# Equivariant Self-homotopy equivalences of product spaces

Gopal Chandra Dutta  ,  Debasis Sen  and  Ajay Singh Thakur  Department of
Mathematics and Statistics
Indian Institute of Technology, Kanpur
Uttar Pradesh 208016
India<EMAIL_ADDRESS>Department of Mathematics and Statistics
Indian Institute of Technology, Kanpur
Uttar Pradesh 208016
India<EMAIL_ADDRESS>Department of Mathematics and Statistics
Indian Institute of Technology, Kanpur
Uttar Pradesh 208016
India<EMAIL_ADDRESS>
###### Abstract.

Let $G$ be a finite group. We study the group of $G$-equivariant self-homotopy
equivalences of product of $G$-spaces. For a product of $n$-spaces, we
represent it as product of $n$-subgroups under the assumption of equivariant
reducibility. Further we describe each factor as a split short exact sequence.
Also we obtain an another kind of factorisation, called $LU$ type
decomposition, as product of two subgroups.

###### Key words and phrases:

Equivariant self homotopy equivalences, $G$-reducibility, product spaces

###### 2010 Mathematics Subject Classification:

Primary 55P10 ; Secondary: 55P15, 55P91, 55Q05.

##  1\. Introduction

For a pointed space $X$, let $\operatorname{Aut}(X)$ denote the set of
homotopy classes of based self homotopy equivalences. This set is a group
under composition of maps [1, 15]. A natural approach for understanding
$\operatorname{Aut}(X)$ would be to start from some decomposition of $X$, for
example as a product or wedge of simpler spaces, and then describe self-
homotopy equivalences of $X$ by in terms of those of the factors. The group
$\operatorname{Aut}(X\times Y)$ was studied in the past by many authors (cf.
[3, 4, 9, 10, 12, 14]). In particular Sawashita considered the group
$\operatorname{Aut}(X\times Y)$ when $X$ and $Y$ are spheres ([12]). We have
projections $p_{X}:X\times Y\to X,~{}~{}p_{Y}:X\times Y\to Y$ and inclusions
$\iota_{X}:X\hookrightarrow X\times Y$ and $\iota_{Y}:Y\hookrightarrow X\times
Y$, determined by the base point. For a map $f:X\times Y\to X\times Y,$ let

$f_{XX}:=p_{X}\circ f\circ\iota_{X}:X\to X,~{}~{}f_{YY}:=p_{Y}\circ
f\circ\iota_{Y}:Y\to Y.$

A self homotopy equivalence $f$ is called _reducible_ if $f_{XX},f_{YY}$ are
self-homotopy equivalences. When self-homotopy equivalences of $X\times Y$ are
reducible, Heath and Booth obtained a split exact sequence

$0\to[X,\operatorname{aut}_{1}(Y)]\to\operatorname{Aut}(X\times
Y)\xrightarrow{}\operatorname{Aut}(X)\times\operatorname{Aut}(Y)\to 0,$

under the condition $[Y,\operatorname{aut}_{1}(X)]=\\{\ast\\}$ ([4, 6]). Here
$\operatorname{aut}_{1}(X)$ is the connected component of $\operatorname{Id}$
of the space of based self homotopy equivalences of $X$. The group
$\operatorname{Aut}(X\times Y)$ has two natural subgroups
$\operatorname{Aut}_{X}(X\times Y)$ and $\operatorname{Aut}_{Y}(X\times Y)$,
which are self-equivalences of $X\times Y$ over $X$ and $Y$ respectively.
Later Pavešić showed that under reducibility (without the condition
$[Y,\operatorname{aut}_{1}(X)]=\\{\ast\\}$), the group
$\operatorname{Aut}(X\times Y)$ can be decomposed as product of the subgroups
$\operatorname{Aut}_{X}(X\times Y)$ and $\operatorname{Aut}_{Y}(X\times Y)$
(see [9, 10]),

(1) $\operatorname{Aut}(X\times Y)=\operatorname{Aut}_{X}(X\times
Y)\cdot\operatorname{Aut}_{Y}(X\times Y).$

Further each of the factors fits into split exact sequences

$0\to[X,\operatorname{aut}_{1}(Y)]\to\operatorname{Aut}_{X}(X\times
Y)\xrightarrow{}\operatorname{Aut}(Y)\to 0,$

and

$0\to[Y,\operatorname{aut}_{1}(X)]\to\operatorname{Aut}_{Y}(X\times
Y)\xrightarrow{}\operatorname{Aut}(X)\to 0.$

The result of Heath and Booth can be deduced as corollary of it. Moreover the
decomposition of Equation 1 was generalised for $n$-product of spaces in terms
of $n$-subgroups. Further a $LU$ kind of decomposition was obtained for higher
products ([11, 10]).

In this paper, we consider equivariant self homotopy equivalences of product
spaces and obtain equivariant versions of the results mentioned about. Let $G$
be a finite group. A based $G$-space is a $G$-space with a point which is
$G$-fixed. The space of based $G$-equivariant maps is denoted by
$\operatorname{map}^{G}(X,Y)$. Let $[X,Y]_{G}$ denote the set of $G$-homotopy
classes of based maps. For a based $G$-space $X$, we denote the space of based
equivariant self-homotopy equivalences of $X$ by $\operatorname{aut}^{G}(X)$.
We denote by $\operatorname{Aut}^{G}(X)$ the set of $G$-homotopy classes of
elements of $\operatorname{aut}^{G}(X)$. The composition of maps induces a
group operation on $\operatorname{Aut}^{G}(X)$. We call it the group of
equivariant self-homotopy equivalences of $X$. This group has been studied by
several authors (cf. [5, 13]). Our goal here is to study the group
$\operatorname{Aut}^{G}(X\times Y)$, where $X,Y$ are based $G$-CW complexes
and also for higher products. For our results we need the condition of
$G$-reducibility.

######  1.1 Definition.

With notations as above, a $G$-self-homotopy equivalence $f$ of $X\times Y$ is
called _$G$ -reducible_ if $f_{XX}\in\operatorname{aut}^{G}(X)$ and
$f_{YY}\in\operatorname{aut}^{G}(Y)$.

As in the non-equivariant case, the group $\operatorname{Aut}^{G}(X\times Y)$
has two natural subgroups $\operatorname{Aut}^{G}_{X}(X\times
Y)=\\{f\in\operatorname{Aut}^{G}(X\times Y):~{}~{}p_{X}\circ f=p_{X}\\}$ and
$\operatorname{Aut}^{G}_{Y}(X\times Y)=\\{f\in\operatorname{Aut}^{G}(X\times
Y):~{}~{}p_{Y}\circ f=p_{Y}\\}$. We show that the product of these two
subgroups is the whole group:

######  Theorem A.

If all $G$-self-homotopy equivalences of $X\times Y$ are $G$-reducible then

$\operatorname{Aut}^{G}(X\times Y)=\operatorname{Aut}^{G}_{X}(X\times
Y)\cdot\operatorname{Aut}^{G}_{Y}(X\times Y).$

(See Theorem 2.7 below.)

Under the assumption of $G$-reducibility, we have a natural map
$\phi:\operatorname{Aut}^{G}(X\times
Y)\to\operatorname{Aut}^{G}(X)\times\operatorname{Aut}^{G}(Y)$ defined by
$\phi(f)=(f_{XX},f_{YY})$. We show that the map $\phi$ gives a split exact
sequence. This is an equivariaint analogue of Heath and Booth [4].

######  Theorem B.

Let $X,Y$ to be pointed connected $G$-CW complexes. If all $G$-self-homotopy
equivalences of $X\times Y$ are $G$-reducible and
$[Y,\operatorname{aut}_{1}(X)]_{G}=\\{\ast\\}$, then we get a split short
exact sequence

$0\to[X,\operatorname{aut}_{1}(Y)]_{G}\to\operatorname{Aut}^{G}(X\times
Y)\xrightarrow[]{\phi}\operatorname{Aut}^{G}(X)\times\operatorname{Aut}^{G}(Y)\to
0.$

(See Theorem 2.9 below.)

We use the above results of two products to generalize for the $n$-product of
spaces ${\bf X}=X_{1}\times\ldots\times X_{n}$, where all $X_{i}$’s are
$G$-spaces. The action of $G$ on ${\bf X}$ is the diagonal action. Let
$\prod_{\hat{i}}:=X_{1}\times\ldots\times\hat{X_{i}}\times\ldots\times X_{n}$,
i.e. $\prod_{\hat{i}}$ denotes the subproduct of ${\bf X}$ obtained by
omitting $X_{i}$. We have a natural projection ${\bf X}\to\prod_{\hat{i}}$.
The set of $G$-homotopy classes of $G$-self-equivalences of ${\bf X}$ over
$\prod_{\hat{i}}$ is denoted by $\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf
X})$. These are subgroups of $\operatorname{Aut}^{G}({\bf X})$.

######  Theorem C.

If all $G$-self-homotopy equivalences of ${\bf X}$ are $G$-reducible, then

$\operatorname{Aut}^{G}({\bf X})=\operatorname{Aut}^{G}_{\prod_{\hat{n}}}({\bf
X})\ldots\operatorname{Aut}^{G}_{\prod_{\hat{1}}}({\bf X}).$

Moreover each $\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf X})$ fits into a
split exact sequence,

$0\to\operatorname{Ker}(\phi_{i})\to\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf
X})\xrightarrow[]{\phi_{i}}\operatorname{Aut}^{G}(X_{i})\to 0,~{}~{}~{}1\leq
i\leq n.$

(See Theorem 3.7 below.)

The group $\operatorname{Aut}^{G}({\bf X})$ can be expressed as the product of
its $n$-subgroups $\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf X})$ (where
$i=1,\ldots,n$), which is difficult from computational viewpoint. Therefore we
develop a new factorization called LU decomposition, which is the following:

######  Theorem D.

If all $G$-self-homotopy equivalences of ${\bf X}$ are $G$-reducible, then

$\operatorname{Aut}^{G}({\bf X})=L^{G}(X_{1},\ldots,X_{n})\cdot
U^{G}(X_{1},\ldots,X_{n}).$

(See Theorem 4.2 below.) The factor $L^{G}(X_{1},\ldots,X_{n})$ has the
property that the induced maps on the homotopy groups can be represented by
lower triangular matrices and similarly by upper triangular matrices for the
other factor.

######  1.2.

Organisation.  The rest of the paper is organised as follows. In Section 2 we
consider product of two spaces. We give some sufficient conditions of
$G$-reducibility and obtain Theorem A and Theorem B. In Section 3, we consider
the case of $n$-products of spaces and obtain Theorem C. In the last Section
4, we obtain the $LU$-decomposition.

Throughout the paper, $G$ will denote a finite group and all spaces are based
$G$-CW complexes. All maps are base point preserving and equivariant.

######  1.3 Acknowledgements.

The first author would like to thank IIT Kanpur for PhD fellowship.

##  2\. $\operatorname{Aut}^{G}(X\times Y)$

We begin this section by recalling some basic facts about $G$-CW complexes. We
refer to [7] for details. Let $X$ be a $G$-space. For a subgroup $H<G$, the
$H$-fixed point set is defined as $X^{H}=\\{x\in X:~{}~{}hx=x,~{}~{}\forall
h\in H\\}$. If $f:X\to Y$ is a $G$-map then $f^{H}:=f|_{X^{H}}:X^{H}\to
Y^{H}$. We know that an equivariant map $f:X\to Y$ between $G$-CW complexes is
$G$-homotopy equivalence if and only if $f^{H}:X^{H}\to Y^{H}$ are ordinary
homotopy equivalences for all subgroups $H$ of $G$.

######  2.1.

$G$-reducibility.  For $G$-CW complexes $X,Y$, an equivariant self-homotopy
equivalence $f$ of $X\times Y$ is $G$-reducible if and only if $f^{H}$ is a
reducible self-homotopy equivalence of $X^{H}\times Y^{H}$ for all subgroups
$H$ of $G$. The following are some examples of when $G$-self homotopy
equivalences of $X\times Y$ are $G$-reducible.

######  2.2 Examples.

Let $X,Y$ be two pointed $G$-CW complexes.

  1. (a)

Assume that for every positive integer $n$ and for every pair of $G$-maps
$f:X\to Y,g:Y\to X$, at least one of the induced maps
$\pi_{n}(f^{H}):\pi_{n}(X^{H})\to\pi_{n}(Y^{H}),\pi_{n}(g^{H}):\pi_{n}(Y^{H})\to\pi_{n}(X^{H})$
is trivial for all $H\leq G$. Then all $G$-self equivalences of $X\times Y$
are $G$-reducible.

To see this, let $h\in\operatorname{Aut}^{G}(X\times Y)$. Therefore
$h^{H}:X^{H}\times Y^{H}\to X^{H}\times Y^{H}$ is an ordinary homotopy
equivalence for all $H\leq G$. By the given condition at least one of the
induced homomorphisms on the homotopy groups of the maps
$(h^{H})_{XY}:Y^{H}\to X^{H},~{}~{}(h^{H})_{YX}:X^{H}\to Y^{H}$ is trivial.
Now from [9, Proposition 2.1], we say that $(h^{H})_{XX},(h^{H})_{YY}$ are
self-homotopy equivalences of $X^{H}$ and $Y^{H}$ respectively for all $H\leq
G$. Therefore $(h_{XX})^{H}$ and $(h_{YY})^{H}$ both are self-homotopy
equivalences of $X^{H}$ and $Y^{H}$ respectively for all $H\leq G$ (since
$(h^{H})_{XX}=(p_{X})^{H}\circ h^{H}\circ(i_{X})^{H}=(p_{X}\circ h\circ
i_{X})^{H}=(h_{XX})^{H}$. Consequently
$h_{XX}\in\operatorname{Aut}^{G}(X),h_{YY}\in\operatorname{Aut}^{G}(Y)$.

  2. (b)

For every positive integer $n$, if at least one of the group
$\pi_{n}(X^{H}),\pi_{n}(Y^{H})$ is trivial for all $H\leq G$, then all
$G$-self equivalences of $X\times Y$ are $G$-reducible.

This follows from the above one.

  3. (c)

If at least one of
$\operatorname{Hom}(\pi_{n}(X^{H}),\pi_{n}(Y^{H})),\operatorname{Hom}(\pi_{n}(Y^{H}),\pi_{n}(X^{H}))$
is trivial $\forall~{}H\leq G,\forall n>0$ then all $G$-self equivalences of
$X\times Y$ are $G$-reducible.

This also immediately follows from the above.

  4. (d)

Let $m,n\geq 2$. Assume that the group $G=\mathbb{Z}_{2}$ acts on $S^{m}$ and
$S^{n}$ by reflection about $m$ and $n$ dimensional hyperplane respectively.
If $m\neq n$, then all $\mathbb{Z}_{2}$-self equivalences of $S^{m}\times
S^{n}$ are $\mathbb{Z}_{2}$-reducible. To see this, observe that,

$(S^{m})^{H}=\begin{cases}S^{m-1},~{}~{}~{}if~{}H=\mathbb{Z}_{2}\\\
S^{m},~{}~{}~{}~{}~{}~{}~{}if~{}~{}H=\\{1\\}.\end{cases}$

If $m<n$, then $[(S^{m})^{H},(S^{n})^{H}]$ is trivial. Therefore the induced
maps between $\pi_{k}(S^{m})^{H}$ and $\pi_{k}(S^{n})^{H}$ are trivial, for
all $H\leq\mathbb{Z}_{2}$ and for all $k>0$. Similarly for $m>n$, we can say
that the induced maps between $\pi_{k}(S^{n})^{H}$ and $\pi_{k}(S^{m})^{H}$
are trivial, for all $H\leq\mathbb{Z}_{2}$ and for all $k>0$. Hence all
$\mathbb{Z}_{2}$-self-homotopy equivalences of $S^{m}\times S^{n}$ are
$\mathbb{Z}_{2}$-reducible.

Recall that, an endomorphism $\phi\colon\Gamma\to\Gamma$ of an additively
written group $\Gamma$ is called _quasi-regular_ if the map
$(\operatorname{Id}-\phi)\colon\Gamma\to\Gamma$ defined by
$\gamma\mapsto\gamma-\phi(\gamma)$, is an automorphism. A self $G$-map
$f\colon X\to X$ is called _quasi-regular_ if the endomorphism of groups
$(f^{H})_{\\#n}\colon\pi_{n}(X^{H})\to\pi_{n}(X^{H})$ are quasi-regular for
all $H\leq G$ and all $n$. An endomorphism $\phi\colon\Gamma\to\Gamma$ is
called _nilpotent_ if $\phi^{k}=0$ for some $k$. For a nilpotent endomorphism
$\phi\colon\Gamma\to\Gamma$ such that $(\operatorname{Id}-\phi)$ is a
homomorphism, then $\phi$ is quasi-regular (cf. [10]).

The following Proposition gives some sufficient conditions for
$G$-reducibility. We denote $f_{XY}=p_{X}\circ f\circ\iota_{Y}:Y\to X$ and
$f_{YX}=p_{Y}\circ f\circ\iota_{X}:X\to Y.$

######  2.3 Proposition.

Let $X,Y$ be based $G$-CW complexes.

  1. (a)

Assume that $f_{YY}\in\operatorname{Aut}^{G}(Y)$ for all
$f\in\operatorname{Aut}^{G}(X\times Y)$. Then all $G$\- self-homotopy
equivalences of $X\times Y$ are $G$-reducible.

  2. (b)

Assume that for any $f\in\operatorname{Aut}^{G}(X\times Y)$ one of the maps
$f_{XY}\circ\widetilde{f}_{YX}:X\to X$ and $f_{YX}\circ\widetilde{f}_{XY}:Y\to
Y$ is quasi-regular. Then all $G$-self-homotopy equivalences of $X\times Y$
are $G$-reducible. Here $\widetilde{f}$ is the $G$-hmotopy inverse of $f$ in
$\operatorname{Aut}^{G}(X\times Y)$.

  3. (c)

Suppose that every self $G$-map of $X$, which can be factored through $Y$ (up
to homotopy), induces a nilpotent endomorphism on $\pi_{n}(X^{H})$ for all
$n,~{}H<G$ (i.e. if $X$ and $Y$ are homotopically distant), then $G$-self-
homotopy equivalences of $X\times Y$ are $G$-reducible.

###### Proof.

  1. (a)

Let $f\in\operatorname{Aut}^{G}(X\times Y)$. Then by assumption
$f_{YY}\in\operatorname{Aut}^{G}(Y)$. We have to show that
$f_{XX}\in\operatorname{Aut}^{G}(X)$ which is equivalent to show that
$(f_{XX})^{H}\in\operatorname{Aut}(X^{H}),~{}~{}\forall H\leq G$. By
definition $(f_{X})^{H}\circ(\iota_{X})^{H}=(f_{XX})^{H}$. Therefore by [9,
Proposition 2.3(b)], it is enough to show
$((f_{X})^{H},(p_{Y})^{H})=(f_{X},p_{Y})^{H}\in\operatorname{Aut}(X^{H}\times
Y^{H})$ for all $H\leq G$.

Since $\operatorname{Aut}^{G}(X\times Y)$ is a group, $f=({f}_{X},{f}_{Y})$
has an inverse
$\bar{f}=(\bar{f}_{X},\bar{f}_{Y})\in\operatorname{Aut}^{G}(X\times Y)$. So
$\bar{f}\circ f=\operatorname{Id}_{X\times Y}$. Now observe that,

$(p_{X},p_{Y})=\operatorname{Id}_{X\times Y}=\bar{f}\circ
f=(\bar{f}_{X},\bar{f}_{Y})\circ(f_{X},f_{Y})=(\bar{f}_{X}\circ(f_{X},f_{Y}),\bar{f}_{Y}\circ(f_{X},f_{Y})).$

So we get,
$\bar{f}_{X}\circ(f_{X},f_{Y})=p_{X};~{}~{}\bar{f}_{Y}\circ(f_{X},f_{Y})=p_{Y}.$
By the given condition, we can say that
$\bar{f}_{YY}\in\operatorname{Aut}^{G}(Y)$, as
$\bar{f}\in\operatorname{Aut}^{G}(X\times Y)$. Hence
$(\bar{f}_{YY})^{H}\in\operatorname{Aut}(Y^{H}),~{}\forall H\leq G$, i.e.
$(\bar{f}_{Y})^{H}\circ(\iota_{Y})^{H}\in\operatorname{Aut}(Y^{H})$ for all
$H\leq G$. Therefore from [9, Proposition 2.3(b)], we have
$(p_{X},\bar{f}_{Y})^{H}=((p_{X})^{H},(\bar{f}_{Y})^{H})\in\operatorname{Aut}(X^{H}\times
Y^{H}),~{}~{}\forall H\leq G$. Hence
$(p_{X},\bar{f}_{Y})\in\operatorname{Aut}^{G}(X\times Y)$. Now,

$(f_{X},p_{Y})=(f_{X},\bar{f}_{Y}\circ(f_{X},f_{Y}))=(p_{X},\bar{f}_{Y})\circ(f_{X},f_{Y})\in\operatorname{Aut}^{G}(X\times
Y).$

Hence $(f_{X},p_{Y})^{H}\in\operatorname{Aut}(X^{H}\times Y^{H})$ for all
$H\leq G$.

  2. (b)

For any map $f\colon X\times Y\to X\times Y$ the induced endomorphism on
homotopy groups
$(f^{H})_{\\#n}\colon{\pi}_{n}(X^{H})\times{\pi}_{n}(Y^{H})\to{\pi}_{n}(X^{H})\times{\pi}_{n}(Y^{H})$
can be represented by the matrix form for all $H\leq G$,

$M_{n}(f^{H}):=\begin{bmatrix}(f^{H}_{XX})_{\\#n}&(f^{H}_{XY})_{\\#n}\\\
(f^{H}_{YX})_{\\#n}&(f^{H}_{YY})_{\\#n}\end{bmatrix}.$

Without any ambiguity we can write $(f^{H})_{XX}=(f_{XX})^{H}=f^{H}_{XX}$ and
so on.

Therefore for any $f\in\operatorname{Aut}^{G}(X\times Y)$ with its inverse
$\widetilde{f}$, we can say that

$M_{n}(f^{H})\cdot
M_{n}(\widetilde{f^{H}})=M_{n}(\operatorname{Id}_{X^{H}\times Y^{H}}).$

This implies that
$(f^{H}_{YX})_{\\#n}\circ(\widetilde{f}^{H}_{XY})_{\\#n}+(f^{H}_{YY})_{\\#n}\circ(\widetilde{f}^{H}_{YY})_{\\#n}=\operatorname{Id}_{\pi_{n}(Y^{H})}$.
So we get,

$(f^{H}_{YY})_{\\#n}\circ(\widetilde{f}^{H}_{YY})_{\\#n}=\operatorname{Id}_{\pi_{n}(Y^{H})}-(f^{H}_{YX})_{\\#n}\circ(\widetilde{f}^{H}_{XY})_{\\#n}=\operatorname{Id}_{\pi_{n}(Y^{H})}-(f_{YX}\circ\widetilde{f}_{XY})^{H}_{\\#n}.$

If $f_{YX}\circ\widetilde{f}_{XY}$ is quasi-regular, then the maps
$\operatorname{Id}_{\pi_{n}(Y^{H})}-(f_{YX}\circ f_{XY})^{H}_{\\#n}$ are
automorphsims for all $n$ and $H\leq G$. Consequently,
$(f^{H}_{YY})_{\\#n}\colon\pi_{n}(Y^{H})\to\pi_{n}(Y^{H})$ are isomorphsim for
all $n$ and all $H\leq G$. By Whitehead theorem we get
$f_{YY}\in\operatorname{Aut}^{G}(Y)$. Using the part (a), we get that the
$G$-self-homotopy equivalences of $X\times Y$ are $G$-reducible.

Similarly, if we take $f_{XY}\circ\widetilde{f}_{YX}$ is quasi-regular, then
using the equality

$(f^{H}_{XY})_{\\#n}\circ(\widetilde{f}^{H}_{YX})_{\\#n}+(f^{H}_{XX})_{\\#n}\circ(\widetilde{f}^{H}_{XX})_{\\#n}=\operatorname{Id}_{\pi_{n}(X^{H})},$

we get the desire result.

  3. (c)

Let $f\colon X\times Y\to X\times Y$ be a self $G$-homotopy equivalence with
$G$-homotopy inverse $\widetilde{f}\colon X\times Y\to X\times Y$. From
part(b) its enough to show that $f_{XY}\circ\widetilde{f}_{YX}$ is quasi-
regular, which is equivalent to show that
$\operatorname{Id}_{\pi_{n}(X^{H})}-(f_{XY}\circ\widetilde{f}_{YX})^{H}_{\\#n}$
are homomorphsims, whenever $(f_{XY}\circ\widetilde{f}_{YX})^{H}_{\\#n}$ are
nilpotent endomorphism for all $n$ and all $H\leq G$. Observe that,
$\operatorname{Id}_{\pi_{n}(X^{H})}-(f_{XY}\circ\widetilde{f}_{YX})^{H}_{\\#n}=(f_{XX})^{H}_{\\#n}\circ(\widetilde{f}_{XX})^{H}_{\\#n}$
are homorphisms for all $n$ and all $H\leq G$. By the given assumption,
$(f_{XY}\circ\widetilde{f}_{YX})_{\\#n}$ are nilpotent natural transformation
for all $n$. Therefore $(f_{XY}\circ\widetilde{f}_{YX})^{H}_{\\#n}$ are
nilpotent endomorphism for all $n$ and all $H\leq G$, so the natural
transformation $(f_{XY}\circ\widetilde{f}_{YX})_{\\#n}$ are quasi-regular
endomorphisms for all $n$. Hence the map $f_{XY}\circ\widetilde{f}_{YX}$ is
quasi-regular and we get the $G$-reducibility of the map $f$.

∎

Recall that we have natural projections $p_{X}:X\times Y\to X$ and
$p_{Y}:X\times Y\to Y$. A self-map of $X\times Y$ such that $p_{X}\circ
f=p_{X}$ looks like $(p_{X},f_{Y})$. We define

$\operatorname{Aut}^{G}_{X}(X\times Y)=\\{f\in\operatorname{Aut}^{G}(X\times
Y)~{}|~{}~{}p_{X}\circ f=p_{X}\\},$ $\operatorname{Aut}^{G}_{Y}(X\times
Y)=\\{f\in\operatorname{Aut}^{G}(X\times Y)~{}|~{}~{}p_{Y}\circ f=p_{Y}\\}.$

We now show that these are subgroups $\operatorname{Aut}^{G}(X\times Y)$ and
obtain a short exact sequence. Recall that $\operatorname{aut}_{1}(X)$ is the
connected component of identity of $\operatorname{aut}(X)$. This has a
$G$-action defined by $(g.f)(x)=gf(g^{-1}x).$

######  2.4 Proposition.

  1. (a)

If $(p_{X},g):X\times Y\to X\times Y$ is a $G$-self-homotopy equivalence, then
there exists a $G$-map $\widetilde{g}:X\times Y\to Y$, such that
$(p_{X},\widetilde{g})$ is a $G$-equivariant homotopy inverse of $(p_{X},g)$.

  2. (b)

The map $(p_{X},g):X\times Y\to X\times Y$ is a $G$-homotopy equivalence if
and only if $g\circ\iota_{Y}\in\operatorname{Aut}^{G}(Y)$.

  3. (c)

$\operatorname{Aut}^{G}_{X}(X\times Y)$ is a subgroup of
$\operatorname{Aut}^{G}(X\times Y)$ whose elements are of the form $(p_{X},g)$
with $g\circ\iota_{Y}\in\operatorname{Aut}^{G}(Y)$.

  4. (d)

If $X$ is a connected $G$-CW complex, then there exists a split exact sequence

$0\to[X,\operatorname{aut}_{1}(Y)]_{G}\to\operatorname{Aut}^{G}_{X}(X\times
Y)\xrightarrow{\phi}\operatorname{Aut}^{G}(Y)\to 0,$

where $\phi(p_{X},g)=g\circ\iota_{Y}$, for all
$(p_{X},g)\in\operatorname{Aut}^{G}_{X}(X\times Y).$

###### Proof.

  1. (a)

Let $(p_{X},g)\in\operatorname{Aut}^{G}(X\times Y)$. Then it has an inverse in
$\operatorname{Aut}^{G}(X\times Y)$, say $(f_{X},f_{Y})$. So,

$\displaystyle(p_{X},g)\circ(f_{X},f_{Y})=\operatorname{Id}_{X\times
Y}=(p_{X},p_{Y})$ $\displaystyle\Rightarrow$
$\displaystyle~{}(f_{X},g\circ(f_{X},f_{Y}))=(p_{X},p_{Y})$
$\displaystyle\Rightarrow$ $\displaystyle~{}f_{X}=p_{X}.$

Therefore $(p_{X},f_{Y})$ is a $G$-equivariant homotopy inverse of
$(p_{X},g)$. Consequently $(p_{X},\widetilde{g})$ is a $G$-equivariant
homotopy inverse of $(p_{X},g)$, where $\widetilde{g}=f_{Y}:X\times Y\to Y$.

  2. (b)

We know that $(p_{X},g):X\times Y\to X\times Y$ is a $G$-equivariant self-
homotopy equivalence, if and only if $(p_{X},g)^{H}:X^{H}\times Y^{H}\to
X^{H}\times Y^{H}$ are self-homotopy equivalences for all $H\leq G$, i.e
$((p_{X})^{H},g^{H})\in\operatorname{Aut}(X^{H}\times Y^{H})$ for all $H\leq
G$. From [9, Proposition 2.3], we say that
$((p_{X})^{H},g^{H})\in\operatorname{Aut}(X^{H}\times Y^{H})$ if and only if
$g^{H}\circ(\iota_{Y})^{H}\in\operatorname{Aut}(Y^{H})$. Therefore
$(p_{X},g):X\times Y\to X\times Y$ is a $G$-equivariant homotopy equivalence,
if and only if
$(g\circ\iota_{Y})^{H}=g^{H}\circ(\iota_{Y})^{H}\in\operatorname{Aut}(Y^{H})$
for all $H\leq G$ i.e. $g\circ\iota_{Y}\in\operatorname{Aut}^{G}(Y)$.

  3. (c)

Follows from part (a) and (b).

  4. (d)

First we prove that $\phi$ is a homomorphism. Let
$(p_{X},g),(p_{X},f)\in\operatorname{Aut}^{G}_{X}(X\times Y)$, then

$\phi((p_{X},g)\circ(p_{X},f))=\phi(p_{X},g\circ(p_{X},f))=g\circ(p_{X},f)\circ\iota_{Y}$
$=g\circ\iota_{Y}\circ f\circ\iota_{Y}=\phi(p_{X},g)\circ\phi(p_{X},f).$

Let us define an another map

$\psi:\operatorname{Aut}^{G}(Y)\to\operatorname{Aut}^{G}_{X}(X\times
Y),~{}~{}\psi(h)=(p_{X},h\circ p_{Y}),$

for all $h\in\operatorname{Aut}^{G}(Y)$. Now
$(\phi\circ\psi)(h)=\phi(p_{X},h\circ p_{Y})=h\circ p_{Y}\circ\iota_{Y}=h$ for
all $h\in\operatorname{Aut}^{G}(Y)$, i.e. $\phi\circ\psi=\operatorname{Id}$.
Therefore $\phi$ is an onto map.

Now we determine the kernel of $\phi.$ Clearly, $(p_{X},f)\in\ker(\phi)$ if
and only if $\phi(p_{X},f)=\operatorname{Id}_{Y}$, i.e
$f\circ\iota_{Y}=\operatorname{Id}_{Y}$, where $f:X\times Y\to Y$ is a
$G$-map. Therefore $(p_{X},f)\in\operatorname{Ker}(\phi)$ corresponds under
the adjoint associativity to a base point preserving map
$\widetilde{f}:X\to\operatorname{map}(Y,Y)$ defined as
$\widetilde{f}(x)(y)=f(x,y)$. Observe that $\widetilde{f}$ sends the base
point of $X$ to $\operatorname{Id}_{Y}$. Since $X$ is a connected, the image
of $\widetilde{f}$ is contained in the component of $\operatorname{map}(Y,Y)$
that contains $\operatorname{Id}_{Y}$, denoted by $\operatorname{aut}_{1}(Y)$.
Again $\widetilde{f}$ is a $G$-equivariant map, because

$(g\cdot\widetilde{f}(x))(y)=g\widetilde{f}(x)(g^{-1}y)=gf(x,g^{-1}y)=f(gx,y)=\widetilde{f}(gx)(y),$

i.e. $\widetilde{f}(gx)=g\cdot\widetilde{f}(x)$. Therefore
$\operatorname{Ker}(\phi)$ can be identified with
$[X,\operatorname{aut}_{1}(Y)]_{G}$, which proves the assertion.

∎

The following corollary follows easily from Proposition 2.4(b).

######  2.5 Corollary.

Assume that all $G$-self-homotopy equivalences of $X\times Y$ are
$G$-reducible. If $f=(f_{X},f_{Y})\in\operatorname{Aut}^{G}(X\times Y)$, then
$(p_{X},f_{Y})\operatorname{Aut}^{G}_{X}(X\times Y)$, and
$(f_{X},p_{Y})\in\operatorname{Aut}^{G}_{Y}(X\times Y)$.

The following is a sufficient condition for an equivariant self-map $f$ of
$X\times Y$ to be a $G$-homotopy equivalence.

######  2.6 Proposition.

Let us assume all $G$-self-homotopy equivalences of $X\times Y$ are
G-reducible. For any $f\in\operatorname{map}^{G}(X\times Y,X\times Y)$, we
have $f\in\operatorname{Aut}^{G}(X\times Y)$ if and only if
$f_{XX}\in\operatorname{Aut}^{G}(X)$ and $f_{YY}\in\operatorname{Aut}^{G}(Y)$.

###### Proof.

The forward implication is trivial.

Conversely, let $f\in\operatorname{map}^{G}(X\times Y,X\times Y)$ such that
$f_{XX}\in\operatorname{Aut}^{G}(X),f_{YY}\in\operatorname{Aut}^{G}(Y)$. By
Proposition 2.4[(a), (b)], we have
$(f_{X},p_{Y})\in\operatorname{Aut}^{G}(X\times Y)$ with inverse
$(\widetilde{f_{X}},p_{Y})\in\operatorname{Aut}^{G}(X\times Y)$. Observe that,

$f=(f_{X},f_{Y})=[p_{X},f_{Y}\circ(f_{X},p_{Y})^{-1}]\circ(f_{X},p_{Y})=[p_{X},f_{Y}\circ(\widetilde{f_{X}},p_{Y})]\circ(f_{X},p_{Y}).$

Moreover,

$(p_{X},f_{Y})\circ(\widetilde{f_{X}},p_{Y})=[p_{X}\circ(\widetilde{f_{X}},p_{Y}),f_{Y}\circ(\widetilde{f_{X}},p_{Y})]\in\operatorname{Aut}^{G}(X\times
Y).$

By applying the $G$-reducibility to the right-hand side, we say that
$f_{Y}\circ(\widetilde{f_{X}},p_{Y})\circ\iota_{Y}\in\operatorname{Aut}^{G}(Y)$.
Therefore from Proposition 2.4(b), we have
$(p_{X},f_{Y}\circ(\widetilde{f_{X}},p_{Y}))\in\operatorname{Aut}^{G}(X\times
Y)$. Consequently,
$f=[p_{X},f_{Y}\circ(\widetilde{f_{X}},p_{Y})]\circ(f_{X},p_{Y})\in\operatorname{Aut}^{G}(X\times
Y)$.

∎

We have seen in Proposition 2.4 that $\operatorname{Aut}^{G}_{X}(X\times Y)$
and $\operatorname{Aut}^{G}_{Y}(X\times Y)$ are two subgroups of
$\operatorname{Aut}^{G}(X\times Y).$ We now show that the product of these
subgroups is the whole group.

######  2.7 Theorem.

Let us asssume all self-homotopy equivalences of $X\times Y$ are
$G$-reducible. Then

$\operatorname{Aut}^{G}(X\times Y)=\operatorname{Aut}^{G}_{X}(X\times
Y)\cdot\operatorname{Aut}^{G}_{Y}(X\times Y).$

###### Proof.

Clearly, $\operatorname{Aut}^{G}_{X}(X\times
Y)\cap\operatorname{Aut}^{G}_{Y}(X\times Y)=(p_{X},p_{Y})$. Since the
intersection of the two subgroups is trivial, therefore if the factorization
of the element in $\operatorname{Aut}^{G}(X\times Y)$ exists, it will be
unique. We have the factorization,

$f=(f_{X},f_{Y})=[p_{X},f_{Y}\circ(f_{X},p_{Y})^{-1}]\circ(f_{X},p_{Y})=[p_{X},f_{Y}\circ(\widetilde{f_{X}},p_{Y})]\circ(f_{X},p_{Y}).$

From Corollary 2.5 we can say that the second factor $(f_{X},p_{Y})$ in the
decomposition of $f$ is in $\operatorname{Aut}^{G}_{Y}(X\times Y)$. Also its
inverse $(\widetilde{f}_{X},p_{Y})\in\operatorname{Aut}^{G}(X\times Y)$. It is
sufficient to show that the first factor is in
$\operatorname{Aut}^{G}_{X}(X\times Y)$. Now

$(f_{X},f_{Y})\circ(\widetilde{f}_{X},p_{Y})=[f_{X}\circ(\widetilde{f}_{X},p_{Y}),f_{Y}\circ(\widetilde{f}_{X},p_{Y})]\in\operatorname{Aut}^{G}(X\times
Y),$

using $G$-reducibility we have,
$f_{Y}\circ(\widetilde{f}_{X},p_{Y})\circ\iota_{Y}\in\operatorname{Aut}^{G}(Y)$.
Therefore by Proposition 2.4(b), we can say that the first factor
$[p_{X},f_{Y}\circ(\widetilde{f_{X}},p_{Y})]\in\operatorname{Aut}^{G}_{X}(X\times
Y)$.

∎

######  2.8 Lemma.

Let Y be a connected $G$-CW complex such that
$[Y,\operatorname{aut}_{1}(X)]_{G}=\\{\ast\\}$. For every $G$-map $h\colon
X\times Y\to X$ such that $h\circ\iota_{X}\in\operatorname{Aut}^{G}(X)$, then
the relation $h\simeq h\circ\iota_{X}\circ p_{X}$ holds.

###### Proof.

From 2.4(d), we have a short exact sequence,

$0\to[Y,\operatorname{aut}_{1}(X)]_{G}\to\operatorname{Aut}^{G}_{Y}(X\times
Y)\xrightarrow[]{\phi}\operatorname{Aut}^{G}(X)\to 0,$

defined as

$\phi(f,p_{Y}):=f\circ\iota_{X}.$

Since $[Y,\operatorname{aut}_{1}(X)]_{G}=\\{\ast\\}$, therefore $\phi$ is an
isomorphism. Let the splitting map of $\phi$ is
$\psi\colon\operatorname{Aut}^{G}(X)\to\operatorname{Aut}^{G}_{Y}(X\times Y)$,
defined as $\psi(g):=(g\circ p_{X},p_{Y})$. Now for any $G$-map $h\colon
X\times Y\to X$ with
$h\circ\iota_{X}\in\operatorname{Aut}^{G}(X)\Rightarrow(h,p_{Y})\in\operatorname{Aut}^{G}_{Y}(X\times
Y)$. Again

$h\circ\iota_{X}\in\operatorname{Aut}^{G}(X)\Rightarrow\psi(h\circ
i_{X})=(h\circ\iota_{X}\circ p_{X},p_{Y})\in\operatorname{Aut}^{G}_{Y}(X\times
Y).$

Since $\phi\big{(}h,p_{Y}\big{)}=\phi\big{(}h\circ\iota_{X}\circ
p_{X},p_{Y}\big{)}$, therefore $(h,p_{Y})\simeq(h\circ\iota_{X}\circ
p_{X},p_{Y})$, otherwise $\phi$ fails to be an isomorphism. Consequently we
get the desired relation $h\simeq h\circ\iota_{X}\circ p_{X}$.

∎

We now obtain a description of $\operatorname{Aut}^{G}(X\times Y)$ in terms of
$\operatorname{Aut}^{G}(X)$ and $\operatorname{Aut}^{G}(Y)$, which is an
equivariant version of Booth and Heath [4, Theorem 2.7].

######  2.9 Theorem.

Let $X$ and $Y$ be connected $G$-CW complexes and all self-homotopy
equivalences of $X\times Y$ are $G$-reducible. If
$[Y,\operatorname{aut}_{1}(X)]_{G}=\\{\ast\\}$, then we have a split exact
sequence

$0\to[X,\operatorname{aut}_{1}(Y)]_{G}\to\operatorname{Aut}^{G}(X\times
Y)\xrightarrow[]{\phi}\operatorname{Aut}^{G}(X)\times\operatorname{Aut}^{G}(Y)\to
0.$

where $\phi(f)=(f_{XX},f_{YY})$.

###### Proof.

First we show that $\phi$ is a homomorphism. Let
$f=(f_{X},f_{Y}),g=(g_{X},g_{Y})\in\operatorname{Aut}^{G}(X\times Y)$, then

$\displaystyle\phi(f\circ g)$
$\displaystyle=\phi\big{(}\big{[}f_{X}\circ(g_{X},g_{Y}),f_{Y}\circ(g_{X},g_{Y})\big{]}\big{)}$
$\displaystyle=\big{[}f_{X}\circ(g_{X},g_{Y})\circ\iota_{X},f_{Y}\circ(g_{X},g_{Y})\circ\iota_{Y}\big{]}$
$\displaystyle=\big{[}f_{X}\circ\iota_{X}\circ
p_{X}\circ(g_{X},g_{Y})\circ\iota_{X},f_{Y}\circ(g_{X}\circ\iota_{X}\circ
p_{X},g_{Y})\circ\iota_{Y}\big{]}~{}~{}~{}(\text{by Lemma }\ref{Lgred})$
$\displaystyle=(f_{X}\circ\iota_{X}\circ
g_{X}\circ\iota_{X},f_{Y}\circ\iota_{Y}\circ g_{Y}\circ\iota_{Y})$
$\displaystyle=(f_{XX}\circ g_{XX},f_{YY}\circ g_{YY})$
$\displaystyle=\phi(f)\circ\phi(g).$

Consider an another map
$\psi:\operatorname{Aut}^{G}(X)\times\operatorname{Aut}^{G}(Y)\to\operatorname{Aut}^{G}(X\times
Y)$, define as

$\psi(h_{1},h_{2}):=(h_{1}\circ p_{X},h_{2}\circ
p_{Y}),~{}~{}~{}~{}~{}~{}\forall(h_{1},h_{2})\in\operatorname{Aut}^{G}(X)\times\operatorname{Aut}^{G}(Y).$

From Proposition 2.6, we can say that the map $\psi$ is well-defined. Clearly
$\phi\circ\psi=\operatorname{Id}_{\operatorname{Aut}^{G}(X)\times\operatorname{Aut}^{G}(Y)}$,
therefore $\phi$ is onto homomorphsim.

Now we determine $\operatorname{Ker}(\phi)$. An element
$f=(f_{X},f_{Y})\in\operatorname{Aut}^{G}(X\times Y)$ is in
$\operatorname{Ker}(\phi)$ if and only if
$f_{XX}=\operatorname{Id}_{X},~{}~{}~{}f_{YY}=\operatorname{Id}_{Y}$.
Therefore $f=(f_{X},f_{Y})\in\operatorname{Ker}(\phi)$ corresponds under
adjoint associativity to the base point preserving maps
$\widetilde{f}_{X}:Y\to\operatorname{map}(X,X)$ and
$\widetilde{f}_{Y}:X\to\operatorname{map}(Y,Y)$, defined as
$\widetilde{f}_{X}(y)(x):=f_{X}(x,y)$ and
$\widetilde{f}_{Y}(x)(y):=f_{Y}(x,y)$ respectively. Since both $X$ and $Y$ are
connected, therefore the images of $\widetilde{f}_{X}$ and $\widetilde{f}_{Y}$
are contained in the component of $\operatorname{map}(X,X)$ and
$\operatorname{map}(Y,Y)$ which contains $\operatorname{Id}_{X}$ and
$\operatorname{Id}_{Y}$ respectively. These components are denoted by
$\operatorname{aut}_{1}(X)$ and $\operatorname{aut}_{1}(Y)$. Again the maps
$\widetilde{f}_{X}$ and $\widetilde{f}_{Y}$ are $G$-maps. Therefore the
$\operatorname{Ker}(\phi)$ can be identified with
$[Y,\operatorname{aut}_{1}(X)]_{G}\times[X,\operatorname{aut}_{1}(Y)]_{G}$.
Since $[Y,\operatorname{aut}_{1}(X)]_{G}=\\{\ast\\}$, we get the required
split exact sequence.

∎

##  3\. $\operatorname{Aut}^{G}(X_{1}\times\ldots\times X_{n})$

Now we use the results of the previous section to generalize for higher
product of based $G$-CW complexes $X_{1},X_{2},\ldots,X_{n}$. Consider

${\bf X}=X_{1}\times\ldots\times X_{n}$

with diagonal $G$-action. For a $G$-map $f=(f_{1},\ldots,f_{n})\colon{\bf
X}\to{\bf X}$, define $f_{ij}\colon X_{j}\to X_{i}$ as the composition
$f_{ij}\colon=p_{i}\circ f\circ\iota_{j}=f_{i}\circ\iota_{j}$, where

$p_{i}\colon{\bf X}\to X_{i},~{}~{}~{}\iota_{j}\colon X_{j}\to{\bf
X},~{}~{}1\leq,i,j\leq n,$

are projections and inclusions respectively. Clearly $p_{i},\iota_{j}$ are
$G$-maps, therefore $f_{ij}$ are $G$-maps and the following diagrams are
commute.

|
---|---
$\textstyle{{\bf
X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{f_{i}}$$\textstyle{{\bf
X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{i}}$$\textstyle{X_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{ij}}$$\scriptstyle{\iota_{j}}$$\textstyle{X_{i}}$

######  3.1 Definition.

A $G$-self-homotopy equivalence $f$ of ${\bf X}$ is said to be _$G$
-reducible_ if the maps $f_{ii}\colon X_{i}\to X_{i}$ are $G$-self-homotopy
equivalences for all $1\leq i\leq n$.

Let $\prod_{\hat{i}}:=X_{1}\times\ldots\times\widehat{X_{i}}\times\ldots\times
X_{n}$, i.e. $\prod_{\hat{i}}$ denote the subproduct of ${\bf X}$ obtained by
omitting $X_{i}$. We have natural projection ${\bf X}\to\prod_{\hat{i}}$. We
denote by $\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf X})$ the set of
$G$-homotopy classes of self-equivalences of ${\bf X}$ over $\prod_{\hat{i}}.$
The following Lemma shows that elements of
$\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf X})$ are of the form
$(p_{1},\ldots,p_{i-1},f_{i},p_{i+1},\ldots,p_{n})$ where
$f_{ii}\in\operatorname{Aut}^{G}(X_{i})$.

######  3.2 Lemma.

Given a $G$-map $f_{i}\colon{\bf X}\to X_{i}$, we have
$f=(p_{1},\ldots,p_{i-1},f_{i},p_{i+1},\ldots,p_{n})\in\operatorname{Aut}^{G}({\bf
X})$ iff $f_{ii}\in\operatorname{Aut}^{G}(X_{i})$ for any $1\leq i\leq n.$
Moreover the inverse of the map $f$ is of the form
$(p_{1},\ldots,p_{i-1},\widetilde{f}_{i},p_{i+1},\ldots,p_{n}),$ for some
$G$-map $\widetilde{f}_{i}:{\bf X}\to X_{i}.$

###### Proof.

Take $\bar{\bf X}=X_{1}\times\ldots\times X_{i-1}\times
X_{i+1}\times\ldots\times X_{n}\times X_{i}$. Consider the permutation map
$\phi:{\bf X}\to\bar{\bf X},$ defined as
$(x_{1},\ldots,x_{n})\mapsto(x_{1},\ldots,x_{i-1},x_{i+1},\ldots,x_{n},x_{i}).$

Clearly $\phi$ is a homeomorphism. Therefore
$(p_{1},\ldots,p_{i-1},f_{i},p_{i+1},\ldots,p_{n})\in\operatorname{Aut}^{G}({\bf
X})$ if and only if
$\phi\circ(p_{1},\ldots,p_{i-1},f_{i},p_{i+1},\ldots,p_{n})\circ\phi^{-1}=(p^{\prime}_{1},\ldots,p^{\prime}_{i-1},p^{\prime}_{i+1},\ldots,p^{\prime}_{n},f^{\prime}_{i})\in\operatorname{Aut}^{G}(\bar{\bf
X}),$ where $p^{\prime}_{j}:\bar{\bf X}\to X_{j}$ are all projection maps for
$j\in\\{1,\ldots,i-1,i+1,\ldots,n\\}$.

|  |  |
---|---|---|---
$\textstyle{{\bf
X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\phi}$$\textstyle{\bar{\bf
X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi\circ
f\circ\phi^{-1}}$$\textstyle{{\bf
X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\textstyle{\bar{\bf
X}}$

Take $\bar{\b{X}}=X_{1}\times\ldots\times X_{i-1}\times
X_{i+1}\times\ldots\times X_{n}$. So
$(p^{\prime}_{\bar{\b{X}}},f^{\prime}_{i})=(p^{\prime}_{1},\ldots,p^{\prime}_{i-1},p^{\prime}_{i+1},\ldots,p^{\prime}_{n},f^{\prime}_{i})$
and $\bar{\bf X}=\bar{\b{X}}\times X_{i}$. From Proposition 2.4(b), we can say
that the map
$(p^{\prime}_{\bar{\b{X}}},f^{\prime}_{i})=(p^{\prime}_{1},\ldots,p^{\prime}_{i-1},p^{\prime}_{i+1},\ldots,p^{\prime}_{n},f^{\prime}_{i})\in\operatorname{Aut}^{G}(\bar{\bf
X})$ if and only if $f^{\prime}_{ii}\in\operatorname{Aut}^{G}(X_{i})$.
Therefore,

$\phi^{-1}\circ(p^{\prime}_{1},\ldots,p^{\prime}_{i-1},p^{\prime}_{i+1},\ldots,p^{\prime}_{n},f^{\prime}_{i})\circ\phi=(p_{1},\ldots,p_{i-1},f_{i},p_{i+1},\ldots,p_{n})\in\operatorname{Aut}^{G}({\bf
X})$ $\iff f_{ii}\in\operatorname{Aut}^{G}(X_{i}),~{}~{}~{}~{}~{}~{}(\because
f^{\prime}_{ii}=f^{\prime}_{i}\circ\iota^{\prime}_{i}=f_{i}\circ\phi^{-1}\circ\iota^{\prime}_{i}=f_{i}\circ\iota_{i}=f_{ii}).$

Since
$(p^{\prime}_{\bar{\b{X}}},f^{\prime}_{i})\in\operatorname{Aut}^{G}(\bar{\bf
X})$, using Proposition 2.4(a), we can say that
$(p^{\prime}_{\bar{\b{X}}},\widetilde{f}^{\prime}_{i})$ is a $G$-homotopy
inverse of $(p^{\prime}_{\bar{\b{X}}},f^{\prime}_{i})$ in
$\operatorname{Aut}^{G}(\bar{\bf X})$. Therefore the map
$\phi^{-1}\circ(p^{\prime}_{\bar{\b{X}}},\widetilde{f}^{\prime}_{i})\circ\phi=(p_{1},\ldots,p_{i-1},\widetilde{f}_{i},p_{i+1},\ldots,p_{n})$
is the $G$-inverse of

$\phi^{-1}\circ(p^{\prime}_{\bar{\b{X}}},f^{\prime}_{i})\circ\phi=(p_{1},\ldots,p_{i-1},f_{i},p_{i+1},\ldots,p_{n})$

in $\operatorname{Aut}^{G}({\bf X})$.

∎

We now deduce that $\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf X})$ are the
subgroups of $\operatorname{Aut}^{G}({\bf X})$, for all $i=1,\ldots,n$.

######  3.3 Corollary.

For a pointed $G$-CW complex ${\bf X}$, the subsets
$\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf X})$ are subgroups of
$\operatorname{Aut}^{G}({\bf X})$, for any $i=1,\ldots,n$.

###### Proof.

Any element of $\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf X})$ is of the
form $(p_{1},\ldots,f_{i},\ldots,p_{n}),$ where
$f_{i}\circ\iota_{i}\in\operatorname{Aut}^{G}(X_{i})$. This set is non-empty
because $(p_{1},\ldots,p_{n})\in\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf
X})$. Let for any two elements
$(p_{1},\ldots,f_{i},\ldots,p_{n}),~{}~{}~{}(p_{1},\ldots,g_{i},\ldots,p_{n})\in\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf
X})$, we have

$(p_{1},\ldots,f_{i},\ldots,p_{n})\circ(p_{1},\ldots,g_{i},\ldots,p_{n})=(p_{1},\ldots,f_{i}\circ(p_{1},\ldots,g_{i},\ldots,p_{n}),\ldots,p_{n}).$

Clearly,

$f_{i}\circ(p_{1},\ldots,g_{i},\ldots,p_{n})\circ\iota_{i}=f_{ii}\circ
g_{ii}\in\operatorname{Aut}^{G}(X_{i}).$

Therefore
$(p_{1},\ldots,f_{i},\ldots,p_{n})\circ(p_{1},\ldots,g_{i},\ldots,p_{n})\in\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf
X})$. From Lemma 3.2, we have,
$(p_{1},\ldots,f_{i},\ldots,p_{n})^{-1}=(p_{1},\ldots,\widetilde{f}_{i},\ldots,p_{n})$,
which is in $\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf X})$. Consequently
$\operatorname{Aut}^{G}_{\prod_{\hat{i}}}({\bf X})$ becomes a subgroup of
$\operatorname{Aut}^{G}({\bf X})$, for any $i=1,\ldots,n$.

∎

The following theorem and its corollary give a sufficient condition for an
equivariant map $f\colon{\bf X}\to{\bf X}$ to be a $G$-self-homotopy
equivalence under the $G$-reducibility assumption.

######  3.4 Theorem.

Assume that all $G$-self-homotopy equivalences of ${\bf X}$ are $G$-reducible
(where each $X_{i}$ are G-CW complexes). Given maps
$f_{i}\in\operatorname{map}^{G}({\bf X},X_{i})$ for $k\leq i\leq n$, we have
$(p_{1},\ldots,p_{k-1},f_{k},\ldots,f_{n})\in\operatorname{Aut}^{G}({\bf X})$
if and only if $f_{ii}\in\operatorname{Aut}^{G}(X_{i})$ for all $k\leq i\leq
n$.

###### Proof.

Let $(p_{1},\ldots,p_{k-1},f_{k},\ldots,f_{n})\in\operatorname{Aut}^{G}({\bf
X})$, therefore by using $G$-reducibility we have
$f_{ii}\in\operatorname{Aut}^{G}(X_{i})$.

Conversely, $k=n$ follows from the above Lemma 3.2. Now we consider the case
$k=n-1.$ Observe that,

$(p_{1},\ldots,p_{n-2},f_{n-1},f_{n})=\Big{[}p_{1},\ldots,p_{n-1},f_{n}\circ(p_{1},\ldots,f_{n-1},p_{n})^{-1}\Big{]}\circ(p_{1},\ldots,f_{n-1},p_{n}).$

By Lemma 3.2, the inverse $(p_{1},\ldots,f_{n-1},p_{n})^{-1}$ is of the form
$(p_{1},\ldots,\widetilde{f}_{n-1},p_{n})\in\operatorname{Aut}^{G}({\bf X}).$
Since $f_{nn}\in\operatorname{Aut}^{G}(X_{n})$, we get by using Lemma 3.2
again,

$\displaystyle(p_{1},\ldots,p_{n-1},f_{n})\circ(p_{1},\ldots,\widetilde{f}_{n-1},p_{n})=\Big{[}p_{1},\ldots,\widetilde{f}_{n-1},f_{n}\circ$
$\displaystyle(p_{1},....,\widetilde{f}_{n-1},p_{n})\Big{]}\in\operatorname{Aut}^{G}({\bf
X}).$

Using $G$-reducibility we have
$f_{n}\circ(p_{1},\ldots,\widetilde{f}_{n-1},p_{n})\circ\iota_{n}\in\operatorname{Aut}^{G}(X_{n})$.
Therefore
$\Big{[}p_{1},\ldots,p_{n-1},f_{n}\circ(p_{1},\ldots,\widetilde{f}_{n-1},p_{n})\Big{]}\in\operatorname{Aut}^{G}({\bf
X})$. Consequently,

$(p_{1},\ldots,p_{n-2},f_{n-1},f_{n})=\Big{[}p_{1},\ldots,p_{n-1},f_{n}\circ(p_{1},\ldots,\widetilde{f}_{n-1},p_{n})\Big{]}\circ(p_{1},\ldots,f_{n-1},p_{n}),$

is in $\operatorname{Aut}^{G}({\bf X})$ whenever
$f_{n-1n-1}\in\operatorname{Aut}^{G}(X_{n-1})$ and
$f_{nn}\in\operatorname{Aut}^{G}(X_{n})$. This completes the proof for
$k=n-1.$

Now consider the case $k=n-2$. We have factorisation,

$\displaystyle(p_{1},\ldots,p_{n-3},f_{n-2},f_{n-1},f_{n})=$
$\displaystyle\Big{[}p_{1},\ldots,p_{n-2},f_{n-1}\circ(p_{1},\ldots,\widetilde{f}_{n-2},p_{n-1},p_{n}),f_{n}\circ(p_{1},\ldots,\widetilde{f}_{n-2},p_{n-1},p_{n})\Big{]}\circ$
$\displaystyle(p_{1},\ldots,f_{n-2},p_{n-1},p_{n}).$

Clearly, the second factor is in $\operatorname{Aut}^{G}({\bf X})$ by Lemma
3.2. To show the first factor is in $\operatorname{Aut}^{G}({\bf X})$, it is
enough to show that (by the previous step $k=n-1$),

$f_{n-1}\circ(p_{1},\ldots,\widetilde{f}_{n-2},p_{n-1},p_{n})\circ\iota_{n-1}\in\operatorname{Aut}^{G}(X_{n-1}),$

and

$f_{n}\circ(p_{1},\ldots,\widetilde{f}_{n-2},p_{n-1},p_{n})\circ\iota_{n}\in\operatorname{Aut}^{G}(X_{n}).$

Since,

$\displaystyle(p_{1},\ldots,p_{n-2},f_{n-1},f_{n})\circ(p_{1},\ldots,p_{n-3},\widetilde{f}_{n-2},p_{n-1},p_{n})=$
$\displaystyle\Big{[}p_{1},\ldots,\widetilde{f}_{n-2},f_{n-1}\circ(p_{1},\ldots,\widetilde{f}_{n-2},p_{n-1},p_{n}),f_{n}\circ(p_{1},\ldots,\widetilde{f}_{n-2},p_{n-1},p_{n})\Big{]}\in\operatorname{Aut}^{G}({\bf
X}).$

Using $G$-reducibility we conclude that,

$f_{n-1}\circ(p_{1},\ldots,p_{n-3},\widetilde{f}_{n-2},p_{n-1},p_{n})\circ\iota_{n-1}\in\operatorname{Aut}^{G}(X_{n-1}),$

and

$f_{n}\circ(p_{1},\ldots,p_{n-3},\widetilde{f}_{n-2},p_{n-1},p_{n})\circ\iota_{n}\in\operatorname{Aut}^{G}(X_{n}).$

Therefore,
$(p_{1},\ldots,p_{n-3},f_{n-2},f_{n-1},f_{n})\in\operatorname{Aut}^{G}({\bf
X})$, whenever $f_{ii}\in\operatorname{Aut}^{G}(X_{i}),~{}~{}~{}~{}\forall
i=n-2,n-1,n$.

Inductively, we can prove that
$(p_{1},\ldots,p_{k-1},f_{k},\ldots,f_{n})\in\operatorname{Aut}^{G}({\bf X})$,
whenever $f_{ii}\in\operatorname{Aut}^{G}(X_{i}),~{}~{}k\leq i\leq n$.

∎

######  3.5 Corollary.

Assume that all $G$-self-homotopy equivalences of ${\bf X}$ are $G$-reducible.
Then an equivariant map $f=(f_{1},\ldots,f_{n})\colon{\bf X}\to{\bf X}$ is
$G$-self homotopy equivalence if $f_{ii}\in\operatorname{Aut}^{G}(X_{i})$ for
all $1\leq i\leq n$.

###### Proof.

Let $f=(f_{1},\ldots,f_{n})\colon{\bf X}\to{\bf X}$ be an equivariant map such
that $f_{ii}\in\operatorname{Aut}^{G}(X_{i})$ for all $1\leq i\leq n.$
Therefore from Theorem 3.4 we have
$(p_{1},f_{2},\ldots,f_{n})\in\operatorname{Aut}^{G}({\bf X})$. Now,

$(f_{1},\ldots,f_{n})=\Big{[}p_{1},f_{2}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n}),\ldots,f_{n}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n})\Big{]}\circ(f_{1},p_{2},\ldots,p_{n}).$

By Lemma 3.2, we have
$(f_{1},p_{2},\ldots,p_{n})\in\operatorname{Aut}^{G}({\bf X})$ with its
inverse $(\widetilde{f}_{1},p_{2},\ldots,p_{n})$. Since,

$(p_{1},f_{2},\ldots,f_{n})\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n})\\\
=\Big{[}\widetilde{f}_{1},f_{2}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n}),\ldots,f_{n}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n})\Big{]}$

in $\operatorname{Aut}^{G}({\bf X})$, using $G$-reducibility we can say that

$f_{j}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n})\circ\iota_{j}\in\operatorname{Aut}^{G}(X_{j}),~{}~{}~{}\forall~{}~{}~{}2\leq
j\leq n.$

Therefore from Theorem 3.4 we have,

$\Big{[}p_{1},f_{2}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n}),\ldots,f_{n}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n})\Big{]}\in\operatorname{Aut}^{G}({\bf
X}).$

Consequently,

$(f_{1},\ldots,f_{n})=\Big{[}p_{1},f_{2}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n}),\ldots,f_{n}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n})\Big{]}\circ(f_{1},p_{2},\ldots,p_{n})$

in $\operatorname{Aut}^{G}({\bf X})$.

∎

######  3.6 Proposition.

Assume that all $G$-self homotopy equivalences of ${\bf X}$ are $G$-reducible.
Let $W=X_{1}\times\cdots\times X_{k}$ and $Z=X_{k+1}\times\cdots\times X_{n}.$
If $f=(f_{W},f_{Z})\in\operatorname{Aut}^{G}({\bf X})$ then
$f_{WW}\in\operatorname{Aut}^{G}(W)$, $f_{ZZ}\in\operatorname{Aut}^{G}(Z)$. In
particular the product spaces $W,Z$ are also $G$-reducible.

###### Proof.

Let $f\in\operatorname{Aut}^{G}({\bf X})$. Therefore
$f=(f_{W},f_{Z})\in\operatorname{Aut}^{G}(W\times Z).$ From Proposition
2.3(a), it is sufficient to show that $f_{ZZ}\in\operatorname{Aut}^{G}(Z)$.

Since $f\in\operatorname{Aut}^{G}({\bf X})$, Theorem 3.4 gives
$(p_{W},f_{Z})\in\operatorname{Aut}^{G}(W\times Z).$ Consider the fiber
sequence,

|  |  |
---|---|---|---
$\textstyle{Z^{H}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{H}_{ZZ}}$$\scriptstyle{(\iota_{Z})^{H}}$$\textstyle{(W\times
Z)^{H}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(p_{W},f_{Z})^{H}}$$\scriptstyle{(p_{W})^{H}}$$\textstyle{W^{H}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Id_{W}}$$\textstyle{Z^{H}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(\iota_{Z})^{H}}$$\textstyle{(W\times
Z)^{H}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(p_{W})^{H}}$$\textstyle{W^{H}}$

Using Five-Lemma corresponding to the long exact sequence of homotopy groups
of the above fiber sequence, we can say that
$f^{H}_{ZZ}\in\operatorname{Aut}(Z^{H})$, for all $H\leq G$ i.e.
$f_{ZZ}\in\operatorname{Aut}^{G}(Z).$

Hence all $G$-self homotopy equivalences of $W\times Z$ are $G$-reducible. ∎

We now show that the group $\operatorname{Aut}^{G}({\bf X})$ is the product of
its subgroups $\operatorname{Aut}_{\prod_{\hat{i}}}^{G}({\bf X})$, where
$i=1,\ldots,n$.

######  3.7 Theorem.

Assume that all $G$-self-homotopy equivalences of ${\bf X}$ are $G$-reducible.
Then,

$\operatorname{Aut}^{G}({\bf X})=\operatorname{Aut}_{\prod_{\hat{n}}}^{G}({\bf
X})\ldots\operatorname{Aut}_{\prod_{\hat{1}}}^{G}({\bf X}).$

Moreover, if each $X_{i}$ are connected then
$\operatorname{Aut}_{\prod_{\hat{i}}}^{G}({\bf X})$ fits into a split exact
sequence

$0\to[\Pi_{\hat{i}},\operatorname{aut}_{1}(X_{i})]_{G}\to\operatorname{Aut}_{\prod_{\hat{i}}}^{G}({\bf
X})\xrightarrow[]{\phi_{i}}\operatorname{Aut}^{G}(X_{i})\to 0,$

defined by

$\phi_{i}\big{(}p_{1},\ldots,f_{i},\ldots,p_{n}\big{)}=f_{i}\circ\iota_{i},~{}~{}~{}\forall
i=1,\ldots,n$

###### Proof.

We first claim that for any $2\leq k\leq n-1$

$\operatorname{Aut}_{X_{1}\times\ldots\times X_{k-1}}^{G}({\bf
X})=\operatorname{Aut}_{X_{1}\times\ldots\times X_{k}}^{G}({\bf
X})\cdot\operatorname{Aut}_{\prod_{\hat{k}}}^{G}({\bf X}).$

Since $\operatorname{Aut}_{X_{1}\times\ldots\times X_{k}}^{G}({\bf
X})\cap\operatorname{Aut}_{\prod_{\hat{k}}}^{G}({\bf
X})=(p_{1},\ldots,p_{n})$, therefore if the factorization exists, it must be
unique. Now,

$\displaystyle(p_{1},\ldots,p_{k-1},f_{k},\ldots,f_{n})=$
$\displaystyle\Big{[}p_{1},\ldots,p_{k},f_{k+1}\circ(p_{1},\ldots,p_{k-1},\widetilde{f}_{k},p_{k+1},\ldots,p_{n}),\ldots,f_{n}\circ(p_{1},\ldots,p_{k-1},\widetilde{f}_{k},p_{k+1},\ldots,p_{n})\Big{]}\circ$
$\displaystyle(p_{1},\ldots,p_{k-1},f_{k},p_{k+1},\ldots,p_{n}).$

Second factor
$(p_{1},\ldots,p_{k-1},f_{k},p_{k+1},\ldots,p_{n})\in\operatorname{Aut}_{\prod_{\hat{k}}}^{G}({\bf
X})$ by Lemma 3.2. Again,

$\displaystyle(p_{1},\ldots,p_{k},f_{k+1},\ldots,f_{n})\circ(p_{1},\ldots,p_{k-1},\widetilde{f}_{k},p_{k+1},\ldots,p_{n})=$
$\displaystyle\Big{[}p_{1},\ldots,\widetilde{f}_{k},f_{k+1}\circ(p_{1},\ldots,p_{k-1},\widetilde{f}_{k},p_{k+1},\ldots,p_{n}),\ldots,f_{n}\circ(p_{1},\ldots,p_{k-1},\widetilde{f}_{k},p_{k+1},\ldots,p_{n})\Big{]}$

in $\operatorname{Aut}^{G}({\bf X}).$ Using $G$-reducibility we have,

$f_{j}\circ(p_{1},\ldots,p_{k-1},\widetilde{f}_{k},p_{k+1},\ldots,p_{n})\circ\iota_{j}\in\operatorname{Aut}^{G}(X_{j}),~{}~{}~{}\forall~{}~{}k+1\leq
j\leq n.$

Therefore the first factor,

$\Big{[}p_{1},\ldots,p_{k},f_{k+1}\circ(p_{1},\ldots,p_{k-1},\widetilde{f}_{k},p_{k+1},\ldots,p_{n}),\ldots,f_{n}\circ(p_{1},\ldots,p_{k-1},\widetilde{f}_{k},p_{k+1},\ldots,p_{n})\Big{]}$

in $\operatorname{Aut}_{X_{1}\times\ldots\times X_{k}}^{G}({\bf X})$, from
Theorem 3.4. Hence we have proved our claim.

Similarly we can prove that,

$\operatorname{Aut}^{G}({\bf X})=\operatorname{Aut}_{X_{1}}^{G}({\bf
X})\cdot\operatorname{Aut}_{\prod_{\hat{1}}}^{G}({\bf X}),$

by using the factorization

$(f_{1},\ldots,f_{n})=\Big{[}p_{1},f_{2}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n}),\ldots,f_{n}\circ(\widetilde{f}_{1},p_{2},\ldots,p_{n})\Big{]}\circ(f_{1},p_{2},\ldots,p_{n}).$

Hence,

$\displaystyle\operatorname{Aut}^{G}({\bf
X})=\operatorname{Aut}_{X_{1}}^{G}({\bf
X})\cdot\operatorname{Aut}_{\prod_{\hat{1}}}^{G}({\bf X})$
$\displaystyle=\operatorname{Aut}_{X_{1}\times X_{2}}({\bf
X})\cdot\operatorname{Aut}_{\prod_{\hat{2}}}^{G}({\bf
X})\cdot\operatorname{Aut}_{\prod_{\hat{1}}}^{G}({\bf X})$
$\displaystyle=\ldots$
$\displaystyle=\operatorname{Aut}_{\prod_{\hat{n}}}^{G}({\bf
X})\ldots\operatorname{Aut}_{\prod_{\hat{1}}}^{G}({\bf X}).$

For the second part of the theorem, we consider the group homomorphism

$\phi_{i}\colon\operatorname{Aut}_{\prod_{\hat{i}}}^{G}({\bf
X})\to\operatorname{Aut}^{G}(X_{i}),~{}~{}\big{(}p_{1},\ldots,f_{i},\ldots,p_{n}\big{)}\mapsto
f_{i}\circ\iota_{i}.$

It has a splitting map

$\psi_{i}\colon\operatorname{Aut}^{G}(X_{i})\to\operatorname{Aut}_{\prod_{\hat{i}}}^{G}({\bf
X})~{}~{}g\mapsto\big{(}p_{1},\ldots,g\circ p_{i},\ldots,p_{n}).$

From Lemma 3.2, we can say that the map $\psi_{i}$ is well-defined. Therefore
we get a split short exact sequence

$0\to\operatorname{Ker}(\phi_{i})\to\operatorname{Aut}_{\prod_{\hat{i}}}^{G}({\bf
X})\xrightarrow[]{\phi_{i}}\operatorname{Aut}^{G}(X_{i})\to 0.$

Since each $X_{i}$ are connected, therefore $\prod_{\hat{i}}$ is connected and
similar to 2.4 (d) we can say that $\operatorname{Ker}(\phi)$ is identified
with $\big{[}\Pi_{\hat{i}},\operatorname{aut}_{1}(X_{i})\big{]}_{G}$. ∎

##  4\. LU-factorization

In the previous section we have described a factorization of
$\operatorname{Aut}^{G}(X_{1}\times\ldots\times X_{n})$ which is symmetric.
However, the number of factors in the decomposition increases with the number
of factors in the topological product, which is inconvenient. In this section
we give a different factorisation with always only two factors, called $LU$
factorisation. It is closely related $LU$ decomposition of matrices. We begin
by introducing some notations. Define maps $u_{k}:{\bf X}\to{\bf X}$ and
$l_{k}:{\bf X}\to{\bf X}$ on the product space ${\bf
X}=X_{1}\times\ldots\times X_{n}$:

$l_{k}(x_{1},\ldots,x_{n}):=(x_{1},\ldots,x_{k},*,\ldots,*),~{}~{}u_{k}(x_{1},\ldots,x_{n}):=(*,\ldots.*,x_{k},\ldots,x_{n}).$

Note that

$l_{j}\circ l_{k}=l_{k}\circ l_{j}=l_{\min\\{j,k\\}},~{}~{}u_{j}\circ
u_{k}=u_{k}\circ u_{j}=u_{\max\\{j,k\\}}.$

Let us define two subsets $L^{G}(X_{1},\ldots,X_{n})$ and
$U^{G}(X_{1},\ldots,X_{n})$ by

$L^{G}(X_{1},\ldots,X_{n}):=\\{f\in\operatorname{Aut}^{G}({\bf
X}):f_{k}=f_{k}\circ l_{k},~{}~{}1\leq k\leq n\\},$

$U^{G}(X_{1},\ldots,X_{n}):=\\{f\in\operatorname{Aut}^{G}({\bf
X}):f_{k}=f_{k}\circ u_{k},~{}~{}f_{kk}=\operatorname{Id}_{X_{k}},~{}~{}1\leq
k\leq n\\}.$

The induced endomorphism on homotopy
$f_{\\#i}^{H}\colon\pi_{i}(X_{1}^{H})\times\ldots\times\pi_{i}(X_{n}^{H})\to\pi_{i}(X_{1}^{H})\times\ldots\times\pi_{i}(X_{n}^{H})$
can be represented as matrix form for all $H\leq G$,

$M_{i}(f^{H}):=\begin{bmatrix}(f^{H}_{11})_{\\#i}&(f^{H}_{12})_{\\#i}&\ldots(f^{H}_{1n})_{\\#i}\\\
(f^{H}_{21})_{\\#i}&(f^{H}_{22})_{\\#i}&\ldots(f^{H}_{2n})_{\\#i}\\\ .\\\ .\\\
.\\\
(f^{H}_{n1})_{\\#i}&(f^{H}_{n2})_{\\#i}&\ldots(f^{H}_{nn})_{\\#i}\end{bmatrix}.$

If $f\in L^{G}(X_{1},\ldots,X_{n})$, then

$f_{kj}=f_{k}\circ\iota_{j}=f_{k}\circ
l_{k}\circ\iota_{j}=0~{}~{}~{}~{}~{}\forall~{}~{}k<j~{}~{}~{}(\because
l_{k}\circ\iota_{j}=0,\forall k<j).$

Therefore the matrix can be written as:

$M_{i}(f^{H})=\begin{bmatrix}(f^{H}_{11})_{\\#i}&0&0&\ldots 0\\\
(f^{H}_{21})_{\\#i}&(f^{H}_{22})_{\\#i}&0&\ldots 0\\\ .\\\ .\\\ .\\\
(f^{H}_{n1})_{\\#i}&(f^{H}_{n2})_{\\#i}&(f^{H}_{n3})_{\\#i}&\ldots(f^{H}_{nn})_{\\#i}\end{bmatrix}.$

Hence the induced map of each element in $L^{G}(X_{1},\ldots,X_{n})$ gives the
lower triangular matrices on homotopy groups of every fixed point spaces.
Similarly, for $g\in U^{G}(X_{1},\ldots,X_{n})$ we have,

$g_{kj}=g_{k}\circ\iota_{j}=g_{k}\circ
u_{k}\circ\iota_{j}=0~{}~{}~{}~{}~{}~{}\forall~{}~{}k>j,~{}~{}~{}~{}~{}(\because
u_{k}\circ\iota_{j}=0,\forall~{}~{}k>j).$

Therefore the matrix can be written as:

$M_{i}(g^{H}):=\begin{bmatrix}1&(g^{H}_{12})_{\\#i}&(g^{H}_{13})_{\\#i}&\ldots(g^{H}_{1n})_{\\#i}\\\
0&1&(g^{H}_{23})_{\\#i}&\ldots(g^{H}_{2n})_{\\#i}\\\ .\\\ .\\\ .\\\
0&0&0&\ldots 1\end{bmatrix}.$

Hence induced map of each element in $U^{G}(X_{1},\ldots,X_{n})$ gives upper
triangular matrices with identity automorphisms on the diagonal entries on
homotopy groups of every fixed point spaces.

As in the non-equivariant case, we have the following chain of inclusions:

$L^{G}(X_{1},\ldots,X_{n})\subseteq L^{G}(X_{1}\times
X_{2},X_{3},\ldots,X_{n})\subseteq\ldots\subseteq L^{G}({\bf
X})=\operatorname{Aut}^{G}({\bf X}),$

$\operatorname{Aut}^{G}({\bf X})\supseteq U^{G}(X_{1},\ldots,X_{n})\supseteq
U^{G}(X_{1}\times X_{2},X_{3},\ldots,X_{n})\supseteq\ldots\supseteq U^{G}({\bf
X})=\\{1\\}.$

Now we show that these two subsets are actually the subgroups of
$\operatorname{Aut}^{G}({\bf X})$.

######  4.1 Proposition.

Assume that all $G$-self-homotopy equivalences of ${\bf X}$ are $G$-reducible.
Then $L^{G}(X_{1},\ldots,X_{n})$ and $U^{G}(X_{1},\ldots,X_{n})$ are subgroups
of $\operatorname{Aut}^{G}({\bf X})$.

###### Proof.

(a) We first consider the case of $L^{G}(X_{1},\ldots,X_{n})$. Let $f,g\in
L^{G}(X_{1},\ldots,X_{n})$. Then $p_{j}\circ f=f_{j}=f_{j}\circ
l_{j};~{}p_{j}\circ g=g_{j}=g_{j}\circ l_{j}$.

For any $j\leq k$, we know that $l_{j}\circ l_{k}=l_{k}\circ l_{j}=l_{j}$.
Now,

$\displaystyle l_{k}\circ g\circ l_{k}$
$\displaystyle=l_{k}\circ(g_{1},\ldots,g_{k},\ldots,g_{n})\circ
l_{k}=l_{k}\circ(g_{1}\circ l_{k},\ldots,g_{k}\circ l_{k},\ldots,g_{n}\circ
l_{k})$ $\displaystyle=(g_{1}\circ l_{k},\ldots,g_{k}\circ
l_{k},*,\ldots,*)=(g_{1}\circ l_{1}\circ l_{k},\ldots,g_{k}\circ l_{k}\circ
l_{k},*,\ldots,*)$ $\displaystyle=(g_{1}\circ l_{1},\ldots,g_{k}\circ
l_{k},*,\ldots,*)=(g_{1},\ldots,g_{k},*,\ldots,*)=l_{k}\circ g.$
$\displaystyle(\because l_{j}\circ l_{k}=l_{\min\\{j,k\\}})$

So,

$p_{k}\circ(f\circ g)=p_{k}\circ f\circ l_{k}\circ g=p_{k}\circ f\circ
l_{k}\circ g\circ l_{k}=p_{k}\circ f\circ g\circ l_{k}=p_{k}\circ(f\circ
g)\circ l_{k}.$

Hence $f,g\in L^{G}(X_{1},\ldots,X_{n})$ implies $f\circ g\in
L^{G}(X_{1},\ldots,X_{n})$. To prove $L^{G}(X_{1},\ldots,X_{n})$ is closed
under formation of inverses, we proceed by induction. Let us first consider
$L^{G}(X_{1},X_{2})$. We can decompose an element $f\in L^{G}(X_{1},X_{2})$ as

$f=(f_{11}\circ p_{1},p_{2})\circ[(f_{11}\circ p_{1},p_{2})^{-1}\circ f].$

We split the proof into several claims.

Claim I: $(f_{11}\circ p_{1},p_{2})^{-1}=(f^{-1}_{11}\circ p_{1},p_{2})$.

Observe that,

$\displaystyle(f_{11}\circ p_{1},p_{2})\circ(f^{-1}_{11}\circ p_{1},p_{2})$
$\displaystyle=[f_{11}\circ p_{1}\circ(f^{-1}_{11}\circ
p_{1},p_{2}),p_{2}]=[f_{11}\circ p_{1}(f^{-1}_{11}\circ p_{1},p_{2}),p_{2}]$
$\displaystyle=(f_{11}\circ f^{-1}_{11}\circ
p_{1},p_{2})=(p_{1},p_{2})=\operatorname{Id}_{X_{1}\times X_{2}}.$

Similarly, $(f^{-1}_{11}\circ p_{1},p_{2})\circ(f_{11}\circ
p_{1},p_{2})=\operatorname{Id}_{X_{1}\times X_{2}}$. Hence the claim is true.

Since $f\in L^{G}(X_{1},X_{2})$ we have $(f_{11}\circ
p_{1},p_{2})\in\operatorname{Aut}^{G}(X_{1}\times X_{2})$, by using
$G$-reducibility and Proposition 2.4. Therefore

$(f^{-1}_{11}\circ p_{1},p_{2})=(f_{11}\circ
p_{1},p_{2})^{-1}\in\operatorname{Aut}^{G}(X_{1}\times X_{2}).$

Again, $f^{-1}_{11}\circ p_{1}=f^{-1}_{11}\circ p_{1}\circ
l_{1},~{}~{}~{}~{}p_{2}=p_{2}\circ l_{2}$. Therefore $(f^{-1}_{11}\circ
p_{1},p_{2})\in L^{G}(X_{1},X_{2}),$ which imply again $(f^{-1}_{11}\circ
p_{1},p_{2})\circ f\in L^{G}(X_{1},X_{2})$. Since,

$\displaystyle p_{1}\circ(f^{-1}_{11}\circ p_{1},p_{2})\circ f$
$\displaystyle=p_{1}\circ(f^{-1}_{11}\circ
p_{1},p_{2})\circ(f_{1},f_{2})=p_{1}\circ[f^{-1}_{11}\circ
p_{1}(f_{1},f_{2}),f_{2}]$ $\displaystyle=p_{1}\circ(f^{-1}_{11}\circ
f_{1},f_{2})=f^{-1}_{11}\circ f_{1}=f^{-1}_{11}\circ f_{11}=p_{1}.$
$\displaystyle~{}~{}~{}~{}~{}~{}(\because f_{1}=f_{1}\circ
l_{1}=f_{1}\circ\iota_{1}).$

Therefore $(f^{-1}_{11}\circ p_{1},p_{2})\circ
f\in\operatorname{Aut}^{G}_{X_{1}}(X_{1}\times X_{2})\subseteq
L^{G}(X_{1}\times X_{2})$. Since $\operatorname{Aut}^{G}_{X_{1}}(X_{1}\times
X_{2})$ is a group and $(f^{-1}_{11}\circ p_{1},p_{2})\circ
f\in\operatorname{Aut}^{G}_{X_{1}}(X_{1}\times X_{2}),$ we have,

$\displaystyle\Rightarrow f^{-1}\circ(f^{-1}_{11}\circ
p_{1},p_{2})^{-1}\in\operatorname{Aut}^{G}_{X_{1}}(X_{1}\times X_{2})\subseteq
L^{G}(X_{1},X_{2})$ $\displaystyle\Rightarrow f^{-1}\circ(f^{-1}_{11}\circ
p_{1},p_{2})^{-1}\circ(f^{-1}_{11}\circ p_{1},p_{2})\in L^{G}(X_{1},X_{2})$
$\displaystyle\Rightarrow f^{-1}\in L^{G}(X_{1},X_{2}).$

Hence $L^{G}(X_{1},X_{2})$ becomes a group.

Now let $f\in L^{G}(X_{1},\ldots,X_{n})\subseteq L^{G}(X_{1}\times
X_{2},\ldots,X_{n})$, we can assume inductively that $f^{-1}\in
L^{G}(X_{1}\times X_{2},\ldots,X_{n})$. Then we get the conditions

$p_{j}\circ f^{-1}=p_{j}\circ f^{-1}\circ l_{j},~{}~{}~{}2\leq j\leq n.$

So it remains to prove that $p_{1}\circ f^{-1}=p_{1}\circ f^{-1}\circ l_{1}$,
which is equivalent to show that $p_{12}\circ f^{-1}\circ\iota_{12}\in
L^{G}(X_{1},X_{2})$. This is because,

$\displaystyle p^{\prime}_{1}\circ(p_{12}\circ
f^{-1}\circ\iota_{12})=p^{\prime}_{1}\circ(p_{12}\circ
f^{-1}\circ\iota_{12})\circ l^{\prime}_{1}$ $\displaystyle\Leftrightarrow$
$\displaystyle p^{\prime}_{1}\circ p_{12}\circ
f^{-1}\circ\iota_{12}=p_{1}\circ f^{-1}\circ l_{1}\circ\iota_{12}$
$\displaystyle\Leftrightarrow$ $\displaystyle p^{\prime}_{1}\circ p_{12}\circ
f^{-1}\circ\iota_{12}\circ p_{12}=p_{1}\circ f^{-1}\circ
l_{1}\circ\iota_{12}\circ p_{12}$ $\displaystyle\Leftrightarrow$
$\displaystyle p^{\prime}_{1}\circ p_{12}\circ f^{-1}\circ l_{12}=p_{1}\circ
f^{-1}\circ l_{1}\circ l_{12}$ $\displaystyle\Leftrightarrow$ $\displaystyle
p^{\prime}_{1}\circ p_{12}\circ f^{-1}=p_{1}\circ f^{-1}\circ
l_{1},~{}~{}~{}~{}~{}(\because f^{-1}\in L^{G}(X_{1}\times
X_{2},\ldots,X_{n}))$ $\displaystyle\Leftrightarrow$ $\displaystyle p_{1}\circ
f^{-1}=p_{1}\circ f^{-1}\circ l_{1}.$

See the commutative diagrams:

|
---|---
$\textstyle{X_{1}\times\ldots\times
X_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{12}}$$\textstyle{X_{1}\times
X_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p^{\prime}_{1}}$$\textstyle{X_{1}}$

|
---|---
$\textstyle{X_{1}\times
X_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{12}}$$\scriptstyle{l^{\prime}_{1}}$$\textstyle{X_{1}\times
X_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{12}}$$\textstyle{X_{1}\times\ldots\times
X_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{l_{1}}$$\textstyle{X_{1}\times\ldots\times
X_{n}}$

and,

|
---|---
$\textstyle{X_{1}\times\ldots\times
X_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{l_{12}}$$\scriptstyle{p_{12}}$$\textstyle{X_{1}\times
X_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{12}}$$\textstyle{X_{1}\times\ldots\times
X_{n}}$

From Proposition 3.6 we have $p_{12}\circ
f\circ\iota_{12}\in\operatorname{Aut}^{G}(X_{1}\times X_{2})$. Now,

$\displaystyle p^{\prime}_{1}\circ(p_{12}\circ f\circ\iota_{12})=p_{1}\circ
f\circ\iota_{12}=p_{1}\circ f\circ l_{1}\circ\iota_{12}=p_{1}\circ
f\circ\iota_{12}\circ l^{\prime}_{1}$
$\displaystyle=p^{\prime}_{1}\circ(p_{12}\circ f\circ\iota_{12})\circ
l^{\prime}_{1},$

and,

$p^{\prime}_{2}\circ(p_{12}\circ
f\circ\iota_{12})=p^{\prime}_{2}\circ(p_{12}\circ f\circ\iota_{12})\circ
l^{\prime}_{2},~{}~{}~{}(\because
l^{\prime}_{2}=\operatorname{Id}_{X_{1}\times X_{2}}).$

Therefore $p_{12}\circ f\circ\iota_{12}\in L^{G}(X_{1},X_{2})$.

Claim II: $(p_{12}\circ f\circ\iota_{12})^{-1}=p_{12}\circ
f^{-1}\circ\iota_{12}$.

Note that,

$\displaystyle(p_{12}\circ f^{-1}\circ\iota_{12})\circ(p_{12}\circ
f\circ\iota_{12})$ $\displaystyle=p_{12}\circ f^{-1}\circ\iota_{12}\circ
p_{12}\circ f\circ\iota_{12}=p_{12}\circ f^{-1}\circ l_{12}\circ
f\circ\iota_{12}$ $\displaystyle=p_{12}\circ f^{-1}\circ
f\circ\iota_{12}=p_{12}\circ\iota_{12}=\operatorname{Id}_{X_{1}\times X_{2}}.$
$\displaystyle~{}~{}~{}~{}~{}(\because f^{-1}\in L^{G}(X_{1}\times
X_{2},\ldots,X_{n}))$

Similarly, $(p_{12}\circ f\circ\iota_{12})\circ(p_{12}\circ
f^{-1}\circ\iota_{12})=\operatorname{Id}_{X_{1}\times X_{2}}$. Therefore
$(p_{12}\circ f^{-1}\circ\iota_{12})=(p_{12}\circ f\circ\iota_{12})^{-1}\in
L^{G}(X_{1}\times X_{2})$. So that $f^{-1}\in L^{G}(X_{1},\ldots,X_{n})$.
Consequently, $L^{G}(X_{1},\ldots,X_{n})$ is closed under taking inverses.
Hence $L^{G}(X_{1},\ldots,X_{n})$ becomes a group.

(b) Now we want to prove that $U^{G}(X_{1},\ldots,X_{n})$ is a group. Let us
first prove that $U^{G}(X_{1},X_{2})$ is a group and then we proceed by
induction. If $f\in U^{G}(X_{1},X_{2})$ then we have $f_{2}=f_{2}\circ u_{2}$
and
$f_{11}=\operatorname{Id}_{X_{1}},~{}~{}~{}f_{22}=\operatorname{Id}_{X_{2}}$.
Now,

$\displaystyle f_{2}=f_{2}\circ u_{2}=f_{2}\circ\iota_{2}\circ
p_{2}=f_{22}\circ p_{2}=p_{2}~{}~{}~{}~{}~{}~{}(\because u_{2}=\iota_{2}\circ
p_{2}).$

This implies that $f\in\operatorname{Aut}^{G}_{X_{2}}(X_{1}\times X_{2})$ (cf.
Proposition 2.4). Thus we have shown that

$U^{G}(X_{1},X_{2})\subseteq\operatorname{Aut}^{G}_{X_{2}}(X_{1},X_{2}).$

From Proposition 2.4, we have a group homomorphsim:

$\phi:\operatorname{Aut}^{G}_{X_{2}}(X_{1}\times
X_{2})\to\operatorname{Aut}^{G}(X_{1}),~{}~{}(f_{1},p_{2})\mapsto
f_{1}\circ\iota_{1}.$

Observe that, $U^{G}(X_{1},X_{2})=\operatorname{Ker}(\phi)$. Hence
$U^{G}(X_{1},X_{2})$ is a group. We now proceed by induction on $n$. Assume
that $U^{G}(X_{1},\ldots,X_{n-1})$ is a group.

Claim III: $f\in U^{G}(X_{1},\ldots,X_{n})\Rightarrow p_{1\ldots(n-1)}\circ
f\circ\iota_{1\ldots(n-1)}\in U^{G}(X_{1},\ldots,X_{n-1})$.

From Propostion 3.6 we have $p_{1\ldots(n-1)}\circ
f\circ\iota_{1\ldots(n-1)}\in\operatorname{Aut}^{G}(X_{1}\times\ldots\times
X_{n-1})$. Now,

$\displaystyle p^{\prime}_{j}\circ p_{1\ldots(n-1)}\circ
f\circ\iota_{1\ldots(n-1)}\circ\iota^{\prime}_{j}$ $\displaystyle=p_{j}\circ
f\circ\iota_{j}=f_{jj}=\operatorname{Id}_{X_{j}},~{}~{}~{}~{}~{}~{}~{}\forall
j=1,\ldots,n-1.$

And,

$\displaystyle p^{\prime}_{j}\circ p_{1\ldots(n-1)}\circ
f\circ\iota_{1\ldots(n-1)}\circ u^{\prime}_{j}=p_{j}\circ f\circ
u_{j}\circ\iota_{1\ldots(n-1)}=p_{j}\circ f\circ\iota_{1\ldots(n-1)}$
$\displaystyle=p^{\prime}_{j}\circ p_{1\ldots(n-1)}\circ
f\circ\iota_{1\ldots(n-1)}~{}~{}~{}~{}\forall j=1,\ldots,n-1.$

Here the maps $p_{j}^{\prime},\iota_{j}^{\prime},u_{j}^{\prime}$ are defined
by the following commutative diagrams:

|
---|---
$\textstyle{X_{1}\times\ldots\times
X_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{j}}$$\scriptstyle{p_{1\ldots(n-1)}}$$\textstyle{X_{1}\times\ldots\times
X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p^{\prime}_{j}}$$\textstyle{X_{j}}$

|
---|---
$\textstyle{X_{1}\times\ldots\times
X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{1\ldots(n-1)}}$$\textstyle{\bf
X}$$\textstyle{X_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{j}}$$\scriptstyle{\iota^{\prime}_{j}}$

|
---|---
$\textstyle{X_{1}\times\ldots\times
X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{1\ldots(n-1)}}$$\scriptstyle{u^{\prime}_{j}}$$\textstyle{X_{1}\times\ldots\times
X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{1\ldots(n-1)}}$$\textstyle{{\bf
X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u_{j}}$$\textstyle{\bf
X}$

Therefore the claim is proved.

From Proposition 2.4, we have the split short exact sequence,

$0\to\operatorname{Ker}(\phi)\xrightarrow[]{\iota}\operatorname{Aut}^{G}_{X_{n}}({\bf
X})\xrightarrow[]{\phi}\operatorname{Aut}^{G}(X_{1}\times\ldots\times
X_{n-1})\to 0,$

where

$\phi(f):=p_{1\ldots(n-1)}\circ
f\circ\iota_{1\ldots(n-1)}=f_{1\ldots(n-1)}\circ\iota_{1\ldots(n-1)},$

and

$\operatorname{Ker}(\phi)=\big{\\{}f\in\operatorname{Aut}^{G}_{X_{n}}({\bf
X}):p_{1\ldots(n-1)}\circ
f\circ\iota_{1\ldots(n-1)}=\operatorname{Id}_{X_{1}\times\ldots\times
X_{n-1}}\big{\\}}.$

Now,

$\psi:\operatorname{Aut}^{G}(X_{1}\times\ldots\times
X_{n-1})\to\operatorname{Aut}^{G}_{X_{n}}({\bf X})$

is the splitting map of $\phi$ defined as $\psi(h):=(h\circ
p_{1\ldots(n-1)},p_{n})$. Observe that,

$f\in U^{G}(X_{1},\ldots,X_{n})\Rightarrow f_{n}=f_{n}\circ u_{n}\text{ and
}~{}~{}f_{nn}=\operatorname{Id}_{X_{n}}.$

Therefore, $f_{n}=f_{n}\circ u_{n}=f_{n}\circ\iota_{n}\circ p_{n}=f_{nn}\circ
p_{n}=p_{n}$. Hence $f\in\operatorname{Aut}^{G}_{X_{n}}({\bf X})$, i.e.
$U^{G}(X_{1},\ldots,X_{n})\subseteq\operatorname{Aut}^{G}_{X_{n}}({\bf X})$.

Claim IV: $\operatorname{Ker}(\phi)=U^{G}(X_{1}\times\ldots\times
X_{n-1},X_{n}).$

Note that,

$\displaystyle f\in U^{G}(X_{1}\times\ldots\times X_{n-1},X_{n})$
$\displaystyle\Rightarrow p_{1\ldots(n-1)}\circ
f\circ\iota_{1\ldots(n-1)}=\operatorname{Id}_{X_{1}\times\ldots\times
X_{n-1}};~{}f_{n}=f_{n}\circ u_{n},~{}f_{nn}=\operatorname{Id}_{X_{n}}$
$\displaystyle\Rightarrow p_{1\ldots(n-1)}\circ
f\circ\iota_{1\ldots(n-1)}=\operatorname{Id}_{X_{1}\times\ldots\times
X_{n-1}};f_{n}=p_{n}$ $\displaystyle\Rightarrow f\in\operatorname{Ker}(\phi).$

Conversely,

$\displaystyle g\in\operatorname{Ker}(\phi)$ $\displaystyle\Rightarrow
p_{1\ldots(n-1)}\circ
g\circ\iota_{1\ldots(n-1)}=\operatorname{Id}_{X_{1}\times...\times
X_{n-1}};~{}~{}g_{n}=p_{n}$ $\displaystyle\Rightarrow p_{1\ldots(n-1)}\circ
g\circ\iota_{1\ldots(n-1)}=\operatorname{Id}_{X_{1}\times...\times
X_{n-1}};~{}~{}g_{n}=g_{n}\circ u_{n},~{}g_{nn}=\operatorname{Id}_{X_{n}}$
$\displaystyle\Rightarrow g\in U^{G}(X_{1}\times\ldots\times X_{n-1},X_{n}).$

So $\operatorname{Ker}(\phi)=U^{G}(X_{1}\times\ldots\times
X_{n-1},X_{n})\subseteq U^{G}(X_{1},\ldots,X_{n})$. Therefore we get the split
short exact sequence,

$0\to U^{G}(X_{1}\times\ldots\times
X_{n-1},X_{n})\xrightarrow[]{\iota}U^{G}(X_{1},\ldots,X_{n})\xrightarrow[]{\phi}U^{G}(X_{1},\ldots,X_{n-1})\to
0.$

Hence $U^{G}(X_{1},\ldots,X_{n})$ is the semi-direct product of
$U^{G}(X_{1}\times\ldots\times X_{n-1},X_{n})$ with
$U^{G}(X_{1},\ldots,X_{n-1}).$ Hence it is a group. ∎

Finally we have the symmetric decomposition of $\operatorname{Aut}^{G}({\bf
X})$ in the following which is an equivariant version of [10, Theorem 5.4]

######  4.2 Theorem.

Assume that all the $G$-self-homotopy equivalences of ${\bf X}$ are
$G$-reducible. Then

$\operatorname{Aut}^{G}({\bf X})=L^{G}(X_{1},\ldots,X_{n})\cdot
U^{G}(X_{1},\ldots,X_{n}).$

###### Proof.

Let $f\in L^{G}(X_{1},\ldots,X_{n})\cap U^{G}(X_{1},\ldots,X_{n}).$ Then

$f_{j}=f_{j}\circ l_{j};~{}~{}f_{j}=f_{j}\circ
u_{j},~{}~{}~{}f_{jj}=\operatorname{Id}_{X_{j}},~{}~{}\forall~{}~{}1\leq j\leq
n.$

Therefore,

$\displaystyle~{}f_{j}=(f_{j}\circ l_{j})\circ u_{j}~{}~{}~{}~{}(\because
f_{j}=f_{j}\circ l_{j})$ $\displaystyle\Rightarrow$
$\displaystyle~{}f_{j}=f_{j}\circ\iota_{j}\circ p_{j}~{}~{}~{}~{}~{}(\because
l_{j}\circ u_{j}=\iota_{j}\circ p_{j})$ $\displaystyle\Rightarrow$
$\displaystyle~{}f_{j}=f_{jj}\circ p_{j}$ $\displaystyle\Rightarrow$
$\displaystyle~{}f_{j}=p_{j}$ $\displaystyle\Rightarrow$
$\displaystyle~{}f=(p_{1},\ldots,p_{n})=\operatorname{Id}_{{\bf X}}.$

Therefore $L^{G}(X_{1},\ldots,X_{n})\cap U^{G}(X_{1},\ldots,X_{n})$ is
trivial.

We begin the proof with $n=2$, then argue by induction. A map
$f\in\operatorname{Aut}^{G}(X_{1}\times X_{2})$ can be decomposed as,

$f=[f\circ(f_{1},p_{2})^{-1}\circ(f_{11}\circ p_{1},p_{2})]\circ[(f_{11}\circ
p_{1},p_{2})^{-1}\circ(f_{1},p_{2})].$

From Proposition 4.1, we have $(f_{11}\circ p_{1},p_{2})\in
L^{G}(X_{1},X_{2})$. For $f\in\operatorname{Aut}^{G}(X_{1}\times X_{2})$ we
have $(f_{1},p_{2})\in\operatorname{Aut}^{G}(X_{1}\times X_{2})$, from
Corollary 2.5. Again

$f\circ(f_{1},p_{2})^{-1}=[f_{1}\circ(f_{1},p_{2})^{-1},f_{2}\circ(f_{1},p_{2})^{-1}].$

Note that $(f_{1},p_{2})\circ(f_{1},p_{2})^{-1}=(p_{1},p_{2})\Rightarrow
f_{1}\circ(f_{1},p_{2})^{-1}=p_{1}$. Therefore

$f\circ(f_{1},p_{2})^{-1}=[p_{1},f_{2}\circ(f_{1},p_{2})^{-1}]\in
L^{G}(X_{1},X_{2}).$

This implies the first factor in the representation of $f$ is in
$L^{G}(X_{1},X_{2})$. On the other hand,

$(f_{11}\circ p_{1},p_{2})^{-1}\circ(f_{1},p_{2})\in U^{G}(X_{1},X_{2}),$

because

$\displaystyle p_{2}\circ(f_{11}\circ p_{1},p_{2})^{-1}\circ(f_{1},p_{2})\circ
u_{2}=$ $\displaystyle~{}p_{2}\circ(f^{-1}_{11}\circ
p_{1},p_{2})\circ(f_{1},p_{2})\circ u_{2}$ $\displaystyle=$
$\displaystyle~{}p_{2}\circ(f^{-1}_{11}\circ p_{1},p_{2})\circ(f_{1}\circ
u_{2},p_{2}\circ u_{2})$ $\displaystyle=$
$\displaystyle~{}p_{2}\circ(f^{-1}_{11}\circ p_{1},p_{2})\circ(f_{1}\circ
u_{2},p_{2})$ $\displaystyle=$ $\displaystyle~{}p_{2}\circ[f^{-1}_{11}\circ
p_{1}\circ(f_{1}\circ u_{2},p_{2}),p_{2}]$ $\displaystyle=$
$\displaystyle~{}p_{2}$ $\displaystyle=$
$\displaystyle~{}p_{2}\circ(f^{-1}_{11}\circ p_{1},p_{2})\circ(f_{1},p_{2}).$

Again,

$\displaystyle p_{1}\circ(f_{11}\circ
p_{1},p_{2})^{-1}\circ(f_{1},p_{2})\circ\iota_{1}=$
$\displaystyle~{}p_{1}\circ(f^{-1}_{11}\circ
p_{1},p_{2})\circ(f_{1}\circ\iota_{1},p_{2}\circ\iota_{1})$ $\displaystyle=$
$\displaystyle~{}p_{1}\circ(f^{-1}_{11}\circ
p_{1},p_{2})\circ(f_{11},p_{2}\circ\iota_{1})$ $\displaystyle=$
$\displaystyle~{}p_{1}\circ[f^{-1}_{11}\circ
p_{1}\circ(f_{11},p_{2}\circ\iota_{1}),p_{2}\circ\iota_{1}]$ $\displaystyle=$
$\displaystyle~{}p_{1}\circ(f^{-1}_{11}\circ f_{11},p_{2}\circ\iota_{1})$
$\displaystyle=$ $\displaystyle~{}f^{-1}_{11}\circ f_{11}$ $\displaystyle=$
$\displaystyle~{}\operatorname{Id}_{X_{1}}.$

And,

$\displaystyle p_{2}\circ(f_{11}\circ
p_{1},p_{2})^{-1}\circ(f_{1},p_{2})\circ\iota_{2}=$
$\displaystyle~{}p_{2}\circ(f^{-1}_{11}\circ
p_{1},p_{2})\circ(f_{1}\circ\iota_{2},p_{2}\circ\iota_{2})$ $\displaystyle=$
$\displaystyle~{}p_{2}\circ(f^{-1}_{11}\circ
p_{1},p_{2})\circ(f_{1}\circ\iota_{2},Id_{X_{2}})$ $\displaystyle=$
$\displaystyle~{}p_{2}\circ[f^{-1}_{11}\circ
p_{1}\circ(f_{1}\circ\iota_{2},Id_{X_{2}}),Id_{X_{2}}]$ $\displaystyle=$
$\displaystyle~{}\operatorname{Id}_{X_{2}}.$

Therefore $(f_{11}\circ p_{1},p_{2})^{-1}\circ(f_{1},p_{2})\in
U^{G}(X_{1},X_{2})$.

Let $f\in\operatorname{Aut}^{G}(X_{1}\times\ldots\times X_{n})$. If we take
$W=X_{1}\times X_{2}$ then $f\in\operatorname{Aut}^{G}(W\times
X_{3}\times\ldots\times X_{n}).$ By induction hypothesis, there exists
$f^{\prime}\in L^{G}(W,X_{3},\ldots,X_{n})$ and $f^{\prime\prime}\in
U^{G}(W,X_{3},\ldots,X_{n})$ such that $f=f^{\prime}\circ f^{\prime\prime}$.
Since $f^{\prime}\in
L^{G}(W,X_{3},\ldots,X_{n})\subseteq\operatorname{Aut}^{G}({\bf X})$ we have,
$p_{12}\circ f^{\prime}\circ\iota_{12}\in\operatorname{Aut}^{G}(W)$ by
Proposition 3.6. Let $\bar{f}=p_{12}\circ f^{\prime}\circ\iota_{12}$, then
$\bar{f}=l\circ\hat{u}$ for some $l\in L^{G}(X_{1},X_{2})$ and $\hat{u}\in
U^{G}(X_{1},X_{2})$. Let

$\bar{u}:=(\hat{u}\circ p_{12},p_{3},\ldots,p_{n})=(\hat{u}_{1}\circ
p_{12},\hat{u}_{2}\circ p_{12},p_{3},\ldots,p_{n}).$

Claim: $\bar{u}\in U^{G}(X_{1},\ldots,X_{n}).$

$\displaystyle\hat{u}_{1}\circ p_{12}\circ\iota_{1}$
$\displaystyle=\hat{u}_{1}\circ\iota^{\prime}_{1}=\hat{u}_{11}=\operatorname{Id}_{X_{1}}~{}~{}~{}(\because
p_{12}\circ\iota_{1}=\iota^{\prime}_{1}).$ $\displaystyle\hat{u}_{2}\circ
p_{12}\circ\iota_{2}=\hat{u}_{2}\circ\iota^{\prime}_{2}=\hat{u}_{22}=\operatorname{Id}_{X_{2}}~{}~{}~{}~{}(\because
p_{12}\circ\iota_{2}=\iota^{\prime}_{2}).$ $p_{{j}}\circ
i_{{j}}=\operatorname{Id}_{X_{j}},~{}~{}3\leq j\leq n.$

And

$\displaystyle\hat{u}_{2}\circ p_{12}\circ u_{2}$
$\displaystyle=\hat{u}_{2}\circ u^{\prime}_{2}\circ
p_{12}~{}~{}~{}~{}~{}~{}(\because p_{12}\circ u_{2}=u^{\prime}_{2}\circ
p_{12})$ $\displaystyle=\hat{u}_{2}\circ
p_{12}~{}~{}~{}(\because\hat{u}=(\hat{u}_{1},\hat{u}_{2})\in
U^{G}(X_{1},X_{2})).$ $p_{j}\circ u_{j}=p_{j},~{}~{}3\leq j\leq n.$

Therefore the claim is proved. Clearly $\bar{u}^{-1}=(u^{-1}\circ
p_{12},p_{3},\ldots,p_{n})\in L^{G}(X_{1}\times X_{2},\ldots,X_{n})$. Hence
$f^{\prime}\circ\bar{u}^{-1}\in L^{G}(X_{1}\times X_{2},\ldots,X_{n})$. Since
$f^{\prime}\in L^{G}(X_{1}\times X_{2},\ldots,X_{n})$, we have,

$p_{12}\circ f^{\prime}=p_{12}\circ f^{\prime}\circ l_{12}=p_{12}\circ
f^{\prime}\circ\iota_{12}\circ
p_{12}~{}~{}~{}~{}~{}~{}(\because\iota_{12}\circ p_{12}=l_{12}).$

To prove $f^{\prime}\circ\bar{u}^{-1}\in L^{G}(X_{1},\ldots,X_{n})$, it is
sufficient to show that
$p_{12}\circ(f^{\prime}\circ\bar{u}^{-1})\circ\iota_{12}\in L^{G}(X_{1}\times
X_{2})$. Now,

$p_{12}\circ f^{\prime}\circ\bar{u}^{-1}\circ\iota_{12}=p_{12}\circ
f^{\prime}\circ\iota_{12}\circ
p_{12}\circ\bar{u}^{-1}\circ\iota_{12}=p_{12}\circ
f^{\prime}\circ\iota_{12}\circ u^{-1}=\bar{f}\circ u^{-1}=l.$

Therefore $p_{12}\circ f^{\prime}\circ\bar{u}^{-1}\circ\iota_{12}\in
L^{G}(X_{1}\times X_{2})$. Consequently we have
$f^{\prime}\circ\bar{u}^{-1}\in L^{G}(X_{1},\ldots,X_{n})$. Clearly,
$\bar{u}\circ f^{\prime\prime}\in U^{G}(X_{1},\ldots,X_{n})$. Hence
$f=(f^{\prime}\circ\bar{u}^{-1})\circ(\bar{u}\circ f^{\prime\prime})$.

∎

Let

$\bar{L}^{G}(X_{1},\ldots,X_{n}):=\\{f\in
L^{G}(X_{1},\ldots,X_{n})|~{}f_{kk}=\operatorname{Id}_{X_{k}},~{}~{}1\leq
k\leq n\\}.$

######  4.3 Proposition.

The group $L^{G}(X_{1},\ldots,X_{n})$ is isomorphic to the semi-direct product
of
$\operatorname{Aut}^{G}(X_{1})\times\ldots\times\operatorname{Aut}^{G}(X_{n})$
with $\bar{L}^{G}(X_{1},\ldots,X_{n})$.

###### Proof.

Consider a map

$\varphi:L^{G}(X_{1},\ldots,X_{n})\to\operatorname{Aut}^{G}(X_{1})\times\ldots\times\operatorname{Aut}^{G}(X_{n}),$

defined as

$f\mapsto(f_{11},\ldots,f_{nn}).$

Now,

$\displaystyle(f\circ g)_{jj}$ $\displaystyle=p_{j}\circ(f\circ
g)\circ\iota_{j}$ $\displaystyle=p_{j}\circ
f\circ(g_{1},\ldots,g_{n})\circ\iota_{j}$ $\displaystyle=p_{j}\circ
f\circ(g_{1}\circ l_{1},\ldots,g_{n}\circ l_{n})\circ\iota_{j}$
$\displaystyle=p_{j}\circ f\circ(g_{1}\circ l_{1}\circ i_{j},\ldots,g_{n}\circ
l_{n}\circ\iota_{j})$ $\displaystyle=p_{j}\circ
f\circ(*,\ldots,*,g_{j}\circ\iota_{j},\ldots,g_{n}\circ\iota_{j})~{}~{}~{}(\because
l_{k}\circ\iota_{j}=*,\forall k<j)$
$\displaystyle=f_{j}\circ(*,\ldots,*,g_{j}\circ\iota_{j},\ldots,g_{n}\circ\iota_{j})$
$\displaystyle=f_{j}\circ
l_{j}\circ(*,\ldots,*,g_{j}\circ\iota_{j},\ldots,g_{n}\circ\iota_{j})$
$\displaystyle=f_{j}\circ(*,\ldots,*,g_{j}\circ\iota_{j},*,\ldots,*)$
$\displaystyle=f_{j}\circ\iota_{j}\circ g_{jj}=f_{jj}\circ g_{jj}.$

Therefore,

$\displaystyle\varphi(f\circ g)=((f\circ g)_{11},\ldots,(f\circ
g)_{nn})=(f_{11}\circ g_{11},\ldots,f_{nn}\circ g_{nn})$
$\displaystyle=(f_{11},\ldots,f_{nn})\circ(g_{11},\ldots,g_{nn})=\varphi(f)\circ\varphi(g).$

So,

$\displaystyle\operatorname{Ker}(\varphi)$ $\displaystyle=\big{\\{}f\in
L^{G}(X_{1},\ldots,X_{n})|~{}~{}f_{jj}=\operatorname{Id}_{X_{j}},~{}\forall~{}~{}1\leq
j\leq n\big{\\}}$ $\displaystyle=\bar{L}^{G}(X_{1},\ldots,X_{n}).$

Now,

$\Psi:\operatorname{Aut}^{G}(X_{1})\times\ldots\times\operatorname{Aut}^{G}(X_{n})\to
L^{G}(X_{1},\ldots,X_{n})$

is the splitting map of $\varphi$ defined as

$\Psi(g_{1},\ldots,g_{n}):=(g_{1}\circ p_{1},\ldots,g_{n}\circ p_{n}).$

Using the Corollary 3.5, we can say that the map $\Psi$ is well-defined.
Therefore we get a split short exact sequence,

$0\to\bar{L}^{G}(X_{1},\ldots,X_{n})\to
L^{G}(X_{1},\ldots,X_{n})\xrightarrow[]{\varphi}\operatorname{Aut}^{G}(X_{1})\times\ldots\times\operatorname{Aut}^{G}(X_{n})\to
0.$

Hence
$L^{G}(X_{1},\ldots,X_{n})\cong(\operatorname{Aut}^{G}(X_{1})\times\ldots\times\operatorname{Aut}^{G}(X_{n}))\rtimes\bar{L}^{G}(X_{1},\ldots,X_{n})$

∎

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