Morita Enveloping Fell Bundles

Abstract

We introduce notions of weak and strong equivalence for non-saturated Fell bundles over locally compact groups and show that every Fell bundle is strongly (resp. weakly) equivalent to a semidirect product Fell bundle for a partial (resp. global) action. Equivalences preserve cross-sectional ${\mathrm {C}}^*$-algebras and amenability. We use this to show that previous results on crossed products and amenability of group actions carry over to Fell bundles.

Appendix A: Tensor Products of Equivalence Bundles

Throughout this section we use the construction of adjoint and tensor product of equivalence bundles of Abadie and Ferraro (2017).

Theorem A.1

If $\mathcal {X}$ and $\mathcal {Y}$ are $\mathcal {A}-\mathcal {B}$ and $\mathcal {B}-\mathcal {C}$-strong equivalence bundles, respectively, then the tensor product bundle $\mathcal {Z}:=\mathcal {X}\otimes _\mathcal {B}\mathcal {Y}$ is (left and right) strongly full. In particular, strong equivalence of Fell bundles is an equivalence relation.

Proof

To show that $\mathcal {Z}$ is (right) full the authors show that given $r,s\in G$, $x_1,x_2\in X_r$, $y_1,y_2\in Y_s$ and $\varepsilon >0$ there exists $\xi _1,\xi _2\in Z_{rs}$ such that $\Vert \langle y_1,\langle x_1,x_2{\rangle _\mathcal {B}}y_2{\rangle _\mathcal {C}}- \langle \xi _1,\xi _2{\rangle _\mathcal {C}}\Vert <\varepsilon$. Now we use that fact to show $\mathcal {Z}$ is strongly full.

Fix $t\in G$, $c\in C_t^*C_t$ and $\varepsilon >0$. Since $\mathcal {Y}$ is strongly full there exists $y_{j,k}\in Y_t$ ($j=1,2$ and $k=1,\ldots ,n$) such that $\Vert c - \sum _{k=1}^n \langle y_{1,k},y_{2,k}{\rangle _\mathcal {C}}\Vert <\varepsilon$. We also know that $\mathcal {X}$ is strongly full, then $X_e$ is a $A_e-B_e$-equivalence bimodule and we can find $x_{j,k}\in X_e$ ($j=1,2$ and $k=1,\ldots ,m$) such that

$$\begin{aligned} \delta :=\left\| c - \sum _{k=1}^n\sum _{l=1}^m \langle y_{1,k}, \langle x_{1,l},x_{2,l}{\rangle _\mathcal {B}}y_{2,k}{\rangle _\mathcal {C}}\right\| <\varepsilon . \end{aligned}$$

From the first paragraph of this proof it follows that we may find $\xi _{p,k,l}\in Z_t$ ($p=1,2$, $k=1,\ldots ,n$, and $l=1,\ldots ,m$) such that $\Vert \langle y_{1,k}, \langle x_{1,l},x_{2,l}{\rangle _\mathcal {B}}y_{2,k}{\rangle _\mathcal {C}}- \langle \xi _{1,k,l}, \xi _{2,k,l}{\rangle _\mathcal {C}}\Vert < \frac{\varepsilon -\delta }{nm}.$ Thus

$$\begin{aligned} \left\| c - \sum _{k=1}^n\sum _{l=1}^m \langle \xi _{k,l,1}, \xi _{k,l,2}{\rangle _\mathcal {C}}\right\| <\varepsilon . \end{aligned}$$

By symmetry, $\mathcal {Z}$ is also left strongly full. Hence strong equivalence is a transitive relation. Regarding the symmetric and reflexive properties of strong equivalence, we leave to the reader the verification of the fact that the adjoint of $\mathcal {X}$, $\tilde{\mathcal {X}}$, is a full $\mathcal {B}- \mathcal {A}$-equivalence bundle and that $\mathcal {B}$, considered as a $\mathcal {B}-\mathcal {B}$-equivalence bundle in the natural way, are left and right strongly full. $\square$

An equivalence module ${}_AX_B$ between the ${{\mathrm {C}}^{*}}$-algebras A and B is usually viewed as an arrow from A to B; here we view it as an arrow from B to A to be consistent with our convention at the beginning of Sect. 2 where we view Fell bundles as actions by equivalences. The composition of arrows is given by inner tensor product. Although the tensor products $(X\otimes _B Y)\otimes _C Z$ and $X\otimes _B (Y\otimes _C Z)$ are not the same object, but they are naturally isomorphic (via a unitary); this gives the associativity of composition. Then we obtain a category with ${\mathrm {C}}^*$-algebras as objects and unitary equivalence classes of equivalence modules as objects. To proceed analogously with Fell bundles we need a notion of unitary operator between equivalence bundles.

Definition A.2

Let $\mathcal {X}$ and $\mathcal {Y}$ be two $\mathcal {A}-\mathcal {B}$-equivalence bundles. A unitary from $\mathcal {X}$ to $\mathcal {Y}$ is an isomorphism of equivalence bundles $\rho :\mathcal {X}\rightarrow \mathcal {Y}$ such that ${}_\mathcal {A}\langle \rho (x),\rho (y){\rangle }= {}_\mathcal {A}\langle x,y{\rangle }$ and $\langle \rho (x),\rho (y){\rangle _\mathcal {B}}= \langle x,y{\rangle _\mathcal {B}}$ (this means that $\rho ^l = {\text {id}}_\mathcal {A}$ and $\rho ^r={\text {id}}_\mathcal {B}$ in the notation of Abadie and Ferraro (2017)). If such an isomorphism exists, we say that $\mathcal {X}$ is unitarily equivalent (or just isomorphic) to $\mathcal {Y}$.

To obtain a category with Fell bundles (over a fixed group G) as objects, isomorphism classes of equivalence bundles as morphisms and the tensor product of Abadie and Ferraro (2017) as composition we need to show that the composition is well defined on the of isomorphism classes and that it is associative. This boils down to the following results.

Proposition A.3

Suppose $\pi :\mathcal {X}_1\rightarrow \mathcal {X}_1$ is a unitary between $\mathcal {A}-\mathcal {B}$-equivalence bundles and $\rho :\mathcal {Y}_1\rightarrow \mathcal {Y}_2$ is a unitary between $\mathcal {B}-\mathcal {C}$-equivalence bundles. Then $\mathcal {X}_1\otimes _\mathcal {B}\mathcal {Y}_1$ is unitarily equivalent to $\mathcal {X}_2\otimes _\mathcal {B}\mathcal {Y}_2$.

Proof

The way the tensor product is constructed is one of the key factors of this proof, so it will be necessary to recall it here. Start by considering the bundle $\mathcal {Z}_j:=\{{X_j}_r\otimes _{B_e}{Y_j}_s\}_{(r,s)\in G\times G}$, $j=1,2$. The topology of that bundle is determined by the set of sections $\Gamma _j:={\text {span}}\{f\boxtimes g:f\in \mathrm {C_c}(\mathcal {X}_j),\ g\in \mathrm {C_c}(\mathcal {Y}_j)\}$ where $f\boxtimes g(r,s)=f(r)\otimes g(s)$. Note that Doran and Fell (1988a, II 13.16) implies the existence of a unique isomorphism of Banach bundles $\mu :\mathcal {Z}_1\rightarrow \mathcal {Z}_2$ such that $\mu (x\otimes y)=\pi (x)\otimes \rho (y)$. Recall that the construction of $\mathcal {X}_j\otimes _\mathcal {B}\mathcal {Y}_j$ is performed using actions of $\mathcal {A}$ and $\mathcal {C}$ on $\mathcal {Z}_j$ and operations $\triangleleft _j:\mathcal {Z}_j\times \mathcal {Z}_j\rightarrow \mathcal {C}$ and $\triangleright _j:\mathcal {Z}_j\times \mathcal {Z}_j\rightarrow \mathcal {A}$ uniquely determined by the identities

$$\begin{aligned} a(x\otimes y)&= (ax)\otimes y&(x\otimes y)c&= x\otimes (yc)\\ (x\otimes y)\triangleleft _j(u\otimes v)&= {}_\mathcal {A}\langle x {}_\mathcal {B}\langle y,v\rangle ,u{\rangle }&(x\otimes y)\triangleright _j(u\otimes v)&= \langle y, \langle x,u{\rangle _\mathcal {B}}u{\rangle _\mathcal {C}}\end{aligned}$$

The reader can easily check that $a\mu (z)=\mu (az)$, $\mu (z)c=\mu (z)$, $\mu (z)\triangleleft _2 \mu (z') = z \triangleleft _1 z'$ and $\mu (z)\triangleright _2 \mu (z') = z \triangleright _2 z'$ for all $z,z'\in \mathcal {Z}_1$, $a\in \mathcal {A}$ and $c\in \mathcal {C}$.

If we think of $\mathcal {Z}_1$ and $\mathcal {Z}_2$ as the same object, then there is nothing else to prove and $\mathcal {X}_1\otimes _\mathcal {B}\mathcal {Y}_1$ is in fact the same as $\mathcal {X}_2\otimes _\mathcal {B}\mathcal {Y}_2$. In other case the next step is to define, for every $t\in G$, ${U_j}_t$ as the reduction of $\mathcal {Z}_j$ to $H^t:=\{(r,s)\in G\times G:rs=t\}$. Then we get an untopologized bundle $\mathcal {U}_j:=\{ {U_j}_t \}_{t\in G}$ and define pre-inner product and actions in the following way:

$$\begin{aligned} {{}_\mathcal {A}^{\mathcal {U}_{j}}\langle } u,v \rangle&:=\iint _{G\times G} u(p,{p^{-1}}r)\triangleleft _j v(q,{q^{-1}}s)\,\mathrm dp\,\mathrm dq; \end{aligned}$$

(A.1)

$$\begin{aligned} \langle u,v {\rangle _\mathcal {C}^{\mathcal {U}_{j}}}&:=\iint _{G\times G} u(p,{p^{-1}}r)\triangleright _j v(q,{q^{-1}}s)\,\mathrm dp\,\mathrm dq; \end{aligned}$$

(A.2)

$$\begin{aligned} au\in {U_j}_{tr}&\text{ by } \text{ the } \text{ formula } (au)(p,{p^{-1}}tr):= au({t^{-1}}p,{p^{-1}}tr) \text{ and } \end{aligned}$$

(A.3)

$$\begin{aligned} uc\in {U_j}_{rt}&\text{ by } \text{ the } \text{ formula } (uc)(p,{p^{-1}}rt):= u(p,{p^{-1}}r)c, \end{aligned}$$

(A.4)

where $u\in {U_j}_r$, $v\in {U_j}_s$, $a\in A_t$ and $c\in C_t$.

It is then clear that the composition with $\mu$ identifies the pre-inner products an actions of $\mathcal {U}_1$ and $\mathcal {U}_2$, for example ${{}_\mathcal {A}^{\mathcal {U}_{1}}\langle } u,v \rangle = {{}_\mathcal {A}^{\mathcal {U}_{2}}\langle } \mu \circ u,\mu \circ v \rangle$. Each fiber ${U_j}_t$ is a seminormed space when considered with the seminorm $\Vert u\Vert := \Vert {{}_\mathcal {A}^{\mathcal {U}_{j}}\langle } u,u \rangle \Vert ^{1/2} = \Vert \langle u,u {\rangle _\mathcal {C}^{\mathcal {U}_{j}}} \Vert ^{1/2}$. The space $[{U_j}_t]$ is defined as the quotient of ${U_j}_t$ by the subspace of zero length vectors, where square brackets are used to represent equivalence classes.

The tensor product $\mathcal {X}_1\otimes _\mathcal {B}\mathcal {Y}_1$ is obtained by completing each fiber of $[\mathcal {U}_j]=\{ [{U_j}_t]\}_{t\in G}$ and a set of continuous sections of this tensor product is given by those of the form $[\xi ]$, for $\xi \in \mathrm {C_c}(\mathcal {Z}_j)$, where $[\xi ](t)=[\xi |_{H^t}]$ and $\xi |_{H^t}$ represents the restriction of $\xi$ to $H^t$.

Note there exists a unique bijective isometry $\mu ^*_t:[{U_1}_t]\rightarrow [{U_2}_t]$ such that $[u]\mapsto [\mu \circ u]$. Then there exists a unique function $\mu ^*:\mathcal {X}_1\otimes _\mathcal {B}\mathcal {Y}_1\rightarrow \mathcal {X}_2\otimes _\mathcal {B}\mathcal {Y}_2$ which is linear and bounded on each fiber and extends each $\mu ^*_t$. Clearly, $\mu ^*$ is an isometry and $\mu ^*\circ [\xi ] = [\mu \circ \xi ]$ for all $\xi \in \mathrm {C_c}(\mathcal {Z}_1)$. In this situation Doran and Fell (1988a, II 13.16) implies $\mu ^*$ is an isomorphism of Banach bundles. Moreover, it preserves the left and right inner products because $\mu$ transforms the inner products and actions of the bundle $\{{U_1}_t\}_{t\in G}$ to those of $\{{U_2}_t\}_{t\in G}$. $\square$

Proposition A.4

Let $\mathcal {X},\ \mathcal {Y}$ and $\mathcal {Z}$ be $\mathcal {A}-\mathcal {B}$, $\mathcal {B}-\mathcal {C}$ and $\mathcal {C}-\mathcal {D}$-equivalence bundles, respectively. Then

• (a) $\mathcal {A}\otimes _\mathcal {A}\mathcal {X}$ is unitarily equivalent to $\mathcal {X}$ and $\mathcal {X}\otimes _\mathcal {B}\mathcal {B}$ to X.

• (b) $\widetilde{\mathcal {X}}\otimes _{\mathcal {A}}\mathcal {X}$ is unitarily equivalent to $\mathcal {B}$ and $\mathcal {X}\otimes _\mathcal {B}\widetilde{\mathcal {X}}$ to $\mathcal {A}$.

• (c) The tensor products $(\mathcal {X}\otimes _\mathcal {B}\mathcal {Y})\otimes _\mathcal {C}\mathcal {Z}$ and $\mathcal {X}\otimes _\mathcal {B}(\mathcal {Y}\otimes _\mathcal {C}\mathcal {Z})$ are unitarily equivalent.

Proof

The proofs of the two claims in (a) are analogous, thus we just prove the first one; the same comment holds for (b).

Let $\mathcal {Z}$ be the bundle constructed from $\mathcal {A}$ and $\mathcal {X}$ ($Z_{(r,s)}=A_r\otimes _{A_e}X_s$) as in the proof of Proposition A.3. We claim that there exists a unique continuous map $\pi :\mathcal {Z}\rightarrow \mathcal {X}$ such that: $\pi ( Z_{(r,s)} )\subset X_{rs}$, $\pi |_{Z_{(r,s)}}$ is linear and $\pi (a\otimes x)=ax$ for all $r,s\in G$, $a\in \mathcal {A}$ and $x\in \mathcal {X}$. First note there exists a unique linear isometry $\pi _{r,s}:Z_{(r,s)}\rightarrow X_{rs}$ sending $a\otimes x$ to ax because, for every $a_1,\ldots ,a_n\in A_r$ and $x_1,\ldots ,x_n$ we have

$$\begin{aligned} \left\langle \sum _{i=1}^n a_i\otimes x_i,\sum _{i=1}^n a_i\otimes x_i\right\rangle =\sum _{i,j=1}^n \langle x_i,a_i^*a_j x_j{\rangle _\mathcal {B}}= \left\langle \sum _{i=1}^n a_ix_i,\sum _{i=1}^n a_ix_i\right\rangle . \end{aligned}$$

If $\pi :\mathcal {Z}\rightarrow \mathcal {A}$ is the unique map extending all the $\pi _{r,s}$, then, using Doran and Fell (1988a, II 13.16), we conclude that $\pi$ is continuous because for all $f\in \mathrm {C_c}(\mathcal {A})$ and $g\in \mathrm {C_c}(\mathcal {X})$, $\pi \circ (f\boxtimes g)$ is continuous.

Now let $\mathcal {U}$ be constructed from $\mathcal {Z}$ as in the proof of Proposition A.3. For $f,u\in \mathrm {C_c}(\mathcal {A})$ and $g,v\in \mathrm {C_c}(\mathcal {X})$ we have

$$\begin{aligned} \langle f\boxtimes g|_{H^r} , u\boxtimes v|_{H^s}\rangle ^\mathcal {U}_\mathcal {B}= \left\langle \int _G \pi ( f\boxtimes g (p,{p^{-1}}r))\,\mathrm dp,\int _G \pi ( u\boxtimes v (q,{q^{-1}}s))\,\mathrm dq \right\rangle _\mathcal {B}. \end{aligned}$$

Since ${\text {span}}\{ f\boxtimes g|_{H^t}:f\in \mathrm {C_c}(\mathcal {A}),\ g\in \mathrm {C_c}(\mathcal {X})\}$ is dense in the inductive limit topology of $U_t$ and $\langle \ , \ \rangle ^\mathcal {U}_\mathcal {B}$ is continuous on each variable (separately) with respect to this topology, we conclude that

$$\begin{aligned} \langle \xi , \eta \rangle ^\mathcal {U}_\mathcal {B}= \left\langle \int _G \pi ( \xi (p,{p^{-1}}r))\,\mathrm dp,\int _G \pi ( \eta (q,{q^{-1}}s))\,\mathrm dq \right\rangle _\mathcal {B}\end{aligned}$$

(A.5)

for all $\xi \in U_r$, $\eta \in U_s$ and $r,s\in G$.

Then we can define a map $\mu :[\mathcal {U}] \rightarrow \mathcal {X}$ such that, for $\xi \in U_r$,

$$\begin{aligned} \mu ([\xi ]) = \int _G\pi ( \xi (p,{p^{-1}}r))\,\mathrm dp . \end{aligned}$$

This map is linear and isometric on each fiber, so it can be (continuously) extended to the closure of each fiber. The resulting extension is a map $\mu :\mathcal {A}\otimes _\mathcal {A}\mathcal {X}\rightarrow \mathcal {X}$. Recall from Abadie and Ferraro (2017) that the topology of $\mathcal {A}\otimes _\mathcal {A}\mathcal {X}$ is constructed using the sections of the form $r\mapsto [\eta |_{H^r}]$, where $\eta \in \mathrm {C_c}(\mathcal {Z})$. Besides, for every $\eta \in \mathrm {C_c}(\mathcal {Z})$, the section $r\mapsto \mu ([\eta |_{H^r}])$ is continuous (this is not immediate but can be proved by standard arguments, see for example the ideas developed in Doran and Fell (1988a, II 15.19). Then Doran and Fell (1988a, II 13.16) implies $\mu$ is continuous. Note Eq. (A.5) implies $\mu ^r={\text {id}}_\mathcal {B}$ and the reader can show that $\mu ^l={\text {id}}_\mathcal {A}$ with analogous computations. Then all we need to do is to show $\mu$ is surjective or, alternatively, that $\mu ([U_r])$ is dense in $X_r$ for all $r\in G$.

Fix $x\in X_r$ and take $a\in A_e$ and $x'\in X_r$ such that $ax'=x$. Now take $f\in \mathrm {C_c}(\mathcal {A})$ and $g\in \mathrm {C_c}(\mathcal {X})$ such that $f(e)=a$ and $g(r)=x'$. Denote I the directed set of compact neighbourhoods of $e\in G$ with respect the usual order: $i\le j$ if $j\subset i$. For each $i\in I$, take a function $\varphi _i\in \mathrm {C_c}(G)^+$ with $\int _G \varphi _i(t)\,\mathrm dt=1$ and ${\text {supp}}(\varphi _i)\subset i$. Then $\lim _i \mu ((\varphi _i f)\boxtimes g|_{H^r})=f(e)g(r) = x$; we conclude that $\mu$ is surjective. This implies that $\mu$ is unitary.

The proof of (b) is very similar to that of (a). We start by constructing a surjective isometry $\pi :\mathcal {Z}\rightarrow \mathcal {B}$, where $\mathcal {Z}=\{ \widetilde{X_{r^{-1}}}\otimes _{A_e}X_s \}_{(r,s)\in G\times G}$. Take $r,s\in G$, $x^j_1,\ldots ,x^j_n\in X_{r^{-1}}$ and $y^j_1,\ldots ,y^j_n\in X_s$ ($j=1,2$). Then note that

$$\begin{aligned} \sum _{i=1}^n \widetilde{x^1_i}\otimes y^1_i \triangleright \sum _{j=1}^n \widetilde{x^2_j}\otimes y^2_j&= \sum _{i,j=1}^n \langle y^1_j,{}_\mathcal {A}\langle x^1_i,x^2_j{\rangle }y^2_j{\rangle _\mathcal {B}}= \sum _{i,j=1}^n \langle y^1_j,x^1_i \langle x^2_j, y^2_j{\rangle _\mathcal {B}}{\rangle _\mathcal {B}}\\&= \left( \sum _{i=1}^n \langle x^1_i,y^1_i{\rangle _\mathcal {B}}\right) ^* \sum _{i=1}^n \langle x^2_i,y^2_i{\rangle _\mathcal {B}}. \end{aligned}$$

Besides, the restriction of $\triangleright$ to $Z_{(r,s)}\times Z_{(r,s)}$ is the inner product of $Z_{(r,s)}$. Then we conclude there exists a unique map $\pi :\mathcal {Z}\rightarrow \mathcal {B}$ such that: $\pi (\widetilde{x}\otimes y)=\langle x,y{\rangle _\mathcal {B}}$, $\pi (Z_{(r,s)})\subset \mathcal {B}_{rs}$ and $\pi |_{Z_{(r,s)}}$ is linear for all $x,y\in \mathcal {X}$ and $r,s\in G$. Moreover, $z\triangleright w = \pi (z)^*\pi (w)$, $\pi (z b)=\pi (zb)$ and $\pi (bz)=b\pi (z)$ for all $z,w\in \mathcal {Z}$ and $b\in \mathcal {B}$. Note also that $\pi$ is continuous because, for $f,g\in \mathrm {C_c}(\mathcal {X})$, we have $\pi \circ \widetilde{f}\boxtimes g(r,s) = \langle f({r^{-1}}),g(s){\rangle _\mathcal {B}}$ and so $\pi \circ \widetilde{f}\boxtimes g$ is continuous.

To complete the proof of (b) it suffices to follow the steps of the proof of (a), using the map $\pi$ we have just constructed instead of the map $\pi$ we used to prove (a).

We now deal with (c). Let $[\mathcal {U}]$ and $[\mathcal {V}]$ be the bundles whose fiber-wise completion gives $\mathcal {X}\otimes _\mathcal {B}\mathcal {Y}$ and $\mathcal {Y}\otimes _\mathcal {C}\mathcal {Z}$, respectively (see the proof of Proposition A.3). For every pair $(f,g)\in \mathrm {C_c}(\mathcal {X})\times \mathrm {C_c}(\mathcal {Y})$ we have a section $[f\boxtimes g]\in \mathrm {C_c}(\mathcal {X}\otimes _\mathcal {B}\mathcal {Y})$ such that $[f\boxtimes g](t) = [f\boxtimes g|_{H^t}]\in [U_t]$. Define $\Gamma _\mathcal {U}$ as the linear span of the sections $[f\boxtimes g]$. Part of the construction of $\mathcal {X}\otimes _\mathcal {B}\mathcal {Y}$ is based on the fact that $\{\xi (t):\xi \in \Gamma _\mathcal {U}\}$ is dense in $[U_t]$ and so in the fiber over t of $\mathcal {X}\otimes _\mathcal {B}\mathcal {Y}$. Of course that the same holds for $\Gamma _\mathcal {V}$.

Fix $r,s \in G$, $f,u\in \mathrm {C_c}(\mathcal {X})$, $g,v\in \mathrm {C_c}(\mathcal {Y})$ and $h,w\in \mathrm {C_c}(\mathcal {Z})$. We want to prove that

$$\begin{aligned} \langle [[f \boxtimes g] \boxtimes h](r) , [[u\boxtimes v] \boxtimes w](s)\rangle _\mathcal {D}= \langle [f\boxtimes [g\boxtimes h]](r) , [u\boxtimes [v\otimes w]](s)\rangle _\mathcal {D}. \end{aligned}$$

(A.6)

To do this first note that, using the definitions of the inner product of tensor products bundles, we obtain

$$\begin{aligned} \langle [[f&\boxtimes g] \boxtimes h](r) , [[u\boxtimes v] \boxtimes w](s) \rangle _\mathcal {D}\\&= \int _{G^2} [f\boxtimes g] \boxtimes h (p,{p^{-1}}r)\triangleright [u\boxtimes v] \boxtimes w (q,{q^{-1}}s)\,\mathrm dp\,\mathrm dq \\&= \int _{G^2} \langle h ({p^{-1}}r), \langle [f\boxtimes g|_{H^p}], [u\boxtimes v|_{H^q}]\rangle _{\mathcal {C}} w ({q^{-1}}s)\rangle _\mathcal {D}\,\mathrm dp\,\mathrm dq \\&= \int _{G^4} \langle h ({p^{-1}}r), \langle g(x^{-1}p) , \langle f(x),u(y)\rangle _\mathcal {B}v(y^{-1}q)\rangle _{\mathcal {C}} w ({q^{-1}}s)\rangle _\mathcal {D}\,\mathrm dx\,\mathrm dy\,\mathrm dp\,\mathrm dq. \end{aligned}$$

Using the substitutions $p\mapsto xp$ and $q\mapsto xq$ in the integrals, we obtain

$$\begin{aligned}&\langle [[f \boxtimes g] \boxtimes h](r) , [[u\boxtimes v] \boxtimes w](s)\rangle _\mathcal {D}\\&\quad = \int _{G^4} \langle h ({p^{-1}}x^{-1} r), \langle g(p) , \langle f(x),u(y)\rangle _\mathcal {B}v(y^{-1}xq)\rangle _{\mathcal {C}} w ({q^{-1}}x^{-1}s)\rangle _\mathcal {D}\,\mathrm dx\,\mathrm dy\,\mathrm dp\,\mathrm dq \\&\quad = \int _{G^4} (g(p)\otimes h({p^{-1}}x^{-1} r)) \triangleright ( \langle f(x),u(y)\rangle _\mathcal {B}v(y^{-1}xq)\otimes w ({q^{-1}}x^{-1}s) ) \,\mathrm dx\,\mathrm dy\,\mathrm dp\,\mathrm dq \\&\quad = \int _{G^2} \langle [g\boxtimes h](x^{-1}r) , \langle f(x),u(y)\rangle _\mathcal {B}[v\otimes w](y^{-1}s)\rangle _\mathcal {D}\,\mathrm dx\,\mathrm dy\\&\quad = \langle [f\boxtimes [g\boxtimes h]](r) , [u\boxtimes [v\otimes w]](s)\rangle _\mathcal {D}. \end{aligned}$$

Using Eq. (A.6) and Doran and Fell (1988a, II 13.16) we can justify the existence of a unique isometric isomorphism of Banach bundles $\mu :(\mathcal {X}\otimes _\mathcal {B}\mathcal {Y})\otimes _\mathcal {C}\mathcal {Z}\rightarrow \mathcal {X}\otimes _\mathcal {B}(\mathcal {Y}\otimes _\mathcal {C}\mathcal {Z})$ such that $\mu ( [[f \boxtimes g] \boxtimes h](r) ) = [f\boxtimes [g\boxtimes h]](r)$. We leave to the reader the verification of the fact that $\mu ( [[f \boxtimes g] \boxtimes h](r) d ) = \mu ([[f \boxtimes g] \boxtimes h](r))d$. After this it is immediate that $\mu (\xi d)=\mu (\xi ) d$ for all $\xi \in (\mathcal {X}\otimes _\mathcal {B}\mathcal {Y})\otimes _\mathcal {C}\mathcal {Z}$ and $d\in \mathcal {D}$. Note Eq. (A.6) implies $\langle \mu (\xi ),\mu (\eta )\rangle _\mathcal {D}= \langle \xi ,\eta \rangle _\mathcal {D}$ for all $\xi ,\eta \in (\mathcal {X}\otimes _\mathcal {B}\mathcal {Y})\otimes _\mathcal {C}\mathcal {Z}$. Then $\mu$ is a morphism of equivalence bundles because

$$\begin{aligned} \mu (\xi \langle \eta ,\zeta \rangle _\mathcal {D}) = \mu (\xi ) \langle \eta ,\zeta \rangle _\mathcal {D}= \mu (\xi ) \langle \mu (\eta ), \mu (\zeta )\rangle _\mathcal {D}. \end{aligned}$$

The identity $\langle \mu (\xi ),\mu (\eta )\rangle _\mathcal {D}= \langle \xi ,\eta \rangle _\mathcal {D}$ tells us that $\mu ^r={\text {id}}_\mathcal {D}$, and we leave to the reader to check that $\mu ^l={\text {id}}_\mathcal {A}$ (the proof of which is analogous to the proof of (A.6)). $\square$

Using the last two propositions one can construct a category $\mathscr {E}_G^w$ (resp. $\mathscr {E}_G^s$) of Fell bundles over G as objects and isomorphism classes of weak (resp. strong) equivalence bundles as arrows. The identity morphism associated to the Fell bundle $\mathcal {A}$ is the isomorphism class of $\mathcal {A}$, $[\mathcal {A}]$. The composition of the arrows $\mathcal {A}{\mathop {\rightarrow }\limits ^{[\mathcal {X}]}}\mathcal {B}$ and $\mathcal {B}{\mathop {\rightarrow }\limits ^{[\mathcal {Y}]}}\mathcal {C}$ is $\mathcal {A}{\mathop {\rightarrow }\limits ^{[\mathcal {X}\otimes _\mathcal {B}\mathcal {Y}]}}\mathcal {C}$. Propositions A.3 and A.4 tell us that we indeed obtain a category with these definitions and, moreover, every arrow is invertible in this category. Only a weak 2-category (or bicategory) can be formed if we do not take isomorphism class, but just the equivalence bundles as arrows. The unitaries introduced in Definition A.2 can be used as 2-arrows for this weak 2-category. This works similarly to the 2-category of ${\mathrm {C}}^*$-algebras with correspondences as arrows introduced in Buss et al. (2013).