gamma=w_{i_{1}}\cdots w_{i_{r}},$

with $r\geqslant 1$, $i_{1},\dots,i_{r}\in\left\\{1,\dots,n\right\\}$,
$i_{\ell}\neq i_{\ell+1}$ and $w_{i_{\ell}}\in
G_{i_{\ell}}\smallsetminus\\{1\\}$. We thus have, for all
$x\in\bigcup_{i}G_{i}\smallsetminus\\{1\\}$,

$x=w_{i_{1}}\cdots w_{i_{r}}xw_{i_{1}}\cdots w_{i_{r}}.$ ($\ast$)

If $i_{1}=i_{r}$, then we can take $j\neq i_{1}$ (because $n\geqslant 2$),
$x\in G_{j}\smallsetminus\\{1\\}$, so that the right-hand side of
$\left(\ast\right)$ is reduced, which is impossible. Hence, $i_{1}\neq i_{r}$.
Since the right-hand side of $\left(\ast\right)$ is not reduced, any choice of
$x\in\bigcup_{i}G_{i}\smallsetminus\\{1\\}$ must satisfy
$x\in\left\\{w_{i_{r}}^{-1},w_{i_{1}}^{-1}\right\\}$. This is impossible
because $\sum_{i}\left(\left|G_{i}\right|-1\right)\geqslant 3$ since
$G\not\cong\mathbb{Z}/2\ast\mathbb{Z}/2$ and $n\geqslant 2$.

Therefore $\gamma=1$, so $\psi_{\left|G_{i}\right.}=\operatorname{id}_{G_{i}}$
for all $i$ and $\psi=\operatorname{id}_{G}$. ∎

Combining this lemma to Proposition 4.6 (which gives an embedding
$\operatorname{Out}(G)\hookrightarrow\operatorname{Aut}(G)/N$), we obtain our
first embedding result.

######  Theorem B.

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product of abelian groups that are
not isomorphic to $\mathbb{Z}$, with $n\geqslant 2$. For each
$i\in\left\\{1,\dots,n\right\\}$, we choose an integer $r_{i}\in\mathbb{Z}$
coprime to $n-1$, in such a way that $r_{i}=r_{j}$ as soon as $G_{i}\cong
G_{j}$. We consider the characteristic subgroup
$N=G^{\prime}G_{1}^{r_{1}}\cdots G_{n}^{r_{n}}$ of $G$. Then there is an
embedding

$\operatorname{Out}(G)\hookrightarrow\operatorname{Out}(N).$

###### Proof.

Proposition 4.6 gives an embedding
$\operatorname{Out}(G)\hookrightarrow\operatorname{Aut}(G)/N$. Lemma 4.7
implies in addition that
$\operatorname{Aut}(G)\rightarrow\operatorname{Aut}\left(N\right)$ is
injective if $G\not\cong\mathbb{Z}/2\ast\mathbb{Z}/2$; it follows that
$\operatorname{Aut}(G)/N\rightarrow\operatorname{Out}(N)$ is injective. By
composition, we get an embedding
$\operatorname{Out}(G)\hookrightarrow\operatorname{Aut}(G)/N\hookrightarrow\operatorname{Out}(N)$.

If $G\cong\mathbb{Z}/2\ast\mathbb{Z}/2$, we have
$\operatorname{Out}(G)\cong\mathbb{Z}/2$. Moreover,
$N=G^{\prime}\cong\mathbb{Z}$ (if $r_{1}$ is even) or $N=G$ (if $r_{1}$ is
odd); in both cases, $\operatorname{Out}(N)\cong\mathbb{Z}/2$, so there is an
embedding $\operatorname{Out}(G)\hookrightarrow\operatorname{Out}(N)$. ∎

If the factors are finite and coprime to $n-1$, we can take
$r_{i}=\left|G_{i}\right|$ for all $i$, so that $N=G^{\prime}$ is a free
subgroup $G$ (by Lemma 4.2). This yields the following corollary.

######  Corollary B.

Let $G=G_{1}\ast\cdots\ast G_{n}$ be a free product of finite abelian groups,
with $n\geqslant 2$. We assume that $n-1$ is coprime to the order
$\left|G_{i}\right|$ of each factor $G_{i}$. Then there exists a free subgroup
$F$ of finite rank and of finite index in $G$ such that there is an embedding

$\operatorname{Out}(G)\hookrightarrow\operatorname{Out}(F).$

In particular, $\operatorname{Out}(G)$ has a faithful free representation.∎

###  4.5 Generators and relations with a free factor

####

We now turn our attention to the case where $G=G_{1}\ast\cdots\ast G_{n}$ is a
free product of freely indecomposable groups that are possibly isomorphic to
$\mathbb{Z}$. We order the factors in such a way that $G_{1}\cong\cdots\cong
G_{d}\cong\mathbb{Z}$, and $G_{d+1},\dots,G_{n}\not\cong\mathbb{Z}$. In other
words,

$G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n},$

with $F_{d}$ free of rank $d$.

A major difference with the case $d=0$ is that, as shown by Grushko’s Theorem,
there is now significantly more flexibility in the free product decomposition
of $G$. In particular, we cannot construct a characteristic subgroup of $G$ as
in Lemma 4.1. Instead, we will work with the characteristic subgroup
$N=G^{\prime}G^{r}$ for $r\in\mathbb{Z}$.

####

Nielsen [18], and more recently Gersten [11], gave presentations of
$\operatorname{Aut}\left(F_{d}\right)$, Fuchs-Rabinovich [9] did the same for
$\operatorname{Aut}\left(G_{d+1}\ast\cdots\ast G_{n}\right)$, as we have seen
above. In fact, in a second article [10], Fuchs-Rabinovich gives a complete
presentation of $\operatorname{Aut}(G)$. The resulting presentation is rather
long, but it generalises the results for the free part and the part without
free factor in a relatively natural way.

The generators of Fuchs-Rabinovich for $\operatorname{Aut}(G)$ are the
following.

######  Notation 4.8.

Let $G=G_{1}\ast\cdots\ast G_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with
$G_{1},\dots,G_{d}$ isomorphic to $\mathbb{Z}$ and equipped with respective
generators $a_{1},\dots,a_{d}$, and $G_{d+1},\dots,G_{n}$ freely
indecomposable not isomorphic to $\mathbb{Z}$.

Factor automorphisms:

For $i\in\left\\{d+1,\dots,n\right\\}$, we consider the group $\Phi_{i}$ of
automorphisms of the factor $G_{i}$ (see Notation 4.3).

Reflections:

For $i\in\left\\{1,\dots,d\right\\}$, we set $\tau_{i}:a_{i}\mapsto
a_{i}^{-1}$. The automorphism $\tau_{i}$ is called the _reflection_ of $G_{i}$
and we have $\Phi_{i}=\left\\{\operatorname{id},\tau_{i}\right\\}$.

Permutation automorphisms:

We consider $\Omega=\left\\{\omega_{ij},\>i\neq j,\>G_{i}\cong G_{j}\right\\}$
the group of permutation automorphisms, defined as in Notation 4.3, with the
additional convention that $\omega_{ij}:a_{i}\mapsto a_{j}$ and $a_{j}\mapsto
a_{i}$ when $i,j\leqslant d$.

Dehn twists:

For $i\in\left\\{1,\dots,n\right\\}$, $j\in\left\\{d+1,\dots,n\right\\}$,
$i\neq j$, we consider $A_{ij}$ the group of Dehn twists of type $ij$ (see
Notation 4.3). If $i\in\left\\{1,\dots,d\right\\}$, we write
$\alpha_{ij}=\alpha_{ij}^{\left(a_{i}\right)}$.

Right transvections:

For $i\in\left\\{1,\dots,n\right\\}$, $j\in\left\\{1,\dots,d\right\\}$, $i\neq
j$, $\gamma_{i}\in G_{i}$, we define

$\displaystyle\rho_{ij}^{\left(\gamma_{i}\right)}:a_{j}\in G_{j}$
$\displaystyle\mapsto a_{j}\gamma_{i},$ $\displaystyle g_{k}\in G_{k}$
$\displaystyle\mapsto g_{k}\quad\textrm{if $k\neq j$}.$

We write $P_{ij}=\left\\{\rho_{ij}^{\left(\gamma_{i}\right)},\>\gamma_{i}\in
G_{i}\right\\}$ and $P=\bigcup_{i\neq j}P_{ij}$. Elements of $P$ are called
_right transvections_. If $i\in\left\\{1,\dots,d\right\\}$, we write
$\rho_{ij}=\rho_{ij}^{\left(a_{i}\right)}$.

Left transvections:

For $i\in\left\\{1,\dots,n\right\\}$, $j\in\left\\{1,\dots,d\right\\}$, $i\neq
j$, $\gamma_{i}\in G_{i}$, we define

$\displaystyle\lambda_{ij}^{\left(\gamma_{i}\right)}:a_{j}\in G_{j}$
$\displaystyle\mapsto\gamma_{i}a_{j},$ $\displaystyle g_{k}\in G_{k}$
$\displaystyle\mapsto g_{k}\quad\textrm{if $k\neq j$}.$

We write
$\Lambda_{ij}=\left\\{\lambda_{ij}^{\left(\gamma_{i}\right)},\>\gamma_{i}\in
G_{i}\right\\}$ and $\Lambda=\bigcup_{i\neq j}\Lambda_{ij}$. Elements of
$\Lambda$ are called _left transvections_. If
$i\in\left\\{1,\dots,d\right\\}$, we write
$\lambda_{ij}=\lambda_{ij}^{\left(a_{i}\right)}$.

Note that we have inverted Bridson and Vogtmann’s notations: we write
$\rho_{ij},\lambda_{ij}$ instead of $\rho_{ji},\lambda_{ji}$ to be consistent
with Fuchs-Rabinovich.

We have a presentation of $\operatorname{Aut}(G)$ as follows.

######  Theorem 4.9 (Fuchs-Rabinovich [10]).

Let $G=G_{1}\ast\cdots\ast G_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$ be a free
product of freely indecomposable groups, with $G_{1}\cong\cdots\cong
G_{d}\cong\mathbb{Z}$ and $G_{d+1},\dots,G_{n}\not\cong\mathbb{Z}$. Then
$\operatorname{Aut}(G)$ is generated by factor automorphisms, reflections,
permutation automorphisms, Dehn twists, left and right transvections.
Moreover, the following relations give a presentation of
$\operatorname{Aut}(G)$:

$\ref{thrm_pres_autg_r1}-\ref{thrm_pres_autg_r11}$

The relations of Theorem 4.4, written for all the indices for which they make
sense.

  1. (12)

Nielsen’s relations [18] of $\operatorname{Aut}\left(F_{d}\right)$, written
for all the indices for which they make sense.

  2. (13)

$\alpha_{ij}^{\left(\gamma_{i}\right)}\tau_{k}=\tau_{k}\alpha_{ij}^{\left(\gamma_{i}\right)}$
for $\gamma_{i}\in G_{i}$, $i,j\neq k$.

  3. (14)

$\alpha_{ij}\tau_{i}=\tau_{i}\alpha_{ij}^{-1}$.

  4. (15)

$\rho_{ij}^{\left(\gamma_{i}\right)}\varphi_{k}=\varphi_{k}\rho_{ij}^{\left(\gamma_{i}\right)}$
for $\gamma_{i}\in G_{i}$, $\varphi_{k}\in\Phi_{k}$, $i,j\neq k$.

  5. (16)

$\rho_{ij}^{\left(\varphi_{i}\gamma_{i}\right)}\varphi_{i}=\varphi_{i}\rho_{ij}^{\left(\gamma_{i}\right)}$
for $\gamma_{i}\in G_{i}$, $\varphi_{i}\in\Phi_{i}$.

  6. (17)

$\rho_{ij}^{\left(\gamma_{i}\right)}\alpha_{k\ell}^{\left(\gamma_{k}\right)}=\alpha_{k\ell}^{\left(\gamma_{k}\right)}\rho_{ij}^{\left(\gamma_{i}\right)}$
for $\gamma_{i}\in G_{i}$, $\gamma_{k}\in G_{k}$, $i,j\neq k,\ell$.

  7. (18)

$\rho_{ij}^{\left(\gamma_{i}\right)}\alpha_{i\ell}^{\left(\gamma_{i}\right)}=\alpha_{i\ell}^{\left(\gamma_{i}\right)}\rho_{ij}^{\left(\gamma_{i}\right)}$
for $\gamma_{i}\in G_{i}$.

  8. (19)

$\rho_{ij}^{\left(\gamma_{i}\right)}\alpha_{j\ell}=\alpha_{j\ell}\rho_{ij}^{\left(\gamma_{i}\right)}\alpha_{i\ell}^{\left(\gamma_{i}\right)}$
for $\gamma_{i}\in G_{i}$.222A misprint in [10] was corrected in [12, § 4].

  9. (20)

$\left(\rho_{kj}^{\left(\gamma_{k}\right)}\right)^{-1}\rho_{ij}^{\left(\gamma_{i}\right)}\rho_{kj}^{\left(\gamma_{k}\right)}\alpha_{ki}^{\left(\gamma_{k}\right)}=\alpha_{ki}^{\left(\gamma_{k}\right)}\rho_{ij}^{\left(\gamma_{i}\right)}$
for $\gamma_{i}\in G_{i}$, $\gamma_{k}\in G_{k}$.

  10. (21)

$\alpha_{ji}\rho_{ij}^{\left(\gamma_{i}\right)}=\lambda_{ij}^{\left(\gamma_{i}\right)}\alpha_{ji}\varphi_{i}$
for $\gamma_{i}\in G_{i}$ with
$\left(\varphi_{i}\right)_{\left|G_{i}\right.}=\operatorname{ad}\left(\gamma_{i}^{-1}\right)$.

  11. (22)

$\tau_{k}\varphi_{i}=\varphi_{i}\tau_{k}$ for $\varphi_{i}\in\Phi_{i}$, $i\neq
k$.

  12. (23)

$\tau_{i}\omega_{k\ell}=\omega_{k\ell}\tau_{i}$ for $i\neq k,\ell$.

######  Remark.

In Theorem 4.9, relations $\ref{thrm_pres_autg_r1}-\ref{thrm_pres_autg_r12}$
are more numerous than the union of relations of
$\operatorname{Aut}\left(F_{d}\right)$ and
$\operatorname{Aut}\left(G_{d+1}\ast\cdots\ast G_{n}\right)$: they include for
example relations involving Dehn twists conjugating one of the factors $G_{i}$
(with $i>d$) by an element of $F_{d}$. However, the choice of notations means
that they are written in the same way as in
$\operatorname{Aut}\left(F_{d}\right)$ or in
$\operatorname{Aut}\left(G_{d+1}\ast\cdots\ast G_{n}\right)$.

As before, we can easily deduce a presentation of $\operatorname{Out}(G)$.

######  Corollary 4.10.

Let $G=G_{1}\ast\cdots\ast G_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$ be a free
product of freely indecomposable groups, with $G_{1}\cong\cdots\cong
G_{d}\cong\mathbb{Z}$ and $G_{d+1},\dots,G_{n}\not\cong\mathbb{Z}$. Then
$\operatorname{Out}(G)$ is generated by factor automorphisms, reflections,
permutation automorphisms, Dehn twists, left and right transvections.
Moreover, relations $\ref{thrm_pres_autg_r1}-\ref{thrm_pres_autg_r23}$ of
Theorem 4.9, together with relation (0) below, give a presentation of
$\operatorname{Out}(G)$:

  1. (0)

$\varphi_{i}\circ\prod_{\begin{subarray}{c}j\leqslant d\\\ j\neq
i\end{subarray}}\rho_{ij}^{\left(\gamma_{i}\right)}\left(\lambda_{ij}^{\left(\gamma_{i}\right)}\right)^{(-1)}\circ\prod_{\begin{subarray}{c}j>d\\\
j\neq i\end{subarray}}\alpha_{ij}^{\left(\gamma_{i}\right)}=1$ for
$\gamma_{i}\in G_{i}$, with
${\varphi_{i}}_{\left|G_{i}\right.}=\operatorname{ad}\left(\gamma_{i}^{-1}\right)$.∎

###  4.6 Embedding results with a free factor

####

We now assume that $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with $F_{d}$
free of rank $d$, $G_{d+1},\dots,G_{n}$ abelian (hence freely indecomposable)
and not isomorphic to $\mathbb{Z}$. We consider the characteristic subgroup
$N=G^{\prime}G^{r}$, where $r\in\mathbb{Z}$ is an integer to be fixed. To
construct a splitting of
$\operatorname{Aut}(G)/N\twoheadrightarrow\operatorname{Out}(G)$, we follow
the same method as in 4.3: for each generator $s\in\operatorname{Out}(G)$, we
pick a correction $\hat{s}\in\operatorname{Aut}(G)/N$ of the standard lift
$s_{0}\in\operatorname{Aut}(G)/N$, in such a way that the corrected lifts
$\hat{s}$ satisfy the relations of $\operatorname{Out}(G)$.

In fact, the generator of $\operatorname{Out}(G)$ in this setting are all of
the same kind as those used by Bridson and Vogtmann [4] when $G=F_{d}$ (for
transvections and reflections) or as those used in 4.3 when
$G=G_{d+1}\ast\cdots\ast G_{n}$ (for factor automorphisms, permutation
automorphisms and Dehn twists). Hence, we can simply take the same corrections
as before. Relations $\ref{thrm_pres_autg_r1}-\ref{thrm_pres_autg_r12}$ of
Theorem 4.9 will be automatically satisfied, and it will suffice to check
$\ref{thrm_pres_autg_r13}-\ref{thrm_pres_autg_r23}$.

######  Proposition 4.11.

Let $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with $F_{d}$ free of rank $d$,
$G_{d+1},\dots,G_{n}$ abelian not isomorphic to $\mathbb{Z}$. We choose an
integer $r\in\mathbb{Z}$ coprime to $n-1$ and we consider the characteristic
subgroup $N=G^{\prime}G^{r}$ of $G$.

Then the projection
$\operatorname{Aut}(G)/N\twoheadrightarrow\operatorname{Out}(G)$ splits.

###### Proof.

We may pick integers $u,t\in\mathbb{Z}$ such that $u(n-1)+tr=1$. We then lift
the generators of $\operatorname{Out}(G)$ as follows:

  * •

For factor automorphisms, permutation automorphisms and Dehn twists, we take
the same corrections as in the proof of Proposition 4.6 (replacing $u_{i}$ by
$u$).

  * •

Right transvections are not corrected (we keep the standard lifts).

  * •

Left transvections are corrected by
$\hat{\lambda}_{ij}^{\left(\gamma_{i}\right)}=\operatorname{ad}\left(\gamma_{i}^{-u}\right)\circ\left(\lambda_{ij}^{\left(\gamma_{i}\right)}\right)_{0}$.

  * •

Reflections are corrected by
$\hat{\tau}_{i}=\operatorname{ad}\left(a_{i}^{u}\right)\circ\left(\tau_{i}\right)_{0}$.

Observe that these corrections of transvections and reflections are an
algebraic restatement of the corrections of Bridson and Vogtmann [4].

It follows that relations $\ref{thrm_pres_autg_r1}-\ref{thrm_pres_autg_r12}$
of Theorem 4.9 are satisfied: their verification would be a rewriting of the
computations of Proposition 4.6 (for
$\ref{thrm_pres_autg_r1}-\ref{thrm_pres_autg_r11}$) and of Bridson and
Vogtmann [4] (for (12)). Moreover, relations
$\ref{thrm_pres_autg_r15}-\ref{thrm_pres_autg_r16}$ are satisfied because they
only involve generators whose lifts are not corrected.

We check that relations $\ref{thrm_pres_autg_r14}-\ref{thrm_pres_autg_r18}$,
(20) and $\ref{thrm_pres_autg_r22}-\ref{thrm_pres_autg_r23}$ are satisfied in
$\operatorname{Aut}(G)$. We give the results of our computations for the other
relations, and we see that they are satisfied in $\operatorname{Aut}(G)/N$:

  1. (13)

$\hat{\alpha}_{ij}^{\left(\gamma_{i}\right)}\hat{\tau}_{k}=\operatorname{ad}\left(\left[\gamma_{i}^{u},a_{k}^{u}\right]\right)\circ\hat{\tau}_{k}\hat{\alpha}_{ij}^{\left(\gamma_{i}\right)}$.

  2. (19)

$\hat{\rho}_{ij}^{\left(\gamma_{i}\right)}\hat{\alpha}_{j\ell}=\operatorname{ad}\left(\left(a_{j}\gamma_{i}\right)^{u}\gamma_{i}^{-u}a_{j}^{-u}\right)\circ\hat{\alpha}_{j\ell}\hat{\rho}_{ij}^{\left(\gamma_{i}\right)}\hat{\alpha}_{i\ell}^{\left(\gamma_{i}\right)}$.

  3. (21)

$\hat{\alpha}_{ji}\hat{\rho}_{ij}^{\left(\gamma_{i}\right)}=\operatorname{ad}\left(a_{j}^{u}\left(\gamma_{i}a_{j}\right)^{-u}\gamma_{i}^{u}\right)\circ\hat{\lambda}_{ij}^{\left(\gamma_{i}\right)}\hat{\alpha}_{ji}\hat{\varphi}_{i}$.

  4. (0)

$\hat{\varphi}_{i}\circ\prod_{\begin{subarray}{c}j\leqslant d\\\ j\neq
i\end{subarray}}\hat{\rho}_{ij}^{\left(\gamma_{i}\right)}\left(\hat{\lambda}_{ij}^{\left(\gamma_{i}\right)}\right)^{(-1)}\circ\prod_{\begin{subarray}{c}j>d\\\
j\neq
i\end{subarray}}\hat{\alpha}_{ij}^{\left(\gamma_{i}\right)}=\operatorname{ad}\left(\gamma_{i}^{u(n-1)-1}\right)=\operatorname{ad}\left(\gamma_{i}^{tr}\right)$.

Hence, $s\mapsto\hat{s}$ gives a splitting of the morphism
$\operatorname{Aut}(G)/N\twoheadrightarrow\operatorname{Out}(G)$. ∎

####

Lemma 4.7, which proved the injectivity of the restriction homomorphism
$\operatorname{Aut}(G)\rightarrow\operatorname{Aut}(N)$, was independent of
the presence of free factors in the free product. Hence, it remains valid in
the present setting, which yields the following result with the same proof as
Theorem B.

######  Theorem C.

Let $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with $n\geqslant 2$, $F_{d}$
free of rank $d$, $G_{d+1},\dots,G_{n}$ abelian not isomorphic to
$\mathbb{Z}$. We choose an integer $r\in\mathbb{Z}$ coprime to $n-1$ and we
consider the characteristic subgroup $N=G^{\prime}G^{r}$ of $G$. Then there is
an embedding

$\operatorname{Out}(G)\hookrightarrow\operatorname{Out}(N).\qed$

######  Remark.

When $d=n$, we get the result of Bridson and Vogtmann [4, Cor. A].

Taking $r=\left|G_{d+1}\right|\cdots\left|G_{n}\right|$ yields a
generalisation of Corollary B.

######  Corollary C1.

Let $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with $n\geqslant 2$, $F_{d}$
free of rank $d$, $G_{d+1},\dots,G_{n}$ finite abelian. We assume that $n-1$
is coprime to the order $\left|G_{i}\right|$ of each factor $G_{i}$. Then
there is a free subgroup $F$ of finite rank and of finite index in $G$ such
that there is an embedding

$\operatorname{Out}(G)\hookrightarrow\operatorname{Out}(F).$

In particular, $\operatorname{Out}(G)$ has a faithful free representation.∎

Observe that, by increasing the rank of the free factor, we can always get to
a situation where $n-1$ is coprime to $\left|G_{i}\right|$ for all $i$. We
thus get our second corollary.

######  Corollary C2.

Let $G=F_{d}\ast G_{d+1}\ast\cdots\ast G_{n}$, with $F_{d}$ free of rank $d$,
$G_{d+1},\dots,G_{n}$ finite abelian. Then there is an integer $k\geqslant 0$
such that $\operatorname{Out}\left(F_{k}\ast G\right)$ has a faithful free
representation.∎

###  4.7 Application to universal Coxeter groups

####

Consider $W_{n}=\left<x_{1},\dots,x_{n}\mid x_{1}^{2},\dots,x_{n}^{2}\right>$.
The group $W_{n}$ is called the _universal Coxeter group_ of rank $n$; it is
the free product of $n$ copies of $\mathbb{Z}/2$, and it is certainly the
simplest example of free product of freely indecomposable groups that are not
isomorphic to $\mathbb{Z}$. It is therefore natural to ask what properties
$\operatorname{Out}\left(W_{n}\right)$ shares with
$\operatorname{Out}\left(F_{n}\right)$.

We focus in particular on free representations of
$\operatorname{Out}\left(W_{n}\right)$. For
$\operatorname{Aut}\left(W_{n}\right)$, Mühlherr [17] showed that there is an
embedding
$\operatorname{Aut}\left(W_{n}\right)\hookrightarrow\operatorname{Aut}\left(F_{n-1}\right)$
for all $n\geqslant 3$. On the contrary, Varghese [21] showed that, if
$n\geqslant 4$ and $m\leqslant n-2$, then any morphism from
$\operatorname{Aut}\left(W_{n}\right)$ to
$\operatorname{Aut}\left(F_{m}\right)$, $\operatorname{Out}\left(F_{m}\right)$
or $GL_{m}\left(\mathbb{Z}\right)$ has finite image.

We turn our attention to the parallel question for outer automorphisms: we
wish to know for which integers $n,m$ there are faithful free representations
$\operatorname{Out}\left(W_{n}\right)\hookrightarrow\operatorname{Out}\left(F_{m}\right)$.
As an application of the results of this text, we have the following
corollary.

######  Corollary 4.12.

If $n$ is even and $m=2^{n-1}(n-2)+1$, then there is an embedding

$\operatorname{Out}\left(W_{n}\right)\hookrightarrow\operatorname{Out}\left(F_{m}\right).$

###### Proof.

By Theorem B (with $r_{i}=2$ for all $i$), there is an embedding
$\operatorname{Out}\left(W_{n}\right)\hookrightarrow\operatorname{Out}\left(W_{n}^{\prime}\right)$.
Let us consider in addition the $W_{n}$-tree $A$ such that $W_{n}\backslash A$
is a tree of group with $n$ vertices labelled by $\mathbb{Z}/2$ and $n-1$
unlabelled edges. Lemma 4.2 implies that $W_{n}^{\prime}$ acts freely on $A$,
so $W_{n}^{\prime}\cong F_{m}$, where $m$ is the genus of the graph
$W_{n}^{\prime}\backslash A$. Observe that
$W_{n}/W_{n}^{\prime}\cong\left(\mathbb{Z}/2\right)^{n}$ and that in $A$,
vertex stabilisers under $W_{n}/W_{n}^{\prime}$ have order $2$ and edge
stabilisers are trivial. Hence, each edge orbit under $W_{n}$ corresponds to
$2^{n}$ edge orbits under $W_{n}^{\prime}$, and each vertex orbit under
$W_{n}$ corresponds to $2^{n-1}$ vertex orbits under $W_{n}^{\prime}$. It
follows that $W_{n}^{\prime}\backslash A$ is a graph with $(n-1)2^{n}$ edges
and $n2^{n-1}$ vertices, hence

$m=\left|E\left(W_{n}^{\prime}\backslash
A\right)\right|-\left|V\left(W_{n}^{\prime}\backslash
A\right)\right|+1=2^{n-1}(n-2)+1.\qed$

We have no reason to believe that this result covers all values of $n,m$ for
which there are embeddings
$\operatorname{Out}\left(W_{n}\right)\hookrightarrow\operatorname{Out}\left(F_{m}\right)$.
To see this, let us examine the case $n=3$, on which Corollary 4.12 does not
say anything. The group $\operatorname{Out}\left(W_{3}\right)$ is simple
enough to be computed explicitly, and we can construct a faithful free
representation by hand.

######  Proposition 4.13.

  1. (i)

$\operatorname{Out}\left(W_{3}\right)\cong W_{3}\rtimes\mathfrak{S}_{3}$,
where $\mathfrak{S}_{3}$ acts on $W_{3}$ by permutation of the generators.

  2. (ii)

There is an embedding
$\operatorname{Out}\left(W_{3}\right)\hookrightarrow\operatorname{Out}\left(F_{4}\right)$.

###### Proof.

  1. (i)

We have a morphism
$\operatorname{Out}\left(W_{3}\right)\rightarrow\mathfrak{S}_{3}$ given by the
action of $W_{3}$ on conjugacy classes of generators; we denote by
$\operatorname{Out}_{0}\left(W_{3}\right)$ its kernel. Hence, there is an
exact sequence

$1\rightarrow\operatorname{Out}_{0}\left(W_{3}\right)\rightarrow\operatorname{Out}\left(W_{3}\right)\rightarrow\mathfrak{S}_{3}\rightarrow
1.$

We can split this exact sequence by mapping
$\left(i\,j\right)\mapsto\omega_{ij}$ (we follow Notation 4.3). Therefore
$\operatorname{Out}\left(W_{3}\right)\cong\operatorname{Out}_{0}\left(W_{3}\right)\rtimes\mathfrak{S}_{3}$.
There remains to determine $\operatorname{Out}_{0}\left(W_{3}\right)$.

It follows from Fuchs-Rabinovich’s presentation of
$\operatorname{Out}\left(W_{3}\right)$ (see Corollary 4.5) that the subgroup
$\operatorname{Out}_{0}\left(W_{3}\right)$ is generated by Dehn twists (in
this case, there is no factor automorphism), with relations (0) and
$\ref{thrm_pres_autg_r4}-\ref{thrm_pres_autg_r5}$ (they are the only ones that
only involve Dehn twists). Each Dehn twist has order $2$. Writing
$\alpha_{ij}=\alpha_{ij}^{\left(x_{i}\right)}$, relation (0) yields

$\alpha_{12}=\alpha_{13},\quad\alpha_{21}=\alpha_{23},\quad\alpha_{31}=\alpha_{32}.$

Relations $\ref{thrm_pres_autg_r4}-\ref{thrm_pres_autg_r5}$ follow, which
gives

$\operatorname{Out}_{0}\left(W_{3}\right)=\left<\alpha_{12},\alpha_{21},\alpha_{31}\mid\alpha_{12}^{2},\alpha_{21}^{2},\alpha_{31}^{2}\right>.$

We therefore see that $\operatorname{Out}_{0}\left(W_{3}\right)\cong W_{3}$,
and the action of $\mathfrak{S}_{3}$ is given by
$\left(1\,2\right)\cdot\alpha_{12}=\alpha_{21}$,
$\left(1\,2\right)\cdot\alpha_{31}=\alpha_{32}=\alpha_{31}$ and
$\left(2\,3\right)\cdot\alpha_{12}=\alpha_{13}=\alpha_{12}$,
$\left(2\,3\right)\cdot\alpha_{31}=\alpha_{21}$. This corresponds to the
action of $\mathfrak{S}_{3}$ by permutation of the generators of $W_{3}$.

  2. (ii)

By (i), it suffices to construct an embedding
$W_{3}\rtimes\mathfrak{S}_{3}\hookrightarrow\operatorname{Out}\left(F_{4}\right)$.
We denote by $x_{1},x_{2},x_{3}$ the generators of $W_{3}$,
$a_{1},\dots,a_{4}$ the generators of $F_{4}$. We write
$\alpha_{ij}=\rho_{ij}\lambda_{ij}^{-1}$ for the Dehn twists in $F_{4}$ and
$\tau_{i}$ for the reflection $a_{i}\mapsto a_{i}^{-1}$ (see Notation 4.8).
For $\sigma\in\mathfrak{S}_{3}$, we define in addition
$\omega_{\sigma}\in\operatorname{Out}\left(F_{4}\right)$ by $a_{i}\mapsto
a_{\sigma(i)}$ for $i\in\left\\{1,2,3\right\\}$ and $a_{4}\mapsto a_{4}$.

We then define
$f:W_{3}\rtimes\mathfrak{S}_{3}\rightarrow\operatorname{Out}\left(F_{4}\right)$
by

$\displaystyle x_{i}\in W_{3}$
$\displaystyle\longmapsto\alpha_{i4}\tau_{i}=\tau_{i}\alpha_{i4}^{-1}\in\operatorname{Out}\left(F_{4}\right),$
$\displaystyle\sigma\in\mathfrak{S}_{3}$
$\displaystyle\longmapsto\omega_{\sigma}\in\operatorname{Out}\left(F_{4}\right).$

We have $f\left(x_{i}\right)^{2}=1$ and
$f\left(\sigma\right)f\left(x_{i}\right)f\left(\sigma\right)^{-1}=f\left(x_{\sigma(i)}\right)$
for all $i\in\left\\{1,2,3\right\\}$ and $\sigma\in\mathfrak{S}_{3}$, so $f$
gives a well-defined group homomorphism
$W_{3}\rtimes\mathfrak{S}_{3}\rightarrow\operatorname{Out}\left(F_{4}\right)$.

We want to show that $f$ is injective. Let
$\gamma=w\sigma\in\operatorname{Ker}f$, with $w\in W_{3}$ and
$\sigma\in\mathfrak{S}_{3}$. Since $f\left(\gamma\right)=1$ in
$\operatorname{Out}\left(F_{4}\right)$, $f(\gamma)$ acts trivially on the
conjugacy classes of the generators $a_{i}$, so $\sigma=1$. We write
$w=x_{i_{1}}\cdots x_{i_{k}}$ in normal form, with $k\geqslant 0$,
$i_{j}\in\left\\{1,2,3\right\\}$, $i_{j+1}\neq i_{j}$. We then have

$\displaystyle f\left(\gamma\right)\cdot a_{i}$ $\displaystyle=a_{i}^{\pm
1}\quad\textrm{if $i\in\left\\{1,2,3\right\\}$}$ $\displaystyle
f\left(\gamma\right)\cdot a_{4}$
$\displaystyle=\operatorname{ad}\left(a_{i_{k}}^{\varepsilon_{k}}\cdots
a_{i_{1}}^{\varepsilon_{1}}\right)\cdot a_{4},$

with $\varepsilon_{1},\dots,\varepsilon_{k}\in\left\\{\pm 1\right\\}$. Since
$f\left(\gamma\right)=1$ in $\operatorname{Out}\left(F_{4}\right)$, it follows
that $a_{i_{k}}^{\varepsilon_{k}}\cdots a_{i_{1}}^{\varepsilon_{1}}=1$. Since
this is a reduced form in $F_{4}$, we must have $k=0$, so $w=1$ and
$\gamma=1$.

Hence,
$f:W_{3}\rtimes\mathfrak{S}_{3}\hookrightarrow\operatorname{Out}\left(F_{4}\right)$
is an embedding.∎

####

This example confirms that out method is not exhaustive: we have been able to
construct embeddings
$\operatorname{Out}\left(W_{n}\right)\hookrightarrow\operatorname{Out}\left(F_{m}\right)$
for some integers $n,m$, but there may exist other values of $n$ for which
$\operatorname{Out}\left(W_{n}\right)$ admits faithful free representations.

In general, the group $\operatorname{Out}(G)$ seems very likely to have
faithful free representations for many free products $G$ for which we cannot
say anything with the method used here.

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Institut Fourier, UMR 5582, Laboratoire de Mathématiques,
Université Grenoble Alpes, CS 40700, 38058 Grenoble cedex 9, France
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