of them similar. We will only deal with the case where $e$ is a red edge and
$f$ is a blue edge. So, let $e$ be a red edge and $f$ a blue edge, with
$s(e)=s(f)=W$ and $r(e)=U$, $r(f)=V$, where $U,V\in{\pazocal P}_{2n}$ and
$W\in{\pazocal P}_{2n+1}$. We then have

(6.1)
$\phi_{n}(\tau_{(e,f)})=\phi_{n}(ef^{*})=\sum_{s(g)=s(h)}gh^{*}=\sum_{s(g)=s(h)}\tau_{(g,h)},$

where $g$ ranges on the set of red edges $g\in R_{s(\overline{e})}$, where
$\overline{e}\in R(e)$, and $h$ ranges on the set of blue edges $h\in
B_{s(\overline{f})}$, where $\overline{f}\in B_{s(f)}$. Note that

$S:=\\{s(g)\mid g\in R_{s(\overline{e})},\overline{e}\in R(e)\\}=\\{s(h)\mid
h\in B_{s(\overline{f})},\overline{f}\in B_{s(f)}\\}$

is exactly the set of vertices appearing in the expression of $\textbf{S}(W)$
(see Lemma 6.1 (ii)), and just as before, this set corresponds to the set of
clopen subsets $\overline{W}\in{\pazocal P}_{2n+3}$ such that
$\overline{W}\subseteq W$. Given $\overline{W}\in{\pazocal P}_{2n+3}$ such
that $\overline{W}\subseteq W$, we have inclusions $\overline{W}\subseteq
W\subseteq\sigma(U)$ and so by Lemma 3.26(ii) there exists a unique
$\overline{U}\in{\pazocal P}_{2n+2}$ such that

(6.2) $\overline{U}\subseteq U\quad\text{ and
}\quad\overline{W}\subseteq\sigma(\overline{U})\subseteq W\subseteq\sigma(U).$

In addition, there is a unique $\overline{V}\in{\pazocal P}_{2n+2}$ such that
$\overline{W}\subseteq\overline{V}$, and since $\overline{W}\subseteq
W\subseteq V$, we must have

(6.3) $\overline{W}\subseteq\overline{V}\subseteq W\subseteq V.$

It follows that each of the terms $gh^{*}$ in the sum appearing in (6.1)
corresponds exactly to a triple $(\overline{U},\overline{W},\overline{V})$,
where $\overline{W}\in{\pazocal P}_{2n+3}$,
$\overline{U},\overline{V}\in{\pazocal P}_{2n+2}$, and (6.2) and (6.3) hold.
We infer that the decomposition (6.1) corresponds, under $\varphi_{n+1}$, to
the decomposition

$\bigsqcup_{\\{\overline{W}\in{\pazocal P}_{2n+3}\mid\overline{W}\subseteq
W\\}}Z((\sigma|_{\overline{U}})^{-1}(\overline{W}),1,0,\overline{W})=Z((\sigma|_{U})^{-1}(W),1,0,W).$

Hence we get that
$(\varphi_{n+1}\circ\phi_{n})(\tau_{(e,f)})=\varphi_{n}(\tau_{(e,f)})$, as
desired. ∎

We now start a study of elements in ${\pazocal A}_{\infty}$.

######  Definition 6.12.

Let $n\geq 0$ and $U,V\in{\pazocal P}_{2n}$, $W\in{\pazocal P}_{2n+1}$. We say
that $(U,W,V)$ is an $n$-triple if $W\subseteq V$ and $W\subseteq\sigma(U)$.
Note that such a triple determines an element $fe^{*}=\tau_{(f,e)}$ in
${\pazocal A}_{2n}$, where $f\in R_{U}$ and $e\in B_{V}$, such that
$\varphi_{n}(fe^{*})=\mathds{1}_{Z((\sigma|_{U})^{-1}(W),1,0,W)}$.

In the following lemma, we describe the images by $\varphi_{n}$ of some
elements of ${\pazocal A}_{2n}$.

######  Lemma 6.13.

Let $n\geq 0$.

  1. i)

Suppose that $x=f_{1}e_{1}^{*}\cdots f_{k}e_{k}^{*}\in{\pazocal A}_{2n}$, with
$f_{i}$ red edges and $e_{i}$ blue edges, such that $s(f_{i})=s(e_{i})=W_{i}$,
$r(f_{1})=U_{1}$ and $r(e_{i})=r(f_{i+1})=U_{i+1}$ for $1\leq i\leq k-1$.
Consider the set

$T_{1}:=(\sigma|_{U_{1}})^{-1}(W_{1}\cap(\sigma|_{U_{2}})^{-1}(W_{2}\cap\cdots\cap(\sigma|_{U_{k-1}})^{-1}(W_{k-1}\cap(\sigma|_{U_{k}})^{-1}(W_{k}))\cdots)).$

Then

$\varphi_{n}(x)=\mathds{1}_{Z(T_{1},k,0,\sigma^{k}(T_{1}))}$

and

$\varphi_{n}(xx^{*})=\mathds{1}_{T_{1}}.$

  2. ii)

Let
$x=(f_{1}e_{1}^{*})\cdots(f_{k-1}e_{k-1}^{*})(f_{k}f_{k+1}^{*})(e_{k+2}f_{k+2}^{*})\cdots(e_{k+l}f_{k+l}^{*})\in{\pazocal
A}_{2n}$, with $e_{i}$ blue edges and $f_{i}$ red edges, such that
$s(e_{i})=s(f_{i})=W_{i}$ for $1\leq i\leq k-1$ and $k+2\leq i\leq k+l$,
$s(f_{k})=s(f_{k+1})=W_{k}=W_{k+1}$, $f_{k}\neq f_{k+1}$, $r(f_{1})=U_{1}$,
$r(e_{i})=r(f_{i+1})=U_{i+1}$ for $1\leq i\leq k-1$,
$r(f_{k+i})=r(e_{k+i+1})=U_{k+i}$ for $1\leq i\leq l-1$ and
$r(f_{k+l})=U_{k+l}$. Let $T_{1}$ be as in i) and set

$T_{2}:=(\sigma|_{U_{k+l}})^{-1}(W_{k+l}\cap(\sigma|_{U_{k+l-1}})^{-1}(W_{k+l-1}\cap\cdots\cap(\sigma|_{U_{k+2}})^{-1}(W_{k+2}\cap(\sigma|_{U_{k+1}})^{-1}(W_{k+1}))\cdots)).$

Then $T:=\sigma^{k}(T_{1})\cap\sigma^{l}(T_{2})\subseteq W_{k}$,
$(\sigma|_{T_{1}})^{k}$ and $(\sigma|_{T_{2}})^{l}$ are injective, and, with
$U:=(\sigma|_{T_{1}})^{-k}(T)$ and $V:=(\sigma|_{T_{2}})^{-l}(T)$, we have

$\varphi_{n}(x)=\mathds{1}_{Z(U,k,l,V)}$

and

$\varphi_{n}(xx^{*})=\mathds{1}_{U}.$

###### Proof.

$i)$ Note that, for $y\in T_{1}$, we have $y\in U_{1}$ and $\sigma(y)\in
W_{1}\subseteq U_{2}$. Moreover $\sigma^{2}(y)=\sigma(\sigma(y))\in
W_{2}\subseteq U_{3}$. In general we have $\sigma^{i}(y)\in W_{i}\subseteq
U_{i+1}$ for $1\leq i\leq k-1$, and $\sigma^{k}(y)\in W_{k}$. Since the
restriction of $\sigma$ to each set $U_{i}$ is injective, we see that
$\sigma^{k}$ is injective on $T_{1}$. We have

$\varphi_{n}(x)=\mathds{1}_{{\pazocal Z}},$

where

${\pazocal Z}=Z((\sigma|_{U_{1}})^{-1}(W_{1}),1,0,W_{1})\cdots
Z((\sigma|_{U_{k}})^{-1}(W_{k}),1,0,W_{k}).$

The elements of the right-hand side of the above equality have the form
$(y_{0},1,y_{1})(y_{1},1,y_{2})\cdots(y_{k-1},1,y_{k})$, where $y_{0}\in
U_{1}$, $y_{1}=\sigma(y_{0})\in W_{1}\subseteq U_{2}$, and in general
$y_{i}=\sigma^{i}(y_{0})\in W_{i}\subseteq U_{i+1}$ for $1\leq i\leq k-1$, and
$\sigma^{k}(y_{0})=y_{k}\in W_{k}$. Then we get that $y_{0}\in T_{1}$ and that

$(y_{0},1,y_{1})\cdots(y_{k-1},1,y_{k})=(y_{0},k,y_{k})\in
Z(T_{1},k,0,\sigma^{k}(T_{1})).$

Conversely, if $(y,k,z)\in Z(T_{1},k,0,\sigma^{k}(T_{1}))$, by the definition
of $T_{1}$ we can write $(y,k,z)$ as a product of elements in
$Z((\sigma|_{U_{i}})^{-1}(W_{i}),1,0,W_{i})$ for all $1\leq i\leq k$.

Now,

$\displaystyle\varphi_{n}(xx^{*})$
$\displaystyle=\varphi_{n}(x)\varphi_{n}(x)^{*}$
$\displaystyle=\mathds{1}_{Z(T_{1},k,0,\sigma^{k}(T_{1}))}\cdot\mathds{1}_{Z(T_{1},k,0,\sigma^{k}(T_{1}))^{-1}}$
$\displaystyle=\mathds{1}_{Z(T_{1},k,0,\sigma^{k}(T_{1}))\cdot
Z(T_{1},k,0,\sigma^{k}(T_{1}))^{-1}}$
$\displaystyle=\mathds{1}_{Z(T_{1},k,0,\sigma^{k}(T_{1}))\cdot
Z(\sigma^{k}(T_{1}),0,k,T_{1})}$ $\displaystyle=\mathds{1}_{T_{1}}.$

The proof of $ii)$ is similar to the proof of $i)$, and is left to the reader.
∎

This result is a first step towards finding a canonical expression for an
element of ${\pazocal A}_{\infty}$. The next step is to show a partial
injectivity result for $\varphi$.

######  Lemma 6.14.

The following statements hold.

  1. i)

Let $x=f_{1}e_{1}^{*}\cdots f_{k}e_{k}^{*}\in{\pazocal A}_{2n}$ be as in Lemma
6.13 (i). Then $\phi_{n,\infty}(x)\neq 0$ if and only if $\varphi_{n}(x)\neq
0$.

  2. ii)

Let
$x=(f_{1}e_{1}^{*})\cdots(f_{k-1}e_{k-1}^{*})(f_{k}f_{k+1}^{*})(e_{k+2}f_{k+2}^{*})\cdots(e_{k+l}f_{k+l}^{*})\in{\pazocal
A}_{2n}$ be as in Lemma 6.13 (ii). Then $\phi_{n,\infty}(x)\neq 0$ if and only
if $\varphi_{n}(x)\neq 0$.

###### Proof.

First, observe that for any term $x$ as in $i)$ or $ii)$, we have
$\phi_{n,\infty}(x)\neq 0$ if and only if $\phi_{n,\infty}(xx^{*})\neq 0$.
This is due to the fact that ${\pazocal A}_{\infty}$ is tame (Theorem 6.7), so
that $\phi_{n,\infty}(x)$ is a partial isometry in ${\pazocal A}_{\infty}$.
Similarly, $\varphi_{n}(x)\neq 0$ if and only if $\varphi_{n}(xx^{*})\neq 0$.

Let us prove $i)$. If $\varphi_{n}(x)\neq 0$ then $\phi_{n,\infty}(x)\neq 0$,
because $\varphi_{n}(x)=\varphi(\phi_{n,\infty}(x))$.

Suppose now that $\phi_{n,\infty}(xx^{*})\neq 0$, so in particular
$\phi_{n,m}(xx^{*})\neq 0$ for all $m>n$. We will show that $T_{1}$ is non-
empty. We proceed by induction on $k$.

For $k=1$, we have $x=f_{1}e_{1}^{*}$ and $xx^{*}=f_{1}f_{1}^{*}$. In this
case $T_{1}=(\sigma|_{U_{1}})^{-1}(W_{1})\neq\emptyset$, because
$W_{1}\neq\emptyset$ and $W_{1}\subseteq\sigma(U_{1})$.

Suppose now that the result is true for all lengths up to $k-1$. Take
$x=f_{1}e_{1}^{*}\cdots f_{k}e_{k}^{*}$. Observe that

$xx^{*}=f_{1}e_{1}^{*}\cdots e_{k-1}^{*}f_{k}f_{k}^{*}e_{k-1}\cdots
e_{1}f_{1}^{*}=(f_{1}e_{1}^{*})yy^{*}(e_{1}^{*}f_{1}),$

where $y=f_{2}e_{2}^{*}\cdots f_{k}e_{k}^{*}$. Recall from Lemma 6.6 that
$\phi_{n,n+k}(xx^{*})=\widetilde{\phi}_{2n,2n+2k}(xx^{*})$ is a sum of a
finite number of distinct vertices in $F_{2n+2k}^{0,0}$, and likewise
$\phi_{n,n+k-1}(yy^{*})=\widetilde{\phi}_{2n,2n+2k-2}(yy^{*})$ is a sum of a
finite number of distinct vertices in $F_{2n+2k-2}^{0,0}$. We will show, using
induction, that any of the open compact sets corresponding to the vertices
appearing in the expression of $\phi_{n,n+k}(xx^{*})$ is contained in $T_{1}$.
Our assumption that $\phi_{n,\infty}(xx^{*})\neq 0$ would then imply that
$T_{1}\neq\emptyset$.

Suppose that $\phi_{n,n+k-1}(yy^{*})=\sum_{v}v$ for distinct vertices $v\in
F_{2n+2k-2}^{0,0}$ and that each of the open compact subsets of $X$
represented by these vertices is contained in

$T^{\prime}:=(\sigma|_{U_{2}})^{-1}(W_{2}\cap(\sigma|_{U_{3}})^{-1}(W_{3}\cap\cdots\cap(\sigma|_{U_{k-1}})^{-1}(W_{k-1}\cap(\sigma|_{U_{k}})^{-1}(W_{k}))\cdots)).$

By Lemma 6.4 (ii), we can write $\phi_{n,n+k-1}(e_{1})=\sum_{g\in N}g$, where
$N$ is a set of blue edges in $F_{2n+2k-2}$ with mutually distinct sources. So
we get

$\displaystyle\phi_{n,n+k-1}(xx^{*})$
$\displaystyle=\phi_{n,n+k-1}(f_{1})\Big{(}\sum_{g\in
N}g^{*}\Big{)}\Big{(}\sum_{v}v\Big{)}\Big{(}\sum_{g\in
N}g\Big{)}\phi_{n,n+k-1}(f_{1}^{*})$
$\displaystyle=\phi_{n,n+k-1}(f_{1})\Big{(}\sum_{g\in
N^{\prime}}s(g)\Big{)}\phi_{n,n+k-1}(f_{1}^{*}),$

where $N^{\prime}$ is a non-empty subset of $N$. Again by Lemma 6.4 (ii), we
can write $\widetilde{\phi}_{2n,2n+2k-1}(f_{1})=\sum_{h\in L}h^{*}$, where $L$
is a subset of red edges in $F_{2n+2k-1}$, whose sources are mutually
disjoint, and such that all edges in $L$ with the same range belong to the
same element of the partition $D$. It follows that

$\widetilde{\phi}_{2n,2n+2k-1}(xx^{*})=\Big{(}\sum_{h\in
L}h^{*}\Big{)}\Big{(}\sum_{g\in N^{\prime}}s(g)\Big{)}\Big{(}\sum_{h\in
L}h\Big{)}=\sum_{h\in L^{\prime}}s(h),$

where $L^{\prime}$ is a subset of $L$. Note that we have
$0\neq\phi_{n,n+k}(xx^{*})=\sum_{h\in L^{\prime}}s(h)$, so
$L^{\prime}\neq\emptyset$. Now take any $h\in L^{\prime}$ and set $Z=s(h)$,
$W=r(h)$. There is $g\in N^{\prime}$ such that $r(h)=s(g)$. Moreover we have
that $r(g)=V$, where $V\subseteq T^{\prime}$ by our induction hypothesis.
Since $g$ is a blue edge we have $W=r(h)=s(g)\subseteq V\subseteq T^{\prime}$,
so that $W\subseteq T^{\prime}$. Since $h$ is a red edge in $F_{2n+2k-1}$, we
must have $Z\subseteq\sigma^{-1}(W)$. On the other hand, since $xx^{*}\leq
f_{1}f_{1}^{*}\leq r(f_{1})=U_{1}$, we have that
$s(h)\leq\phi_{n,n+k}(xx^{*})\leq\phi_{n,n+k}(r(f_{1}))$, and we conclude that
$Z\subseteq U_{1}$. Similarly $e_{1}^{*}yy^{*}e_{1}\leq s(e_{1})=W_{1}$ and
thus
$s(g)\leq\phi_{n,n+k-1}(e_{1}^{*}yy^{*}e_{1})\leq\phi_{n,n+k-1}(s(e_{1}))$
implies that $W\subseteq W_{1}$. We thus have that

$W\subseteq W_{1}\cap T^{\prime}\qquad\text{and}\qquad Z\subseteq
U_{1}\cap\sigma^{-1}(W)\subseteq U_{1}\cap\sigma^{-1}(W_{1}\cap T^{\prime}),$

so we conclude that $Z\subseteq(\sigma|_{U_{1}})^{-1}(W_{1}\cap
T^{\prime})=T_{1}$, as desired.

The similar proof of $ii)$ is left to the reader. ∎

We can now state our result on canonical forms.

######  Theorem 6.15.

Each element $a\in{\pazocal A}_{\infty}$ can be represented as
$a=\phi_{n,\infty}(b)$ for some $n\geq 0$ and $b\in{\pazocal A}_{2n}$ such
that $b$ is a finite linear combination of the following types of elements:

  1. (A)

Vertices $p_{U}$, with $U\in{\pazocal P}_{2n}$.

  2. (B)

Terms of the form $x=f_{1}e_{1}^{*}\cdots f_{k}e_{k}^{*}$ as in Lemma 6.13
(i), with $\varphi_{n}(x)\neq 0$.

  3. (C)

Adjoints of the terms from (B).

  4. (D)

Terms of the form
$x=(f_{1}e_{1}^{*})\cdots(f_{k-1}e_{k-1}^{*})(f_{k}f_{k+1}^{*})(e_{k+2}f_{k+2}^{*})\cdots(e_{k+l}f_{k+l}^{*})$
as in Lemma 6.13 (ii), where $k,l\geq 1$ and with $\varphi_{n}(x)\neq 0$. When
$k=l=1$, the corresponding term must be interpreted as $f_{1}f_{2}^{*}$, where
$f_{1},f_{2}$ are distinct red edges with the same source.

Note that the family (D) is closed under adjoints.

###### Proof.

Let $a$ be a nonzero element of ${\pazocal A}_{\infty}$. There exists $m\geq
0$ and $c\in{\pazocal A}_{2m}$ such that $\phi_{m,\infty}(c)=a$. The element
$c$ will be a linear combination of vertices $p_{U}$, for $U\in{\pazocal
P}_{2m}$, and terms which are products of elements $ef^{*}$ , where $e$ and
$f$ are red or blue edges such that $s(e)=s(f)\in F_{2m}^{0,1}$.

Observe that a term $ee^{*}$, where $e$ is either a blue or a red edge, can be
simplified by applying $\phi_{m}$, since $\phi_{m}(ee^{*})$ is a sum of
distinct vertices in ${\pazocal A}_{2m+2}$ by Lemma 6.6, so of the form (A).
We can thus assume that we do not have terms of the form $ee^{*}$ in our
product. Moreover, since $s(e_{1})=s(e_{2})$ implies $e_{1}=e_{2}$ for
$e_{1},e_{2}$ blue edges, we may, in view of the above, assume that no terms
$e_{1}e_{2}^{*}$, with both $e_{1}$ and $e_{2}$ blue edges, appear in the
product.

Suppose that we have a product $(fe^{*})(gh^{*})$ such that $e,g$ are both
blue edges. By our assumptions, we necessarily have that both $f$ and $h$ are
red edges. Then we get that

$(fe^{*})(gh^{*})=\delta_{e,g}(fh^{*}).$

So either the product is $0$ or it can be simplified to a product containing a
term $fh^{*}$, with both $f$ and $h$ red edges. By the previous paragraph, we
can assume that $f\neq h$. We can thus simplify these terms.

Suppose now that we have a product $(ef^{*})(gh^{*})$ such that $f,g$ are both
red edges. Then we get, as before,

$(ef^{*})(gh^{*})=\delta_{f,g}(eh^{*}).$

If both $e$ and $h$ are red edges, then we can assume that they satisfy $e\neq
h$ (otherwise we can reduce the length of the word by applying $\phi_{m}$). If
$e$ and $h$ are both blue edges, then either $eh^{*}=0$ or $e=h$. In the
latter case, we can simplify by applying $\phi_{m}$. If $e$ is blue and $h$ is
red the product will be either $0$ or of the form $eh^{*}$, and similarly when
$e$ is red and $h$ is blue. Hence these terms can be simplified too.

In conclusion, the only products that cannot be simplified are those where all
the factors are of the form $e_{i}f_{i}^{*}$, where all $e_{i}$ are blue and
all $f_{i}$ are red (terms of the form (C)), or all $e_{i}$ are red and all
$f_{i}$ are blue (terms of the form (B)), and those of the form (D). ∎

We are now ready to show the main result of this section.

######  Theorem 6.16.

With the previous notation, we have that the map

$\varphi\colon{\pazocal A}_{\infty}\to A_{K}({\pazocal G}(X,\sigma))$

is a graded $*$-isomorphism.

###### Proof.

In view of Proposition 6.11, it only remains to show that $\varphi$ is
bijective. We first show it is surjective. For this, it is enough to check
that all the characteristic functions of sets of the form $Z(U,k,l,V)$ belong
to the image of $\varphi$. Suppose therefore that $U$ and $V$ are open compact
subsets of $X$ such that $\sigma^{k}|_{U}$ and $\sigma^{l}|_{V}$ are injective
and that $\sigma^{k}(U)=\sigma^{l}(V)$. Observe that

$Z(U,k,l,V)=Z(U,k,0,\sigma^{k}(U))\cdot Z(V,l,0,\sigma^{l}(V))^{-1},$

so that it suffices to show that $Z(U,k,0,\sigma^{k}(U))$ belongs to the image
of $\varphi$.

There is some $n\geq 0$ such that $U$ can be written as a disjoint union of
sets in ${\pazocal P}_{2n}$. Hence we can assume that $U\in{\pazocal P}_{2n}$.
We claim that

(6.4) $U=\bigsqcup_{(W_{1},\dots,W_{k})\in{\pazocal
B}}(\sigma|_{U_{1}})^{-1}(W_{1}\cap(\sigma|_{U_{2}})^{-1}(W_{2}\cap\cdots\cap(\sigma|_{U_{k-1}})^{-1}(W_{k-1}\cap(\sigma|_{U_{k}})^{-1}(W_{k}))\cdots)),$

where

${\pazocal B}=\\{(W_{1},\dots,W_{k})\in{\pazocal P}_{2n+1}^{k}\mid
W_{i}\subseteq\sigma(U_{i})\text{ and }W_{i}\subseteq U_{i+1}\text{ for }1\leq
i\leq k,\text{ for unique }U_{i}\text{'s}\text{ with }U_{1}=U\\}.$

First, we observe that $T\cap T^{\prime}=\emptyset$, where $T$ and
$T^{\prime}$ are sets as in the right-hand side of (6.4) corresponding to
different choices $(W_{1},\dots,W_{k})$ and
$(W_{1}^{\prime},\dots,W_{k}^{\prime})$. Indeed, suppose that $T\cap
T^{\prime}\neq\emptyset$ and take $x\in T\cap T^{\prime}$. Then $x\in U_{1}$
and $\sigma(x)\in W_{1}\cap W_{1}^{\prime}$. Since $W_{1},W_{1}^{\prime}$
belong to the partition ${\pazocal P}_{2n+1}$ of $X$, we obtain that
$W_{1}=W_{1}^{\prime}$. This in turn implies that $U_{2}=U_{2}^{\prime}$. Now
note that $\sigma^{2}(x)\in W_{2}\cap W_{2}^{\prime}$ and thus
$W_{2}=W_{2}^{\prime}$ and $U_{3}=U_{3}^{\prime}$. Continuing in this way, we
obtain that $(W_{1},\dots,W_{k})=(W_{1}^{\prime},\dots,W_{k}^{\prime})$.

Take now $x\in U=U_{1}$. We have

$\sigma(U_{1})=\bigsqcup_{\\{W\in{\pazocal P}_{2n+1}\mid
W\subseteq\sigma(U_{1})\\}}W,$

so that there is a unique $W_{1}\in{\pazocal P}_{2n+1}$ such that
$\sigma(x)\in W_{1}$. Then there is a unique $U_{2}\in{\pazocal P}_{2n}$ such
that $W_{1}\subseteq U_{2}$. Since $\sigma(x)\in U_{2}$, by the same argument
we get $W_{2}\in{\pazocal P}_{2n+1}$ such that $W_{2}\subseteq\sigma(U_{2})$
and $\sigma^{2}(x)=\sigma(\sigma(x))\in W_{2}$. In this way, we build a family
of sets $W_{1},\dots,W_{k}$ and $U_{2},\dots,U_{k}$ satisfying the specified
conditions and such that

$x\in(\sigma|_{U_{1}})^{-1}(W_{1}\cap(\sigma|_{U_{2}})^{-1}(W_{2}\cap\cdots\cap(\sigma|_{U_{k-1}})^{-1}(W_{k-1}\cap(\sigma|_{U_{k}})^{-1}(W_{k}))\cdots)).$

This shows our claim.

Note that each $(U_{i},W_{i},U_{i+1})$ is an $n$-triple in the sense of
Definition 6.12, hence it gives rise to an element $f_{i}e_{i}^{*}$ in
${\pazocal A}_{2n}$, where $f_{i}\in R_{U_{i}}$, $e_{i}\in B_{U_{i+1}}$ and
$\varphi_{n}(f_{i}e_{i}^{*})=\mathds{1}_{Z((\sigma|_{U_{i}})^{-1}(W_{i}),1,0,W_{i})}$.
Therefore we see that each sequence $(W_{1},W_{2},\dots,W_{k})\in{\pazocal
P}_{2n+1}^{k}$ satisfying the conditions stated above, and such that the
corresponding set $T$ is non-empty, determines an element
$x=f_{1}e_{1}^{*}\cdots f_{k}e_{k}^{*}$ of the form (B) from Theorem 6.15,
with $\varphi_{n}(x)=\mathds{1}_{Z(T,k,0,\sigma^{k}(T))}$ by Lemma 6.13. Using
this observation and (6.4), the surjectivity of $\varphi$ follows.

Finally we show that $\varphi$ is injective. Since $\varphi$ is a graded
homomorphism, it is enough to show that it is injective in each graded
component. Suppose that we have a nonzero homogeneous element, say $a$, in
${\pazocal A}_{\infty}$. By Theorem 6.15, there exists some $n\geq 0$ and an
element $b\in{\pazocal A}_{2n}$ which is a linear combination of terms of the
forms (A), (B), (C), (D) such that $a=\phi_{n,\infty}(b)$. Of course we can
assume that $\phi_{n,\infty}$ does not annihilate any of these terms, which by
Lemma 6.14 implies that $\varphi_{n}$ also does not annihilate any of them. It
is sufficient to show that the compact open bisections corresponding to two
distinct terms of the same degree are disjoint. We will treat only the case
where the two terms are of the form (D), i.e. satisfy the conditions in Lemma
6.13 (ii). The cases where other types of terms arise are treated in a similar
way.

Suppose that we have two terms

$\alpha=(f_{1}e_{1}^{*})\cdots(f_{k-1}e_{k-1}^{*})(f_{k}f_{k+1}^{*})(e_{k+2}f_{k+2}^{*})\cdots(e_{k+l}f_{k+l}^{*}),$
$\beta=f^{\prime}_{1}(e^{\prime}_{1})^{*}\cdots
f^{\prime}_{k^{\prime}-1}(e^{\prime}_{k^{\prime}-1})^{*}f^{\prime}_{k^{\prime}}(f^{\prime}_{k^{\prime}+1})^{*}e^{\prime}_{k^{\prime}+2}(f^{\prime}_{k^{\prime}+2})^{*}\cdots
e^{\prime}_{k^{\prime}+l^{\prime}}(f^{\prime}_{k^{\prime}+l^{\prime}})^{*}$

in ${\pazocal A}_{2n}$ as in Lemma 6.13 (ii), and such that
$k-l=k^{\prime}-l^{\prime}$. We will assume that $k^{\prime}\geq k$, so that
$k^{\prime}=k+r$ and $l^{\prime}=l+r$ for some $r\geq 0$. Let

$W_{1},\dots,W_{k}=W_{k+1},W_{k+2},\dots,W_{k+l}\quad\text{and}\quad
U_{1},\dots,U_{k},U_{k+1},\dots,U_{k+l}$

be the sets associated to $\alpha$, and

$W_{1}^{\prime},\dots,W^{\prime}_{k^{\prime}}=W^{\prime}_{k^{\prime}+1},W^{\prime}_{k^{\prime}+2},\dots,W^{\prime}_{k^{\prime}+l^{\prime}}\quad\text{and}\quad
U^{\prime}_{1},\dots,U^{\prime}_{k^{\prime}},U^{\prime}_{k^{\prime}+1},\dots,U^{\prime}_{k^{\prime}+l^{\prime}}$

be the sets associated to $\beta$. Let $T_{1}$, $T_{2}$, $T$,
$U=(\sigma|_{T_{1}})^{-k}(T)$ and $V=(\sigma|_{T_{2}})^{-l}(T)$ be the sets
associated to $\alpha$ defined in Lemma 6.13 (ii), and let $T_{1}^{\prime}$,
$T_{2}^{\prime}$, $T^{\prime}$,
$U^{\prime}=(\sigma|_{T^{\prime}_{1}})^{-k^{\prime}}(T^{\prime})$ and
$V^{\prime}=(\sigma|_{T^{\prime}_{2}})^{-l^{\prime}}(T^{\prime})$ be the ones
corresponding to $\beta$. Suppose, by way of contradiction, that

$Z(U,k,l,V)\cap Z(U^{\prime},k^{\prime},l^{\prime},V^{\prime})\neq\emptyset,$

and let $(x,k-l,y)$ be an element in this intersection. Then $x\in U\cap
U^{\prime}$, $y\in V\cap V^{\prime}$, and $\sigma^{k}(x)=\sigma^{l}(y)$, which
automatically implies $\sigma^{k^{\prime}}(x)=\sigma^{l^{\prime}}(y)$ because
$k^{\prime}=k+r$ and $l^{\prime}=l+r$ with $r\geq 0$. Since $U\cap
U^{\prime}\neq\emptyset$, necessarily $U_{1}=U_{1}^{\prime}$, and also
$U_{k+l}=U^{\prime}_{k^{\prime}+l^{\prime}}$ because $V\cap
V^{\prime}\neq\emptyset$. Now observe that $\sigma(x)\in W_{1}\cap
W_{1}^{\prime}$, so that $W_{1}=W_{1}^{\prime}$ and $U_{2}=U_{2}^{\prime}$.
Then we get $\sigma^{2}(x)\in W_{2}\cap W_{2}^{\prime}$, so
$W_{2}=W_{2}^{\prime}$ and $U_{3}=U_{3}^{\prime}$, and so on up to
$U_{k}=U^{\prime}_{k}$ and $W_{k}=W_{k}^{\prime}$. Starting at the other end,
we have $\sigma(y)\in W_{k+l}\cap W^{\prime}_{k^{\prime}+l^{\prime}}$, so that
$W_{k+l}=W^{\prime}_{k^{\prime}+l^{\prime}}$ and
$U_{k+l-1}=U^{\prime}_{k^{\prime}+l^{\prime}-1}$. Then since $\sigma^{2}(y)\in
W_{k+l-1}\cap W^{\prime}_{k^{\prime}+l^{\prime}-1}$, we get
$W_{k+l-1}=W^{\prime}_{k^{\prime}+l^{\prime}-1}$ and
$U_{k+l-2}=U^{\prime}_{k^{\prime}+l^{\prime}-2}$. Continuing in this way, we
get that $W_{k+l-i}=W^{\prime}_{k^{\prime}+l^{\prime}-i}$ and
$U_{k+l-i}=U^{\prime}_{k^{\prime}+l^{\prime}-i}$ for all $0\leq i\leq l-1$. We
now distinguish two cases.

  1. $\cdot$

In case $r=0$ we conclude that all sets $W_{i}$ and all sets $U_{i}$ agree
with $W_{i}^{\prime}$ and $U_{i}^{\prime}$ respectively, so that
$\alpha=\beta$.

  2. $\cdot$

If $r>0$, then observe that

$\sigma^{k}(x)=\sigma^{l}(y)\in W_{k}^{\prime}\cap
W^{\prime}_{k^{\prime}+l^{\prime}-l+1},$

so that
$W_{k}^{\prime}=W^{\prime}_{k^{\prime}+l^{\prime}-l+1}=W^{\prime}_{k+2r+1}$.
Therefore $U^{\prime}_{k+1}=U^{\prime}_{k+2r}$. Since
$\sigma^{k+1}(x)=\sigma^{l+1}(y)\in W^{\prime}_{k+1}\cap W^{\prime}_{k+2r}$,
we get that $W^{\prime}_{k+1}=W^{\prime}_{k+2r}$ and thus
$U^{\prime}_{k+2}=U^{\prime}_{k+2r-1}$. Continuing in this way, we see that
$W^{\prime}_{k+r-1}=W^{\prime}_{k+r+2}$ and thus
$U^{\prime}_{k+r}=U^{\prime}_{k+r+1}$, that is, we obtain that
$U^{\prime}_{k^{\prime}}=U^{\prime}_{k^{\prime}+1}$. But since
$f^{\prime}_{k^{\prime}}\neq f^{\prime}_{k^{\prime}+1}$ and
$s(f^{\prime}_{k^{\prime}})=s(f^{\prime}_{k^{\prime}+1})$, we necessarily have
that $U^{\prime}_{k^{\prime}}=r(f^{\prime}_{k^{\prime}})\neq
r(f^{\prime}_{k^{\prime}+1})=U^{\prime}_{k^{\prime}+1}$, and we obtain a
contradiction. Therefore the case where $r>0$ is not possible.

The conclusion is that, whenever $Z(U,k,l,V)\cap
Z(U^{\prime},k^{\prime},l^{\prime},V^{\prime})\neq\emptyset$ holds, we
necessarily have $\alpha=\beta$, as desired. The proof is complete. ∎

We consider now the situation for the corresponding $C^{*}$-algebras. Writing
$C_{n}=C^{*}(F_{n},D^{n})$, we have that, by definition, $C_{n}$ is the
enveloping $C^{*}$-algebra of $L_{n}$. Hence we get a sequence of
$*$-homomorphisms $C_{n}\to C_{n+1}$, also denoted by $\tilde{\phi}_{n}$. Set
$C_{\infty}=\varinjlim(C_{n},\tilde{\phi}_{n})$, where the above inductive
limit is the colimit of $\\{(C_{n},\tilde{\phi}_{n})\\}$ in the cateory of
$C^{*}$-algebras. We now set

${\pazocal B}_{2n}:=p_{2n}C_{2n}p_{2n}\quad\text{ and }\quad{\pazocal
B}_{\infty}=\varinjlim({\pazocal B}_{2n},\phi_{n}),$

where, as before, $\phi_{n}$ is the restriction of
$\tilde{\phi}_{2n+1}\circ\tilde{\phi}_{2n}$ to ${\pazocal B}_{2n}$.

We can now state our main result for $C^{*}$-algebras.

######  Theorem 6.17.

With the above notation, there is a unique $*$-isomorphism
$\overline{\varphi}\colon{\pazocal B}_{\infty}\to C^{*}({\pazocal
G}(X,\sigma))$ such that the following diagram

(6.5) $\textstyle{{\pazocal
A}_{\infty}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\kern-20.0pt{\varphi}}$$\scriptstyle{\kern-20.0pt{\cong}}$$\textstyle{A_{{\mathbb{C}}}({\pazocal
G}(X,\sigma))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\pazocal
B}_{\infty}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\kern-20.0pt{\overline{\varphi}}}$$\scriptstyle{\kern-20.0pt{\cong}}$$\textstyle{C^{*}({\pazocal
G}(X,\sigma))}$

is commutative.

###### Proof.

It is a simple matter to check that $C_{\infty}$ is the enveloping
$C^{*}$-algebra of $L_{\infty}$.

Just as in Lemma 6.8, we have a natural $*$-isomorphism
$\overline{\lambda}\colon{\pazocal B}_{\infty}\to pC_{\infty}p$, such that the
following diagram

(6.6) $\textstyle{{\pazocal
A}_{\infty}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\kern-7.0pt{\lambda}}$$\scriptstyle{\kern-7.0pt{\cong}}$$\textstyle{pL_{\infty}p\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\pazocal
B}_{\infty}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\kern-7.0pt{\overline{\lambda}}}$$\scriptstyle{\kern-7.0pt{\cong}}$$\textstyle{pC_{\infty}p}$

is commutative. Since there is a partial isometry $s\in L_{\infty}$ such that
$ss^{*}=p$ and $s^{*}s=1-p$ (take for instance
$s=\widetilde{\phi}_{0,\infty}(\sum_{e\in B}e)$, where $B$ is the set of blue
edges in $F^{1,0}$), it follows that $L_{\infty}\cong pL_{\infty}p\otimes
M_{2}({\mathbb{C}})$, and $C_{\infty}\cong pC_{\infty}p\otimes
M_{2}({\mathbb{C}})$ in a compatible way. Since $C_{\infty}$ is the enveloping
$C^{*}$-algebra of $L_{\infty}$, it follows that $pC_{\infty}p$ is the
enveloping $C^{*}$-algebra of $pL_{\infty}p$. We infer from (6.6) that
${\pazocal B}_{\infty}$ is the enveloping $C^{*}$-algebra of ${\pazocal
A}_{\infty}$. The result follows from this and Theorem 6.16, because
$C^{*}({\pazocal G}(X,\sigma))$ is the enveloping $C^{*}$-algebra of
$A_{{\mathbb{C}}}({\pazocal G}(X,\sigma))$ (see Definition 2.4). ∎

###  6.2. Functoriality properties

In this section we show that the Deaconu-Renault construction is indeed a
functor from LHomeo to AHGrd, where AHGrd is the category whose objects are
all the second countable, ample Hausdorff groupoids, and whose morphisms are
the continuous groupoid homomorphisms. We further show that, whenever
$\psi\colon Y\to X$ is the universal continuous $(E,C)$-equivariant map
induced by an $(E,C)$-structure on $Y$, there are induced $*$-homomorphisms at
the levels of Steinberg algebras and $C^{*}$-algebras of the corresponding
groupoids, respectively.

The following lemma, which gives the mentioned functoriality, may be well-
known to experts.

######  Lemma 6.18.

Let $(X,\sigma)$ and $(Y,\rho)$ be objects in LHomeo, and let $\psi\colon X\to
Y$ be an equivariant continuous map. Then the induced map
$\psi_{*}\colon{\pazocal G}(X,\sigma)\to{\pazocal G}(Y,\rho)$ defined by

$\psi_{*}(x,m-n,y)=(\psi(x),m-n,\psi(y))$

for $(x,m-n,y)\in{\pazocal G}(X,\sigma)$, is a morphism in AHGrd. Hence the
Deaconu-Renault groupoid construction can be extended to a functor
$\emph{{DR}}\colon\emph{{LHomeo}}\to\emph{{AHGrd}}$.

###### Proof.

It is clear that $\psi_{*}$ is well-defined and an algebraic morphism of
groupoids. Thus we only need to show that $\psi_{*}$ is continuous.

Observe that the restriction of $\psi_{*}$ to the space of units $X$ agrees
with $\psi$, hence it is continuous. Let $\alpha=(x,m-n,y)\in{\pazocal
G}(X,\sigma)$, and let $U$ be an open compact subset of ${\pazocal G}(Y,\rho)$
containing $\psi_{*}(\alpha)$. Since $\rho^{i}$ is a local homeomorphism for
each $i\geq 0$, we can find open neighborhoods $U_{1}$ of $\psi(x)$ and
$U_{2}$ of $\psi(y)$ such that $\rho^{m}|_{U_{1}}$ and $\rho^{n}|_{U_{2}}$ are
injective. Replacing $U$ with $U\cap Z(U_{1},m,n,U_{2})$, we can assume that
$U\subseteq Z(U_{1},m,n,U_{2})$, and in particular that $U$ is an open compact
bisection. Note that $s(U)\subseteq U_{2}$ and $r(U)\subseteq U_{1}$.

By continuity of $\psi$, we can take open neighborhoods $U_{1}^{\prime}$ and
$U_{2}^{\prime}$ of $x$ and $y$ in $X$, respectively, such that
$\psi(U_{1}^{\prime})\subseteq r(U)$ and $\psi(U_{2}^{\prime})\subseteq s(U)$.
Then $Z(U_{1}^{\prime},m,n,U_{2}^{\prime})$ is an open subset of ${\pazocal
G}(X,\sigma)$, and clearly $\alpha\in Z(U_{1}^{\prime},m,n,U_{2}^{\prime})$.
Now if $\gamma\in Z(U_{1}^{\prime},m,n,U_{2}^{\prime})$, then
$r(\psi_{*}(\gamma))\in r(U)$, and since $U$ is contained in the bisection
$Z(U_{1},m,n,U_{2})$ we conclude that $\psi_{*}(\gamma)\in U$. This shows that

$\psi_{*}(Z(U_{1}^{\prime},m,n,U_{2}^{\prime}))\subseteq U,$

which gives the continuity of $\psi_{*}$. The last statement follows
straightforwardly. ∎

######  Notation 6.19.

Let $(E,C)$ be an $l$-diagram, or a layer in an $l$-diagram, and let $v$ be a
vertex in $E$ . We will denote by $s_{\mathrm{red}}^{-1}(v)$ the set of red
edges $f$ in $E^{1}$ such that $s(f)=v$.

If ${\pazocal G}$ is a groupoid and $x\in{\pazocal G}^{0}$, we denote by
${\pazocal G}x$ the set of elements $\gamma\in{\pazocal G}$ such that
$s(\gamma)=x$.

######  Lemma 6.20.

Let $(Y,\rho)\in\emph{{LHomeo}}$ and let $(E,C)$ be a generalized finite shift
graph. Suppose that we have an $(E,C)$-structure on $Y$, and let $\psi\colon
Y\to X$ be the unique $(E,C)$-equivariant continuous map, where $(X,\sigma)$
is the generalized finite shift associated to $(E,C)$. Then for $y\in Y$, the
map $\psi_{*}$ from Lemma 6.18 induces a bijection from ${\pazocal
G}(Y,\rho)y$ onto ${\pazocal G}(X,\sigma)\psi(y)$.

###### Proof.

We start by fixing the notation. We will denote the natural $(E,C)$-structure
on $X$ by $(\\{H_{e}\\}_{e\in B},\\{H_{f}\\}_{f\in R})$, where $B$ and $R$ are
the sets of blue and red edges of $E$, respectively. For $f\in R$, we have a
homeomorphism $\theta_{f}\colon H_{f}\to H_{e}$, where $e$ is the unique blue
edge such that $s(e)=s(f)$, and where $\theta_{f}$ is just the restriction of
$\sigma$ to $H_{f}$. We will also denote by $(\\{Z_{e}\\}_{e\in
B},\\{Z_{f}\\}_{f\in R})$ the open compact sets corresponding to the
$(E,C)$-structure on $Y$, and $\eta_{f}\colon Z_{f}\to Z_{e}$ for the
corresponding homeomorphisms, which are the restrictions of $\rho$ to $Z_{f}$.
We recall from Definition 5.22 that we have decompositions $Y=\bigsqcup_{e\in
B}Z_{e}=\bigsqcup_{f\in R}Z_{f}$ such that, for every $v\in E^{0,0}$, we have
$\bigsqcup_{e\in B_{v}}Z_{e}=\bigsqcup_{f\in R_{v}}Z_{f}$, and, for each $f\in
R$, $\rho|_{Z_{f}}$ is injective and $\rho(Z_{f})=Z_{e}$, where $e$ is the
unique blue edge such that $s(e)=s(f)$.

We use the description of the map $\psi\colon Y\to X$ given in Remark 5.24. It
is clear that, given $y\in Y$, the map $\psi_{*}$ induces a well-defined map
${\pazocal G}(Y,\rho)y\to{\pazocal G}(X,\sigma)\psi(y)$. Observe that

${\pazocal G}(Y,\rho)y=\\{(z,m-n,y)\in{\pazocal
G}(Y,\rho)\mid\rho^{m}(z)=\rho^{n}(y)\\}.$

Moreover, when $\rho^{m}(z)=\rho^{n}(y)$ and $n,m>0$, we can obviously assume
that $\rho^{m-1}(z)\neq\rho^{n-1}(y)$.

We now give a description of the elements $(z,m-n,y)$ as above. Suppose that
$y\in Y$ and that $\rho^{m}(z)=\rho^{n}(y)$, where $n,m>0$ and
$\rho^{m-1}(z)\neq\rho^{n-1}(y)$ (the cases where $n=0$ or $m=0$ are simpler
and left to the reader). Let $g_{1}\in R$ be the unique red edge such that
$y\in Z_{g_{1}}$. Then we have $\eta_{g_{1}}(y)=\rho(y)\in Z_{g_{2}}$ for a
unique $g_{2}\in R$, and continuing in this way we find a sequence
$g_{1},\dots,g_{n}$ of red edges such that $\rho^{i}(y)\in Z_{g_{i+1}}$ for
$0\leq i\leq n-1$. Now
$\rho^{n}(y)=\rho(\rho^{n-1}(y))\in\rho(Z_{g_{n}})=Z_{e}$, where $e$ is the
unique blue edge such that $s(e)=s(g_{n})$. Set $e_{m}:=e$ for notational
convenience. We also have $\rho(\rho^{m-1}(z))=\rho^{m}(z)=\rho^{n}(y)$, and
hence there exists a unique $f_{m}\in s_{\mathrm{red}}^{-1}(s(e_{m}))$ such
that $\rho^{m-1}(z)\in Z_{f_{m}}$. Note that $f_{m}\neq g_{n}$ because
$\rho^{m-1}(z)\neq\rho^{n-1}(y)$. Now $\rho(\rho^{m-2}(z))\in Z_{e_{m-1}}$ for
a unique blue edge $e_{m-1}$, and so $\rho^{m-2}(z)\in Z_{f_{m-1}}$ for a
unique red edge $f_{m-1}\in s_{\mathrm{red}}^{-1}(s(e_{m-1}))$. In this way,
we get unique sequences of blue edges $e_{1},\dots,e_{m}$ and red edges
$f_{1},\dots,f_{m}$, such that $f_{i}\in s_{\mathrm{red}}^{-1}(s(e_{i}))$ for
$1\leq i\leq m$, and $\rho^{i}(z)\in Z_{f_{i+1}}$ for $0\leq i\leq m-1$. Note
that the sequence of red edges $f_{1},\dots,f_{m},g_{1},\dots,g_{n}$ satisfies
the conditions given in Construction 5.19, so that we may consider the element
$a_{f_{1}}^{-1}\cdots a_{f_{m}}^{-1}a_{g_{n}}\cdots
a_{g_{1}}\in\mathbb{F}(A)$.

Now observe that

(6.7)
$\eta_{f_{1}}^{-1}\circ\cdots\circ\eta_{f_{m}}^{-1}\circ\eta_{g_{n}}\circ\cdots\circ\eta_{g_{1}}(y)=z,$

so that certainly
$y\in\text{Dom}(\eta_{f_{1}}^{-1}\circ\cdots\circ\eta_{f_{m}}^{-1}\circ\eta_{g_{n}}\circ\cdots\circ\eta_{g_{1}})$.
Conversely, given a sequence $f_{1},\dots,f_{m},g_{1},\dots,g_{n}$ satisfying
the conditions given in Construction 5.19 and such that
$y\in\text{Dom}(\eta_{f_{1}}^{-1}\circ\cdots\circ\eta_{f_{m}}^{-1}\circ\eta_{g_{n}}\circ\cdots\circ\eta_{g_{1}})$,
the element $z$ defined by equation (6.7) satisfies that
$\rho^{m}(z)=\rho^{n}(y)$ and that $\rho^{m-1}(z)\neq\rho^{n-1}(y)$, thus
$(z,m-n,y)\in{\pazocal G}(Y,\rho)$.

We now have, for $(z,m-n,y)\in{\pazocal G}(Y,\rho)y$ and corresponding red
edges $f_{1},\dots,f_{m},g_{1},\dots,g_{n}$ as above,

$a_{f_{1}}^{-1}\cdots a_{f_{m}}^{-1}a_{g_{n}}\cdots a_{g_{1}}\in\psi(y)$

and, by the $(E,C)$-equivariance of $\psi$,

$\psi(z)=\theta_{f_{1}}^{-1}\circ\cdots\circ\theta_{f_{m}}^{-1}\circ\theta_{g_{n}}\circ\cdots\circ\theta_{g_{1}}(\psi(y)).$

To show that $\psi_{*}|_{{\pazocal G}(Y,\rho)y}$ is surjective onto ${\pazocal
G}(X,\sigma)\psi(y)$, observe that the above argument applied to $(X,\sigma)$
gives that any element in ${\pazocal G}(X,\sigma)\psi(y)$ is of the form

$(\theta_{f_{1}}^{-1}\circ\cdots\circ\theta_{f_{m}}^{-1}\circ\theta_{g_{n}}\circ\cdots\circ\theta_{g_{1}}(\psi(y)),m-n,\psi(y)),$

where $a_{f_{1}}^{-1}\cdots a_{f_{m}}^{-1}a_{g_{n}}\cdots
a_{g_{1}}\in\psi(y)$. Then by the description of $\psi$ in Remark 5.24 we have
that

$y\in\text{Dom}(\eta_{f_{1}}^{-1}\circ\cdots\circ\eta_{f_{m}}^{-1}\circ\eta_{g_{n}}\circ\cdots\circ\eta_{g_{1}}),$

and then
$\psi_{*}(z,m-n,y)=(\theta_{f_{1}}^{-1}\circ\cdots\circ\theta_{f_{m}}^{-1}\circ\theta_{g_{n}}\circ\cdots\circ\theta_{g_{1}}(\psi(y)),m-n,\psi(y))$,
where $z$ is given by equation (6.7). So we get the surjectivity.

Finally, to show that $\psi_{*}|_{{\pazocal G}(Y,\rho)y}$ is injective,
suppose that

$\displaystyle\psi_{*}(z_{1},m-n,y)$
$\displaystyle=(\theta_{f_{1}}^{-1}\circ\cdots\circ\theta_{f_{m}}^{-1}\circ\theta_{g_{n}}\circ\cdots\circ\theta_{g_{1}}(\psi(y)),m-n,\psi(y))$
$\displaystyle=(\theta_{f_{1}^{\prime}}^{-1}\circ\cdots\circ\theta_{f_{m^{\prime}}^{\prime}}^{-1}\circ\theta_{g_{n^{\prime}}^{\prime}}\circ\cdots\circ\theta_{g_{1}^{\prime}}(\psi(y)),m^{\prime}-n^{\prime},\psi(y))$
$\displaystyle=\psi_{*}(z_{2},m^{\prime}-n^{\prime},y),$

for sequences of red edges $f_{1},\dots,f_{m},g_{1},\dots,g_{n}$ and
$f_{1}^{\prime},\dots,f_{m^{\prime}}^{\prime},g_{1}^{\prime},\dots,g_{n^{\prime}}^{\prime}$
satisfying the conditions in Construction 5.19, and which correspond to
$(z_{1},m-n,y),(z_{2},m^{\prime}-n^{\prime},y)\in{\pazocal G}(Y,\rho)y$
respectively. We have $m-n=m^{\prime}-n^{\prime}$ and we may assume that
$m^{\prime}\geq m$, so that setting $m^{\prime}=m+k$ we have $k\geq 0$ and
$n^{\prime}=n+k$. We emphasize that the conditions in Construction 5.19
include in particular that $f_{m}\neq g_{n}$ and that
$f^{\prime}_{m^{\prime}}\neq g^{\prime}_{n^{\prime}}$.

Setting $\xi=\psi(y)$, we have

$\theta_{f_{1}}^{-1}\circ\cdots\circ\theta_{f_{m}}^{-1}\circ\theta_{g_{n}}\circ\cdots\circ\theta_{g_{1}}(\xi)=\theta_{f_{1}^{\prime}}^{-1}\circ\cdots\circ\theta_{f_{m^{\prime}}^{\prime}}^{-1}\circ\theta_{g_{n^{\prime}}^{\prime}}\circ\cdots\circ\theta_{g_{1}^{\prime}}(\xi).$

Hence
$\xi\in\text{Dom}(\theta_{g_{1}})\cap\text{Dom}(\theta_{g^{\prime}_{1}})=H_{g_{1}}\cap
H_{g_{1}^{\prime}}$ and so $g_{1}=g_{1}^{\prime}$. Now we get

$\theta_{f_{1}}^{-1}\circ\cdots\circ\theta_{f_{m}}^{-1}\circ\theta_{g_{n}}\circ\cdots\circ\theta_{g_{2}}(\xi_{1})=\theta_{f_{1}^{\prime}}^{-1}\circ\cdots\circ\theta_{f_{m^{\prime}}^{\prime}}^{-1}\circ\theta_{g_{n^{\prime}}^{\prime}}\circ\cdots\circ\theta_{g_{2}^{\prime}}(\xi_{1}),$

where $\xi_{1}=\theta_{g_{1}}(\xi)=\sigma(\xi)$. The same argument gives that
$g_{2}=g_{2}^{\prime}$. Proceeding in this way, we obtain that
$g_{i}=g_{i}^{\prime}$ for $1\leq i\leq n$. Set
$\xi_{n}=\theta_{g_{n}}\circ\cdots\circ\theta_{g_{1}}(\xi)$. Then we have

$\xi^{\prime}:=\theta_{f_{1}}^{-1}\circ\cdots\circ\theta_{f_{m}}^{-1}(\xi_{n})=\theta_{f_{1}^{\prime}}^{-1}\circ\cdots\circ\theta_{f^{\prime}_{m}}^{-1}\circ\theta_{f^{\prime}_{m+1}}^{-1}\circ\cdots\circ\theta_{f_{m+k}^{\prime}}^{-1}\circ\theta_{g_{n+k}^{\prime}}\circ\cdots\circ\theta_{g_{n+1}^{\prime}}(\xi_{n}).$

Hence $\xi^{\prime}\in H_{f_{1}}\cap H_{f_{1}^{\prime}}$, and thus
$f_{1}=f_{1}^{\prime}$. Applying the same argument to
$\theta_{f_{1}}(\xi^{\prime})$, we get that $f_{2}=f_{2}^{\prime}$, and
continuing in this way, that $f_{i}=f_{i}^{\prime}$ for $1\leq i\leq m$.
Therefore, we get

$\xi_{n}=\theta_{f_{m}}\circ\cdots\circ\theta_{f_{1}}(\xi^{\prime})=\theta_{f^{\prime}_{m+1}}^{-1}\circ\cdots\circ\theta_{f_{m+k}^{\prime}}^{-1}\circ\theta_{g_{n+k}^{\prime}}\circ\cdots\circ\theta_{g_{n+1}^{\prime}}(\xi_{n}).$

If $k>0$, then we have

$\theta_{f_{m+k}^{\prime}}\circ\cdots\circ\theta_{f_{m+1}^{\prime}}(\xi_{n})=\theta_{g_{n+k}^{\prime}}\circ\cdots\circ\theta_{g_{n+1}^{\prime}}(\xi_{n}),$

and proceeding as above we get $g_{n+i}^{\prime}=f_{n+i}^{\prime}$ for $1\leq
i\leq k$. In particular we have
$f^{\prime}_{m^{\prime}}=f_{m+k}^{\prime}=g_{n+k}^{\prime}=g^{\prime}_{n^{\prime}}$,
which is a contradiction with the condition $f^{\prime}_{m^{\prime}}\neq
g^{\prime}_{n^{\prime}}$. Therefore $k=0$, and thus we have $m^{\prime}=m$,
$n^{\prime}=n$, $f_{i}=f_{i}^{\prime}$ for $1\leq i\leq m$, and
$g_{j}=g_{j}^{\prime}$ for $1\leq j\leq n$. Hence we obtain that
$(z_{1},m-n,y)=(z_{2},m^{\prime}-n^{\prime},y)$, as desired. ∎

######  Theorem 6.21.

Let $(Y,\rho)\in\emph{{LHomeo}}$ and let $(E,C)$ be a generalized finite shift
graph. Suppose that we have an $(E,C)$-structure on $Y$, and let $\psi\colon
Y\to X$ be the unique $(E,C)$-equivariant continuous map, where $(X,\sigma)$
is the generalized finite shift associated to $(E,C)$. Then the associated
groupoid homomorphism

$\psi_{*}\colon{\pazocal G}(Y,\rho)\to{\pazocal G}(X,\sigma)$

given in Lemma 6.18 induces a $K$-algebra $*$-homomorphism

$\psi^{*}\colon A_{K}({\pazocal G}(X,\sigma))\to A_{K}({\pazocal G}(Y,\rho))$

by the rule $\psi^{*}(f)=f\circ\psi_{*}$ for $f\in A_{K}({\pazocal
G}(X,\sigma))$.

###### Proof.

We first show that $\psi_{*}$ is proper. For this it is enough to shown that
$\psi_{*}^{-1}(Z(U,m,n,V))$ is compact, where $U$ and $V$ are open compact
subsets of $X$ such that $\sigma^{m}|_{U}$ and $\sigma^{n}|_{V}$ are
injective, and $\sigma^{m}(U)=\sigma^{n}(V)$.

For such open compact subset $Z(U,m,n,V)$ of ${\pazocal G}(X,\sigma)$, we
claim that the set $Z(\psi^{-1}(U),m,n,\psi^{-1}(V))$ is a bisection
homeomorphic to $\psi^{-1}(V)$, and thus a compact set. For this, we show that
the source map $s$ induces a bijection

$Z(\psi^{-1}(U),m,n,\psi^{-1}(V))\to\psi^{-1}(V).$

Let $y\in\psi^{-1}(V)$. Then $\psi(y)\in V$ and since
$\sigma^{m}(U)=\sigma^{n}(V)$, there is $x\in U$ such that
$\sigma^{m}(x)=\sigma^{n}(\psi(y))$. Hence $(x,m-n,\psi(y))\in{\pazocal
G}(X,\sigma)\psi(y)$. Now by the proof of surjectivity in Lemma 6.20, there is
$z\in Y$ such that $\psi_{*}(z,m-n,y)=(x,m-n,\psi(y))$. But then $(z,m-n,y)\in
Z(\psi^{-1}(U),m,n,\psi^{-1}(V))$ and $s(z,m-n,y)=y$, showing surjectivity.

To show injectivity, suppose that $(z_{1},m-n,y),(z_{2},m-n,y)$ are two
elements in $Z(\psi^{-1}(U),m,n,\psi^{-1}(V))$. Then,
$(\psi(z_{1}),m-n,\psi(y)),(\psi(z_{2}),m-n,\psi(y))\in Z(U,m,n,V)$, and thus
$\sigma^{m}(\psi(z_{1}))=\sigma^{n}(\psi(y))=\sigma^{m}(\psi(z_{2}))$. But
$\sigma^{m}|_{U}$ is injective, so $\psi(z_{1})=\psi(z_{2})$. By Lemma 6.20,
we conclude that $z_{1}=z_{2}$.

We have completed the proof that $\psi_{*}$ is proper. Hence, since it is also
continuous by Lemma 6.18, we obtain that $\psi^{*}$ is a well-defined linear
map. We have to show that $\psi^{*}$ preserves the convolution product. For
$f,g\in A_{K}({\pazocal G}(X,\sigma))$ and an element
$\gamma=(z,m-n,y)\in{\pazocal G}(Y,\rho)$, we have

$\displaystyle(\psi^{*}(f)*\psi^{*}(g))(\gamma)$
$\displaystyle=\sum_{\alpha\in{\pazocal
G}(Y,\rho)y}\psi^{*}(f)(\gamma\alpha^{-1})\psi^{*}(g)(\alpha)$
$\displaystyle=\sum_{\alpha\in{\pazocal
G}(Y,\rho)y}f(\psi_{*}(\gamma)\psi_{*}(\alpha)^{-1})g(\psi_{*}(\alpha))$
$\displaystyle=\sum_{\beta\in{\pazocal
G}(X,\sigma)\psi(y)}f(\psi_{*}(\gamma)\beta^{-1})g(\beta)=\psi^{*}(f*g)(\gamma),$

where we have used Lemma 6.20 in the third equality. We thus have
$\psi^{*}(f)*\psi^{*}(g)=\psi^{*}(f*g)$. It is straightforward to check that
$\psi^{*}$ is a $*$-map. ∎

We also have a similar result for groupoid $C^{*}$-algebras.

######  Theorem 6.22.

Let $(Y,\rho)\in\emph{{LHomeo}}$ and $(X,\sigma)$ and $\psi$ be as in Theorem
6.21. Then the groupoid homomorphism $\psi_{*}\colon{\pazocal
G}(Y,\rho)\to{\pazocal G}(X,\sigma)$ induces a $*$-homomorphism

$\psi^{*}\colon C^{*}({\pazocal G}(X,\sigma))\to C^{*}({\pazocal G}(Y,\rho))$

such that $\psi^{*}(f)=f\circ\psi_{*}$ for $f\in C_{c}({\pazocal
G}(X,\sigma))$.

###### Proof.

As in the proof of Theorem 6.21, we see that $\psi^{*}$ gives a
$*$-homomorphism $C_{c}({\pazocal G}(X,\sigma))\to C_{c}({\pazocal
G}(Y,\rho))$. Since $C^{*}({\pazocal G})$ is the enveloping $C^{*}$-algebra of
$C_{c}({\pazocal G})$ for any second-countable locally compact Hausdorff étale
groupoid ${\pazocal G}$ (see [55, Theorem 9.2.3 and Lemma 9.2.4]), we conclude
the result. Alternatively, we may also use that $C^{*}({\pazocal G})$ is the
enveloping $C^{*}$-algebra of $A_{{\mathbb{C}}}({\pazocal G})$, and use
Theorem 6.21. ∎

###  6.3. Algebras as colimits of generalized finite shift algebras

We now consider the situation treated at the end of Section 5, so that
$(X,\sigma)\in\textsf{LHomeo}$, $(F,D)$ is a fixed $l$-diagram for
$(X,\sigma)$, and we set $(E(n),C(n))=(F_{2n},D^{2n})$ for all $n\geq 0$.
Recall that $(E(n),C(n))$ is a generalized finite shift graph, with associated
generalized finite shift $(X_{n},\sigma_{n})$. We denote by $(F(n),D(n))$ the
canonical resolution of $(E(n),C(n))$, which is an $l$-diagram for
$(X_{n},\sigma_{n})$. By Theorem 5.26, we have
$(X,\sigma)=\varprojlim(X_{n},\sigma_{n})$, and the inverse limit maps
$\psi_{n}\colon X\to X_{n}$ and the connecting maps $\psi_{n,n+1}\colon
X_{n+1}\to X_{n}$ are universal maps induced by the corresponding
$(E(n),C(n))$-structures. It therefore follows from Theorems 6.21 and 6.22
that we have directed systems

$(A_{K}({\pazocal G}(X_{n},\sigma_{n})),\psi_{n,n+1}^{*})\quad\text{ and
}\quad(C^{*}({\pazocal G}(X_{n},\sigma_{n})),\psi_{n,n+1}^{*})$

together with compatible $*$-homomorphisms

$A_{K}({\pazocal
G}(X_{n},\sigma_{n}))\overset{\psi_{n}^{*}}{\longrightarrow}A_{K}({\pazocal
G}(X,\sigma))\quad\text{ and }\quad C^{*}({\pazocal
G}(X_{n},\sigma_{n}))\overset{\psi_{n}^{*}}{\longrightarrow}C^{*}({\pazocal
G}(X,\sigma)).$

We show in this section that indeed $(A_{K}({\pazocal
G}(X,\sigma)),\psi_{n}^{*})$ and $(C^{*}({\pazocal
G}(X,\sigma)),\psi_{n}^{*})$ are the colimits of the above directed systems in
the categories of $*$-algebras and $C^{*}$-algebras, respectively.

We introduce some useful notation. Recall the definitions of the $*$-algebra
${\pazocal A}_{\infty}$ and the $C^{*}$-algebra ${\pazocal B}_{\infty}$
associated to the $l$-diagram $(F,D)$ from Subsection 6.1.

Given a generalized finite shift graph $(E,C)$, we identify $LV_{K}(E,C)$ with
the corner $VL_{K}(E,C)V$, where $V=\sum_{v\in E^{0,0}}v\in L_{K}(E,C)$, see
Proposition 6.10. Thinking of $V$ as a projection in $C^{*}(E,C)$, we set
$C^{*}V(E,C):=VC^{*}(E,C)V$.

######  Notation 6.23.

Let $(E,C)$ be a generalized finite shift graph. We denote by $LV^{{\rm
ab}}_{K}(E,C)$ the universal tame $*$-algebra associated to the canonical
generating set of partial isometries of $LV_{K}(E,C)$. Similarly ${\pazocal
O}V(E,C)$ will denote the universal tame $C^{*}$-algebra associated to
$C^{*}V(E,C)$. By using [5, Lemma 4.3], we will identify $LV^{{\rm
ab}}_{K}(E,C)$ with the algebra ${\pazocal A}_{\infty}$ of the canonical
resolution of $(E,C)$. Moreover, the arguments in [5, Lemma 4.3] apply
verbatim to the $C^{*}$-algebra counterparts, therefore we will also identify
${\pazocal O}V(E,C)$ with the ${\pazocal B}_{\infty}$ $C^{*}$-algebra
associated to the canonical resolution of $(E,C)$.

Applying Theorem 6.16 to $(F(n),D(n))$ we obtain a canonical $*$-isomorphism

$\varphi^{(n)}\colon LV^{{\rm ab}}_{K}(E(n),C(n))={\pazocal
A}_{\infty}^{(n)}\longrightarrow A_{K}({\pazocal G}(X_{n},\sigma_{n})),$

where $LV^{{\rm ab}}_{K}(E(n),C(n))={\pazocal
A}_{\infty}^{(n)}=\varinjlim_{i}p_{2i}^{(n)}L_{K}(F(n)_{2i},D(n)^{2i})p_{2i}^{(n)}$
is the ${\pazocal A}_{\infty}$ algebra corresponding to $(F(n),D(n))$ (see
Notation 6.23). In particular, since the first layer of $(F(n),D(n))$ is
$(E(n),C(n))$, we have the $*$-homomorphism

$\varphi^{(n)}_{0}\colon LV_{K}(E(n),C(n))\to A_{K}({\pazocal
G}(X_{n},\sigma_{n}))$

which satisfies

(6.8) $\varphi^{(n)}_{0}=\varphi^{(n)}\circ\pi_{n},$

where $\pi_{n}\colon LV_{K}(E(n),C(n))\to LV^{{\rm ab}}_{K}(E(n),C(n))$ is the
canonical projection map.

In a completely analogous fashion, using Theorem 6.17, we get $*$-isomorphisms

$\overline{\varphi}^{(n)}\colon{\pazocal O}V(E(n),C(n))={\pazocal
B}_{\infty}^{(n)}\longrightarrow C^{*}({\pazocal G}(X_{n},\sigma_{n}))$

satisfying corresponding properties.

We can now obtain the following result, which links the two developments in
Subsections 6.1 and 6.2.

######  Theorem 6.24.

Let $(X,\sigma)\in\emph{{LHomeo}}$ and let $(F,D)$ be an $l$-diagram for
$(X,\sigma)$. With the above notation, for each $n\geq 0$ there exists a
unique $*$-homomorphism $\phi_{n}^{\mathrm{ab}}\colon LV^{{\rm
ab}}_{K}(E(n),C(n))\to LV^{{\rm ab}}_{K}(E(n+1),C(n+1))$ such that the
following diagram

(6.9)
$\textstyle{LV_{K}(E(n),C(n))\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{n}}$$\scriptstyle{\phi_{n}}$$\textstyle{LV^{{\rm
ab}}_{K}(E(n),C(n))\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{n}^{\rm
ab}}$$\scriptstyle{\kern 5.0pt{\varphi^{(n)}}}$$\scriptstyle{\kern
5.0pt{\cong}}$$\textstyle{A_{K}({\pazocal
G}(X_{n},\sigma_{n}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{n,n+1}^{*}}$$\textstyle{LV_{K}(E(n+1),C(n+1))\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{n+1}}$$\scriptstyle{\phi_{n+1,\infty}}$$\textstyle{LV^{{\rm
ab}}_{K}(E(n+1),C(n+1))\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\kern
10.0pt{\varphi^{(n+1)}}}$$\scriptstyle{\kern
10.0pt{\cong}}$$\scriptstyle{\phi_{n+1,\infty}^{{\rm
ab}}}$$\textstyle{A_{K}({\pazocal
G}(X_{n+1},\sigma_{n+1}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{n+1}^{*}}$$\textstyle{{\pazocal
A}_{\infty}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\rm
id}}$$\textstyle{{\pazocal
A}_{\infty}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\kern-15.0pt{\varphi}}$$\scriptstyle{\kern-15.0pt{\cong}}$$\textstyle{A_{K}({\pazocal
G}(X,\sigma))}$

is commutative, where $\pi_{n}\colon LV_{K}(E(n),C(n))\to LV^{{\rm
ab}}_{K}(E(n),C(n))$ are the canonical quotient maps. Hence we get
$*$-isomorphisms

(6.10) $A_{K}({\pazocal G}(X,\sigma))\cong{\pazocal A}_{\infty}\cong\varinjlim
LV^{{\rm ab}}_{K}(E(n),C(n))\quad\text{ and }\quad A_{K}({\pazocal
G}(X,\sigma))\cong\varinjlim A_{K}({\pazocal G}(X_{n},\sigma_{n})).$

Similarly, we have $*$-isomorphisms

(6.11) $C^{*}({\pazocal G}(X,\sigma))\cong{\pazocal
B}_{\infty}\cong\varinjlim{{\pazocal O}V}(E(n),C(n))\quad\text{ and }\quad
C^{*}({\pazocal G}(X,\sigma))\cong\varinjlim C^{*}({\pazocal
G}(X_{n},\sigma_{n})).$

###### Proof.

Since $\varphi^{(n)}$ and $\varphi^{(n+1)}$ are isomorphisms, there is a
unique $*$-homomorphism

$\phi_{n}^{{\rm ab}}\colon LV^{{\rm ab}}_{K}(E(n),C(n))\to LV^{{\rm
ab}}_{K}(E(n+1),C(n+1))$

making the upper right square in (6.9) commutative.

By (6.8), in order to show that the upper left square in (6.9) is commutative,
it suffices to show the identity
$\varphi_{0}^{(n+1)}\circ\phi_{n}=\psi_{n,n+1}^{*}\circ\varphi_{0}^{(n)}$. To
do so, we have to fix some notation.

For $u\in E(n)^{0}$, we will denote by $Z^{(n)}(u)$ the corresponding cylinder
set in $X_{n}$. Let $(G,D)$ be the $l$-diagram obtained by placing the two
layers $(F_{2n},D^{2n})=(E(n),C(n))$ and $(F_{2n+1},D^{2n+1})$ on top of
$(F(n+1),D(n+1))$, and denote by ${\pazocal P}_{i}^{G}$, $i\geq 0$, the
corresponding partitions of $X_{n+1}$. Note that the first three layers of
$(G,D)$ are identical to
$(F_{2n},D^{2n})\cup(F_{2n+1},D^{2n+1})\cup(F_{2n+2},D^{2n+2})$, so in
particular we have a natural bijection

$\gamma\colon E(n)^{0,0}\cup E(n)^{0,1}\cup E(n+1)^{0,0}\cup
E(n+1)^{0,1}\longrightarrow{\pazocal P}^{G}_{0}\cup{\pazocal
P}^{G}_{1}\cup{\pazocal P}^{G}_{2}\cup{\pazocal P}^{G}_{3}.$

Note that $\gamma(u)=Z^{(n+1)}(u)$ for $u\in E(n+1)^{0}$. Using the definition
of $\psi_{n,n+1}$, we get, for $u\in E(n)^{0,0}$, that
$\psi_{n,n+1}^{-1}(Z^{(n)}(u))=\gamma(u)$, so setting $U:=\gamma(u)$, we have

$\psi_{n,n+1}^{-1}(Z^{(n)}(u))=U=\bigsqcup_{\\{U^{\prime}\in{\pazocal
P}^{G}_{2}\mid U^{\prime}\subseteq U\\}}U^{\prime}.$

Similarly, for $w\in E(n)^{0,1}$, we have
$\psi_{n+1,n}^{-1}(Z^{(n)}(w))=\bigsqcup_{W^{\prime}}W^{\prime}$ where the
union is extended to all $W^{\prime}\in{\pazocal P}^{G}_{3}$ such that
$W^{\prime}\subseteq W$, where $W:=\gamma(w)\in{\pazocal P}^{G}_{1}$. It
follows that, for $u\in E(n)^{0,0}$, and with $U:=\gamma(u)$, we have

$\displaystyle\psi_{n,n+1}^{*}(\varphi_{0}^{(n)}(u))$
$\displaystyle=\psi_{n,n+1}^{*}(\mathds{1}_{Z^{(n)}(u)})$
$\displaystyle=\mathds{1}_{Z^{(n)}(u)}\circ\psi_{n,n+1}$
$\displaystyle=\mathds{1}_{\psi_{n,n+1}^{-1}(Z^{(n)}(u))}$
$\displaystyle=\sum_{\\{U^{\prime}\in{\pazocal P}^{G}_{2}\mid
U^{\prime}\subseteq U\\}}\mathds{1}_{U^{\prime}}$
$\displaystyle=\varphi_{0}^{(n+1)}(\mathbf{S}(u))=\varphi_{0}^{(n+1)}(\phi_{n}(u)).$

Similarly
$\psi_{n,n+1}^{*}(\varphi_{0}^{(n)}(w))=\varphi_{0}^{(n+1)}(\phi_{n}(w))$ for
all $w\in E(n)^{0,1}$.

Let now $f\in R_{u}$, $e\in B_{v}$, where $e,f\in E(n)^{1}$, $u,v\in
E(n)^{0,0}$, and $s(e)=s(f)=:w\in E(n)^{0,1}$. Using Lemma 6.1 and the
definition of $\phi_{n}$, we have

$\phi_{n}(fe^{*})=\sum_{(f^{\prime},e^{\prime})\in
A}f^{\prime}(e^{\prime})^{*},$

where $A$ is the set of all pairs $(f^{\prime},e^{\prime})\in E(n+1)^{1}\times
E(n+1)^{1}$ such that $f^{\prime}\in R_{u^{\prime}}$, $e^{\prime}\in
B_{v^{\prime}}$, $u^{\prime}=r(f^{\prime})\leq\mathbf{S}(u)$,
$v^{\prime}=r(e^{\prime})\leq\mathbf{S}(v)$ and
$w^{\prime}:=s(f^{\prime})=s(e^{\prime})\leq\mathbf{S}(w)$. Each pair
$(f^{\prime},e^{\prime})\in A$ corresponds exactly to a triple
$(U^{\prime},W^{\prime},V^{\prime})$ such that
$U^{\prime},V^{\prime}\in{\pazocal P}^{G}_{2}$ and $W^{\prime}\in{\pazocal
P}^{G}_{3}$, with $W^{\prime}\subseteq\sigma_{n+1}(U^{\prime})$ and
$W^{\prime}\subseteq V^{\prime}$, with $U^{\prime}=\gamma(u^{\prime})\subseteq
U=\gamma(u)$, $V^{\prime}=\gamma(v^{\prime})\subseteq V:=\gamma(v)$ and
$W^{\prime}=\gamma(w^{\prime})\subseteq W=\gamma(w)$. Moreover, observe that

$\varphi^{(n+1)}_{0}(f^{\prime}(e^{\prime})^{*})=\mathds{1}_{Z((\sigma|_{U^{\prime}})^{-1}(W^{\prime}),1,0,W^{\prime})}\in
A_{K}({\pazocal G}(X_{n+1},\sigma_{n+1})).$

We denote by $B$ the family of all such triples
$(U^{\prime},W^{\prime},V^{\prime})$. Using this, we compute

$\displaystyle\psi_{n,n+1}^{*}(\varphi_{0}^{(n)}(fe^{*}))$
$\displaystyle=\psi_{n,n+1}^{*}(\mathds{1}_{Z((\sigma_{n}|_{Z^{(n)}(u)})^{-1}(Z^{(n)}(w)),1,0,Z^{(n)}(w))})$
$\displaystyle=\mathds{1}_{Z((\sigma_{n}|_{Z^{(n)}(u)})^{-1}(Z^{(n)}(w)),1,0,Z^{(n)}(w))}\circ(\psi_{n,n+1})_{*}$
$\displaystyle=\mathds{1}_{({\psi_{n+1,n}}_{*})^{-1}Z((\sigma_{n}|_{Z^{(n)}(u)})^{-1}(Z^{(n)}(w)),1,0,Z^{(n)}(w))}$
$\displaystyle=\mathds{1}_{(Z((\sigma_{n+1}|_{U})^{-1}(W)),1,0,W)}$
$\displaystyle=\sum_{(U^{\prime},W^{\prime},V^{\prime})\in
B}\mathds{1}_{Z((\sigma_{n+1}|_{U^{\prime}})^{-1}(W^{\prime}),1,0,W^{\prime})}$
$\displaystyle=\sum_{(f^{\prime},e^{\prime})\in
A}\varphi_{0}^{(n+1)}(f^{\prime}(e^{\prime})^{*})=\varphi_{0}^{(n+1)}(\phi_{n}(fe^{*})).$

Since the family of partial isometries $fe^{*}$, with $f\in R_{u}$, $e\in
B_{v}$, $u,v\in E(n)^{0,0}$, generates $LV_{K}(E(n),C(n))$ as a $*$-algebra,
we get that
$\psi_{n,n+1}^{*}\circ\varphi_{0}^{(n)}=\varphi_{0}^{(n+1)}\circ\phi_{n}$, and
hence $\phi_{n}^{\text{ab}}\circ\pi_{n}=\pi_{n+1}\circ\phi_{n}$, so that the
upper left square of (6.9) is commutative.

Using that $\varphi$ is an isomorphism, we see that there is a unique
$*$-homomorphism

$\phi_{n+1,\infty}^{\text{ab}}\colon LV^{{\rm
ab}}_{K}(E(n+1),C(n+1))\longrightarrow{\pazocal A}_{\infty}$

making the lower right square of (6.9) commutative. Now a computation similar
to the above gives that

$\psi_{n+1}^{*}\circ\varphi_{0}^{(n+1)}=\varphi_{n+1}=\varphi\circ\phi_{n+1,\infty},$

and so the lower left square of (6.9) is also commutative.

Now recall that, by its definition, $({\pazocal A}_{\infty},\phi_{n,\infty})$
is the colimit of the directed system

$(LV_{K}(E(n),C(n)),\phi_{n}).$

Since all the maps $\pi_{n}$ are surjective, the commutativity of (6.9) gives
first that the maps $\phi_{n,\infty}^{{\rm ab}}$ are compatible with the
transition maps $\phi_{n}^{{\rm ab}}$, that is, $\phi_{n,\infty}^{{\rm
ab}}=\phi_{n+1,\infty}^{{\rm ab}}\circ\phi_{n}^{{\rm ab}}$, and second that
$({\pazocal A}_{\infty},\phi_{n,\infty}^{{\rm ab}})$ is the colimit of the
directed system

$(LV^{{\rm ab}}_{K}(E(n),C(n)),\phi_{n}^{{\rm ab}}).$

This gives the isomorphism ${\pazocal A}_{\infty}\cong\varinjlim
LV_{K}^{\text{ab}}(E(n),C(n))$. Using the commutativity of the right squares
of (6.9) and the fact that $\varphi$ and all $\varphi^{(n)}$ are isomorphisms,
we obtain (6.10).

The proof of (6.11) uses a diagram of $C^{*}$-algebras obtained by taking
$C^{*}$-completions in diagram (6.9). We leave the routine details to the
reader. This concludes the proof. ∎

## Acknowledgments

The authors would like to thank the anonymous referee for her/his many
suggestions, which have improved the exposition of the paper.

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