# Cohomology groups invariant under continuous orbit equivalence

Yongle Jiang  Institute of Mathematics, Polish Academy of Sciences, Warsaw,
00-656, Poland<EMAIL_ADDRESS>
###### Abstract.

By the work of Brodzki-Niblo-Nowak-Wright and Monod, topological amenability
of a continuous group action can be characterized using uniformly finite
homology groups or bounded cohomology groups associated to this action. We
show that (certain variations of) these groups are invariants for
topologically free actions under continuous orbit equivalence.

###### Key words and phrases:

Continuous orbit equivalence; uniformly finite homology; bounded cohomology

##  1\. Introduction

We continue our study of continuous orbit equivalence introduced by Li [li_0].
In this paper, we focus on certain (co)homology groups for a continuous group
action. Let us first review the introduction of these (co)homology groups.

Let $G$ be a countable discrete group. There are two remarkable
characterizations of amenablility of $G$. One, given by Johnson-Ringrose
[johnson], says that $G$ is amenable if and only if the first bounded
cohomology group with coefficients in $\ell^{1}_{0}(G)^{**}$ vanishes, i.e.
$H_{b}^{1}(G,\ell^{1}_{0}(G)^{**})=0$, where $\ell^{1}_{0}(G)$ denotes the
augmentation ideal, i.e. kernel of the summation map from $\ell^{1}(G)$ to
$\mathbb{R}$. By contrast, Block and Weinberger [bw] described amenability in
terms of non-vanishing of the 0-dimensional uniformly finite homology of $G$,
i.e. $H_{0}^{uf}(G,\mathbb{R})\neq 0$. For a short, unified proof of these two
results, see [bnw]. In particular, it was observed there that
$H_{*}^{uf}(G,\mathbb{R})=H_{*}(G,\ell^{\infty}(G))$ if $G$ is finitely
generated.

The notion of amenable actions of groups acting on topological spaces
generalizes the concept of amenability and appears in many areas of
mathematics. For example, a group acts amenably on a point if and only if it
is amenable, while every hyperbolic group acts amenably on its Gromov
boundary. For more on amenable actions, see [AR, gk, hr, oz].

Parallel with these characterizations, people also found two similar
characterizations for the amenablity of group actions in the topological
sense. To do this, the key step is to find appropriate coefficient modules
associated to a continuous action $G\curvearrowright X$.

More precisely, Brodzki, Niblo, Nowak and Wright considered the standard
module $W_{0}(G,X)$ and its submodule $N_{0}(G,X):=C(X,\ell^{1}_{0}(G))$. Note
that $W_{0}(G,X)^{*}$ and $N_{0}(G,X)^{**}$ reduce to $\ell^{\infty}(G)$ and
$\ell^{1}_{0}(G)^{**}$ respectively when $X$ is a point. Naturally, they
considered the bounded cohomology groups with coefficients in
$N_{0}(G,X)^{**}$ [bnnw2] and also the uniformly finite homology of an action,
$H^{uf}_{n}(G\curvearrowright X)$ as the group homology with coefficients in
$W_{0}(G,X)^{*}$, i.e. $H^{uf}_{n}(G\curvearrowright
X):=H_{n}(G,W_{0}(G,X)^{*})$ [bnnw]. They succeeded in characterizing
amenability of actions using these (co)homology groups, generalizing the above
results of Johson-Ringrose and Block-Weinberger for group case. A similar
approach was also taken by Monod in [monod].

In [bw] (see also [li]), among other results, $H_{*}^{uf}(G,\mathbb{R})$ is
shown to be an invariant for groups under coarse equivalence, i.e. quasi-
isometry if the groups are finitely generated. Hence it is natural to ask
whether the above (co)homology groups for actions are also invariants under
some “coarse equivalence” for actions.

In this paper, we show this is indeed possible if we take “coarse equivalence”
to be “continuous orbit equivalence”, and actions are assumed to be
topologically free.

Let us recall the definition of continuous orbit equivalence, for known
results on this notion and its connection to geometric group theory, see
[cohen, cj1, cj2, j, li_0, li].

Let $G\curvearrowright X$ and $H\curvearrowright Y$ be two actions by
homeomorphisms, where $G$, $H$ are countable discrete groups and $X$, $Y$ are
compact Hausdorff spaces. Following [li_0], we say the two actions are
_continuous orbit equivalent_ (abbreviated as COE) if there are homeomorphisms
$\phi:X\approx Y$, $\psi:Y\approx X$ and continuous maps $c:G\times X\to H$,
$c^{\prime}:H\times Y\to G$ such that $\phi(gx)=c(g,x)\phi(x)$ and
$\psi(hy)=c^{\prime}(h,y)\psi(y)$ hold for all $g\in G$, $h\in H$, $x\in X$
and $y\in Y$. If these two actions are _topologically free_ , i.e. points with
trivial stabilizers are dense, then both $c$ and $c^{\prime}$ are cocycles
[li_0, Lemma 2.8]. Recall that $c:G\times X\to H$ is a _cocycle_ if
$c(g_{1}g_{2},x)=c(g_{1},g_{2}x)c(g_{2},x)$ for all $g_{1},g_{2}\in G$ and all
$x\in X$.

Now, we can state our main theorems. Note that all acting groups are assumed
to be countable discrete and spaces are assumed to be compact Hausdorff.

######  Theorem 1.1.

Let $G\curvearrowright X$ and $H\curvearrowright Y$ be topologically free
actions. If these two actions are COE, then $H^{uf}_{0}(G\curvearrowright
X)\cong H^{uf}_{0}(H\curvearrowright Y)$ and $H_{n}(G,N_{0}(G,X)^{*})\cong
H_{n}(H,N_{0}(H,Y)^{*})$ for all $n\geq 0$.

######  Theorem 1.2.

Let $G\curvearrowright X$ and $H\curvearrowright Y$ be topologically free
actions. If these two actions are COE, then $H^{p}_{b}(G,N_{0}(G,X)^{**})\cong
H^{p}_{b}(H,N_{0}(H,Y)^{**})$ for all $p\geq 0$.

The paper is organized as follows.

Besides the introduction, the paper contains five other sections. In Section
2, we review the definitions of group (co)homology and certain coefficient
modules. In Section 3, we state several lemmas on basic properties of the
orbit cocycles and observe that the coefficient modules are closed under
restriction maps. These are the crucial ingredients for constructing bijective
(co)chain maps later. The next two sections will take up the majority of our
paper: Section 4 proves Theorem 1.1 and Section 5 proves Theorem 1.2. The
proof of these two theorems uses the same idea: we construct maps between
(co)chain complexes directly and then use the whole sections to verify these
maps are bijective (co)chain maps. Detailed proof for lower dimensional cases
will be included to illustrate the main ideas. We conclude with several
remarks in Section 6 to discuss corollaries of the theorems and other related
questions.

Following the convention in [bnnw2, bnnw, monod], all Banach spaces are
assumed to be real.

##  2\. Preliminaries

###  2.1. Group (co)homology

We briefly recall the definition of group homology and (bounded) group
cohomology using bar resolutions, see [li, §4.3], [brown, Chapter III, §1] and
[monod_book].

Let $G$ be a group and $V$ be a $\mathbb{Z}G$-module. Let
$(C_{*}(V),\partial_{*})$ be the chain complex
$\dots\overset{\partial_{3}}{\rightarrow}C_{2}(V)\overset{\partial_{2}}{\rightarrow}C_{1}(V)\overset{\partial_{1}}{\rightarrow}C_{0}(V)$
with $C_{0}(V)=V$, $C_{n}(V)=C_{f}(G^{n},V)$, where $C_{f}$ stands for maps
with finite support, and
$\partial_{n}=\sum_{i=0}^{n}(-1)^{i}\partial_{n}^{(i)}$, where

$\displaystyle\partial_{n}^{(0)}(f)(g_{1},\dots,g_{n-1})$
$\displaystyle=\sum_{g_{0}\in G}g_{0}^{-1}f(g_{0},g_{1},\dots,g_{n-1}),$
$\displaystyle\partial_{n}^{(i)}(f)(g_{1},\dots,g_{n-1})$
$\displaystyle=\sum_{\begin{subarray}{c}g,\bar{g}\in G\\\
g\bar{g}=g_{i}\end{subarray}}f(g_{1},\dots,g_{i-1},g,\bar{g},g_{i+1},g_{n-1})~{}\text{for}~{}1\leq
i\leq n-1,$ $\displaystyle\partial_{n}^{(n)}(f)(g_{1},\dots,g_{n-1})$
$\displaystyle=\sum_{g_{n}\in G}f(g_{1},\dots,g_{n-1},g_{n}).$

Then we define $H_{n}(G,V):=Ker(\partial_{n})/Im(\partial_{n+1})$.

Let $(C^{*}(V),\partial^{*})$ be the cochain complex
$C^{0}(V)\overset{\partial^{0}}{\rightarrow}C^{1}(V)\overset{\partial^{1}}{\rightarrow}C^{2}(V)\overset{\partial^{2}}{\rightarrow}\dots$
with $C^{0}(V)=V$, $C^{n}(V)=C(G^{n},V)$ for all $n\geq 1$, and
$\partial^{n}=\sum_{i=0}^{n+1}(-1)^{i}\partial_{(i)}^{n}$, where

$\displaystyle\partial_{(0)}^{n}(f)(g_{0},\dots,g_{n})$
$\displaystyle=g_{0}f(g_{1},\dots,g_{n}),$
$\displaystyle\partial_{(i)}^{n}(f)(g_{0},\dots,g_{n})$
$\displaystyle=f(g_{0},\cdots,g_{i-1}g_{i},\dots,g_{n})~{}\text{for}~{}1\leq
i\leq n,$ $\displaystyle\partial_{(n+1)}^{n}(f)(g_{0},\dots,g_{n})$
$\displaystyle=f(g_{0},\dots,g_{n-1}).$

Then we define $H^{n}(G,V):=Ker(\partial^{n})/Im(\partial^{n-1})$.

If $V$ is a Banach $G$-module, we may replace $C^{*}(V)$ by $C^{*}_{b}(V)$,
where $C^{n}_{b}(V):=C_{b}(G^{n},V)$ and $C_{b}$ stands for uniformly bounded
maps, then the group cohomology we get is the bounded cohomology group,
written as $H^{n}_{b}(G,V)$.

###  2.2. Coefficient modules

The module $V$ we are interested in are certain dual or double dual Banach
spaces, i.e. $N_{0}(G,X)^{*}$, $W_{0}(G,X)^{*}$ and $N_{0}(G,X)^{**}$
associated to a continuous action $G\curvearrowright X$. Let us recall their
definition below [bnnw, bnnw2].

The space $C(X,\ell^{1}(G))$ of continuous $\ell^{1}(G)$-valued functions on
$X$ is equipped with the sup-$\ell^{1}$ norm

$||\xi||=\sup_{x\in X}\sum_{g\in G}|\xi_{x}(g)|.$

Here, for $\xi\in C(X,\ell^{1}(G))$, we write $\xi_{x}(g):=\xi(x)(g)$. We
remind the reader that we are using a different notation from the one in
[bnnw, bnnw2], where $\xi_{g}(x)$ was used.

In this notation, the Banach space $C(X,\ell^{1}(G))$ is equipped with a
natural action of $G$,

$(g\cdot\xi)_{x}(h):=\xi_{g^{-1}x}(g^{-1}h),$

for each $g,h\in G$ and $x\in X$.

The summation map on $\ell^{1}(G)$ induces a continuous map
$\sigma:C(X,\ell^{1}(G))\to C(X)$, where $C(X)$ is equipped with the
$\ell^{\infty}$ norm. The space $N_{0}(G,X)$ is defined to be the pre-image
$\sigma^{-1}(0)$ which we identify as $C(X,\ell^{1}_{0}(G))$. Then we define
$W_{0}(G,X):=\sigma^{-1}(\mathbb{R})$, where $\mathbb{R}$ is regarded as
constant functions on $X$. Note that $W_{0}(G,X)=N_{0}(G,X)\oplus\mathbb{R}$,
where $\mathbb{R}$ is identified as constant $\mathbb{R}\delta_{e}$-valued
functions on $X$, here $e$ is the neutral element in $G$. Obviously, both
$N_{0}(G,X)$ and $W_{0}(G,X)$ are invariant under the above $G$-action since
$\sigma$ is $G$-equivariant. Also recall that if $V$ is a Banach $G$-module,
then the Banach dual space $V^{*}$ is also equipped with a $G$-module
structure by requiring $(g\phi)(\xi)=\phi(g^{-1}\xi)$ for $g\in G$, $\phi\in
V^{*}$ and $\xi\in V$.

##  3\. Properties of orbit cocycles and restriction invariant property

In this section, we state several lemmas proving basic properties of orbit
cocycles and observe that the module $N_{0}(G,X)$ is restriction invariant.
These give us the hint on how to define bijective (co)chain maps between two
(co)chain complexes later.

From now on, we use the notations introduced when defining COE, i.e.
$\phi:X\approx Y$ and $\psi:Y\approx X$ are homeomorphisms; $c:G\times X\to H$
and $c^{\prime}:H\times Y\to G$ are maps satisfying certain identities.
Concerning COE between topologically free actions, the following property will
be used frequently.

######  Lemma 3.1.

[li_0, Lemma 2.10] Let $G\curvearrowright X$ and $H\curvearrowright Y$ be
topologically free actions that are COE. Then $c^{\prime}(c(g,x),\phi(x))=g$
for all $g\in G$, $x\in X$. Similarly, $c(c^{\prime}(h,y),\psi(y))=h$ for all
$h\in H$, $y\in Y$.

Clearly, this implies that for all $x\in X$, $G\ni g\mapsto c(g,x)\in H$ is a
bijection.

The starting point for our proof is the following lemma.

######  Lemma 3.2.

Let $G\curvearrowright X$ and $H\curvearrowright Y$ be topologically free
actions which are COE. Then there exists a (linear) map $\pi$ from
$C(Y,\ell^{1}(H))$ onto $C(X,\ell^{1}(G))$ which is an isometry, and which
moreover has the following properties: $\pi(W_{0}(H,Y))=W_{0}(G,X)$,
$\pi(N_{0}(H,Y))=N_{0}(G,X)$ and $\pi(\mathbb{R})=\mathbb{R}$, where
$\mathbb{R}$ is identified as constant $\mathbb{R}\delta_{e}$-valued functions
on $X$.

###### Proof.

We define $\pi:C(Y,\ell^{1}(H))\to C(X,\ell^{1}(G))$ by setting
$\pi(\xi)=\xi^{\prime}$, where

$\xi^{\prime}_{x}(g):=\xi_{\phi(x)}(c(g,g^{-1}x)).$

First, we check $\pi$ is a well-defined isometry.

Note that for all $x\in X$, $G\ni g\mapsto c(g^{-1},x)^{-1}=c(g,g^{-1}x)\in H$
is a bijection. A calculation shows that
$\sum_{g}\xi^{\prime}_{x}(g)=\sum_{h}\xi_{\phi(x)}(h)$ for all $x\in X$ and
hence $||\xi||=||\xi^{\prime}||$. So $\pi$ is an isometry.

Now we check $\xi^{\prime}\in C(X,\ell^{1}(G))$. For any given $x\in X$ and
$\epsilon>0$, we need to find $\mathcal{U}_{x}$, a neighborhood of $x$, such
that if $x^{\prime}\in\mathcal{U}_{x}$, then $\sum_{g\in
G}|\xi^{\prime}_{x}(g)-\xi^{\prime}_{x^{\prime}}(g)|<\epsilon$.

Since $\xi\in C(Y,\ell^{1}(H))$, there exists a small neighborhood
$\mathcal{V}_{\phi(x)}$ containing $\phi(x)$ such that if
$y^{\prime}\in\mathcal{V}_{\phi(x)}$, then $\sum_{h\in
H}|\xi_{y^{\prime}}(h)-\xi_{\phi(x)}(h)|<\epsilon/4$.

Take a large finite set $F\subseteq H$ such that $\sum_{h\not\in
F}|\xi_{\phi(x)}(h)|<\epsilon/4$. This implies that if
$y^{\prime}\in\mathcal{V}_{\phi(x)}$, then

$\sum_{h\not\in F}|\xi_{y^{\prime}}(h)|\leq\sum_{h\not\in
F}|\xi_{y^{\prime}}(h)-\xi_{\phi(x)}(h)|+\sum_{h\not\in
F}|\xi_{\phi(x)}(h)|\leq\epsilon/2.$

Now, we take $\mathcal{U}_{x}$ to be any sufficiently small neighborhood of
$x$, which is contained in $\psi(\mathcal{V}_{\phi(x)})$ and satisfies the
property that for any $x^{\prime}\in\mathcal{U}_{x}$, we have
$c(g^{-1},x)=c(g^{-1},x^{\prime})$ for all $g\in K$, where $K:=\\{g\in
G:~{}c(g^{-1},x)^{-1}\in F\\}$ is a finite set.

Take any $x^{\prime}\in\mathcal{U}_{x}$, note that $g\in K$ iff
$c(g^{-1},x)^{-1}\in F$ iff $c(g^{-1},x^{\prime})^{-1}\in F$. Moreover,
$x^{\prime}\in\mathcal{U}_{x}$ implies
$\phi(x^{\prime})\in\mathcal{V}_{\phi(x)}$. Hence,

$\displaystyle\sum_{g\in G}|\xi^{\prime}_{x}(g)-\xi^{\prime}_{x^{\prime}}(g)|$
$\displaystyle=\sum_{g\in
K}|\xi^{\prime}_{x}(g)-\xi^{\prime}_{x^{\prime}}(g)|+\sum_{g\not\in
K}|\xi^{\prime}_{x}(g)-\xi^{\prime}_{x^{\prime}}(g)|$
$\displaystyle\leq\sum_{h\in
F}|\xi_{\phi(x)}(h)-\xi_{\phi(x^{\prime})}(h)|+\sum_{h\not\in
F}|\xi_{\phi(x)}(h)|+\sum_{h\not\in F}|\xi_{\phi(x^{\prime})}(h)|$
$\displaystyle\leq\epsilon/4+\epsilon/4+\epsilon/2=\epsilon.$

Second, it is easy to check $\pi(W_{0}(H,Y))\subseteq
W_{0}(G,X),\pi(N_{0}(H,Y))\subseteq N_{0}(G,X)$ and
$\pi({\mathbb{R}})\subseteq{\mathbb{R}}$ hold. To see $\subseteq$ is really
$=$, we need to find the inverse of $\pi$.

By symmetry, we can define another isometry $L:C(X,\ell^{1}(G))\to
C(Y,\ell^{1}(H))$ by setting $L(\eta)=\eta^{\prime}$, where

$\eta^{\prime}_{y}(h):=\eta_{\psi(y)}({c^{\prime}(h,h^{-1}y)}).$

It is straightforward to check that $L$ is the inverse of $\pi$. Indeed, this
boils down to check that $c(c^{\prime}(h,y),\psi(y))=h$ and
$c^{\prime}(c(g,x),\phi(x))=g$, which hold by Lemma 3.1. Then using $L$, we
know $\pi$ maps the three subspaces onto the corresponding ones. ∎

### Observation

In the proof of the main theorems, we will use the following observation,
which shows that $N_{0}(G,X)$ is restriction invariant.

Under the assumptions in the main theorems, for any given $g\in G$, let us
list the elements in the finite set $c(g^{-1},X):=\\{c(g^{-1},x)|~{}x\in X\\}$
as $h_{1}^{-1},\dots,h_{n}^{-1}$. Then define $X_{i}:=\\{x\in
X:c(g^{-1},x)=h_{i}^{-1}\\}$. Clearly, $X=\sqcup_{i}X_{i}$ and each $X_{i}$ is
clopen since $c$ is continuous. Note that
$N_{0}(G,X)=C(X,\ell^{1}_{0}(G))=\oplus_{i}C(X_{i},\ell^{1}_{0}(G))$. For each
$\xi^{\prime}\in C(X,\ell^{1}_{0}(G))$, let $\xi^{\prime}|_{X_{i}}$ be the
$i$-th component with respect to the above decomposition, i.e.
$(\xi^{\prime}|_{X_{i}})_{g}(x)=\xi^{\prime}_{g}(x)$ if $x\in X_{i}$ and zero
otherwise. Clearly, $\xi^{\prime}|_{X_{i}}\in C(X,\ell^{1}_{0}(G))$. For each
$\tau\in N_{0}(G,X)^{*}$, define $\tau_{i}\in N_{0}(G,X)^{*}$ by setting
$\tau_{i}(\xi^{\prime})=\tau(\xi^{\prime}|_{X_{i}})$. Then, we have that
$\xi^{\prime}=\sum_{i}\xi^{\prime}|_{X_{i}}$ and $\tau=\sum_{i}\tau_{i}$. Note
that the above decomposition is not $G$-equivariant, i.e.
$g\tau_{i}\neq(g\tau)_{i}$. Moreover, for any clopen subset $X_{0}$ of $X$,
$\xi|_{X_{0}}\in N_{0}(G,X)$ if $\xi\in N_{0}(G,X)$, hence $\tau|_{X_{0}}$ is
also well-defined. (By convention, $\xi^{\prime}|_{\emptyset}:=0$ and
$\tau|_{\emptyset}:=0$.)

Below we record two more lemmas that we use frequently later to simplify the
proof. The first lemma tells us the map $\pi$ is equivariant in a local sense,
and the second one tells us how the composition of $\pi$ and $L$ behaves under
restrictions.

######  Lemma 3.3.

Let $g\in G$, $h\in H$ and $\xi\in N_{0}(H,Y)$. If $X_{0}\subseteq\\{x\in
X:~{}c(g^{-1},x)=h^{-1}\\}$ is clopen, then
$(\pi(h\xi))|_{X_{0}}=(g\pi(\xi))|_{X_{0}}=g(\pi(\xi)|_{g^{-1}X_{0}})$, where
$\pi$ is the map constructed in Lemma 3.2.

###### Proof.

We may assume $X_{0}\neq\emptyset$; otherwise, the equality holds trivially as
both side equals zero. Take any $x\in X$ and $g^{\prime}\in G$, we get

$\displaystyle[(\pi(h\xi))|_{X_{0}}]_{x}(g^{\prime})$
$\displaystyle=\begin{cases}(h\xi)_{\phi(x)}(c(g^{\prime},g^{\prime-1}x)),~{}\mbox{if
$x\in X_{0}$}\\\ 0,~{}\mbox{otherwise}\end{cases}$
$\displaystyle=\begin{cases}\xi_{h^{-1}\phi(x)}(h^{-1}c(g^{\prime},g^{\prime-1}x)),~{}\mbox{if
$x\in X_{0}$}\\\ 0,~{}\mbox{otherwise}\end{cases}$
$\displaystyle=\begin{cases}\xi_{\phi(g^{-1}x)}(c(g^{-1}g^{\prime},g^{\prime-1}x)),~{}\mbox{if
$x\in X_{0}$}\\\ 0,~{}\mbox{otherwise}\end{cases}$
$\displaystyle=\begin{cases}[\pi(\xi)]_{g^{-1}x}(g^{-1}g^{\prime}),~{}\mbox{if
$x\in X_{0}$}\\\ 0,~{}\mbox{otherwise}\end{cases}$
$\displaystyle=\begin{cases}[g(\pi(\xi))]_{x}(g^{\prime}),~{}\mbox{if $x\in
X_{0}$}\\\ 0,~{}\mbox{otherwise}\end{cases}$
$\displaystyle=[(g\pi(\xi))|_{X_{0}}]_{x}(g^{\prime}).$

Hence, $\pi(h\xi)|_{X_{0}}=(g\pi(\xi))|_{X_{0}}$ holds. It is easy to check
$(g\pi(\xi))|_{X_{0}}=g(\pi(\xi)|_{g^{-1}X_{0}})$ holds by definition. ∎

######  Lemma 3.4.

Let $X_{0}$ and $Y_{0}$ be clopen subsets of $X$ and $Y$ respectively. If
$\eta\in N_{0}(G,X)$ and $\xi\in N_{0}(H,Y)$, then
$\pi(L(\eta)|_{Y_{0}})|_{X_{0}}=\eta|_{X_{0}\cap\psi(Y_{0})}$ and
$L(\pi(\xi)|_{X_{0}})|_{Y_{0}}=\xi|_{Y_{0}\cap\phi(X_{0})}$.

###### Proof.

We will check the first equality below, the second one can be checked
similarly.

Let $\xi=L(\eta)|_{Y_{0}}\in N_{0}(H,Y)$. Take any $g\in G$ and $x\in X$, then

$\displaystyle\quad(\pi(\xi)|_{X_{0}})_{x}(g)$
$\displaystyle=\begin{cases}\xi_{\phi(x)}(c(g,g^{-1}x)),\quad\mbox{if $x\in
X_{0}$}\\\ 0,\quad\mbox{otherwise}\end{cases}$ $\displaystyle\quad(\mbox{def.
of restriction map and $\pi$})$
$\displaystyle=\begin{cases}\eta_{x}(c^{\prime}(c(g,g^{-1}x),c(g,g^{-1}x)^{-1}\phi(x))),\quad\mbox{if
$x\in X_{0}\cap\psi(Y_{0})$}\\\ 0,\quad\mbox{otherwise}\end{cases}$
$\displaystyle\quad(\mbox{def. of $\xi$ and $L$})$
$\displaystyle=\begin{cases}\eta_{x}(g),\quad\mbox{if $x\in
X_{0}\cap\psi(Y_{0})$}\\\ 0,\quad\mbox{otherwise}\end{cases}$
$\displaystyle\quad(\mbox{cocycle identity and Lemma \ref{lemma: xin's
lemma}})$ $\displaystyle=(\eta|_{X_{0}\cap\psi(Y_{0})})_{x}(g).$

Hence, $(\pi(L(\eta)|_{Y_{0}}))|_{X_{0}}=\eta|_{X_{0}\cap\psi(Y_{0})}$. ∎

### Notations

Here we record some notations used in the context.

  * •

$\phi:X\approx Y$ and $\psi:Y\to X$ are the homemophisms in the definition of
COE.

  * •

$c:G\times X\to H$ and $c^{\prime}:H\times Y\to X$ denote the orbit cocycles.

  * •

$\pi$ is the isometry from $N_{0}(H,Y)$ to $N_{0}(G,X)$ as defined in Lemma
3.2.

  * •

$L$ is the inverse map of $\pi$ as defined in Lemma 3.2.

  * •

$c(g,X)$ denotes the finite set $\\{c(g,x):x\in X\\}$.

  * •

$c^{\prime}(h,Y)$ denotes the finite set $\\{c^{\prime}(h,y):y\in Y\\}$.

  * •

$X_{g,h}$ denotes the set $\\{x\in X:c(g^{-1},x)=h^{-1}\\}$. For later use, we
usually simplify $X_{g_{i},h}$ (resp. $X_{g,h_{j}}$) to $X_{i}$ (resp.
$X_{j}$) if the variable $h$ (resp. $g$) is fixed or clear from the context.

  * •

$\eta$ denotes an element in $N_{0}(G,X)$ or $N_{0}(G,X)^{**}$ depending on
the context.

  * •

$\xi$ denotes an element in $N_{0}(H,Y)$ or $N_{0}(H,Y)^{**}$ depending on the
context.

  * •

$\tau$ denotes an element in $N_{0}(G,X)^{*}$.

  * •

$\nu$ denotes an element in $N_{0}(H,Y)^{*}$.

##  4\. Proof of Theorem 1.1

By [bnnw, Corollary 10], we know that $H_{0}^{uf}(G\curvearrowright
X)\cong\mathbb{R}\oplus H_{0}(G,N_{0}(G,X)^{*})$ or $H_{0}(G,N_{0}(G,X)^{*})$
depending on whether the action is amenable or not. Therefore, to prove
Theorem 1.1, it suffices to prove the second part, i.e.
$H_{n}(G,N_{0}(G,X)^{*})\cong H_{n}(H,N_{0}(H,Y)^{*})$ for all $n\geq 0$.

We will construct a (linear) map $S_{n}:C_{f}(G^{n},N_{0}(G,X)^{*})\to
C_{f}(H^{n},N_{0}(H,Y)^{*})$ and check $\partial_{n}S_{n}=S_{n-1}\partial_{n}$
for each $n$, i.e. $S_{n}$ is a chain map. Moreover, we will show it is an
isomorphism by finding the inverse map $T_{n}$. Clearly, this will induce an
isomorphism between homology groups and hence finish the proof.

We present the detailed proof when $n=0,1$ to illustrate the main ideas. The
proof of the general case will be given after the proof of these two cases.

###  4.1. Case $n=0$

Clearly, the map $\pi:N_{0}(H,Y)\to N_{0}(G,X)$ as used in Lemma 3.2 induces a
map, denoted by $\pi^{*}$, on the dual spaces.

Define $S_{0}:=\pi^{*}$.

Let $f\in C_{f}(G,N_{0}(G,X)^{*})$, we define $S_{1}(f)\in
C_{f}(H,N_{0}(H,Y)^{*})$ by setting $S_{1}(f)=f^{\prime}$, where

$f^{\prime}(h_{1})(\xi):=\sum_{j}f(g_{j})(\pi(\xi)|_{X_{j}}).$

Here $\xi\in N_{0}(H,Y)$, $c^{\prime}(h_{1},Y):=\\{c^{\prime}(h_{1},y):y\in
Y\\}:=\\{g_{j}~{}|~{}j\\},X_{j}=\\{x\in X:~{}c(g_{j}^{-1},x)=h_{1}^{-1}\\}$.
Strictly speaking, it is better to use notation $X_{g_{j},h_{1}}$ for $X_{j}$,
but we often simplify the subscripts if no confusion arises.

Clearly, $S_{0},S_{1}$ are well-defined. Indeed, to see $S_{1}$ is well-
defined, observe that if $supp(f)=F$, then $supp(S_{1}(f))\subseteq\cup_{g\in
F}c(g,X)$, which is finite once $F$ is a finite set.

Step 1: we check $\partial_{1}S_{1}=S_{0}\partial_{1}$.

Let $f\in C_{f}(G,N_{0}(G,X)^{*})$ and $f^{\prime}=S_{1}(f)$. Recall
$\partial_{1}(f^{\prime})=\sum_{h_{1}\in
H}h_{1}^{-1}f^{\prime}(h_{1})-\sum_{h_{1}\in H}f^{\prime}(h_{1})$. We need to
show $\partial_{1}S_{1}(f)=S_{0}\partial_{1}(f)$; equivalently,

$\sum_{h_{1}\in H}h_{1}^{-1}f^{\prime}(h_{1})-\sum_{h_{1}\in
H}f^{\prime}(h_{1})=\sum_{g\in G}\pi^{*}(g^{-1}f(g))-\sum_{g\in
G}\pi^{*}(f(g)).$

It suffices to check

(1) $\displaystyle\sum_{h_{1}\in H}h_{1}^{-1}f^{\prime}(h_{1})$
$\displaystyle=\sum_{g\in G}\pi^{*}(g^{-1}f(g)),$ (2)
$\displaystyle\sum_{h_{1}\in H}f^{\prime}(h_{1})$ $\displaystyle=\sum_{g\in
G}\pi^{*}(f(g)).$

To check (2), take any $\xi\in N_{0}(H,Y)$, then

$\displaystyle\sum_{h_{1}}f^{\prime}(h_{1})(\xi)$
$\displaystyle=\sum_{h_{1}}\sum_{j}f(g_{j})(\pi(\xi)|_{X_{j}})=\sum_{g}\sum_{h_{1},h_{1}\in
c(g,X)}f(g)(\pi(\xi)|_{X_{g,h_{1}}})$
$\displaystyle=\sum_{g}f(g)(\pi(\xi))=\sum_{g}\pi^{*}(f(g))(\xi).$

The 2nd last equality holds since $X=\sqcup_{h_{1},h_{1}\in
c(g,X)}X_{g,h_{1}}$ for every $g$.

Similarly, let us check (1) holds below.

$\displaystyle\sum_{h_{1}}(h_{1}^{-1}f^{\prime}(h_{1}))(\xi)$
$\displaystyle=\sum_{h_{1}}f^{\prime}(h_{1})(h_{1}\xi)=\sum_{h_{1}}\sum_{j}f(g_{j})(\pi(h_{1}\xi)|_{X_{j}})$
$\displaystyle=\sum_{g}\sum_{h_{1},h_{1}\in
c(g,X)}f(g)(\pi(h_{1}\xi)|_{X_{g,h_{1}}})$
$\displaystyle=\sum_{g}\sum_{h_{1},h_{1}\in
c(g,X)}f(g)((g\pi(\xi))|_{X_{g,h_{1}}})~{}\mbox{(by Lemma \ref{lem: module
structure under restriction})}$
$\displaystyle=\sum_{g}f(g)((g\pi(\xi))|_{\sqcup_{h_{1}\in
c(g,X)}X_{g,h_{1}}})$ $\displaystyle=\sum_{g}f(g)(g\pi(\xi))~{}\mbox{(as
$X=\sqcup_{h_{1}\in c(g,X)}X_{g,h_{1}}$)}$
$\displaystyle=\sum_{g}\pi^{**}(g^{-1}f(g))(\xi).$

Step 2: $S_{0}$ is an isomorphism.

Recall in the proof of Lemma 3.2, there exists some map $L:N_{0}(G,X)\to
N_{0}(H,Y)$ defined by setting $L(\eta)=\eta^{\prime}$, where
$\eta^{\prime}_{h}(y):=\eta_{\psi(y)}({c^{\prime}(h,h^{-1}y)})$. By symmetry,
it also induces a well-defined map $T_{0}:=L^{*}:N_{0}(H,Y)^{*}\to
N_{0}(G,X)^{*}$. Clearly, $\pi^{*}L^{*}=id$ and $L^{*}\pi^{*}=id$.

###  4.2. Case $n=1$

Write $c^{\prime}(h_{0},Y)=\\{g_{i}|~{}i\\}$,
$c^{\prime}(h_{1},Y)=\\{g_{j}|~{}j\\}$,
$c^{\prime}(h_{2},Y)=\\{g_{k}|~{}k\\}$, $c^{\prime}(h,Y)=\\{g_{s}|~{}s\\}$ and
$c^{\prime}(\bar{h},Y)=\\{g_{t}|~{}t\\}$. In this subsection, we will use the
following notations.

$\displaystyle X_{i}$ $\displaystyle:=\\{x:c(g_{i}^{-1},x)=h_{0}^{-1}\\},\quad
X_{j}:=\\{x:c(g_{j}^{-1},x)=h_{1}^{-1}\\},$ $\displaystyle X_{k}$
$\displaystyle:=\\{x:c(g_{k}^{-1},x)=h_{2}^{-1}\\},\quad
X_{s}:=\\{x:c(g_{s}^{-1},x)=h^{-1}\\},$ $\displaystyle X_{t}$
$\displaystyle:=\\{x:c(g_{t}^{-1},x)=\bar{h}^{-1}\\}.$

Strictly speaking, it is better to use the notation
$X_{g_{i},h_{0}},X_{g_{j},h_{1}},X_{g_{k},h_{2}},X_{g_{s},h},X_{g_{t},\bar{h}}$
for $X_{i},X_{j},X_{k},X_{s},X_{t}$ respectively, but we often simplify the
subscripts as the variables $h_{0},h_{1},h_{2},h,\bar{h}$ are usually fixed in
the context.

Let $\theta\in C_{f}(G^{2},N_{0}(G,X)^{*})$, we define
$S_{2}:C_{f}(G^{2},N_{0}(G,X)^{*})\to C_{f}(H^{2},N_{0}(H,Y)^{*})$ by setting
$\theta^{\prime}=S_{2}\theta$, where for any $\xi\in N_{0}(H,Y)$,

$\theta^{\prime}(h_{0},h_{1})(\xi):=\sum_{i,j}\theta(g_{i},g_{j})(\pi(\xi)|_{g_{i}X_{j}\cap
X_{i}}).$

Note that $S_{2}$ is well-defined. Indeed, observe that

$supp(\theta^{\prime})\subseteq\bigcup_{(g_{i},g_{j})\in
supp(\theta)}c(g_{i},X)\times c(g_{j},X).$

Therefore, $supp(\theta)$ is finite implies $supp(\theta^{\prime})$ is finite.

Write $f=\partial_{2}\theta$ and $f^{\prime}=S_{1}f$, from the definition of
$\partial_{2}$, we know

$f(g_{1})=\sum_{g_{0}}g_{0}^{-1}\theta(g_{0},g_{1})-\sum_{g,\bar{g},g\bar{g}=g_{1}}\theta(g,\bar{g})+\sum_{g_{2}}\theta(g_{1},g_{2}).$

Step 1: we check $S_{1}\partial_{2}=\partial_{2}S_{2}$.

Evaluate both sides at $\theta$ and use the above notations, we are left to
show the following identity holds.

$f^{\prime}(h_{1})=\sum_{h_{0}}h_{0}^{-1}\theta^{\prime}(h_{0},h_{1})-\sum_{\begin{subarray}{c}h,\bar{h}\\\
h\bar{h}=h_{1}\end{subarray}}\theta^{\prime}(h,\bar{h})+\sum_{h_{2}}\theta^{\prime}(h_{1},h_{2}).$

Let $\xi\in N_{0}(H,Y)$, a calculation shows:

$\displaystyle\quad f^{\prime}(h_{1})(\xi)$
$\displaystyle=\sum_{j}f(g_{j})(\pi(\xi)|_{X_{j}})$
$\displaystyle=\sum_{j}[\sum_{g}\theta(g,g_{j})(g(\pi(\xi)|_{X_{j}}))-\sum_{\begin{subarray}{c}g,\bar{g}\\\
g\bar{g}=g_{j}\end{subarray}}\theta(g,\bar{g})(\pi(\xi)|_{X_{j}})+\sum_{g}\theta(g_{j},g)(\pi(\xi)|_{X_{j}})].$
$\displaystyle\quad[\sum_{h_{0}}h_{0}^{-1}\theta^{\prime}(h_{0},h_{1})-\sum_{\begin{subarray}{c}h,\bar{h}\\\
h\bar{h}=h_{1}\end{subarray}}\theta^{\prime}(h,\bar{h})+\sum_{h_{2}}\theta^{\prime}(h_{1},h_{2})](\xi)$
$\displaystyle=\sum_{h_{0}}\theta^{\prime}(h_{0},h_{1})(h_{0}\xi)-\sum_{\begin{subarray}{c}h,\bar{h}\\\
h\bar{h}=h_{1}\end{subarray}}\theta^{\prime}(h,\bar{h})(\xi)+\sum_{h_{2}}\theta^{\prime}(h_{1},h_{2})(\xi)$
$\displaystyle=\sum_{h_{0}}\sum_{i,j}\theta(g_{i},g_{j})(\pi(h_{0}\xi)|_{g_{i}X_{j}\cap
X_{i}})-\sum_{\begin{subarray}{c}h,\bar{h}\\\
h\bar{h}=h_{1}\end{subarray}}\sum_{s,t}\theta(g_{s},g_{t})(\pi(\xi)|_{g_{s}X_{t}\cap
X_{s}})$
$\displaystyle\quad{}+\sum_{h_{2}}\sum_{j,k}\theta(g_{j},g_{k})(\pi(\xi)|_{g_{j}X_{k}\cap
X_{j}}).$

Comparing the two expressions above, we just need to prove the corresponding
terms are equal, i.e.

(3) $\displaystyle\sum_{j}\sum_{g}\theta(g,g_{j})(g(\pi(\xi)|_{X_{j}}))$
$\displaystyle=\sum_{h_{0}}\sum_{i,j}\theta(g_{i},g_{j})(\pi(h_{0}\xi)|_{g_{i}X_{j}\cap
X_{i}}),$ (4) $\displaystyle\sum_{j}\sum_{\begin{subarray}{c}g,\bar{g}\\\
g\bar{g}=g_{j}\end{subarray}}\theta(g,\bar{g})(\pi(\xi)|_{X_{j}})$
$\displaystyle=\sum_{\begin{subarray}{c}h,\bar{h}\\\
h\bar{h}=h_{1}\end{subarray}}\sum_{s,t}\theta(g_{s},g_{t})(\pi(\xi)|_{g_{s}X_{t}\cap
X_{s}}),$ (5)
$\displaystyle\sum_{j}\sum_{g}\theta(g_{j},g)(\pi(\xi)|_{X_{j}})$
$\displaystyle=\sum_{h_{2}}\sum_{j,k}\theta(g_{j},g_{k})(\pi(\xi)|_{g_{j}X_{k}\cap
X_{j}}).$

To check (3), apply Lemma 3.3 to get $\pi(h_{0}\xi)|_{g_{i}X_{j}\cap
X_{i}}=g_{i}(\pi(\xi)|_{X_{j}\cap g_{i}^{-1}X_{i}})$.

Then, RHS of (3) =
$\sum_{j}\sum_{h_{0},i}\theta(g_{i},g_{j})(g_{i}(\pi(\xi)|_{X_{j}\cap
g_{i}^{-1}X_{i}}))$.

Now it suffices to prove for every $j$,

$\sum_{h_{0},i}\theta(g_{i},g_{j})(g_{i}(\pi(\xi)|_{X_{j}\cap
g_{i}^{-1}X_{i}}))=\sum_{g}\theta(g,g_{j})(g(\pi(\xi)|_{X_{j}})).$

First, recall that $g_{i}^{-1}X_{i}=\\{x:c(g_{i},x)=h_{0}\\}$, which we denote
by $Z_{g_{i},h_{0}}$.

Clearly, for every $g\in G$, $X=\sqcup_{h_{0}\in c(g,X)}Z_{g,h_{0}}$, so

$\displaystyle\sum_{h_{0},i}\theta(g_{i},g_{j})(g_{i}(\pi(\xi)|_{X_{j}\cap
g_{i}^{-1}X_{i}}))$ $\displaystyle=\sum_{g}\sum_{h_{0}\in
c(g,X)}\theta(g,g_{j})(g(\pi(\xi)|_{X_{j}\cap Z_{g,h_{0}}}))$
$\displaystyle=\sum_{g}\theta(g,g_{j})(g(\pi(\xi)|_{X_{j}})).$

To check (4), first, observe that we have a bijection/reordering between the
two index sets:

$\displaystyle\left\\{(h,\bar{h},g_{s},g_{t})\begin{tabular}[]{|l}$h\bar{h}=h_{1}$\\\
$g_{s}\in c^{\prime}(h,Y),g_{t}\in c^{\prime}(\bar{h},Y)$\\\
$g_{s}X_{g_{t},\bar{h}}\cap X_{g_{s},h}\neq\emptyset$\end{tabular}\right\\}$

and

$\displaystyle\left\\{(g_{j},g,\bar{g},h,\bar{h})\begin{tabular}[]{|l}$h\bar{h}=h_{1},g\bar{g}=g_{j}$\\\
$g_{j}\in c^{\prime}(h_{1},Y),h\in c(g,X),\bar{h}\in c(\bar{g},X)$\\\
$gX_{\bar{g},\bar{h}}\cap X_{g,h}\neq\emptyset$\end{tabular}\right\\}.$

Indeed, one can define a bijection as follows: $g_{s}\mapsto
g,g_{t}\mapsto\bar{g},h\mapsto h,\bar{h}\mapsto\bar{h}$.

Then we have the following.

RHS of (4) $\displaystyle=\sum_{\begin{subarray}{c}h,\bar{h}\\\
h\bar{h}=h_{1}\end{subarray}}\sum_{s,t}\theta(g_{s},g_{t})(\pi(\xi)|_{g_{s}X_{g_{t},\bar{h}}\cap
X_{g_{s},h}})$ $\displaystyle=\sum_{j}\sum_{\begin{subarray}{c}g,\bar{g}\\\
g\bar{g}=g_{j}\end{subarray}}\sum_{\begin{subarray}{c}h,\bar{h},h\bar{h}=h_{1}\\\
h\in c(g,X),\bar{h}\in
c(\bar{g},X)\end{subarray}}\theta(g,\bar{g})(\pi(\xi)|_{gX_{\bar{g},\bar{h}}\cap
X_{g,h}})$ $\displaystyle\quad(\mbox{use the above bijection to do change of
variables})$ $\displaystyle=\sum_{j}\sum_{\begin{subarray}{c}g,\bar{g}\\\
g\bar{g}=g_{j}\end{subarray}}\theta(g,\bar{g})(\pi(\xi)|_{X_{j}})=\text{LHS of
\eqref{eq7}}.$

The 2nd last equality holds since for every $g,\bar{g}$ with $g\bar{g}=g_{j}$,

$X_{j}=\bigsqcup_{\begin{subarray}{c}h,\bar{h},h\bar{h}=h_{1}\\\ h\in
c(g,X),\bar{h}\in c(\bar{g},X)\end{subarray}}(gX_{\bar{g},\bar{h}}\cap
X_{g,h}),$

which can be checked easily.

To check (5), observe for every $j$,

$\displaystyle\quad\sum_{h_{2},k}\theta(g_{j},g_{k})(\pi(\xi)|_{g_{j}X_{k}\cap
X_{j}})$
$\displaystyle=\sum_{h_{2},k}\theta(g_{j},g_{k})(\pi(\xi)|_{g_{j}X_{g_{k},h_{2}}\cap
X_{j}})~{}(\mbox{as $X_{k}=X_{g_{k},h_{2}}$ by convention})$
$\displaystyle=\sum_{g}\sum_{\begin{subarray}{c}h_{2}\\\ h_{2}\in
c(g,X)\end{subarray}}\theta(g_{j},g)(\pi(\xi)|_{g_{j}X_{g,h_{2}}\cap X_{j}})$
$\displaystyle=\sum_{g}\theta(g_{j},g)(\pi(\xi)|_{X_{j}}).$

The last equality holds since $X=\sqcup_{h_{2}~{}s.t.~{}h_{2}\in
c(g,X)}g_{j}X_{g,h_{2}}$ for every $g$.

Step 2: $S_{1}$ is an isomorphism.

Recall $L:N_{0}(G,X)\to N_{0}(H,Y)$ is the inverse of $\pi$. Then we may
define a map $T_{1}:C_{f}(H,N_{0}(H,Y)^{*})\to C_{f}(G,N_{0}(G,X)^{*})$ by
setting $T_{1}(f^{\prime})=f^{\prime\prime}$, where

$f^{\prime\prime}(g_{1})(\eta):=\sum_{j}f^{\prime}(h_{j})(L(\eta)|_{Y_{j}}).$

Here $\eta\in N_{0}(G,X)$ and $Y_{j}:=\\{y\in
Y:~{}c^{\prime}(h_{j}^{-1},y)=g_{1}^{-1}\\}$.

Let $f\in C_{f}(G,N_{0}(G,X)^{*})$, write $f^{\prime}=S_{1}f$,
$f^{\prime\prime}=T_{1}f^{\prime}$. We check $T_{1}S_{1}=id$. Equivalently,
$f^{\prime\prime}=f$. A calculation shows

$\displaystyle\quad f^{\prime\prime}(g_{1})(\eta)$
$\displaystyle=\sum_{j}f^{\prime}(h_{j})(L(\eta)|_{Y_{j}})~{}(\mbox{def. of
$T_{1}$})$
$\displaystyle=\sum_{j}\sum_{i}f(g_{i})(\pi(L(\eta)|_{Y_{j}})|_{X_{g_{i},h_{j}}})~{}(\mbox{def.
of $S_{1}$})$
$\displaystyle=\sum_{j,i}f(g_{i})(\eta|_{X_{g_{i},h_{j}}\cap\psi(Y_{j})}).~{}(\mbox{Lemma
\ref{lem: composition of pi and L under restrictions}})$

Recall by our notation, $X_{g_{i},h_{j}}:=\\{x\in
X:c(g_{i}^{-1},x)=h_{j}^{-1}\\}$. Now, it is easy to check that
$X_{g_{i},h_{j}}\cap\psi(Y_{j})=\emptyset$ unless $g_{i}=g_{1}$. And
$X_{g_{1},h_{j}}\cap\psi(Y_{j})=X_{g_{1},h_{j}}$ by Lemma 3.1.

Hence,
$f^{\prime\prime}(g_{1})(\eta)=\sum_{j}f(g_{1})(\eta|_{X_{g_{1},h_{j}}})=f(g_{1})(\eta)$.

So $T_{1}S_{1}=id$. By symmetry, $S_{1}T_{1}=id$.

###  4.3. General case

Based on the definition of $S_{i}$ for $i=0,1$, it is natural to consider the
following general formula for $S_{n}$.

For all $n\geq 2$, let $\theta\in C_{f}(G^{n},N_{0}(G,X)^{*})$, define
$S_{n}(\theta)=\theta^{\prime}\in C_{f}(H^{n},N_{0}(H,Y)^{*})$ by setting for
every $\xi\in N_{0}(H,Y)$,

(6)
$\displaystyle{}\theta^{\prime}(h_{0},\dots,h_{n-1})(\xi):=\sum_{t_{0},\dots,t_{n-1}}\theta(g_{t_{0}},\dots,g_{t_{n-1}})(\pi(\xi)|_{[t_{0},\ldots,t_{n-1}]}).$

Here, $c^{\prime}(h_{i}^{-1},Y)=\\{g_{t_{i}}^{-1}:t_{i}\in T_{i}\\}$ for some
finite set $T_{i}$ for all $i=0,\cdots,n-1$,

(7)
$\displaystyle\begin{split}X_{t_{i}}&:=\\{x:c(g_{t_{i}}^{-1},x)=h_{i}^{-1}\\},\\\
[t_{0},\ldots,t_{n-1}]&:=X_{t_{0}}\cap
g_{t_{0}}X_{t_{1}}\cap\dots\cap(g_{t_{0}}\cdots
g_{t_{n-2}})X_{t_{n-1}}.\end{split}$

Strictly speaking, the index set for the above $\sum$ in (6) should be
$(g_{t_{0}},\ldots,g_{t_{n-1}})$, but we simplify it to
$(t_{0},\ldots,t_{n-1})$ for convenience. Again, we will switch to the
notation $X_{g_{t_{i}},h_{i}}$ once using $X_{t_{i}}$ causes confusion. We
also reserve the notation $g_{i}$ for an arbitrary element in $G$.

Clearly, $S^{n}$ is well-defined. Indeed, observe that

$supp(\theta^{\prime})\subseteq\bigcup_{(g_{t_{0}},\cdots,g_{t_{n-1}})\in
supp(\theta)}c(g_{t_{0}},X)\times\cdots\times c(g_{t_{n-1}},X).$

Hence, $supp(\theta)$ is finite implies $supp(\theta^{\prime})$ is finite.

Our goal is to check that $\partial_{n}S_{n}=S_{n-1}\partial_{n}$ holds and
$S_{n}$ is a bijection for each $n\geq 2$. This will finish the proof of
Theorem 1.1.

Step 1: We claim $\partial_{n}S_{n}=S_{n-1}\partial_{n}$.

Take any $\theta\in C_{f}(G^{n},N_{0}(G,X)^{*})$, write
$\theta^{\prime}=S_{n}(\theta)$ and $f=\partial_{n}\theta$. Fix any
$(h_{1},\ldots,h_{n-1})\in H^{n-1}$ and any $\xi\in N_{0}(H,Y)$.

Now, we calculate both
$[\partial_{n}S_{n}(\theta)](h_{1},\ldots,h_{n-1})(\xi)$ and
$[S_{n-1}\partial_{n}(\theta)](h_{1},\ldots,h_{n-1})(\xi)$ and show they are
equal.

First, we have the following calculation.

$\displaystyle\quad[\partial_{n}S_{n}(\theta)](h_{1},\ldots,h_{n-1})$
$\displaystyle=[\partial_{n}\theta^{\prime}](h_{1},\ldots,h_{n-1})$
$\displaystyle=\sum_{h_{0}\in
H}h_{0}^{-1}\theta^{\prime}(h_{0},\ldots,h_{n-1})+\sum_{i=1}^{n-1}(-1)^{i}\sum_{\begin{subarray}{c}h,\bar{h}\in
H\\\
h\bar{h}=h_{i}\end{subarray}}\theta^{\prime}(h_{1},\ldots,h_{i-1},h,\bar{h},h_{i+1},\ldots,h_{n-1})$
$\displaystyle\quad{}+(-1)^{n}\sum_{h_{n}\in
H}\theta^{\prime}(h_{1},\ldots,h_{n}).$

Evaluate both sides of the above at the element $\xi$, then plug in the
definition of $\theta^{\prime}$, i.e. (6) into the RHS of the above, we deduce
that

(8)
$\displaystyle\begin{split}&\quad\quad[\partial_{n}S_{n}(\theta)](h_{1},\ldots,h_{n-1})(\xi)\\\
&=\sum_{h_{0}\in
H}\sum_{t_{0},\ldots,t_{n-1}}\big{\langle}\theta(g_{t_{0}},\ldots,g_{t_{n-1}}),\pi(h_{0}\xi)|_{[t_{0},\ldots,t_{n-1}]}\big{\rangle}\\\
&\quad+\sum_{i=1}^{n-1}(-1)^{i}\sum_{\begin{subarray}{c}h,\bar{h}\in H\\\
h\bar{h}=h_{i}\end{subarray}}\sum_{\begin{subarray}{c}t_{1},\ldots,t_{i-1},s\\\
k,t_{i+1},\ldots,t_{n-1}\end{subarray}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{i-1}},g_{s},g_{k},g_{t_{i+1}},\ldots,g_{t_{n-1}}),\\\
&\qquad\qquad\qquad\qquad\pi(\xi)|_{[t_{1},\ldots,t_{i-1},s,k,t_{i+1},\ldots,t_{n-1}]}\big{\rangle}\\\
&\quad+(-1)^{n}\sum_{h_{n}\in
H}\sum_{t_{1},\ldots,t_{n}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{n}}),\pi(\xi)|_{[t_{1},\ldots,t_{n}]}\big{\rangle}.\end{split}$

Here, we use $\langle-,-\rangle$ to denote the evaluation, i.e. evaluating the
first entry (a function) at the second entry (a variable). Let us recall some
notations used here:

$\displaystyle X_{s}:=\\{x:c(g_{s}^{-1},x)=h^{-1}\\},\quad
X_{k}:=\\{x:c(g_{k}^{-1},x)=\bar{h}^{-1}\\}.$ (9)
$[t_{1},\ldots,t_{i-1},s,k,t_{i+1},\ldots,t_{n-1}]:=\\\ X_{t_{1}}\cap
g_{t_{1}}X_{t_{2}}\cap\cdots\cap(g_{t_{1}}\cdots
g_{t_{i-2}})X_{t_{i-1}}\cap(g_{t_{1}}\cdots g_{t_{i-1}})X_{s}\\\
\cap(g_{t_{1}}\cdots g_{t_{i-1}}g_{s})X_{k}\cap\cdots\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g_{s}g_{k}g_{t_{i+1}}\cdots g_{t_{n-2}})X_{t_{n-1}}.$

Second, another calculation tells us the following.

(10)
$\displaystyle\begin{split}&\quad[S_{n-1}\partial_{n}\theta](h_{1},\ldots,h_{n-1})(\xi)\\\
&=[S_{n-1}f](h_{1},\ldots,h_{n-1})(\xi)\\\
&=\sum_{t_{1},\ldots,t_{n-1}}f(g_{t_{1}},\ldots,g_{t_{n-1}})(\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}).\quad(\mbox{def.
of $S_{n-1}$})\end{split}$

Now, recall that $f=\partial_{n}\theta$ and the definition of $\partial_{n}$
from Section 2, we get

$\displaystyle\quad f(g_{t_{1}},\ldots,g_{t_{n-1}})$
$\displaystyle=\sum_{g_{0}\in
G}g_{0}^{-1}\theta(g_{0},g_{t_{1}},\ldots,g_{t_{n-1}})$
$\displaystyle\quad+\sum_{i=1}^{n-1}(-1)^{i}\sum_{\begin{subarray}{c}g,\bar{g}\in
G\\\
g\bar{g}=g_{t_{i}}\end{subarray}}\theta(g_{t_{1}},\ldots,g_{t_{i-1}},g,\bar{g},g_{t_{i+1}},\ldots,g_{t_{n-1}})$
$\displaystyle\quad+(-1)^{n}\sum_{g_{n}\in
G}\theta(g_{t_{1}},\ldots,g_{t_{n-1}},g_{n}).$

Evaluate both sides of the above at $\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}$ and
combine with (10), we get

(11)
$\displaystyle\begin{split}&\quad[S_{n-1}\partial_{n}\theta](h_{1},\ldots,h_{n-1})(\xi)\\\
&=\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{0}\in G}\langle
g_{0}^{-1}\theta(g_{0},g_{t_{1}},\ldots,g_{t_{n-1}}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}\rangle\\\
&\quad+\sum_{t_{1},\ldots,t_{n-1}}\sum_{i=1}^{n-1}(-1)^{i}\sum_{\begin{subarray}{c}g,\bar{g}\in
G\\\
g\bar{g}=g_{t_{i}}\end{subarray}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{i-1}},g,\bar{g},g_{t_{i+1}},\ldots,g_{t_{n-1}}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}\big{\rangle}\\\
&\quad+\sum_{t_{1},\ldots,t_{n-1}}(-1)^{n}\sum_{g_{n}\in
G}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{n-1}},g_{n}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}\big{\rangle}.\end{split}$

We aim to show (8)=(11). To do this, we just need to compare all the
corresponding summands and prove they are equal, i.e. it suffices to check the
following hold.

(a)

$\sum_{h_{0}\in
H}\sum_{t_{0},\ldots,t_{n-1}}\big{\langle}\theta(g_{t_{0}},\ldots,g_{t_{n-1}}),\pi(h_{0}\xi)|_{[t_{0},\ldots,t_{n-1}]}\big{\rangle}\\\
=\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{0}\in G}\langle
g_{0}^{-1}\theta(g_{0},g_{t_{1}},\ldots,g_{t_{n-1}}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}\rangle.$

(b) For any $1\leq i\leq n-1$,

$\sum_{\begin{subarray}{c}h,\bar{h}\in H\\\
h\bar{h}=h_{i}\end{subarray}}\sum_{\begin{subarray}{c}t_{1},\ldots,t_{i-1},s\\\
k,t_{i+1},\ldots,t_{n-1}\end{subarray}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{i-1}},g_{s},g_{k},g_{t_{i+1}},\ldots,g_{t_{n-1}}),\pi(\xi)|_{[t_{1},\ldots,t_{i-1},s,k,t_{i+1},\ldots,t_{n-1}]}\big{\rangle}\\\
=\sum_{t_{1},\ldots,t_{n-1}}\sum_{\begin{subarray}{c}g,\bar{g}\in G\\\
g\bar{g}=g_{t_{i}}\end{subarray}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{i-1}},g,\bar{g},g_{t_{i+1}},\ldots,g_{t_{n-1}}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}\big{\rangle}.$

(c)

$\sum_{h_{n}\in
H}\sum_{t_{1},\ldots,t_{n}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{n}}),\pi(\xi)|_{[t_{1},\ldots,t_{n}]}\big{\rangle}\\\
=\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{n}\in
G}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{n-1}},g_{n}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}\big{\rangle}.$

We prove (a), (b), (c) below respectively.

###### Proof of (a).

By Lemma 3.3,
$\pi(h_{0}\xi)|_{[t_{0},\ldots,t_{n-1}]}=g_{t_{0}}(\pi(\xi)|_{g_{t_{0}}^{-1}[t_{0},\ldots,t_{n-1}]}).$
Then,

(12) $\displaystyle\begin{split}&\quad\quad\mbox{Top expression in (a)}\\\
&=\quad\sum_{h_{0}\in
H}\sum_{t_{0},\ldots,t_{n-1}}\big{\langle}\theta(g_{t_{0}},\ldots,g_{t_{n-1}}),\pi(h_{0}\xi)|_{[t_{0},\ldots,t_{n-1}]}\big{\rangle}\\\
&=\quad\sum_{h_{0}\in
H}\sum_{t_{0},\ldots,t_{n-1}}\big{\langle}\theta(g_{t_{0}},\ldots,g_{t_{n-1}}),g_{t_{0}}(\pi(\xi)|_{g_{t_{0}}^{-1}[t_{0},\ldots,t_{n-1}]})\big{\rangle}\\\
&=\quad\sum_{h_{0}\in
H}\sum_{t_{0},\ldots,t_{n-1}}\big{\langle}g_{t_{0}}^{-1}\theta(g_{t_{0}},\ldots,g_{t_{n-1}}),\pi(\xi)|_{g_{t_{0}}^{-1}X_{t_{0}}\cap[t_{1},\ldots,t_{n-1}]}\big{\rangle}\\\
&\quad(\mbox{def. of dual module and by (\ref{eq for lem 1.1: def of
[t0,...,tn]}),
$g_{t_{0}}^{-1}[t_{0},\ldots,t_{n-1}]=g_{t_{0}}^{-1}X_{t_{0}}\cap[t_{1},\ldots,t_{n-1}]$})\\\
&=\quad\sum_{t_{1},\ldots,t_{n-1}}\sum_{h_{0}\in
H}\sum_{t_{0}}\big{\langle}g_{t_{0}}^{-1}\theta(g_{t_{0}},\ldots,g_{t_{n-1}}),\pi(\xi)|_{g_{t_{0}}^{-1}X_{t_{0}}\cap[t_{1},\ldots,t_{n-1}]}\big{\rangle}\\\
&\quad(\mbox{switching sums is possible as $t_{1},\ldots,t_{n-1}$ and $h_{0}$
are independent})\\\ &=\quad\sum_{t_{1},\ldots,t_{n-1}}\sum_{h_{0}\in
H}\sum_{\begin{subarray}{c}g_{t_{0}}\\\ g_{t_{0}}\in
c^{\prime}(h_{0},Y)\end{subarray}}\big{\langle}g_{t_{0}}^{-1}\theta(g_{t_{0}},\ldots,g_{t_{n-1}}),\pi(\xi)|_{g_{t_{0}}^{-1}X_{g_{t_{0}},h_{0}}\cap[t_{1},\ldots,t_{n-1}]}\big{\rangle}\\\
&\quad(\mbox{notation convention, check the explanation given when defing
$S_{n}$, i.e. (\ref{eq for thm 1.1: def of S_n})})\\\
&=\quad\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{0}\in G}\sum_{h_{0}\in
c(g_{0},X)}\big{\langle}g_{0}^{-1}\theta(g_{0},\ldots,g_{t_{n-1}}),\pi(\xi)|_{g_{0}^{-1}X_{g_{0},h_{0}}\cap[t_{1},\ldots,t_{n-1}]}\big{\rangle}.\end{split}$

The last equality holds for the following reason:

The map $h_{0}\mapsto h_{0},g_{t_{0}}\mapsto g_{0}$ induces a bijection
between the two index sets: $\\{(h_{0},g_{t_{0}}):h_{0}\in H,g_{t_{0}}\in
c^{\prime}(h_{0},Y)\\}$ and $\\{(g_{0},h_{0}):g_{0}\in G,h_{0}\in
c(g_{0},X)\\}$. This explains the change of the index sets under the two
rightmost sums. So under the above bijection,
$g_{t_{0}}^{-1}X_{g_{t_{0}},h_{0}}$ is replaced by
$g_{0}^{-1}X_{g_{0},h_{0}}$.

Now, recall $x\in g_{0}^{-1}X_{g_{0},h_{0}}$ iff $g_{0}x\in X_{g_{0},h_{0}}$
iff $c(g_{0},x)=h_{0}$. Therefore, for each $g_{0}\in G$, $X=\sqcup_{h_{0}\in
c(g_{0},X)}g_{0}^{-1}X_{g_{0},h_{0}}$.

Using this fact, we can continue to simplify (12) to the following
expressions.

The top expression in (a)
$\displaystyle=\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{0}\in
G}\big{\langle}g_{0}^{-1}\theta(g_{0},\ldots,g_{t_{n-1}}),\pi(\xi)|_{(\sqcup_{h_{0}\in
c(g_{0},X)}g_{0}^{-1}X_{g_{0},h_{0}})\cap[t_{1},\ldots,t_{n-1}]}\big{\rangle}$
$\displaystyle=\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{0}\in
G}\big{\langle}g_{0}^{-1}\theta(g_{0},\ldots,g_{t_{n-1}}),\pi(\xi)|_{X\cap[t_{1},\ldots,t_{n-1}]}\big{\rangle}$
$\displaystyle=\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{0}\in
G}\big{\langle}g_{0}^{-1}\theta(g_{0},\ldots,g_{t_{n-1}}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}\big{\rangle}$
$\displaystyle=\quad\mbox{The bottom expression in (a)}.\qed$

###### Proof of (b).

The proof relies on the following two facts:

(Fact 1) For every fixed $t_{1},\ldots,t_{i-1},t_{i+1},\ldots,t_{n-1}$, the
map $t_{j}\mapsto t_{j}$ ($\forall j\neq i$), $s\mapsto g$, $k\mapsto\bar{g}$
induces a bijection between the index sets:

$\displaystyle\left\\{(t_{1},\ldots,t_{i-1},t_{i+1},\ldots,t_{n-1},s,k)~{}\begin{tabular}[]{|l}$h\bar{h}=h_{i}$\\\
$g_{s}\in c^{\prime}(h,Y),g_{k}\in c^{\prime}(\bar{h},Y)$\\\
$[t_{1},\ldots,t_{i-1},s,k,t_{i+1},\ldots,t_{n-1}]\neq\emptyset$\end{tabular}\right\\}$

and

$\displaystyle\left\\{(t_{1},\ldots,t_{n-1},g,\bar{g})~{}\begin{tabular}[]{|l}$g\bar{g}=g_{t_{i}},h\bar{h}=h_{i}$\\\
$h\in c(g,X),\bar{h}\in c(\bar{g},X),g_{t_{i}}\in c^{\prime}(h_{i},Y)$\\\
$[t_{1},\ldots,t_{i-1},g,\bar{g},t_{i+1},\ldots,t_{n-1}]\neq\emptyset$\end{tabular}\right\\}.$

Indeed, just remember that the index $s$ stands for $g_{s}$, $k$ stands for
$g_{k}$, $t_{j}$ stands for $g_{t_{j}}$ and $g_{t_{i}}$ is determined by $g$
and $\bar{g}$.

Here, $[t_{1},\ldots,t_{i-1},g,\bar{g},t_{i+1},\ldots,t_{n-1}]$ is the image
of $[t_{1},\ldots,t_{i-1},s,k,t_{i+1},\ldots,t_{n-1}]$ under the bijection as
in Fact 1. More precisely, from the def. (9), we know

(13) $[t_{1},\ldots,t_{i-1},g,\bar{g},t_{i+1},\ldots,t_{n-1}]=X_{t_{1}}\cap
g_{t_{1}}X_{t_{2}}\cap\cdots\cap(g_{t_{1}}\cdots g_{t_{i-2}})X_{t_{i-1}}\\\
\cap(g_{t_{1}}\cdots g_{t_{i-1}})X_{g}\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g)X_{\bar{g}}\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g\bar{g})X_{t_{i+1}}\\\ \cap(g_{t_{1}}\cdots
g_{t_{i-1}}g\bar{g}g_{t_{i+1}})X_{t_{i+2}}\cap\cdots\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g\bar{g}g_{t_{i+1}}\cdots g_{t_{n-2}})X_{t_{n-1}}.$

We remind the reader in the above expression,
$X_{g}=X_{g,h}=\\{x:c(g^{-1},x)=h^{-1}\\}$ and
$X_{\bar{g}}=\\{x:c(\bar{g}^{-1},x)=\bar{h}^{-1}\\}$.

(Fact 2) For all $g,\bar{g}\in G$ satisfying $g\bar{g}=g_{t_{i}}$, we have

$\displaystyle\bigsqcup_{\begin{subarray}{c}h,\bar{h},h\bar{h}=h_{i}\\\ h\in
c(g,X),\bar{h}\in c(\bar{g},X)\end{subarray}}(g_{t_{1}}\cdots
g_{t_{i-1}})X_{g,h}\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g)X_{\bar{g},\bar{h}}\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g\bar{g})X_{t_{i+1}}$ $\displaystyle\quad{}=(g_{t_{1}}\cdots
g_{t_{i-1}})X_{t_{i}}\cap(g_{t_{1}}\cdots g_{t_{i}})X_{t_{i+1}}.$

Recall
$X_{t_{i+1}}=X_{g_{t_{i+1}},h_{i+1}}=\\{x:c(g_{t_{i+1}}^{-1},x)=h_{i+1}^{-1}\\}$,
$X_{g,h}=\\{x:c(g^{-1},x)=h^{-1}\\}$ and
$X_{\bar{g},\bar{h}}=\\{x:c(\bar{g}^{-1},x)=\bar{h}^{-1}\\}$. Then Fact 2 is
easy to verify using cocycle identities.

Now, we can prove (b) as follows.

By Fact 1,

(14) $\displaystyle\begin{split}&\quad\mbox{The top expression in (b)}\\\
&=\sum_{t_{1},\ldots,t_{n-1}}\sum_{\begin{subarray}{c}g,\bar{g}\in G\\\
g\bar{g}=g_{t_{i}}\end{subarray}}\sum_{\begin{subarray}{c}h,\bar{h},h\bar{h}=h_{i}\\\
h\in c(g,X),\bar{h}\in
c(\bar{g},X)\end{subarray}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{i-1}},g,\bar{g},g_{t_{i+1}},\ldots,g_{t_{n-1}}),\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pi(\xi)|_{[t_{1},\ldots,t_{i-1},g,\bar{g},t_{i+1},\ldots,t_{n-1}]}\big{\rangle}.\end{split}$

Since $g\bar{g}=g_{t_{i}}$, we can rewrite (13) as

(15) $[t_{1},\ldots,t_{i-1},g,\bar{g},t_{i+1},\ldots,t_{n-1}]=X_{t_{1}}\cap
g_{t_{1}}X_{t_{2}}\cap\cdots\cap(g_{t_{1}}\cdots g_{t_{i-2}})X_{t_{i-1}}\\\
\cap\underline{(g_{t_{1}}\cdots g_{t_{i-1}})X_{g,h}\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g)X_{\bar{g},\bar{h}}\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g\bar{g})X_{t_{i+1}}}\\\ \cap(g_{t_{1}}\cdots
g_{t_{i-1}}g_{t_{i}}g_{t_{i+1}})X_{t_{i+2}}\cap\cdots\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g_{t_{i}}g_{t_{i+1}}\cdots g_{t_{n-2}})X_{t_{n-1}}.$

Now, observe that the underlined part above is exactly the one appeared in
Fact 2, and the rest pieces in $(\ref{eq for thm 1.1: final form for
[t1,...,g, g bar,...]})$ are exactly the ones appeared in the definition of
$[t_{1},\ldots,t_{n-1}]$, which is recalled below

$[t_{1},\ldots,t_{n-1}]=X_{t_{1}}\cap
g_{t_{1}}X_{t_{2}}\cap\cdots\cap(g_{t_{1}}\cdots g_{t_{i-2}})X_{t_{i-1}}\\\
\cap\underline{(g_{t_{1}}\cdots g_{t_{i-1}})X_{t_{i}}\cap(g_{t_{1}}\cdots
g_{t_{i}})X_{t_{i+1}}}\\\ \cap(g_{t_{1}}\cdots
g_{t_{i-1}}g_{t_{i}}g_{t_{i+1}})X_{t_{i+2}}\cap\cdots\cap(g_{t_{1}}\cdots
g_{t_{i-1}}g_{t_{i}}g_{t_{i+1}}\cdots g_{t_{n-2}})X_{t_{n-1}}.$

Apply the above observation, we can continue the computation in (14) as
follows.

First, we can move

$\sum_{\begin{subarray}{c}h,\bar{h},h\bar{h}=h_{i}\\\ h\in c(g,X),\bar{h}\in
c(\bar{g},X)\end{subarray}}$

inside $\langle-,-\rangle$ and put it before the 2nd entry, i.e.
$\pi(\xi)|_{[t_{1},\ldots,t_{i-1},g,\bar{g},t_{i+1},\ldots,t_{n-1}]}$, to get

$\displaystyle\sum_{\begin{subarray}{c}h,\bar{h},h\bar{h}=h_{i}\\\ h\in
c(g,X),\bar{h}\in
c(\bar{g},X)\end{subarray}}\pi(\xi)|_{[t_{1},\ldots,t_{i-1},g,\bar{g},t_{i+1},\ldots,t_{n-1}]}=\pi(\xi)|_{\mathcal{J}}.$

Here,

$\displaystyle\mathcal{J}$
$\displaystyle:=\bigsqcup_{\begin{subarray}{c}h,\bar{h},h\bar{h}=h_{i}\\\ h\in
c(g,X),\bar{h}\in
c(\bar{g},X)\end{subarray}}[t_{1},\ldots,t_{i-1},g,\bar{g},t_{i+1},\ldots,t_{n-1}]$
$\displaystyle=[t_{1},\ldots,t_{n-1}].\qquad(\mbox{by Fact ~{}2~{}+~{}(\ref{eq
for thm 1.1: final form for [t1,...,g, g bar,...]})})$

Put all the above information together, we finally arrive at the following.

The top expression in (b)
$\displaystyle=\sum_{t_{1},\ldots,t_{n-1}}\sum_{\begin{subarray}{c}g,\bar{g}\in
G\\\
g\bar{g}=g_{t_{i}}\end{subarray}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{i-1}},g,\bar{g},g_{t_{i+1}},\ldots,g_{t_{n-1}}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}\big{\rangle}$
$\displaystyle=\mbox{The bottom expression in (b)}.$

This finishes the proof of (b). ∎

###### Proof of (c).

The top expression in (c)
$\displaystyle=\sum_{t_{1},\ldots,t_{n-1}}\sum_{h_{n}\in
H}\sum_{t_{n}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{n}}),\pi(\xi)|_{[t_{1},\ldots,t_{n}]}\big{\rangle}$
$\displaystyle=\sum_{t_{1},\ldots,t_{n-1}}\sum_{h_{n}\in
H}\sum_{t_{n}}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{n}}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]\cap(g_{t_{1}}\cdots
g_{t_{n-1}})X_{t_{n}}}\big{\rangle}$ (def. of $[t_{1},\ldots,t_{n}]$)
$\displaystyle=\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{n}\in G}\sum_{h_{n}\in
c(g_{n},X)}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{n-1}},g_{n}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]\cap(g_{t_{1}}\cdots
g_{t_{n-1}})X_{g_{n},h_{n}}}\big{\rangle}.$

The last equality holds for the following reason.

Recall here
$X_{t_{n}}=X_{g_{t_{n}},h_{n}}=\\{x:c(g_{t_{n}}^{-1},x)=h_{n}^{-1}\\}$, so
$X_{t_{n}}$ is replaced by $X_{g_{n},h_{n}}$ under the bijective map
$t_{n}\mapsto g_{n}$, $h_{n}\mapsto h_{n}$ between the index sets

$\\{(h_{n},t_{n}):t_{n}\in c^{\prime}(h_{n},Y),h_{n}\in H\\}$

and

$\\{(g_{n},h_{n}):h_{n}\in c(g_{n},X),g_{n}\in G\\}.$

Then, since

$X=\bigsqcup_{h_{n}\in c(g_{n},X)}(g_{t_{1}}\cdots
g_{t_{n-1}})X_{g_{n},h_{n}},$

we can continue the computation of the above expression as follows.

The top expression in (c)
$\displaystyle=\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{n}\in
G}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{n-1}},g_{n}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]\cap\sqcup_{h_{n}\in
c(g_{n},X)}(g_{t_{1}}\cdots g_{t_{n-1}})X_{g_{n},h_{n}}}\big{\rangle}$
$\displaystyle=\sum_{t_{1},\ldots,t_{n-1}}\sum_{g_{n}\in
G}\big{\langle}\theta(g_{t_{1}},\ldots,g_{t_{n-1}},g_{n}),\pi(\xi)|_{[t_{1},\ldots,t_{n-1}]}\big{\rangle}$
$\displaystyle=\mbox{The bottom expression in (c).}\qed$

Step 2: $S_{n}$ is a bijection.

By symmetry, we define a (linear) map $T_{n}:C_{f}(H^{n},N_{0}(H,Y)^{*})\to
C_{f}(G^{n},N_{0}(G,X)^{*})$ by setting
$T_{n}(\theta^{\prime})=\theta^{\prime\prime}$, where

(16)
$\displaystyle\theta^{\prime\prime}(g_{0},\ldots,g_{n-1})(\eta):=\sum_{s_{0},\ldots,s_{n-1}}\theta^{\prime}(h_{s_{0}},\ldots,h_{s_{n-1}})(L(\eta)|_{[s_{0},\ldots,s_{n-1}]}).$

Here, $\eta\in N_{0}(G,X)$, $Y_{s_{i}}:=\\{y\in
Y:~{}c^{\prime}(h_{s_{i}}^{-1},y)=g_{i}^{-1}\\}$ and

(17) $\displaystyle[s_{0},\ldots,s_{n-1}]:=Y_{s_{0}}\cap
h_{s_{0}}Y_{s_{1}}\cap\cdots\cap(h_{s_{0}}\cdots h_{s_{n-2}})Y_{s_{n-1}}.$

If we can prove the following claim, then by symmetry, we also have
$S_{n}T_{n}=id$, and hence $S_{n}$ is a bijection.

Claim: $T_{n}S_{n}=id$.

###### Proof of the claim.

Take any $\theta\in C_{f}(G^{n},N_{0}(G,X)^{*})$, $(g_{0},\ldots,g_{n-1})\in
G^{n}$ and any $\eta\in N_{0}(G,X)$, let $\theta^{\prime}=S_{n}(\theta)$ and
$\theta^{\prime\prime}=T_{n}(\theta^{\prime})$.

We aim to show $\theta=\theta^{\prime\prime}$, i.e.
$\theta^{\prime\prime}(g_{0},\ldots,g_{n-1})(\eta)=\theta(g_{0},\ldots,g_{n-1})(\eta)$.

From the definitions of $S_{n}$ and $T_{n}$, i.e. (6) and (16), we deduce that

(18)
$\displaystyle\begin{split}&\quad\theta^{\prime\prime}(g_{0},\ldots,g_{n-1})(\eta)\\\
&=\sum_{s_{0},\ldots,s_{n-1}}\theta^{\prime}(h_{s_{0}},\ldots,h_{s_{n-1}})(L(\eta)|_{[s_{0},\ldots,s_{n-1}]})\\\
&=\sum_{s_{0},\ldots,s_{n-1}}\sum_{t_{0},\ldots,t_{n-1}}\theta(g_{t_{0}},\ldots,g_{t_{n-1}})(\pi(L(\eta)|_{[s_{0},\ldots,s_{n-1}]})|_{[t_{0},\ldots,t_{n-1}]}).\end{split}$

Since there are many variables here, let us recall some notations used above.

$[s_{0},\ldots,s_{n-1}]$ is defined as in (17), and def. (7) in our context is
the following.

(19) $\displaystyle[t_{0},\ldots,t_{n-1}]=X_{g_{t_{0}},h_{s_{0}}}\cap
g_{t_{0}}X_{g_{t_{1}},h_{s_{1}}}\cap\cdots\cap(g_{t_{0}}\ldots
g_{t_{n-2}})X_{g_{t_{n-1}}h_{s_{n-1}}},$

where $X_{g_{t_{i}},h_{s_{i}}}=\\{x:c(g_{t_{i}}^{-1},x)=h_{s_{i}}^{-1}\\}.$

To continue the computation, we need the following facts.

(Fact 3)
$\pi(L(\eta)|_{[s_{0},\ldots,s_{n-1}]})|_{[t_{0},\ldots,t_{n-1}]}=\eta|_{[t_{0},\ldots,t_{n-1}]\cap\psi([s_{0},\ldots,s_{n-1}])}.$

This is clear by Lemma 3.4.

(Fact 4) $[t_{0},\ldots,t_{n-1}]\cap\psi([s_{0},\ldots,s_{n-1}])=\emptyset$
unless $g_{t_{i}}=g_{i}$ for all $0\leq i\leq n-1$. When these conditions
hold, the intersection equals

$X_{g_{0},s_{0}}\cap g_{0}X_{g_{1},s_{1}}\cap\cdots\cap(g_{0}\cdots
g_{n-2})X_{g_{n-1},s_{n-1}},$

where $X_{g_{i},s_{i}}=\\{x:c(g_{i}^{-1},x)=h_{s_{i}}^{-1}\\}$.

This fact can be checked using (17), (19) and Lemma 3.1.

Now, we can continue the computation of (18) as follows.

$\displaystyle\quad\theta^{\prime\prime}(g_{0},\ldots,g_{n-1})(\eta)$
$\displaystyle\overset{Fact~{}3}{=}\sum_{s_{0},\ldots,s_{n-1}}\sum_{t_{0},\ldots,t_{n-1}}\theta(g_{t_{0}},\ldots,g_{t_{n-1}})(\eta|_{[t_{0},\ldots,t_{n-1}]\cap\psi([s_{0},\ldots,s_{n-1}])})$
$\displaystyle\overset{Fact~{}4}{=}\sum_{s_{0},\ldots,s_{n-1}}\theta(g_{0},\ldots,g_{n-1})(\eta|_{X_{g_{0},s_{0}}\cap
g_{0}X_{g_{1},s_{1}}\cap\cdots\cap(g_{0}\cdots g_{n-2})X_{g_{n-1},s_{n-1}}})$
$\displaystyle=\theta(g_{0},\ldots,g_{n-1})(\eta|_{\bigsqcup_{s_{0},\ldots,s_{n-1}}(X_{g_{0},s_{0}}\cap
g_{0}X_{g_{1},s_{1}}\cap\cdots\cap(g_{0}\cdots g_{n-2})X_{g_{n-1},s_{n-1}})})$
$\displaystyle=\theta(g_{0},\ldots,g_{n-1})(\eta).$

The last equality holds as

$X=\bigsqcup_{s_{0},\ldots,s_{n-1}}(X_{g_{0},s_{0}}\cap
g_{0}X_{g_{1},s_{1}}\cap\cdots\cap(g_{0}\cdots g_{n-2})X_{g_{n-1},s_{n-1}}),$

which can be checked directly. ∎

##  5\. Proof of Theorem 1.2

The idea for the proof is similar to the previous one: For each $n$, we will
construct a (linear) map $S^{n}$ from $C_{b}(H^{n},N_{0}(H,Y)^{**})$ to
$C_{b}(G^{n},N_{0}(G,X)^{**})$ and check directly $S^{n}$ is a cochain map,
i.e. $\partial^{n-1}S^{n-1}=S^{n}\partial^{n-1}$ holds. Then, we use symmetry
to find the inverse of $S^{n}$. Since the proof of general case uses
cumbersome notations, we include the proof of the initial cases ($n=1,2,3$) to
illustrate the main ideas.

###  5.1. Case $n=1$

Recall that $C_{b}(H^{0},N_{0}(H,Y)^{**})=N_{0}(H,Y)^{**}$. We define
$S^{0}=\pi^{**}$ and $S^{1}(f)=f^{\prime}$, where

$f^{\prime}(g)(\tau):=\sum_{i}\langle\pi^{**}(f(h_{i})),\tau_{i}\rangle.$

Here, $f\in C_{b}(H,N_{0}(H,Y)^{**})$, $\tau\in N_{0}(G,X)^{*}$ and
$\tau_{i}=\tau|_{X_{i}}$. Moreover, we write $X_{i}=\\{x\in
X:~{}c(g^{-1},x)=h_{i}^{-1}\\}$, where $c(g,X)=\\{h_{i}|~{}i\\}$.

Step 1: check $S^{1}$ is well-defined, i.e. $\sup_{g\in
G}||f^{\prime}(g)||<\infty$.

Since $\langle\pi^{**}(f(h_{i})),\tau_{i}\rangle=f(h_{i})(\tau_{i}\pi)$, it
suffices to show that $\sum_{i}||\tau_{i}||/||\tau||$ is bounded (and
independent of the choice of $g$, $\\{X_{i}\\}$ and $n:=\\#c(g,X)$). This is
done by the following lemma.

######  Lemma 5.1.

Let $\tau\in N_{0}(G,X)^{*}$ and $X=\sqcup_{i}X_{i}$ be a finite clopen
partition. Then $\sum_{i}||\tau_{i}||\leq||\tau||$.

###### Proof.

For any $\epsilon>0$, take $a_{i}\in N_{0}(G,X)$ such that $||a_{i}||=1$ and
$||\tau_{i}||\leq|\tau_{i}(a_{i})|+\epsilon/n$ for all $i$.

Then
$\sum_{i}||\tau_{i}||\leq\sum_{i}|\tau_{i}(a_{i})|+\epsilon=\sum_{i}\tau(a_{i}|_{X_{i}}\cdot\lambda_{i})+\epsilon=\tau(\sum_{i}(a_{i}|_{X_{i}}\cdot\lambda_{i}))+\epsilon$,
where $\lambda_{i}\in\\{\pm 1\\}$ is the sign of $\tau_{i}(a_{i})$.

Observe that $\sum_{i}(a_{i}|_{X_{i}}\cdot\lambda_{i})\in N_{0}(G,X)$ and
$||\sum_{i}(a_{i}|_{X_{i}}\cdot\lambda_{i})||\leq 1$. Hence,
$\sum_{i}||\tau_{i}||\leq||\tau||+\epsilon$ for all $\epsilon>0$. So,
$\sum_{i}||\tau_{i}||\leq||\tau||$. ∎

Step 2: check $\partial^{0}S^{0}=S^{1}\partial^{0}$.

Take any $\xi\in N_{0}(H,Y)^{**}$, $g\in G$ and $\tau\in N_{0}(G,X)^{*}$. We
do computation as follows.

$\displaystyle(S^{1}\partial^{0}\xi)(g)(\tau)$
$\displaystyle=\sum_{i}\langle\pi^{**}((\partial^{0}\xi)(h_{i})),\tau_{i}\rangle\qquad(\mbox{def.
of $S^{1}$})$
$\displaystyle=\sum_{i}\langle(\partial^{0}\xi)(h_{i}),\tau_{i}\pi\rangle\qquad(\mbox{def.
of $\pi^{**}$})$ $\displaystyle=\sum_{i}\langle
h_{i}\xi,\tau_{i}\pi\rangle\qquad(\mbox{def. of $\partial^{0}$})$
$\displaystyle=\sum_{i}\langle\xi,h_{i}^{-1}(\tau_{i}\pi)\rangle.\qquad(\mbox{def.
of dual action})$ $\displaystyle(\partial^{0}S^{0}\xi)(g)(\tau)$
$\displaystyle=[g(S^{0}\xi)](\tau)\qquad(\mbox{def. of $\partial^{0}$})$
$\displaystyle=(S^{0}\xi)(g^{-1}\tau)\qquad(\mbox{def. of dual action})$
$\displaystyle=\langle\pi^{**}\xi,g^{-1}\tau\rangle\qquad(\mbox{$S^{0}=\pi^{**}$})$
$\displaystyle=\langle\xi,(g^{-1}\tau)\pi\rangle\qquad(\mbox{def. of
$\pi^{**}$})$
$\displaystyle=\sum_{i}\langle\xi,(g^{-1}\tau_{i})\pi\rangle.\qquad(\mbox{$\tau=\sum_{i}\tau_{i}$})$

Therefore, to show the above two equations are equal, it suffices to check
that for each $i$, we have $h_{i}^{-1}(\tau_{i}\pi)=(g^{-1}\tau_{i})\pi$.
Equivalently, we need to check that
$\tau(\pi(h_{i}\xi)|_{X_{i}})=\tau([g(\pi(\xi))]|_{X_{i}})$ for each $\xi\in
N_{0}(H,Y)$, which is clear by Lemma 3.3.

Clearly, $S^{0}$ is a bijection by Lemma 3.2.

###  5.2. Case $n=2$

Now, we construct a (linear) map $S^{2}:C_{b}(H^{2},N_{0}(H,Y)^{**})\to
C_{b}(G^{2},N_{0}(G,X)^{**})$.

Write

(20) $\displaystyle\begin{split}X_{i}&=\\{x:c(g_{0}^{-1},x)=h_{i}^{-1}\\},\\\
X_{j}&=\\{x:c(g_{1}^{-1},x)=h_{j}^{-1}\\},\\\
X_{k}&=\\{x:c((g_{0}g_{1})^{-1},x)=h_{k}^{-1}\\},\\\
X_{s}&=\\{x:c((g_{1}g_{2})^{-1},x)=h_{s}^{-1}\\},\\\
X_{l}&=\\{x:c(g_{2}^{-1},x)=h_{l}^{-1}\\}.\end{split}$

Given $f\in C_{b}(H^{2},N_{0}(H,Y)^{**})$, we define $S^{2}(f)=f^{\prime}$,
where for each $\tau\in N_{0}(G,X)^{*}$,

$f^{\prime}(g_{0},g_{1})(\tau):=\sum_{i,j}\langle
f(h_{i},h_{j}),\tau|_{g_{0}X_{j}\cap X_{i}}\pi\rangle.$

Step 1: $S^{2}$ is well-defined.

To show $\sup_{(g_{0},g_{1})\in G^{2}}||f^{\prime}(g_{0},g_{1})||<\infty$,
apply Lemma 5.1 to $X=\sqcup_{i,j}g_{0}X_{j}\cap X_{i}$.

Step 2: check $\partial^{1}S^{1}=S^{2}\partial^{1}$.

Take any $f\in C_{b}(H,N_{0}(H,Y)^{**})$, write $f^{\prime}=S^{1}f$. For any
$g_{1},g_{2}\in G$ and $\tau\in N_{0}(G,X)^{*}$, we have

$\displaystyle\quad(S^{2}\partial^{1}f)(g_{1},g_{2})(\tau)$
$\displaystyle=\sum_{j,l}\langle(\partial^{1}f)(h_{j},h_{l}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle\qquad(\mbox{def. of $S^{2}$})$
$\displaystyle=\sum_{j,l}\langle
h_{j}f(h_{l})-f(h_{j}h_{l})+f(h_{j}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle\qquad(\mbox{def. of $\partial^{1}$})$
$\displaystyle=\sum_{j,l}\langle h_{j}f(h_{l}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle-\sum_{j,l}\langle f(h_{j}h_{l}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle+\sum_{j,l}\langle f(h_{j}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle.$ $\displaystyle\quad(\partial^{1}S^{1}f)(g_{1},g_{2})(\tau)$
$\displaystyle=(g_{1}f^{\prime}(g_{2})-f^{\prime}(g_{1}g_{2})+f^{\prime}(g_{1}))(\tau)\qquad(\mbox{def.
of $\partial^{1}$})$
$\displaystyle=\sum_{l}\langle\pi^{**}(f(h_{l})),(g_{1}^{-1}\tau)_{l}\rangle-\sum_{s}\langle\pi^{**}(f(h_{s})),\tau_{s}\rangle+\sum_{j}\langle\pi^{**}(f(h_{j})),\tau_{j}\rangle.$

By comparing the two expressions above, it suffices to show each corresponding
terms are equal, i.e.

(21) $\displaystyle\sum_{j,l}\langle h_{j}f(h_{l}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle$
$\displaystyle=\sum_{l}\langle\pi^{**}(f(h_{l})),(g_{1}^{-1}\tau)_{l}\rangle,$
(22) $\displaystyle\sum_{j,l}\langle f(h_{j}h_{l}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle$
$\displaystyle=\sum_{s}\langle\pi^{**}(f(h_{s})),\tau_{s}\rangle,$ (23)
$\displaystyle\sum_{j,l}\langle f(h_{j}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle$
$\displaystyle=\sum_{j}\langle\pi^{**}(f(h_{j})),\tau_{j}\rangle.$

To check (21), it suffices to show for each $l$,

$\sum_{j}\langle h_{j}f(h_{l}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle=\langle\pi^{**}(f(h_{l})),(g_{1}^{-1}\tau)_{l}\rangle.$

Equivalently, we check $\sum_{j}h_{j}^{-1}(\tau|_{g_{1}X_{l}\cap
X_{j}}\pi)=(g_{1}^{-1}\tau)_{l}\pi$ holds.

Take any $\xi\in N_{0}(H,Y)$, observe that

$\displaystyle\sum_{j}h_{j}^{-1}(\tau|_{g_{1}X_{l}\cap X_{j}}\pi)(\xi)$
$\displaystyle=\sum_{j}(\tau|_{g_{1}X_{l}\cap
X_{j}}\pi)(h_{j}\xi)\qquad(\mbox{def. of dual action})$
$\displaystyle=\sum_{j}\tau(\pi(h_{j}\xi)|_{g_{1}X_{l}\cap X_{j}})$
$\displaystyle=\sum_{j}\tau((g_{1}\pi(\xi))|_{g_{1}X_{l}\cap
X_{j}}))\qquad(\mbox{Lemma \ref{lem: module structure under restriction}})$
$\displaystyle=\tau((g_{1}\pi(\xi))|_{\sqcup_{j}g_{1}X_{l}\cap X_{j}}))$
$\displaystyle=\tau((g_{1}\pi(\xi))|_{g_{1}X_{l}})\qquad(\mbox{$X=\sqcup_{j}X_{j}$})$
$\displaystyle=\tau(g_{1}(\pi(\xi)|_{X_{l}}))$
$\displaystyle=(g_{1}^{-1}\tau)_{l}\pi(\xi).$

Hence, $\sum_{j}h_{j}^{-1}(\tau|_{g_{1}X_{l}\cap
X_{j}}\pi)=(g_{1}^{-1}\tau)_{l}\pi$ holds.

To check (22), observe that by the cocycle identity, $X_{s}=\sqcup_{(j,l)\in
I_{s}}(g_{1}X_{l}\cap X_{j})$ holds, where
$I_{s}:=\\{(j,l):h_{s}=h_{j}h_{l}\\}$.

LHS of (22) $\displaystyle=\sum_{s}\sum_{(j,l)\in I_{s}}\langle
f(h_{j}h_{l}),\tau|_{g_{1}X_{l}\cap X_{j}}\pi\rangle~{}(\mbox{as
$g_{1}X_{l}\cap X_{j}\neq\emptyset$ only if $(j,l)\in I_{s}$ for some $s$})$
$\displaystyle=\sum_{s}\sum_{(j,l)\in I_{s}}\langle
f(h_{s}),\tau|_{g_{1}X_{l}\cap X_{j}}\pi\rangle$
$\displaystyle=\sum_{s}\langle f(h_{s}),\tau|_{\sqcup_{(j,l)\in
I_{s}}g_{1}X_{l}\cap X_{j}}\pi\rangle$ $\displaystyle=\sum_{s}\langle
f(h_{s}),\tau|_{X_{s}}\rangle~{}(\mbox{as $X_{s}=\sqcup_{(j,l)\in
I_{s}}(g_{1}X_{l}\cap X_{j})$})$ $\displaystyle=\sum_{s}\langle
f(h_{s}),\tau_{s}\pi\rangle=\mbox{RHS of (\ref{eq 2 for n=1, thm 1.2})}.$

To check (23), it suffices to show for each $j$, $\sum_{l}\langle
f(h_{j}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle=\langle\pi^{**}(f(h_{j})),\tau_{j}\rangle$. Observe that

$\displaystyle\sum_{l}\langle f(h_{j}),\tau|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle=\langle f(h_{j}),\tau|_{\sqcup_{l}g_{1}X_{l}\cap
X_{j}}\pi\rangle=\langle
f(h_{j}),\tau|_{X_{j}}\pi\rangle=\langle\pi^{**}(f(h_{j})),\tau_{j}\rangle.$

Step 3: check that $S^{1}$ is a bijection.

For this purpose, we define a map $T^{1}:C_{b}(G,N_{0}(G,X)^{**})\to
C_{b}(H,N_{0}(H,Y)^{**})$ by setting $T^{1}(z)=z^{\prime}$, where
$z^{\prime}(h)(\nu):=\sum_{i}\langle L^{**}(z(g_{i})),\nu_{i}\rangle$. Here
$\nu\in N_{0}(H,Y)^{*}$,
$\nu_{i}=\nu|_{Y_{i}},Y_{i}=\\{y:c^{\prime}(h^{-1},y)=g_{i}^{-1}\\}$.

Now, we check $T^{1}S^{1}=id$ holds. $S^{1}T^{1}=id$ can be checked similarly
by symmetry.

Take any $f\in C_{b}(H,N_{0}(H,Y)^{**})$, $h\in H$ and $\nu\in
N_{0}(H,Y)^{*}$, write $f^{\prime}=S^{1}f$.

$\displaystyle(T^{1}S^{1}f)(h)(\nu)$ $\displaystyle=\sum_{i}\langle
L^{**}(f^{\prime}(g_{i})),\nu_{i}\rangle=\sum_{i}\langle
f^{\prime}(g_{i}),\nu_{i}L\rangle$
$\displaystyle=\sum_{i}\sum_{j}\langle\pi^{**}(f(h_{j})),(\nu_{i}L)_{j}\rangle$
(Here, $(\nu_{i}L)_{j}:=(\nu_{i}L)|_{X_{g_{i},h_{j}}}$,
$X_{g_{i},h_{j}}=\\{x:c(g_{i}^{-1},x)=h_{j}^{-1}\\}.$)
$\displaystyle=\sum_{i}\sum_{j}\langle f(h_{j}),(\nu_{i}L)_{j}\pi\rangle.$

Now, we claim that $(\nu_{i}L)_{j}\pi=\nu|_{\phi(X_{g_{i},h_{j}})\cap Y_{i}}$.
To see this, take any $\xi\in N_{0}(H,Y)$, then

$\displaystyle(\nu_{i}L)_{j}\pi(\xi)$
$\displaystyle=(\nu_{i}L)(\pi(\xi)|_{X_{g_{i},h_{j}}})=\nu([L(\pi(\xi)|_{X_{g_{i},h_{j}}})]|_{Y_{i}})$
$\displaystyle=\nu(\xi|_{\phi(X_{g_{i},h_{j}})\cap Y_{i}})\qquad(\mbox{Lemma
\ref{lem: composition of pi and L under restrictions}})$
$\displaystyle=(\nu|_{\phi(X_{g_{i},h_{j}})\cap Y_{i}})(\xi).$

Note that $\phi(X_{g_{i},h_{j}})\cap Y_{i}=Y_{i}$ if $h_{j}=h$ and $\emptyset$
otherwise. Therefore,

$(T^{1}S^{1}f)(h)(\nu)=\sum_{i}\sum_{j}\langle
f(h_{j}),(\nu_{i}L)_{j}\pi\rangle=\sum_{i}\langle f(h),\nu_{i}\rangle=\langle
f(h),\sum_{i}\nu_{i}\rangle=f(h)(\nu).$

So, $T^{1}S^{1}=id$. By symmetry, $S^{1}T^{1}=id$.

###  5.3. Case $n=3$

Now, we construct a (linear) map $S^{3}:C_{b}(H^{3},N_{0}(H,Y)^{**})\to
C_{b}(G^{3},N_{0}(G,X)^{**})$. For each $f\in C_{b}(H^{3},N_{0}(H,Y)^{**})$,
define $S^{3}(f)=f^{\prime}$, where

$\displaystyle f^{\prime}(g_{0},g_{1},g_{2})(\tau):=\sum_{i,j,l}\langle
f(h_{i},h_{j},h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle.$

Here, $\tau\in N_{0}(H,Y)^{*}$ and we still use (20) for the definition of
$X_{i},X_{j},X_{k},X_{l},X_{s}$.

Step 1: check $S^{3}$ is well-defined.

The proof is similar to the one when showing $S^{1}$ is well-defined.

Step 2: check $S^{3}\partial^{2}=\partial^{2}S^{2}$.

Take any $f\in C_{b}(H^{3},N_{0}(H,Y)^{**})$, write $f^{\prime}=S^{2}f$, we
have

$\displaystyle\quad(S^{3}\partial^{2}f)(g_{0},g_{1},g_{2})(\tau)$
$\displaystyle=\sum_{i,j,l}\langle(\partial^{2}f)(h_{i},h_{j},h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap
g_{0}X_{j}\cap X_{i}}\pi\rangle$ $\displaystyle=\sum_{i,j,l}\langle
h_{i}f(h_{j},h_{l})-f(h_{i}h_{j},h_{l})+f(h_{i},h_{j}h_{l})-f(h_{i},h_{j}),\tau|_{(g_{0}g_{1})X_{l}\cap
g_{0}X_{j}\cap X_{i}}\pi\rangle$ $\displaystyle=\sum_{i,j,l}\langle
h_{i}f(h_{j},h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle-\sum_{i,j,l}\langle
f(h_{i}h_{j},h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle+\sum_{i,j,l}\langle
f(h_{i},h_{j}h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle-\sum_{i,j,l}\langle
f(h_{i},h_{j}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap X_{i}}\pi\rangle$
$\displaystyle=:①-②+③-④.$

Here, $①,②,③$ and $④$ denote the corresponding terms appeared above.

$\displaystyle\quad(\partial^{2}S^{2}f)(g_{0},g_{1},g_{2})(\tau)$
$\displaystyle=(g_{0}f^{\prime}(g_{1},g_{2})-f^{\prime}(g_{0}g_{1},g_{2})+f^{\prime}(g_{0},g_{1}g_{2})-f^{\prime}(g_{0},g_{1}))(\tau)$
$\displaystyle=\sum_{j,l}\langle
f(h_{j},h_{l}),(g_{0}^{-1}\tau)|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle-\sum_{k,l}\langle f(h_{k},h_{l}),\tau|_{g_{0}g_{1}X_{l}\cap
X_{k}}\pi\rangle$ $\displaystyle+\sum_{i,s}\langle
f(h_{i},h_{s}),\tau|_{g_{0}X_{s}\cap X_{i}}\pi\rangle-\sum_{i,j}\langle
f(h_{i},h_{j}),\tau|_{g_{0}X_{j}\cap X_{i}}\pi\rangle$
$\displaystyle=:⑤-⑥+⑦-⑧.$

Now, in order to prove $①-②+③-④=⑤-⑥+⑦-⑧$, it suffices to check the
corresponding terms are equal, i.e. $①=⑤$, $②=⑥$, $③=⑦$ and $④=⑧$. More
precisely, we want to prove the following equalities hold.

(24) $\displaystyle\sum_{i,j,l}\langle
h_{i}f(h_{j},h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\sum_{j,l}\langle
f(h_{j},h_{l}),(g_{0}^{-1}\tau)|_{g_{1}X_{l}\cap X_{j}}\pi\rangle,$ (25)
$\displaystyle\sum_{i,j,l}\langle
f(h_{i}h_{j},h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\sum_{k,l}\langle
f(h_{k},h_{l}),\tau|_{g_{0}g_{1}X_{l}\cap X_{k}}\pi\rangle,$ (26)
$\displaystyle\sum_{i,j,l}\langle
f(h_{i},h_{j}h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\sum_{i,s}\langle
f(h_{i},h_{s}),\tau|_{g_{0}X_{s}\cap X_{i}}\pi\rangle,$ (27)
$\displaystyle\sum_{i,j,l}\langle f(h_{i},h_{j}),\tau|_{(g_{0}g_{1})X_{l}\cap
g_{0}X_{j}\cap X_{i}}\pi\rangle$ $\displaystyle=\sum_{i,j}\langle
f(h_{i},h_{j}),\tau|_{g_{0}X_{j}\cap X_{i}}\pi\rangle.$

To check (24), it suffices to check for each $j,l$,

(28) $\displaystyle\sum_{i}\langle
h_{i}f(h_{j},h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle=\langle f(h_{j},h_{l}),(g_{0}^{-1}\tau)|_{g_{1}X_{l}\cap
X_{j}}\pi\rangle.$

Observe that

(29) $\displaystyle h_{i}^{-1}(\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi)=(g_{0}^{-1}\tau)|_{g_{1}X_{l}\cap X_{j}\cap g_{0}^{-1}X_{i}}\pi.$

Indeed, take any $\xi\in N_{0}(H,Y)$, and write $X_{0}:=(g_{0}g_{1})X_{l}\cap
g_{0}X_{j}\cap X_{i}$, we have
$(h_{i}^{-1}(\tau|_{X_{0}}\pi))\xi=\tau(\pi(h_{i}\xi)|_{X_{0}})\overset{Lem.\ref{lem:
module structure under
restriction}}{=}\tau(g_{0}(\pi(\xi)|_{g_{0}^{-1}X_{0}}))=[(g_{0}^{-1}\tau)|_{g_{0}^{-1}X_{0}}\pi](\xi)$.

Therefore,

LHS of (28) $\displaystyle=\sum_{i}\langle
h_{i}f(h_{j},h_{l}),\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\sum_{i}\langle
f(h_{j},h_{l}),h_{i}^{-1}(\tau|_{(g_{0}g_{1})X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi)\rangle$ $\displaystyle\overset{(\ref{middle eq for thm 1.2,
n=2})}{=}\sum_{i}\langle f(h_{j},h_{l}),(g_{0}^{-1}\tau)|_{g_{1}X_{l}\cap
X_{j}\cap g_{0}^{-1}X_{i}}\pi)\rangle$ $\displaystyle=\langle
f(h_{j},h_{l}),(g_{0}^{-1}\tau)|_{\sqcup_{i}(g_{1}X_{l}\cap X_{j}\cap
g_{0}^{-1}X_{i})}\pi\rangle$ $\displaystyle=\langle
f(h_{j},h_{l}),(g_{0}^{-1}\tau)|_{g_{1}X_{l}\cap X_{j}}\pi\rangle$
$\displaystyle=\mbox{RHS of (\ref{eq9})}.$

To check (25), observe $X_{k}=\sqcup_{(i,j)\in I_{k}}g_{0}X_{j}\cap X_{i}$,
where $I_{k}=\\{(i,j):h_{k}=h_{i}h_{j}\\}$. So,

RHS of (25) $\displaystyle=\sum_{k,l}\sum_{(i,j)\in I_{k}}\langle
f(h_{i}h_{j},h_{l}),\tau|_{g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\sum_{l}\sum_{k}\sum_{(i,j)\in I_{k}}\langle
f(h_{i}h_{j},h_{l}),\tau|_{g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\sum_{l,i,j}\langle
f(h_{i}h_{j},h_{l}),\tau|_{g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\mbox{LHS of (\ref{eq 2= eq 6 in thm 1.2
n=2})}.$

Here, the 2nd last equality holds as $g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}\neq\emptyset$ only if $(i,j)\in I_{k}$ for some $k$.

To check (26), use $X_{s}=\sqcup_{(j,l)\in I_{s}}g_{1}X_{l}\cap X_{j}$, where
$I_{s}=\\{(j,l):h_{s}=h_{j}h_{l}\\}$ to deduce

RHS of (26) $\displaystyle=\sum_{i,s}\langle
f(h_{i},h_{s}),\tau|_{g_{0}X_{s}\cap X_{i}}\pi\rangle$
$\displaystyle=\sum_{i,s}\sum_{(j,l)\in I_{s}}\langle
f(h_{i},h_{j}h_{l}),\tau|_{g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\sum_{k,l}\sum_{(i,j)\in I_{k}}\langle
f(h_{i},h_{j}h_{l}),\tau|_{g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\sum_{l,i,j}\langle
f(h_{i},h_{j}h_{l}),\tau|_{g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}}\pi\rangle$ $\displaystyle=\mbox{LHS of (\ref{eq 3= eq 7 in thm 1.2
n=2})}.$

The 3rd last equality holds since we have a bijection between the index sets:

$\\{(i,s,j,l):h_{s}=h_{j}h_{l},g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}\neq\emptyset\\}$

and

$\\{(k,l,i,j):h_{k}=h_{i}h_{j},g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}\neq\emptyset\\}.$

And the 2nd last equality holds as $g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap
X_{i}\neq\emptyset$ only if $(i,j)\in I_{k}$ for some $k$.

To check (27), just observe $g_{0}X_{j}\cap
X_{i}=\sqcup_{l}(g_{0}g_{1}X_{l}\cap g_{0}X_{j}\cap X_{i}).$

Step 3: $S^{2}$ is a bijection.

Define $T^{2}:C_{b}(G^{2},N_{0}(G,X)^{**})\to C_{b}(H^{2},N_{0}(H,Y)^{**})$ by
setting $T^{2}(f^{\prime})=f^{\prime\prime}$, where

$f^{\prime\prime}(h_{0},h_{1})(\nu)=\sum_{i,j}\langle
f^{\prime}(g_{i},g_{j}),\nu|_{h_{0}Y_{j}\cap Y_{i}}L\rangle.$

Here $\nu\in N_{0}(H,Y)^{*}$, and

$\displaystyle Y_{i}$ $\displaystyle:=\\{y\in
Y;c^{\prime}(h_{0}^{-1},y)=g_{i}^{-1}\\},$ $\displaystyle Y_{j}$
$\displaystyle:=\\{y\in Y;c^{\prime}(h_{1}^{-1},y)=g_{j}^{-1}\\}.$

Now, let us check $T^{2}S^{2}=id$.

Take any $f\in C_{b}(H^{2},N_{0}(H,Y)^{**})$, $\nu\in N_{0}(H,Y)^{*}$, let
$f^{\prime}=S^{2}f$ and $f^{\prime\prime}=T^{2}f^{\prime}$, we aim to show
$f^{\prime\prime}(h_{0},h_{1})(\nu)=f(h_{0},h_{1})(\nu)$.

A calculation shows

$\displaystyle f^{\prime\prime}(h_{0},h_{1})(\nu)$
$\displaystyle=\sum_{i,j}\sum_{s,t}\langle
f(h_{s},h_{t}),(\nu|_{h_{0}Y_{j}\cap Y_{i}}L)|_{g_{i}X_{g_{j},h_{t}}\cap
X_{g_{i},h_{s}}}\pi\rangle$ $\displaystyle=\sum_{i,j}\sum_{s,t}\langle
f(h_{s},h_{t}),\nu|_{h_{0}Y_{j}\cap Y_{i}\cap\phi(g_{i}X_{g_{j},h_{t}}\cap
X_{g_{i},h_{s}})}\rangle.$

Here, $X_{g_{i},h_{s}}=\\{x\in X:c(g_{i}^{-1},x)=h_{s}^{-1}\\}$ and
$X_{g_{j},h_{t}}=\\{x\in X:c(g_{j}^{-1},x)=h_{t}^{-1}\\}$.

One can check that $h_{0}Y_{j}\cap Y_{i}\cap\phi(g_{i}X_{g_{j},h_{t}}\cap
X_{g_{i},h_{s}})=\emptyset$ unless $(h_{s},h_{t})=(h_{0},h_{1})$. If this
condition holds, then $h_{0}Y_{j}\cap Y_{i}\cap\phi(g_{i}X_{g_{j},h_{1}}\cap
X_{g_{i},h_{0}})=h_{0}Y_{j}\cap Y_{i}$.

Hence,

$\displaystyle f^{\prime\prime}(h_{0},h_{1})(\nu)=\sum_{i,j}\langle
f(h_{0},h_{1}),\nu|_{h_{0}Y_{j}\cap Y_{i}}\rangle=f(h_{0},h_{1})(\nu).$

###  5.4. General case

We’ll define a map $S^{n}:C_{b}(H^{n},N_{0}(H,Y)^{**})\to
C_{b}(G^{n},N_{0}(G,X)^{**})$ and check it is a bijective cochain map. Hence,
it induces an isomorphism between the two bounded cohomology groups.

We introduce some notations which will be used in this subsection.

Let $g_{0},\ldots,g_{n}$ be elements in $G$. Write

(30)
$\displaystyle\begin{split}c(g_{j}^{-1},X)&:=\\{h_{t_{j}}^{-1}|~{}t_{j}\\},\\\
X_{t_{j}}&:=\\{x\in X:c(g_{j}^{-1},x)=h_{t_{j}}^{-1}\\},\\\
c((g_{i}g_{i+1})^{-1},X)&:=\\{h_{s_{i}}^{-1}|~{}s_{i}\\},\\\
X_{s_{i}}&:=\\{x\in X:c((g_{i}g_{i+1})^{-1},x)=h_{s_{i}}^{-1}\\}.\end{split}$

Note that $X_{s_{i}}=\sqcup_{(t_{i},t_{i+1})\in\Delta_{i}}X_{t_{i}}\cap
g_{i}X_{t_{i+1}}$, where
$\Delta_{i}:=\\{(t_{i},t_{i+1}):h_{s_{i}}=h_{t_{i}}h_{t_{i+1}}\\}$.

We remind the reader once again that $X_{t_{j}}$, $X_{s_{i}}$ and $\Delta_{i}$
are just simplified versions for the notations $X_{g_{j},h_{t_{j}}}$,
$X_{g_{i}g_{i+1},h_{s_{i}}}$ and $\Delta_{s_{i}}$ respectively, which we will
switch to if necessary.

Let $f\in C_{b}(H^{n},N_{0}(H,Y)^{**})$, we define $S^{n}(f):=f^{\prime}\in
C_{b}(G^{n},N_{0}(G,X)^{**})$, where

$f^{\prime}(g_{0},\dots,g_{n-1})(\tau):=\sum_{t_{0},\dots,t_{n-1}}\langle
f(h_{t_{0}},\dots,h_{t_{n-1}}),\tau|_{[t_{0},\ldots,t_{n-1}]}\circ\pi\rangle$

for all $g_{i}\in G$ and $\tau\in N_{0}(G,X)^{*}$.

Here,

(31) $\displaystyle[t_{0},\ldots,t_{n-1}]:=X_{t_{0}}\cap g_{0}X_{t_{1}}\cap
g_{0}g_{1}X_{t_{2}}\cap\dots\cap(g_{0}\cdots g_{n-2})X_{t_{n-1}}.$

We split the proof into several steps.

Step 1: $S^{n}$ is well-defined.

Apply Lemma 5.1 to $X=\sqcup_{t_{0},\ldots,t_{n-1}}[t_{0},\ldots,t_{n-1}]$.

Step 2: $S^{n}$ is a cochain map, i.e.
$S^{n+1}\partial^{n}=\partial^{n}S^{n}$.

Fix $f\in C_{b}(H^{n},N_{0}(H,Y)^{**})$, let $f^{\prime}=S^{n}f$. We aim to
show $S^{n+1}\partial^{n}f=\partial^{n}f^{\prime}$.

Take any $g_{0},\ldots,g_{n}$ in $G$, and any $\tau\in N_{0}(G,X)^{*}$. We
compute as follows.

First,

(32)
$\displaystyle\begin{split}&\qquad\partial^{n}f^{\prime}(g_{0},\ldots,g_{n})(\tau)\\\
&=[g_{0}f^{\prime}(g_{1},\ldots,g_{n})+\sum_{i=1}^{n}(-1)^{i}f^{\prime}(g_{0},\ldots,g_{i-1}g_{i},\ldots,g_{n})+(-1)^{n+1}f^{\prime}(g_{0},\ldots,g_{n-1})](\tau)\\\
&=\sum_{t_{1},\ldots,t_{n}}\langle
f(h_{t_{1}},\ldots,h_{t_{n}}),(g_{0}^{-1}\tau)|_{[t_{1},\ldots,t_{n}]}\pi\rangle\\\
&\quad+\sum_{i=1}^{n}(-1)^{i}\sum_{\begin{subarray}{c}t_{0},\ldots,t_{i-2},\\\
s_{i-1},t_{i+1},\ldots,t_{n}\end{subarray}}\langle
f(h_{t_{0}},\ldots,h_{t_{i-2}},h_{s_{i-1}},h_{t_{i+1}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{i-2},s_{i-1},t_{i+1},\ldots,t_{n}]}\pi\rangle\\\
&\quad+(-1)^{n+1}\sum_{t_{0},\ldots,t_{n-1}}\langle
f(h_{t_{0}},\ldots,h_{t_{n-1}}),\tau|_{[t_{0},\ldots,t_{n-1}]}\pi\rangle.\end{split}$

Recall here, $c(g_{i-1}g_{i},X)=\\{h_{s_{i-1}}|~{}s_{i-1}\\}$,
$X_{s_{i-1}}=\\{x:c((g_{i-1}g_{i})^{-1},x)=h_{s_{i-1}}^{-1}\\}$ and

(33) $[t_{0},\ldots,t_{i-2},s_{i-1},t_{i+1},\ldots,t_{n}]=X_{t_{0}}\cap
g_{0}X_{t_{1}}\cap\cdots\cap(g_{0}\cdots g_{i-2})X_{s_{i-1}}\\\
\cap(g_{0}\cdots g_{i-2}g_{i-1}g_{i})X_{t_{i+1}}\cap\cdots\cap(g_{0}\dots
g_{n-1})X_{t_{n}}.$

Second,

(34)
$\displaystyle\begin{split}&\qquad(S^{n+1}\partial^{n}f)(g_{0},\ldots,g_{n})(\tau)\\\
&=\sum_{t_{0},\ldots,t_{n}}\langle\partial^{n}(f)(h_{t_{0}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{n}]}\pi\rangle\\\
&=\sum_{t_{0},\ldots,t_{n}}\langle
h_{t_{0}}f(h_{t_{1}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{n}]}\pi\rangle\\\
&\quad+\sum_{t_{0},\ldots,t_{n}}\langle\sum_{i=1}^{n}(-1)^{i}f(h_{t_{0}},\ldots,h_{t_{i-1}}h_{t_{i}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{n}]}\pi\rangle\\\
&\quad+\sum_{t_{0},\ldots,t_{n}}\langle(-1)^{n+1}f(h_{t_{0}},\ldots,h_{t_{n-1}}),\tau|_{[t_{0},\ldots,t_{n}]}\pi\rangle.\end{split}$

By comparing (34) with (32), it suffices to show that the corresponding terms
are equal, i.e. we want to establish the following identities.

(a)

$\displaystyle\sum_{t_{1},\ldots,t_{n}}\langle
f(h_{t_{1}},\ldots,h_{t_{n}}),(g_{0}^{-1}\tau)|_{[t_{1},\ldots,t_{n}]}\pi\rangle=\sum_{t_{0},\ldots,t_{n}}\langle
h_{t_{0}}f(h_{t_{1}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{n}]}\pi\rangle.$

(b) For each $1\leq i\leq n$,

$\sum_{\begin{subarray}{c}t_{0},\ldots,t_{i-2},\\\
s_{i-1},t_{i+1},\ldots,t_{n}\end{subarray}}\langle
f(h_{t_{0}},\ldots,h_{t_{i-2}},h_{s_{i-1}},h_{t_{i+1}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{i-2},s_{i-1},t_{i+1},\ldots,t_{n}]}\pi\rangle\\\
=\sum_{t_{0},\ldots,t_{n}}\langle
f(h_{t_{0}},\ldots,h_{t_{i-1}}h_{t_{i}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{n}]}\pi\rangle.$

(c)

$\displaystyle\sum_{t_{0},\ldots,t_{n-1}}\langle
f(h_{t_{0}},\ldots,h_{t_{n-1}}),\tau|_{[t_{0},\ldots,t_{n-1}]}\pi\rangle=\sum_{t_{0},\ldots,t_{n}}\langle(-1)^{n+1}f(h_{t_{0}},\ldots,h_{t_{n-1}}),\tau|_{[t_{0},\ldots,t_{n}]}\pi\rangle.$

###### Proof of (a).

By Lemma 3.4,
$(g_{0}^{-1}\tau)|_{g_{0}^{-1}X_{t_{0}}\cap[t_{1},\ldots,t_{n}]}\pi=h_{t_{0}}^{-1}(\tau|_{[t_{0},\ldots,t_{n}]}\pi)$
holds.

Indeed, evaluate both sides at any $\xi$ in $N_{0}(H,Y)$, this boils down to
check $X_{t_{0}}\cap g_{0}[t_{1},\ldots,t_{n}]=[t_{0},\ldots,t_{n}]$, which is
clear from the definition of $[t_{0},\ldots,t_{n}].$

Using this identity, we deduce

RHS of (a) $\displaystyle=\sum_{t_{0},\ldots,t_{n}}\langle
f(h_{t_{1}},\ldots,h_{t_{n}}),h_{t_{0}}^{-1}(\tau|_{[t_{0},\ldots,t_{n}]}\pi)\rangle$
$\displaystyle=\sum_{t_{0},\ldots,t_{n}}\langle
f(h_{t_{1}},\ldots,h_{t_{n}}),(g_{0}^{-1}\tau)|_{g_{0}^{-1}X_{t_{0}}\cap[t_{1},\ldots,t_{n}]}\pi\rangle$
$\displaystyle=\sum_{t_{1},\ldots,t_{n}}\langle
f(h_{t_{1}},\ldots,h_{t_{n}}),(g_{0}^{-1}\tau)|_{\sqcup_{t_{0}}g_{0}^{-1}X_{t_{0}}\cap[t_{1},\ldots,t_{n}]}\pi\rangle$
$\displaystyle=\sum_{t_{1},\ldots,t_{n}}\langle
f(h_{t_{1}},\ldots,h_{t_{n}}),(g_{0}^{-1}\tau)|_{[t_{1},\ldots,t_{n}]}\pi\rangle$
$\displaystyle=\mbox{LHS of (a)}.\qed$

###### Proof of (b).

From $X_{s_{i-1}}=\sqcup_{(t_{i-1},t_{i})\in\Delta_{i-1}}X_{t_{i-1}}\cap
g_{i-1}X_{t_{i}}$, we deduce that

(35)
$\displaystyle[t_{0},\ldots,t_{i-2},s_{i-1},t_{i+1},\ldots,t_{n}]=\sqcup_{(t_{i-1},t_{i})\in\Delta_{i-1}}[t_{0},\ldots,t_{n}].$

Indeed, just compare definition (31) and (33).

Hence,

$\displaystyle\quad\sum_{s_{i-1}}\langle
f(h_{t_{0}},\ldots,h_{t_{i-2}},h_{s_{i-1}},h_{t_{i+1}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{i-2},s_{i-1},t_{i+1},\ldots,t_{n}]}\pi\rangle$
$\displaystyle\overset{(\ref{eq10})}{=}\sum_{s_{i-1}}\sum_{(t_{i-1},t_{i})\in\Delta_{i-1}}\langle
f(h_{t_{0}},\ldots,h_{t_{i-2}},h_{t_{i-1}}h_{t_{i}},h_{t_{i+1}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{n}]}\pi\rangle$
$\displaystyle=\sum_{t_{i-1},t_{i}}\langle
f(h_{t_{0}},\ldots,h_{t_{i-2}},h_{t_{i-1}}h_{t_{i}},h_{t_{i+1}},\ldots,h_{t_{n}}),\tau|_{[t_{0},\ldots,t_{n}]}\pi\rangle.$

The last equality holds since $[t_{0},\ldots,t_{n}]\neq\emptyset$ only if
$(t_{i-1},t_{i})\in\Delta_{i-1}$ for some $s_{i-1}$. (Recall
$\Delta_{i-1}=\Delta_{s_{i-1}}$ for simplicity.)

Therefore, (b) holds by taking sum over
$t_{0},\ldots,t_{i-2},t_{i+1},\ldots,t_{n}$ on both sides of the above
equality. ∎

###### Proof of (c).

As $[t_{0},\ldots,t_{n}]=[t_{0},\ldots,t_{n-1}]\cap(g_{0}\cdots
g_{n-1})X_{t_{n}}$, we get

RHS of (c) $\displaystyle=\sum_{t_{0},\ldots,t_{n}}\langle
f(h_{t_{0}},\ldots,h_{t_{n-1}}),\tau|_{[t_{0},\ldots,t_{n-1}]\cap(g_{0}\cdots
g_{n-1})X_{t_{n}}}\pi\rangle$
$\displaystyle=\sum_{t_{0},\ldots,t_{n-1}}\langle
f(h_{t_{0}},\ldots,h_{t_{n-1}}),\tau|_{[t_{0},\ldots,t_{n-1}]\cap\sqcup_{t_{n}}(g_{0}\cdots
g_{n-1})X_{t_{n}}}\pi\rangle$
$\displaystyle=\sum_{t_{0},\ldots,t_{n-1}}\langle
f(h_{t_{0}},\ldots,h_{t_{n-1}}),\tau|_{[t_{0},\ldots,t_{n-1}]}\pi\rangle$
$\displaystyle=\mbox{LHS of (c)}.\qed$

Step 3: $S^{n}$ is a bijection.

To check $S^{n}$ is a bijection, consider the natural inverse map $T^{n}$
defined by symmetry as follows.

For $f^{\prime}\in C_{b}(G^{n},N_{0}(G,X)^{**})$, define
$f^{\prime\prime}:=T(f^{\prime})$ as follows:

$\displaystyle
f^{\prime\prime}(h_{0},\ldots,h_{n-1})(\nu):=\sum_{i_{0},\ldots,i_{n-1}}\langle
f^{\prime}(g_{i_{0}},\ldots,g_{i_{n-1}}),\nu|_{[i_{0},\ldots,i_{n-1}]}\circ
L\rangle.$

Here, $\nu\in N_{0}(H,Y)^{*}$, $[i_{0},\ldots,i_{n-1}]:=Y_{i_{0}}\cap
h_{0}Y_{i_{1}}\cap\cdots\cap(h_{0}\cdots h_{n-2})Y_{i_{n-1}}$ and
$Y_{i_{j}}:=\\{y:c^{\prime}(h_{j}^{-1},y)=g_{i_{j}}^{-1}\\}$.

Let $f\in C_{b}(H^{n},N_{0}(H,Y)^{**})$, write $f^{\prime}=S^{n}f$ and
$f^{\prime\prime}=T^{n}f^{\prime}$. We aim to show $T^{n}S^{n}=id$, i.e.
$f^{\prime\prime}=f$. Take any $h_{0},\ldots,h_{n-1}$ in $H$ and any $\nu$ in
$N_{0}(H,Y)^{*}$, then

(36) $\displaystyle\begin{split}&\quad
f^{\prime\prime}(h_{0},\ldots,h_{n-1})(\nu)\\\
&=\sum_{i_{0},\ldots,i_{n-1}}\langle
f^{\prime}(g_{i_{0}},\ldots,g_{i_{n-1}}),\nu|_{[i_{0},\ldots,i_{n-1}]}\circ
L\rangle\\\ &=\sum_{i_{0},\ldots,i_{n-1}}\sum_{t_{0},\ldots,t_{n-1}}\langle
f(h_{t_{0}},\ldots,h_{t_{n-1}}),(\nu|_{[i_{0},\ldots,i_{n-1}]}\circ
L)|_{[t_{0},\ldots,t_{n-1}]}\pi\rangle.\end{split}$

Here, $[t_{0},\ldots,t_{n-1}]:=X_{g_{i_{0}},h_{t_{0}}}\cap
g_{i_{0}}X_{g_{i_{1}},h_{t_{1}}}\cap\cdots\cap(g_{i_{0}}\cdots
g_{i_{n-2}})X_{g_{i_{n-1}},h_{t_{n-1}}}$ and
$X_{g_{i_{j}},h_{t_{j}}}:=\\{x:c(g_{i_{j}}^{-1},x)=h_{t_{j}}^{-1}\\}$.

To continue the proof, we notice the following facts.

(Fact 5) $(\nu|_{[i_{0},\ldots,i_{n-1}]}\circ
L)|_{[t_{0},\ldots,t_{n-1}]}\pi=\nu|_{[i_{0},\ldots,i_{n-1}]\cap\phi([t_{0},\ldots,t_{n-1}])}$.

To prove this fact, take any $\xi\in N_{0}(H,Y)$, $h\in H$ and $y\in Y$. Let
$\xi^{\prime}=\pi(\xi)$ and $\eta=\xi^{\prime}|_{[t_{0},\ldots,t_{n-1}]}$. We
aim to prove $((\nu|_{[i_{0},\ldots,i_{n-1}]}\circ
L)|_{[t_{0},\ldots,t_{n-1}]}\pi)(\xi)=(\nu|_{[i_{0},\ldots,i_{n-1}]\cap\phi([t_{0},\ldots,t_{n-1}])})(\xi)$;
equivalently,
$\nu((L\circ\eta)|_{[i_{0},\ldots,i_{n-1}]})=\nu(\xi|_{{[i_{0},\ldots,i_{n-1}]\cap\phi([t_{0},\ldots,t_{n-1}])}})$.
This is clear by Lemma 3.4.

(Fact 6) $[i_{0},\ldots,i_{n-1}]\cap\phi([t_{0},\ldots,t_{n-1}])=\emptyset$
unless $h_{t_{j}}=h_{j}$ for all $0\leq j\leq n-1$. When these conditions
hold, the intersection equals $[i_{0},\ldots,i_{n-1}]$.

This fact can be directly checked using cocycle identity and Lemma 3.1.

Now we can continue the computation in (36) as follows.

$\displaystyle\begin{split}&\quad
f^{\prime\prime}(h_{0},\ldots,h_{n-1})(\nu)\\\
&\overset{Fact~{}5}{=}\sum_{i_{0},\ldots,i_{n-1}}\sum_{t_{0},\ldots,t_{n-1}}\langle
f(h_{t_{0}},\ldots,h_{t_{n-1}}),\nu|_{[i_{0},\ldots,i_{n-1}]\cap\phi([t_{0},\ldots,t_{n-1}])}\rangle\\\
&\overset{Fact~{}6}{=}\sum_{i_{0},\ldots,i_{n-1}}\langle
f(h_{0},\ldots,h_{n-1}),\nu|_{[i_{0},\ldots,i_{n-1}]}\rangle\\\ &=\langle
f(h_{0},\ldots,h_{n-1}),\nu\rangle.\end{split}$

Therefore, $f^{\prime\prime}=f$ and the proof is finished.

##  6\. Concluding remarks

We end this paper with several remarks addressing various issues related to
our main theorems.

(1) It is routine to check that the isomorphism $S$ constructed in the proof
preserves the Johnson classes [bnnw2, Definition 20] and the fundamental
classes [bnnw, Definition 3] for the two actions, which were introduced to
study amenability of group actions. Since the triviality of the Johnson class
[bnnw2, Theorem 1] and the nontriviality of the fundamental class [bnnw,
Theorem 9] are used to characterize the amenability of an action, we see that
for two topologically free actions, topologically amenability is an invariant
property under continuous orbit equivalence.

(2) Under the same assumptions as in our theorems, it is natural to ask
whether $H^{uf}_{n}(G\curvearrowright X)\cong H^{uf}_{n}(H\curvearrowright Y)$
for $n\geq 1$ holds. For this question, our method does not work. Indeed, we
use crucially the decomposition of elements in $N_{0}(G,X)$ as a finite sum of
elements in $N_{0}(G,X)$ with respect to a finite partition of $X$ into clopen
subsets. The summands do not belong to $W_{0}(G,X)$ anymore if we decompose an
element in $W_{0}(G,X)$. This may be interpreted as saying $W_{0}(G,X)$ is not
res-invariant (w.r.t. $X$), following [li, Definition 4.1] in spirit or
reflecting the fact $W_{0}(G,X)$ is not a $(G,X)$-module in the sense of
[monod].

In fact, the following suggests the above isomorphism may fail in general.
First, by [bnnw, Corollary 5], we know that for a topologically amenable
action $G\curvearrowright X$, $H^{uf}_{n}(G\curvearrowright X)\cong
H_{n}(G,\mathbb{R})\oplus H_{n}(G,N_{0}(G,X)^{*})$ holds. Then, we consider
two continuous actions which are both topologically amenable and topologically
free, e.g. left translation actions on the Stone-C̆ech compactifications of
free groups with different rank, say $F_{2}$ and $F_{3}$. As $F_{2}$ is quasi-
isometric to $F_{3}$, we know they are also bilipschitz equivalent by [whyte];
equivalently (see [li, Corollary 2.21]), $F_{2}\curvearrowright\beta
F_{2}\overset{coe}{\sim}F_{3}\curvearrowright\beta F_{3}$. On the one hand,
$H_{1}(F_{k},\mathbb{R})\cong\mathbb{R}^{k}$ for all $k\geq 1$ implies
$H_{1}(F_{2},\mathbb{R})\not\cong H_{1}(F_{3},\mathbb{R})$, but on the other
hand, $H_{1}(F_{2},N_{0}(F_{2},\beta F_{2})^{*})\cong
H_{1}(F_{3},N_{0}(F_{3},\beta F_{3})^{*})$ by our main theorems. This suggests
that for the above example, one expects these two actions have non-isomorphic
$H_{1}^{uf}$.

Nevertheless, under certain assumptions, we can still have some positive
result.

######  Corollary 6.1.

Let $G\curvearrowright X$ and $H\curvearrowright Y$ be topologically free
actions which are COE. If $G\curvearrowright X$ is topologically amenable and
both $G$ and $H$ are finitely generated torsion free nilpotent groups with
finite cohomological dimension, then $H_{i}^{uf}(G\curvearrowright X)\cong
H_{i}^{uf}(H\curvearrowright Y)$ for all $i\geq 0$.

###### Proof.

First, observe that $H\curvearrowright Y$ is also topologically amenable. This
is clear by previous Remark (1). One can also check this using [bnnw,
Definition 7] and the map $\pi$ defined in Lemma 3.2.

Then by [bnnw, Corollary 5], for topologically amenable actions, we have
$H^{uf}_{i}(G\curvearrowright X)\cong H_{i}(G,\mathbb{R})\oplus
H_{i}(G,N_{0}(G,X)^{*})$. The same holds for the action $H\curvearrowright Y$.

By Theorem 1.1, it suffices to show that $H_{i}(G,\mathbb{R})\cong
H_{i}(H,\mathbb{R})$.

Observe that COE between the above two actions implies $G$ and $H$ are quasi-
isometric. Indeed, just fix any $x\in X$, it is easy to check $g\in G\to H\ni
c(g,x)$ is a quasi-isometry using Lemma 3.1.

Now, as $G$ and $H$ are quasi-isometric nilpotent groups, we deduce
$H^{i}(G,\mathbb{R})\cong H^{i}(H,\mathbb{R})$ by [sauer, Theorem 1.5].
Moreover, $cd(G)=cd(H)$ by [sauer, Theorem 1.2] or [li, Corollary 4.42]. The
proof is finished by noticing that for finitely generated torsion free
nilpotent groups, they are orientable Poincaré duality groups, which implies
that $H^{i}(G,\mathbb{R})\cong H_{n-i}(G,\mathbb{R})$, where $n=cd(G)$. And
the same holds for $H$. (For the above assertion and the definition of
orientable Poincaré duality groups, see Section 10, Chapter VIII in [brown],
in particular, [brown, Example 1, p. 222].) ∎

(3) (Co)homology groups associated to many coefficient modules are proved to
be invariants under COE for two topologically free actions in [li, Theorem
3.1, 3.5], but as far as we can see, the (co)homology groups considered in our
paper are not covered by these theorems. It seems plausible one may also prove
our theorems using the method in [li], i.e. try to interpret the
(co)homologies for $G$ as (co)homologies for the transformation groupoid, but
one may need to extend unitary representations of étale locally compact
groupoid ([li, §3.2], [renault]) to linear isometric representations on Banach
spaces. In fact, for two COE topologically free actions $G\curvearrowright X$
and $H\curvearrowright Y$, it may be possible to show there is a one to one
correspondence between $(G,X)$-modules of type $M$ in the sense of [monod],
say $E$, and $(H,Y)$-modules of type $M$, say $F$, and under this
correspondence, $H_{b}^{*}(G,E^{*})\cong H_{b}^{*}(H,F^{*})$. Theorem 1.2 may
be thought of as an evidence for this.

Acknowledgement: We thank Prof. Piotr Nowak for helpful discussion related to
this paper. We are also very grateful to the anonymous referee. He/She
provided us with many valuable suggestions which greatly improved the
readability of the paper.

## References