Proof of Lemma 2.1

Condition (DE9) is just a restatement of other properties. To be precise, $M(x)$ is a right inner product $B(\sigma (x))$ -module in the sense of [Reference Raeburn and Williams24, Definition 2.1] because of Conditions (DE3)–(DE6), and it is a Hilbert $B(\sigma (x))$ -module because we assumed that the norm with respect to which $M(x)$ is complete coincides with the norm induced by the inner product (Assumption (DE7)). Fullness is exactly Assumption (DE8).

Condition (DE10) follows from (DE5) and (DE4) combined, and (DE11) follows directly from (DE2).

(DE13) Fix $x\in X$ and $m,m_{i}\in M(x)$ . If $b\in B(h)$ for some $h\in \sigma (x)\mathcal {H}$ , then

Since multiplication in $\mathscr {B}$ is bilinear, we conclude that

for all $m\in M(x)$ . Choosing , it follows from (DE6) that $m=0$ , meaning that is linear in the first component. A similar computation, using (DE10), proves linearity in the second component and also (DE14), using associativity of the multiplication of $\mathscr {B}$ .

(DE15) We have

where $(\dagger )$ follows from (DE4) and (DE10).

(DE16) Suppose $(m_{i },b_{i })$ is a net in that converges to $(m,b)$ ; in particular, converges to in X and converges to in $\mathcal {H}$ . Now choose a continuous section $\mu \in \Gamma _0 (X;\mathscr {M})$ of $\mathscr {M}$ with . Since $\mu $ and the right $\mathcal {H}$ -action on X are continuous, we have . Using (DE4), (DE10), and sesquilinearity of the inner product, we have

By continuity of the involution and multiplication of $\mathscr {B}$ and by continuity of the inner product on $\mathscr {M}$ , the above converges to

which, again by (DE4) and (DE10), is $0$ . Since

by (DE7), it thus follows from [Reference Duwenig and Li4, Lemma A.5] that converges in the total space of $\mathscr {M}$ to the zero-element of the Banach space . Since converges to and since limits are unique, this implies , as claimed.

The proof of (DE12) is verbatim that for [Reference Duwenig and Li4, Lemma 4.7] (after translating from the left to the right).

Proof First, we will do some sanity checks: Condition (1) makes sense because the actions on X commute, meaning that and live over the same fibre of $\mathscr {M}$ by (DE1). Next, let us check that Condition (3) makes sense. Since is defined on and on , we can evaluate the shown inner products. By (DE11), we have , and since , we therefore have that $m_{1}$ can be acted on by on the right; a similar argument shows that $m_{3}$ can be acted on by on the left. Lastly, note that Conditions (DE1) and (DE2) combined with Equation (1.4) show that the elements in question are indeed living in the same fibre of $\mathscr {M}$ .

To see that $\mathscr {M}$ is an equivalence, we have to check Properties (FE1), (FE2), and (FE3) in [Reference Duwenig and Li4, Definition 2.11]. By assumption, $\mathscr {M}$ is an upper semi-continuous Banach bundle over a groupoid equivalence, and as explained in Lemma 2.1, and are actions in the sense of [Reference Duwenig and Li4, Definition 2.10]. Since they are assumed to commute, we therefore get Condition (FE1). For (FE2), note that the inner products are assumed to be sesquilinear in the correct sense. Furthermore,

1. (FE2.a) is satisfied by (DE2);

2. (FE2.b) is satisfied by (DE5);

3. (FE2.c) is satisfied by (DE4); and

4. (FE2.d) is exactly Assumption (3).

Lastly, for (FE3), we must check that each $M(x)$ is an imprimitivity bimodule between $A(\rho (x))$ and $B(\sigma (x))$ . We know by (DE9) that it is a full left- and right-Hilbert $\mathrm {C}^{*}$ -module. By Assumption (3), the left- and right-inner products are compatible and the left-action is by $B(\sigma (x))$ -adjointable operators; it remains to show that the right-action is by $A(\rho (x))$ -adjointable operators. For $n_{i}\in M(x)$ , we have

This proves that the element of $A(\rho (x))$ “kills” all of $M(x)$ . By Assumption (2), a is an adjointable operator with respect to the $B(\sigma (x))$ -valued right-inner product. It therefore follows from [Reference Raeburn and Williams24, Remark 2.29] that $a^*a$ and hence a is $0$ , i.e., , so b is an $A(\rho (x))$ -adjointable operator.

Proof Consider

Recall that $\mathscr {M}$ has enough continuous sections, so that for each $m\in M(x)$ , there exists $\mu \in \Gamma _0(X;\mathscr {M})$ such that $\mu (x)=m$ . In particular, it follows that $ \{\gamma (x,y^{\mathrm {op}}) : \gamma \in \Gamma \} $ is dense in $K(x,y^{\mathrm {op}})$ . If we can show for an arbitrary element $\gamma =\sum _{i=1}^{k}\mu _{i}\otimes \nu ^{\mathrm {op}}_{i}$ of $\Gamma $ that $(x,y^{\mathrm {op}})\mapsto \left \| \gamma (x,y^{\mathrm {op}}) \right \|$ is upper semi-continuous, then the claim follows from [Reference Hofmann12, Corollary 3.7] (see also [Reference Fell and Doran9, p. 13.18], [Reference Williams29, Theorem C.25], [Reference Duwenig and Li4, Remark 2.7]). We have by Equation (3.2)

It was shown in [Reference Raeburn and Williams24, Lemma 2.65] that the $k\times k$ -matrix with $i,j$ -entry is positive for each $x\in X$ , so there exists a matrix $(b_{i,j}(x))_{i,j}$ over $B(\sigma (x))$ such that . Note that we may choose each $b_{i,j}\colon x\mapsto b_{i,j}(x)$ to be a continuous section of the pullback bundle $\sigma ^*(\mathscr {B})$ . For $1\leq l \leq k$ , we let . By continuity of , of each $\nu _{i}$ and $b_{il}$ , and of summation in $\mathscr {M}$ , each $\nu _{l}$ is continuous. Moreover,

Since the $\mathscr {B}$ -valued inner product on $\mathscr {M}$ , summation in $\mathscr {B}$ , and each $\nu _{l}$ are continuous, and the norm on $\mathscr {B}$ is upper semi-continuous, we conclude that $(x,y^{\mathrm {op}})\mapsto \left \| \gamma (x,y^{\mathrm {op}}) \right \|$ is upper semi-continuous.

Proof We construct $\Psi _{h}$ on each of the fibres . Consider the map

Since it is multilinear, it descends to a linear map with domain $[M(x) \odot B(h)] \odot M(y)^{\mathrm {op}} $ , where $\odot $ denotes the algebraic tensor product. Because of the $B(u)$ -balancing in the codomain, the map descends to a map with domain $[M(x) \odot _{B(u)} B(h)] \odot M(y)^{\mathrm {op}}$ . Since for any $b_v\in B(v)$ , it further descends to a map with domain $[M(x) \odot _{B(u)} B(h)] \odot _{B(v)} M(y)^{\mathrm {op}}$ . It is easy to check that that map is isometric (when $M(x) \odot _{B(u)} B(h)$ is given the norm of the balanced tensor product of Hilbert modules), and therefore extends to $[M(x) \otimes _{u} B(h)] \otimes _{v} M(y)^{\mathrm {op}}$ . Now, Assumption (DE8) implies that every element of is the limit of sums of elements of the form ; in other words, the linear map determined by has dense range. We conclude the existence of $\Psi _{h}$ . The remaining properties are easy to check.

Proof To see that $\Psi $ is an isomorphism, we will invoke Lemma B.2 and [Reference Duwenig and Li4, Proposition A.8]. Because of Lemma 3.1 and our comment preceding Lemma 3.4, the set

$$\begin{align*}\Gamma = \{ (h,x,y^{\mathrm{op}} )\mapsto (h,\mu(x)\otimes \nu(y)^{\mathrm{op}} ) : \mu,\nu\in \Gamma_0(X;\mathscr{M}) \} \end{align*}$$

is a collection of continuous sections of . As the fibre of over a given $(h,x,y^{\mathrm {op}} )$ is just $K(x,y^{\mathrm {op}})$ and since the linear span of $\{{\mu (x)\otimes \nu (y)}^{\mathrm {op}}:{\mu ,\nu \in \Gamma _0(X;\mathscr {M})}\}$ is dense in $K(x,y^{\mathrm {op}})$ , we conclude that the linear span of $\{{\gamma (h,x,y^{\mathrm {op}})}:{\gamma \in \Gamma }\}$ is likewise dense in the fibre of . We claim that, given any element $\gamma \colon (h,x,y^{\mathrm {op}} )\mapsto (h,\mu (x)\otimes \nu (y)^{\mathrm {op}} )$ of $\Gamma $ , the section of is continuous. Indeed, it suffices to check that, if

then

Fix $\epsilon>0$ . As mentioned earlier, Assumption (DE8) implies that , so we may take finitely many elements $m_{i}\in M(x)$ and $b_{i}\in B(h)$ such that

(3.5)

For each $1\leq i\leq k$ , fix a section $\mu _{i}$ of $\mathscr {M}$ and $\tau _{i}$ of $\mathscr {B}$ with $\mu _{i}(x)=m_i$ respectively $\tau _{i}(h)=b_{i}$ . Since addition in $\mathscr {M}$ and the right $\mathscr {B}$ -action are continuous, the net converges to in M. By [Reference Duwenig and Li4, Lemma A.3, $(i) \Longrightarrow (iii)$ ], we therefore have

(3.6)

Note that

and likewise without subscript- $\lambda $ ’s. Thus, by continuity of $\mu _{i},\nu ,\tau _{i}$ , of the involution in $\mathscr {B}$ , of the right $\mathscr {B}$ -action, and of addition in $\mathscr {K}$ , combined with Remark 3.2, this implies that

(3.7) $$ \begin{align} \Psi_{h_{\lambda}}(\zeta_{\lambda}\otimes \nu(y_{\lambda})^{\mathrm{op}}) \to \Psi_{h}(\zeta \otimes \nu(y)^{\mathrm{op}}). \end{align} $$

Since $\Psi _{h}$ is isometric ( $\Psi $ 2), we then have for large enough $\lambda $

where $(*)$ follows from (3.6) and continuity of $\nu $ . The same computation without subscripts yields

As $\epsilon $ was arbitrary, Lemma B.1 states that this combined with (3.7) implies

Because each $\Psi _{h}$ is linear, isometric, and surjective, the claim now follows from Lemma B.2 and [Reference Duwenig and Li4, Proposition A.8].

Proof The proof that $\mathbin {\texttt {R}}$ is closed is verbatim that given for [Reference Duwenig and Li4, Lemma 6.2], where it is also explained why $\mathbin {\texttt {R}}$ is an equivalence relation.

To show that the quotient map Q is open, the proof of [Reference Duwenig and Li4, Proposition 6.8] goes through for , even though $\mathscr {M}$ was assumed to be an equivalence and $\mathcal {H}$ was assumed to be étale in [Reference Duwenig and Li4]. In fact, étaleness was only needed to be allowed to invoke [Reference Duwenig and Li4, Lemma 6.10], but that lemma’s proof holds as long as the range map of $\mathcal {H}$ is open (which we assumed here).

Proof The proof of [Reference Duwenig and Li4, Lemma 6.5] for goes through, even though $\mathscr {M}$ was assumed to be an equivalence in that lemma.

Proof If $\xi ,\eta \in K(x,y^{\mathrm {op}})$ , then since the $\mathcal {H}$ -action on X is free, is the only possible element that allows $\xi \mathbin {\texttt {R}}\eta $ , in which case $\Psi _{h}(\eta )=\eta $ (since h is a unit) and hence $\xi =\eta $ . Thus, $Q_{(x,y^{\mathrm {op}})}$ is injective. For surjectivity, note that Q is surjective, so any element a of $A(g)$ can be written as $a=Q(\xi )$ for $\xi \in K$ with . In particular, there exists $h\in \mathcal {H}$ with , so that $\Psi _{h}(\xi ) \in K(x,y^{\mathrm {op}})$ and $a=Q(\xi )=Q_{(x,y^{\mathrm {op}})}(\Psi _{h}(\xi ))$ .

Since $p=p_{\mathscr {A}}$ is continuous, is closed in A, and so continuity of Q implies continuity of $Q_{(x,y^{\mathrm {op}})}$ . To see that $Q_{(x,y^{\mathrm {op}})}$ is closed (and hence a homeomorphism), we follow the idea in the proof of [Reference Duwenig and Li4, Lemma 6.7]: Let $F\subseteq K(x,y^{\mathrm {op}})$ be a closed set. To show that $Q(F)\subseteq A(g)$ is closed, it suffices to show that it is closed in A, i.e., that is closed in K. So assume $\xi _{\lambda }\to \xi $ in K with $Q(\xi _{\lambda })\in Q(F)$ ; we must prove that $Q(\xi )\in Q(F)$ . Since $Q(\xi _{\lambda })\in F$ , there exist $\eta _{\lambda }\in F$ with $\eta _{\lambda }\mathbin {\texttt {R}}\xi _{\lambda }$ . Since $F\subseteq K(x,y^{\mathrm {op}})$ , this means there exist $h_{\lambda }\in \sigma (x)\mathcal {H} $ with and $\Psi _{h_{\lambda }}(\xi _{\lambda })=\eta _{\lambda }$ . Since $\xi _{\lambda }\to \xi $ , we have

Since the $\mathcal {H}$ -action on X is proper, it follows from [Reference Williams28, Proposition 2.17] that (a subnet of) $\{h_{\lambda }\}_{\lambda }$ converges to, say, h; since X is Hausdorff, it follows that . Since $\Psi $ is jointly continuous, we conclude that (a subnet of) $\eta _{\lambda } = \Psi _{h_{\lambda }}(\xi _{\lambda })$ converges to $\Psi _{h}(\xi )$ in K. Since $\eta _{\lambda }\in F$ and F is closed in $K(x,y^{\mathrm {op}})$ and hence also in K, we have $\Psi _{h}(\xi )\in F$ . Since $\xi \mathbin {\texttt {R}}\Psi _{h}(\xi )$ , this means that , as needed.

Proof That $\mathscr {A}$ is an upper semi-continuous Banach bundle follows from an application of [Reference Duwenig and Li4, Proposition 6.13]; here, we need that both and $Q\colon {K\to A}$ are open quotient maps (Lemma 3.5) and that $Q_{(x,y^{\mathrm {op}})}$ is surjective (Lemma 3.7) and a linear isometry (by definition of the linear structure on each fibre of $\mathscr {A}$ ).

That the given section is continuous follows since is open, $\mu \otimes \nu ^{\mathrm {op}}$ is a continuous section of $\mathscr {K}$ (Lemma 3.1), and Q is continuous.

Proof To see that the map $\texttt {Flip}$ exists on each fibre, fix and let $u=\sigma (x)=\sigma (y)$ . Recall that $K(x, y^{\mathrm {op}})$ is isomorphic to $\mathbb {K}_{B(u)} (M(y),M(x))$ as bi-Hilbert $\mathbb {K}_{B(u)} (M(x))-\mathbb {K}_{B(u)} (M(y))$ -bimodules. With this identification, the restriction of $\texttt {Flip}$ to this fibre is given by or in other words, $\texttt {Flip}$ is the adjoint map $T\mapsto T^*$ . In particular, $\texttt {Flip}$ is fibrewise antilinear and isometric. Its continuity is built into the topology on $\mathscr {K}$ : If $\mu ,\nu $ are continuous sections of $\mathscr {M}$ , then $\mu \otimes \nu ^{\mathrm {op}}$ and $\nu \otimes \mu ^{\mathrm {op}}$ are continuous sections of $\mathscr {K}$ . Since $\texttt {Flip}$ transforms the former into the latter, it is continuous and open by an application of [Reference Duwenig and Li4, Propositions A.7 and A.8].Footnote ⁵

Proof The existence of $\Phi _{x,y}$ follows from the following, well-studied isomorphisms of bi-Hilbert $A-A$ -modules for any full right-Hilbert A-module $\mathbf {Y}$ : Firstly, the map $\mathbf {Y}^{\mathrm {op}}\otimes _{\mathbb {K}}\mathbf {Y} \to A$ determined by , and secondly, $\mathbf {Y} \otimes _{A} A \to \mathbf {Y}$ determined by $\mathbf {y}\otimes a\mapsto \mathbf {y}\cdot a$ . In our situation, $A=B(u)$ where $u=\sigma (x)=\sigma (y)$ , and we use first $M(y)$ and then $M(x)$ to take the rôle of $\mathbf {Y}$ . To be precise, if we write $\mathbb {K}$ for $\mathbb {K}_{B(u)} (M(y))$ , then

$$\begin{align*}K(x,y^{\mathrm{op}})\times M(y) \twoheadrightarrow \big( M(x)\otimes_{u}M(y)^{\mathrm{op}} \big) \otimes_{\mathbb{K}} M(y) \cong M(x)\otimes_{u} B(u) \cong M(x). \end{align*}$$

The claim about the norm is also well known, but we add it here for completion. Since $\Phi _{x,y}$ is continuous, it suffices to prove the claim for $\xi =\sum _{i=1}^{\ell } m_{i}\otimes n_{i}^{\mathrm {op}}$ , a sum of elementary tensors. We have

where the norm on the right-hand side is the operator norm on $\mathbb {K}_{B(u)}(M(x))$ . Note that acts by $\mathbb {K}_{B(u)}(M(x))$ -adjointable operators, so that

Now, by (DE4). Since $k,n_{i}\in M(y)$ , we further know that . Recall that is a positive element of $\mathbb{K}_{B(u)}(M(y))$ , so we can write it as $T^*T$ . In particular,

Since $\left \| \sum _{i} m_{i}\otimes (Tn_{i})^{\mathrm {op}} \right \|\leq \left \| 1\otimes T \right \|\left \| \sum _{i} m_{i}\otimes n_{i}^{\mathrm {op}} \right \|$ and $\left \| 1\otimes T \right \|=\left \| T \right \|=\left \| k \right \|$ , we conclude that

$$\begin{align*}\left\| \Phi_{x,y}(\xi, k) \right\| \leq \left\| \xi \right\|\left\| k \right\|, \end{align*}$$

as claimed.

Proof of Lemma 3.12

The existence of $U_{y}$ follows from observations of general Hilbert $\mathrm {C}^{*}$ -modules similar to those following Lemma 1.5: If $\mathbf {X,Y,Z}$ are full right-Hilbert $\mathrm {C}^{*}$ -modules over a $\mathrm {C}^{*}$ -algebra A, then the maps

(3.16) $$ \begin{align} (\mathbf{X}\otimes_{A}\mathbf{Y}^{\mathrm{op}}) &\otimes_{\mathbb{K}_{A}(\mathbf{Y})} (\mathbf{Y}\otimes_{A}\mathbf{Z}^{\mathrm{op}}) &\longrightarrow& & \mathbf{X}\otimes_{A}\mathbf{Z}^{\mathrm{op}} &&\longrightarrow& & \mathbb{K}_{A}(\mathbf{Z},\mathbf{X}) \end{align} $$

determined by

are isomorphisms of bi-Hilbert $\mathbb {K}_{A}(\mathbf {X})-\mathbb {K}_{A}(\mathbf {Z})$ -modules. The above isomorphism can be precomposed with the universal bilinear map $\mathbf {X}\otimes _{A}\mathbf {Y}^{\mathrm {op}} \times \mathbf {Y}\otimes _{A}\mathbf {Z}^{\mathrm {op}} \to {(\mathbf {X}\otimes _{A}\mathbf {Y}^{\mathrm {op}})} \otimes _{\mathbb {K}_{A}(\mathbf {Y})} {(\mathbf {Y}\otimes _{A}\mathbf {Z}^{\mathrm {op}})}$ . We apply this to the case where $\mathbf {Y}=M(y)$ and $A=B(u)$ . Condition (U1) is therefore by construction.

(U2) Since the maps in (3.16) are isomorphisms of modules (in particular, they are isometric), we have $ \left \| U_{y}(\xi ,\eta ) \right \|=\left \| \xi \otimes \eta \right \|$ for $\xi \in K(x,y^{\mathrm {op}})$ and $\eta \in K(y,z^{\mathrm {op}})$ , which is known (or easily shown) to be bounded by $\left \| \xi \right \|\left \| \eta \right \|$ . In particular, $U_{y}$ is not just linear but also continuous on each fibre. For the second claim, it therefore suffices to consider one of the dense elements $\xi = \sum _{i}m_{i}\otimes n_{i}^{\mathrm {op}}$ in which case $\texttt {Flip}(\xi ) = \sum _{j}n_{j}\otimes m_{j}^{\mathrm {op}}$ and

(3.17)

Again, since the maps in (3.16) are isometric, we see that

which equals $\left \| \xi \right \|^2$ by Equation (3.2).

(U3) The properties of and of imply that

whenever $ m ,n_{1},n_{2}\in M$ and $b\in B$ are chosen such that the left (and hence each) side of the above equations makes sense. Likewise, we have For $h\in u\mathcal {H}$ , suppose we are given elements . Then

Because of continuity and (bi)linearity of $\Psi _{h}$ and of , we conclude that , as claimed. One likewise checks that (U4) holds for for elementary tensors, and again uses (bi)linearity and continuity of $U_{y}$ and of $\texttt {Flip}$ to deduce the claim.

Proof Since $\Phi _{x,y}$ lands in $M(x)$ by construction, $\Phi $ covers $(x,y^{\mathrm {op}},y) \mapsto x$ . To see that $\Phi $ is jointly continuous, assume that we are given a convergent net $(\xi _{\lambda },k_{\lambda })\to (\xi , k)$ in , and let $q (\xi _{\lambda },k_{\lambda }) = (x_{\lambda },y_{\lambda }^{\mathrm {op}}, y_{\lambda })$ and $q (\xi ,k) = (x,y^{\mathrm {op}},y)$ ; note that $x_{\lambda }\to x$ and $y_{\lambda }\to y$ in X. To prove that $\Phi (\xi _{\lambda },k_{\lambda })\to \Phi (\xi , k)$ , fix $\delta>0$ ; we must find $\kappa \in \Gamma _0(X;\mathscr {M})$ with $\left \| \Phi (\xi , k) - \kappa (x) \right \|<\delta $ and $\left \| \Phi (\xi _{\lambda },k_{\lambda }) - \kappa (x_{\lambda }) \right \|<\delta $ for large $\lambda $ [Reference Duwenig and Li4, Lemma A.3].

Because of how the topology on $\mathscr {K}$ is defined and because $\xi _{\lambda }\to \xi $ in $\mathscr {K}$ , we can find finitely many nets $m_{j,\lambda }\to m_{j}$ and $n_{j,\lambda } \to n_{j}$ in M such that

(3.18) $$ \begin{align} \left\| \xi - \sum_{j=1}^{\ell} m_{j}\otimes n_{j}^{\mathrm{op}} \right\| < \frac{\delta}{2(\left\| k \right\|+1)} , \quad \left\| \xi_{\lambda} - \sum_{j=1}^{\ell} m_{j,\lambda}\otimes n_{j,\lambda}^{\mathrm{op}} \right\| < \frac{\delta}{2(\left\| k \right\|+1)}. \end{align} $$

Since is jointly continuous, this implies that for each j, in B. Since is jointly continuous, this in turn implies that

That means that we can find a section $\kappa $ of $\mathscr {M}$ with

(3.19)

for all $\lambda \geq \lambda _0$ . Let $\lambda $ be large enough such that we also have $|\left \| k_{\lambda } \right \|-\left \| k \right \|| < 1.$ Linearity of $\Phi $ in the first component yields:

and the exact same computation with subscript- $\lambda $ ’s yields $\left \| \Phi (\xi _{\lambda },k_{\lambda }) -\kappa (x_{\lambda }) \right \| < \delta $ for large $\lambda $ , which proves the claim.

Proof To see that U is jointly continuous, assume that $\{(\xi _{\lambda },\eta _{\lambda })\}_{\lambda }$ is a net in that converges to $(\xi ,\eta )$ . Let , which converges to , and let , which converges to . Fix an arbitrary $\delta \in (0,1)$ ; by [Reference Duwenig and Li4, Lemma A.3], we should find a section with

$$\begin{align*}\left\| U\big( \xi , \eta \big) - \kappa(x,z^{\mathrm{op}}) \right\|<4\delta \text{ and } \left\| U\big( \xi_{\lambda} , \eta_{\lambda} \big) - \kappa(x_{\lambda},z_{\lambda}^{\mathrm{op}}) \right\|<4\delta \end{align*}$$

for large $\lambda $ . Because of how the topology on $\mathscr {K}$ is defined and because $\xi _{\lambda }\to \xi $ , $\eta _{\lambda }\to \eta $ in $\mathscr {K}$ , there exist elements

$$\begin{align*}\zeta_{\lambda} \in M(x_{\lambda})\odot M(y_{\lambda})^{\mathrm{op}} \text{ converging to } \zeta \in M(x)\odot M(y)^{\mathrm{op}} \text{ in } K \end{align*}$$

and elements

$$\begin{align*}\chi_{\lambda} \in M(y_{\lambda})\odot M(z_{\lambda})^{\mathrm{op}} \text{ converging to } \chi \in M(y)\odot M(z)^{\mathrm{op}} \text{ in } K \end{align*}$$

such that

(3.21) $$ \begin{align} \left\| \xi - \zeta \right\| , \left\| \xi_{\lambda} - \zeta_{\lambda} \right\| < \frac{\delta}{\left\| \eta \right\|+1}, &&\text{ and }& \left\| \eta - \chi \right\|, \left\| \eta_{\lambda} - \chi_{\lambda} \right\| < \frac{\delta}{\left\| \xi \right\|+1} \end{align} $$

for large $\lambda $ . Here, it is important to point out that $\odot $ refers to the algebraic tensor product, so each of the elements considered above is a finite sum of elementary tensors. Note that

(3.22) $$ \begin{align} U(m\otimes n^{\mathrm{op}}, k\otimes \ell^{\mathrm{op}}) = \Phi (m\otimes n^{\mathrm{op}}, k) \otimes \ell^{\mathrm{op}}, \end{align} $$

so using bilinearity of U and continuity of $\Phi $ , the fact that $\zeta _{\lambda }\to \zeta $ and $\chi _{\lambda }\to \chi $ in K and that addition and scalar multiplication is continuous on $\mathscr {K}$ , imply that

$$\begin{align*}U( \zeta_{\lambda}, \chi_{\lambda}) \to U(\zeta,\chi) \end{align*}$$

for these particular sums of elementary tensors $ \zeta _{\lambda }, \chi _{\lambda }$ . This convergence means that we can find a section $\kappa $ of $\mathscr {K}$ with

(3.23) $$ \begin{align} \left\| U( \zeta_{\lambda}, \chi_{\lambda}) - \kappa(x_{\lambda},z_{\lambda}^{\mathrm{op}}) \right\|<\delta \text{ and } \left\| U( \zeta,\chi) - \kappa(x,z^{\mathrm{op}}) \right\|<\delta. \end{align} $$

Combining this with Condition (U2), we conclude for large $\lambda $ :

$$ \begin{align*} \left\| U\big( \xi , \eta \big) - \kappa(x,z^{\mathrm{op}}) \right\| &\leq \left\| U\big( \xi , [\eta-\chi]\big) \right\| + \left\| U\big( [\xi-\zeta] , \chi\big) \right\| + \left\| U\big(\zeta,\chi\big) - \kappa(x,z^{\mathrm{op}}) \right\|\\& < \left\| \xi \right\|\, \left\| \eta-\chi \right\| + \left\| \xi-\zeta \right\| \,\left\| \chi \right\| +\delta \quad\text{ by (3.23), (U2)}\\&< 3\delta \qquad\qquad\qquad\qquad\qquad\qquad\text{by (3.21)}. \end{align*} $$

We likewise get for $\lambda $ large enough such that $|\left \| \xi \right \|-\left \| \xi _{\lambda } \right \||$ and $ |\left \| \eta \right \|-\left \| \eta _{\lambda } \right \||$ are bounded by $1$ ,

$$ \begin{align*} \left\| U\big( \xi_{\lambda} , \eta_{\lambda} \big) - \kappa(x_{\lambda},z_{\lambda}^{\mathrm{op}}) \right\| &<\left\| \xi_{\lambda} \right\|\, \left\| \eta_{\lambda}-\chi_{\lambda} \right\| + \left\| \xi_{\lambda}-\zeta_{\lambda} \right\| \,\left\| \chi_{\lambda} \right\| +\delta \\ &\overset{(3.21)}{<} (\left\| \xi \right\|+1)\, \frac{\delta}{\left\| \xi \right\|+1} + \frac{\delta}{\left\| \eta \right\|+1} \,\left\| \chi_{\lambda} \right\| +\delta. \end{align*} $$

Note that

$$ \begin{align*} \left\| \chi_{\lambda} \right\| \leq \left\| \chi_{\lambda} - \eta_{\lambda} \right\| + \left\| \eta_{\lambda} \right\| \leq \left\| \chi_{\lambda} - \eta_{\lambda} \right\| + \left\| \eta \right\|+1 \leq \frac{\delta}{\left\| \xi \right\|+1} + \left\| \eta \right\|+1 , \end{align*} $$

so that

$$ \begin{align*} \frac{\delta}{\left\| \eta \right\|+1} \,\left\| \chi_{\lambda} \right\| \leq \frac{\delta}{\left\| \eta \right\|+1} \, \left[\frac{\delta}{\left\| \xi \right\|+1} + \left\| \eta \right\|+1\right] \leq 2\delta. \end{align*} $$

We conclude that

$$ \begin{align*} \left\| U( \xi_{\lambda} , \eta_{\lambda} ) - \kappa(x_{\lambda},z_{\lambda}^{\mathrm{op}}) \right\| < (\left\| \xi \right\|+1)\, \frac{\delta}{\left\| \xi \right\|+1} + 2\delta +\delta < 4\delta, \end{align*} $$

as needed.

For (3.20), it likewise suffices to prove the claims for elementary tensors $\mu =\ell \otimes m_{1}^{\mathrm {op}}$ , $\xi =m_{2}\otimes n_{1}^{\mathrm {op}}$ , and $\eta =n_{2}\otimes k^{\mathrm {op}}$ . The computations make use of $\mathscr {B}$ -linearity of and are left to the reader.

Proof of Lemma 3.15

Since , the map is well defined. Since $U_{y}$ and $\Psi _{h}$ are linear, linearity of follows directly from the definition of the fibrewise Banach space structure on $\mathscr {A}$ .

To see that it is continuous, we will invoke openness of the quotient map Q: Suppose $\{\chi _{\lambda }\}_{\lambda }$ is a net in which converges to $\chi $ . Since it suffices to show that a subnet of converges to , we can without loss of generality assume that the entire net $\{\chi _{\lambda }\}_{\lambda }$ lifts to a net $\{(\xi _{\lambda }, \eta _{\lambda })\}_{\lambda }$ in $K\times K$ that converges to a lift $(\xi ,\eta )$ of $\chi $ (here, we have made use of [Reference Williams28, Proposition 1.1 (Fell’s criterion)] for the open map Q). Denote $q_{\mathscr {K}}(\xi _{\lambda })=(x_{\lambda }, y_{\lambda }^{\mathrm {op}})$ . Since , we may let $h_{\lambda },h\in \mathcal {H}$ be the unique elements such that $q_{\mathscr {K}}\big (\Psi _{h_{\lambda }}(\eta _{\lambda })\big )=(y_{\lambda }, z_{\lambda }^{\mathrm {op}})$ and $q_{\mathscr {K}}\big (\Psi _{h}(\eta )\big )=(y, z^{\mathrm {op}})$ for some $z_{\lambda },z\in X$ . Since $\xi _{\lambda }\to \xi $ , it follows that $y_{\lambda }\to y$ . Since $\eta _{\lambda }\to \eta $ and $q_{\mathscr {K}}\big (\Psi _{h_{\lambda }}(\eta _{\lambda })\big )=(y_{\lambda }, z_{\lambda }^{\mathrm {op}})$ , we further have that . Since both $y_{\lambda }$ and converge, it follows from properness and freeness of the $\mathcal {H}$ -action on X that a subnet of $\{h_{\lambda }\}_{\lambda }$ converges to h; again, without loss of generality we can assume that the entire net converges. By continuity of $\Psi $ , we conclude that $ \Psi _{h_{\lambda }}(\eta _{\lambda })\to \Psi _{h} (\eta ) $ . Since U is jointly continuous by Lemma 3.14, we therefore have

$$\begin{align*}U \big( \xi_{\lambda} , \Psi_{h_{\lambda}}(\eta_{\lambda}) \big) \longrightarrow U\big( \xi , \Psi_{h} (\eta) \big) \end{align*}$$

in K, which suffices since Q is continuous.

Proof We will check the conditions as stated in [Reference Duwenig and Li4, Definition 2.9]. It is clear that multiplication is bilinear and that is conjugate linear and self-inverse; this takes care of Conditions (F2), (F6), and (F8). For the following, fix , and let $h\in \mathcal {H}$ be the unique element such that $q_{\mathscr {K}}(\xi )=(x,y^{\mathrm {op}})$ and $q_{\mathscr {K}}(\Psi _{h}(\eta ))=(y,z^{\mathrm {op}})$ , meaning that .

(F1) Since $U_{y}\big ( \xi , \Psi _{h}(\eta )\big )\in K(x,z^{\mathrm {op}})$ and since $p_{\mathscr {A}}([\xi ])=[q_{\mathscr {K}}(\xi )]$ , we get the first and last equality in the following computation:

$$\begin{align*}p_{\mathscr{A}} ([\xi]\cdot[\eta]) = [x,z^{\mathrm{op}}] = [x,y^{\mathrm{op}}]\cdot [y,z^{\mathrm{op}}] = p_{\mathscr{A}} ([\xi])\cdot p_{\mathscr{A}}([\eta]). \end{align*}$$

(F3) Associativity of the multiplication can be shown using uniqueness of h, the second identity in (3.20), commutativity of Diagram (3.15), and the fact that $\Psi _{hh'}=\Psi _{h}\circ \Psi _{h'}$ ; the details are left as an exercise.

(F4) The definition of the norm on the fibres of $\mathscr {A}$ implies that

$$\begin{align*}\| [\xi]\cdot[\eta] \| = \left\| U_{y}\big( \xi , \Psi_{h}(\eta)\big) \right\| \overset{\textrm{(U2)}}{\leq} \left\| \xi \right\|\, \left\| \Psi_{h}(\eta) \right\| \overset{(\Psi 2 )}{=} \left\| \xi \right\| \, \left\|\eta \right\| = \left\| [\xi] \right\| \, \left\|[\eta] \right\|, \end{align*}$$

proving that multiplication is norm-decreasing.

(F5) Since $p_{\mathscr {A}}([\xi ]^*)= [q_{\mathscr {K}}(\texttt {Flip}(\xi ))]$ and since in $\mathcal {G}$ , we have that maps $A(g)$ to .

(F7) We must show that $([\xi ]\cdot [\eta ])^*=[\eta ]^*\cdot [\xi ]^*$ . We have

which is exactly $[\eta ]^*\cdot [\xi ]^*$ .

(F9) It is clear that $\|[\xi ]\| = \|[\xi ]^*\|$ . To show $\|[\xi ]\cdot [\xi ]^*\| =\|[\xi ]\|^2$ , note that has $q_{\mathscr {K}}(\eta )=(y,x^{\mathrm {op}})$ ; in particular, is the unique element of $\mathcal {H}$ with $\texttt {sec}_{\mathscr {K}}(\xi ) = \texttt {fir}_{\mathscr {K}}\big (\Psi _{u}(\eta )\big )$ . Since $\Psi _{u}$ is the identity, we conclude that

$$\begin{align*}[\xi]\cdot[\xi]^* = \big[ U_{y} \big(\xi, \texttt{Flip}(\xi))\big], \end{align*}$$

which has norm equal to $\left \| \xi \right \|^2=\left \| [\xi ] \right \|^2$ by (U2), as needed.

(F10) To see that $[\xi ]\cdot [\xi ]^* \geq 0$ , we again invoke [Reference Raeburn and Williams24, Lemma 2.65]: The matrix with $i,j$ -entry is positive, so there exists a matrix $(b_{i,j})_{i,j}$ over $B(u)$ such that . In particular, the $B(u)$ -balancing in $M(x)\otimes _{u}M(y)^{\mathrm {op}}$ allows us to write

If we let , then this combined with (3.17) shows that

$$ \begin{align*} [\xi]\cdot [\xi]^* = \big[ U_{y} \big(\xi, \texttt{Flip}(\xi))\big] = \sum_{l} \left[ \mathbf{x}_{l}\otimes \mathbf{x}_{l}^{\mathrm{op}} \right]. \end{align*} $$

Since the element $\mathbf {x}_{l}\otimes \mathbf {x}_{l}^{\mathrm {op}}$ of $K(x,x^{\mathrm {op}})$ is positive (it corresponds to the positive operator in $\mathbb {K}_{B(u)}(M(x))$ ), and since the algebra $A([x,x^{\mathrm {op}}])$ is *-isomorphic to $K(x,x^{\mathrm {op}})$ (see Remark 3.16), the claim follows.

Lastly, to see that $\mathscr {A}$ is saturated, we must show that the linear span of elements of the form $[\xi ]\cdot [\eta ]$ for $[\xi ]\in A([x,y^{\mathrm {op}}])$ and $[\eta ]\in A([y,z^{\mathrm {op}}])$ is dense in $A([x,z^{\mathrm {op}}])$ . When we consider that $A([x,y^{\mathrm {op}}])\cong M(x)\otimes _{u}M(y)^{\mathrm {op}}$ and that these isomorphisms respect the multiplicative and linear structure we defined on $\mathscr {A}$ , then the claim follows directly from the first isomorphism in Equation (3.16).

Proof of Proposition 4.1

Write $p_{\mathscr {A}}([\xi ])=[x_{1},x_{2}^{\mathrm {op}}]$ . Since , we have $s_{\mathcal {G}}([x_{1},x_{2}^{\mathrm {op}}])=\rho _{\mathscr {M}} (m)=[y,y^{\mathrm {op}}]$ , which means that there exists a unique $h\in \mathcal {H}$ such that . Now that the second component is fixed as $y^{\mathrm {op}}$ , the first component is also uniquely determined, and satisfies $\sigma (x)=s_{\mathcal {H}}(h)$ . In other words, we have argued that the equality $s_{\mathscr {A}}([\xi ])=\rho _{\mathscr {M}} (m)$ implies that there exists a unique representative $\Psi _{h}(\xi )\in K$ of the $\mathbin {\texttt {R}}$ -equivalence class $[\xi ]$ for which $q_{\mathscr {K}}(\Psi _{h}(\xi ))$ has $q_{\mathscr {M}}(m)^{\mathrm {op}}$ as its second component: $\Psi _{h}(\xi )\in M(x)\otimes _{\sigma (x)} M(y^{\mathrm {op}})$ . We can now invoke Lemma 3.11 to conclude that is a well-defined element of $M(x)$ .

(LA1) Follows from .

(LA2) We compute

so that bilinearity and continuity of $\Phi $ implies

for any appropriate $\eta \in K$ . The claim follows.

Proof The map is clearly well defined. Property (LIP1) follows immediately from Equation (1.3).

(LIP2) Sesquilinearity is obvious. For continuity, it suffices to check that the map , $(m,n)\mapsto m\otimes n^{\mathrm {op}}$ , is continuous since the quotient map ${\mathscr {K}\to \mathscr {A}}$ is continuous. But said continuity is built into the definition of $\mathscr {K}$ ’s topology; see Lemma 3.1.

(LIP3) Follows from the definition of the involution ${}^*$ on $\mathscr {A}$ .

(LIP4) Recall from Proposition 4.1 that

where $y=q_{\mathscr {M}}(m_{1})$ and where is the unique element such that $q_{\mathscr {K}}(\Psi _{h}(\xi ))=(x,y^{\mathrm {op}})$ , so that

so (LIP4) holds.

(LIP5) Let $y=q_{\mathscr {M}}(m_{3})$ and let $h\in \mathcal {H}$ and $x\in X$ be the unique elements such that ${q_{\mathscr {K}}(\Psi _{h}(m_{1}\otimes m_{2}^{\mathrm {op}}))}={(x,y^{\mathrm {op}})}$ . Because of (3.4), we know that and . In particular, we can approximate $m_{1}$ by for finitely many $n_{i}\in M(x)$ and $b_{i}\in B(h)$ . In the following final computation, the first instance of $\approx $ is valid because $\Phi _{x,y}$ and $\Psi _{h}$ are continuous and linear (see ( $\Psi $ 2)), and the second because the right $\mathscr {B}$ -action on $\mathscr {M}$ is continuous by (DE16).

Proof of Corollary 4.3

We first check that $\mathscr {M}$ is a left $\mathscr {A}$ -demi-equivalence. In the following lists, all references are either to Proposition 4.1 or to Proposition 4.2.

1. (DE1) was shown in (LA1);

2. (DE2) was shown in (LIP1);

3. (DE3) was shown in (LIP2);

4. (DE4) was shown in (LIP4); and

5. (DE5) was shown in (LIP3).

For the remaining conditions necessary for a demi-equivalence that we have not checked previously, note that each $M(x)$ is a full right-Hilbert $\mathrm {C}^{*}$ -module by (DE9), and so it is an imprimitivity bimodule between $M(x)\otimes _{u} M(x)^{\mathrm {op}}$ and $B(u)$ for . All structure with which we equipped the bundle $\mathscr {A}$ is compatible with the homeomorphism $A(\rho (x))\cong M(x)\otimes _{u} M(x)^{\mathrm {op}}$ in Lemma 3.7; in other words, with the restriction to $M(x)$ of the given $\mathscr {A}$ -action and $\mathscr {A}$ -inner product on $\mathscr {M}$ , we recover exactly the structure that $M(x)$ naturally carries. In particular:

1. (DE6): We have in the $\mathrm {C}^{*}$ -algebra $A(\rho _{\mathscr {M}}(m))$ , and only if $m=0$ .

2. (DE7): By [Reference Raeburn and Williams24, Lemma 2.30], the norm of in $B(\sigma (x))$ coincides with the norm of $m\otimes m^{\mathrm {op}}$ in $M(x)\otimes _{u}M(x)^{\mathrm {op}}$ which, by definition, is exactly the norm of in $A(\rho (x))$ . Since $\mathscr {M}$ is a right $\mathscr {B}$ -demi-equivalence, the norm agrees with the norm that the upper semi-continuous Banach bundle $\mathscr {M}$ already carries, which is therefore in turn identical to the norm .

3. (DE8): For each $x\in X$ , the linear span of is dense in $A(\rho (x))$ , since $M(x)$ is an imprimitivity bimodule.

That $A(\rho (x))\cong M(x)\otimes _{u}M(x)^{\mathrm {op}}$ further implies that Assumption (2) of Proposition 2.3 is satisfied. Since Assumptions (1) and (3) were shown in (LA2) and (LIP5), respectively, we can conclude that $\mathscr {M}$ is an equivalence by Proposition 2.3.

Proof of Proposition 5.1

By Corollary 4.3, we may without loss of generality assume that $\mathscr {A}=\mathscr {M}\otimes _{\mathscr {B}}\mathscr {M}^{\mathrm {op}}$ and $\mathcal {G}=X\times _{\mathcal {H}}X^{\mathrm {op}}$ . In that setting, is exactly $[q_{\mathscr {M}}(m),q_{\mathscr {M}}(n)] \in \mathcal {G}$ ; see Lemma 1.3. We also remind the reader that , is known to be an isomorphism of topological groupoids. It is therefore clear that $\mathscr {A}'\cong \omega ^*(\mathscr {A}')$ , and so we may without loss of generality assume that $\mathcal {G}'=\mathcal {G}$ (and hence $\omega =\operatorname {id}$ and $\mathscr {A}' = \omega ^*(\mathscr {A}')$ ). As before, we will denote the anchor maps of X by and . Recall that via .

To construct $\Omega $ , we will first construct a surjective bundle map $ \tilde {\Omega } \colon \mathscr {K} \to \mathscr {A}' $ covering the quotient map . We will then show that $\tilde {\Omega }$ is constant on $\mathbin {\texttt {R}}$ -equivalence classes (where $\mathbin {\texttt {R}}$ is as in Lemma 3.5), so that $\tilde {\Omega }$ induces a bundle map $\Omega $ from the quotient $\mathscr {M}\otimes _{\mathscr {B}}\mathscr {M}^{\mathrm {op}} = Q(\mathscr {K})$ to $\mathscr {A}'$ covering the identity map $\mathcal {G}\to \mathcal {G}$ . After we have shown that $\Omega $ is injective, we will then prove the conditions in [Reference Duwenig and Li4, Propositions A.7 and A.8] to deduce that $\Omega $ is open and continuous and hence the claimed isomorphism of Fell bundles.

We define $\tilde {\Omega }$ fibrewise. Fix and $x,y\in Xu$ , let , and consider

(5.1)

By the assumption on the left $\mathscr {A}'$ -valued inner product on $\mathscr {M}$ , this map is bilinear. Moreover, by [Reference Duwenig and Li4, Corollary 4.6]Footnote ⁸ , we have for all $b\in B(u)$ , meaning that the map in (5.1) descends to a linear map

$$\begin{align*}\tilde{\Omega}_{(x,y^{\mathrm{op}})}\colon M(x)\odot_{B(u)}M(y)^{\mathrm{op}} \to A'(g), \end{align*}$$

determined by

where $M(x)\odot _{B(u)}M(y)^{\mathrm {op}}$ denotes the balanced algebraic tensor product.

We claim that $\tilde {\Omega }_{(x,y^{\mathrm {op}})}$ has dense range. Since $\mathscr {A}'$ is a Fell bundle, $A'(g)$ is an -imprimitivity bimodule, so by [Reference Raeburn and Williams24, Proposition 2.33 (Hewitt-Cohen)], any element of $ A'(g)$ can be written as for some $a_{g}\in A'(g)$ and some , where the product is the multiplication in the Fell bundle $\mathscr {A}'$ . By assumption, $M(y)$ is a -imprimitivity bimodule. In particular, is dense in , meaning we can approximate by linear combinations of elements of the form for $n',n\in M(y)$ . Since and , we conclude that the arbitrary element of $ A'(g)$ can be approximated by linear combinations of elements of the form for $m\in M(x),n\in M(y)$ , as claimed.

Since $\mathscr {M}$ is an equivalence, $M(x)$ is an imprimitivity bimodule between and $B(\sigma (x))=B(u)$ , so that there exists a canonical isomorphism of $\mathrm {C}^{*}$ -algebras. In particular, the norm with which we equip $M(x)\odot _{B(u)}M(y)^{\mathrm {op}}$ , is unambiguously given by Equation (3.2) and, using the $*$ -isomorphism, can be rewritten to

Using the properties [Reference Duwenig and Li4, Definition 2.9, (F9)] and [Reference Duwenig and Li4, Definition 2.11, (FE2.b), (FE2.c), (FE2.d)], we therefore see that each $ \tilde {\Omega }_{(x,y^{\mathrm {op}})}$ is isometric and hence extends to a map on the completion $K(x,y^{\mathrm {op}})$ of $M(x)\odot _{B(u)}M(y)^{\mathrm {op}}$ . All in all, we get a fibrewise linear map

determined by

covering the quotient map .

Next, fix , $h\in u\mathcal {H} v$ , and $x\in Xu$ , $y\in Xv$ . For an arbitrary but fixed element of , we have

Since $\tilde {\Omega }_{(x,y^{\mathrm {op}})}(\Psi _{h}(\xi ))=\tilde {\Omega }_{(x,y^{\mathrm {op}})}(\xi )$ for all $\xi $ in the algebraic tensor product, we conclude that $\tilde {\Omega }$ descends to a bundle map

covering the identity map $\mathcal {G}\to \mathcal {G}$ . We must now show that this map is a continuous open bijection.

Since $\Omega $ covers the identity map, surjectivity of $\Omega $ follows from the fact that each $\tilde {\Omega }_{(x,y^{\mathrm {op}})}$ has dense range, and injectivity of $\Omega $ follows since each $\tilde {\Omega }_{(x,y^{\mathrm {op}})}$ is isometric. To see that $\Omega $ is continuous, suppose that we have a convergent net in $\mathscr {A}=\mathscr {M}\otimes _{\mathscr {B}}\mathscr {M}^{\mathrm {op}}$ , say $[\xi _{\lambda }]\to [\xi ]$ ; by $(2)\implies (1)$ in [Reference Munkres23, Theorem 18.1], it suffices to show that a subnet of $\{\Omega ([\xi _{\lambda }])\}_{\lambda }$ converges to $\Omega ([\xi ])$ . Since the quotient map $Q\colon \mathscr {K}\to \mathscr {A}$ is open (Lemma 3.5), we can without loss of generality assume that $\xi _{\lambda }\to \xi $ in $\mathscr {K}$ . Let $q_{\mathscr {K}}(\xi _{\lambda })=(x_{\lambda },y_{\lambda }^{\mathrm {op}})$ with limit $q_{\mathscr {K}}(\xi )=(x,y)$ in . We now invoke the definition of the topology on $\mathscr {K}$ (Lemma 3.1; see also [Reference Duwenig and Li4, Lemma A.3]): for any given $\epsilon>0$ , we can find finitely many continuous sections $\mu _{i},\nu _{i}$ of $\mathscr {M}$ and a $\lambda _0$ such that for all $\lambda \geq \lambda _0$ , we have

(5.2) $$ \begin{align} \left\| \xi_{\lambda} - \sum_{i=1}^{k}\mu_{i}(x_{\lambda})\otimes \nu_{i}(y_{\lambda})^{\mathrm{op}} \right\| < \epsilon \text{ and } \left\| \xi - \sum_{i=1}^{k}\mu_{i}(x)\otimes \nu_{i}(y)^{\mathrm{op}} \right\|<\epsilon. \end{align} $$

The norm on each fibre $A([x,y^{\mathrm {op}}])$ of $\mathscr {A}$ is defined as the norm on one of its lifts under Q, meaning that Q restricted to $M(x)\otimes _{u}M(y)^{\mathrm {op}}$ is isometric. Since $\Omega $ is likewise isometric, (5.2) implies

(5.3) $$ \begin{align} \begin{aligned} \left\| \Omega([\xi_{\lambda}]) - \Omega\left(\left[\sum\nolimits_{i}\mu_{i}(x_{\lambda})\otimes \nu_{i}(y_{\lambda})^{\mathrm{op}}\right]\right) \right\| < \epsilon&\text{ and } \\ \left\| \Omega([\xi]) - \Omega\left(\left[\sum\nolimits_{i}\mu_{i}(x)\otimes \nu_{i}(y)^{\mathrm{op}}\right]\right) \right\|<\epsilon &. \end{aligned} \end{align} $$

By definition of $\Omega $ ,

By assumption on $\mathscr {M}$ , is continuous, so that continuity of $\mu _{i},\nu _{i}$ and of addition in $\mathscr {A}$ implies

Combining (5.3) with Lemma B.1, we conclude that $\Omega ([\xi _{\lambda }])\to \Omega ([\xi ])$ , so $\Omega $ is continuous. Lastly, since $\Omega $ is isometric, it follows from [Reference Duwenig and Li4, Proposition A.8] that $\Omega $ is open as well.

Proof sketch

First, let us make sure that the objects we have written down are not nonsense: By Theorem 1.6, all three demi-equivalences are equivalences between their respective imprimitivity Fell bundles and $\mathscr {B}$ . As explained in [Reference Muhly and Williams22, Example 6.7], their opposites are then likewise equivalences, just in the other direction. It was shown in [Reference Duwenig and Li4] that equivalences can be concatenated: $\mathscr {M} \otimes _{\mathscr {B}}\mathscr {N}^{\mathrm {op}}$ is an equivalence from the imprimitivity Fell bundle of $\mathscr {M}$ to $\mathscr {A}$ , and so forth. In particular, both the left- and the right-hand side of the alleged isomorphism is an equivalence between the imprimitivity Fell bundle of $\mathscr {M}$ and that of $\mathscr {K}$ .

Since the balanced tensor product of imprimitivity bimodule is associative, it is easy to see that the same is true for $\otimes _{\mathscr {B}}$ of Fell bundle equivalences. In particular,

$$ \begin{align*} (\mathscr{M} \otimes_{\mathscr{B}}\mathscr{N}^{\mathrm{op}})\otimes_{\mathscr{A}} (\mathscr{N}\otimes_{\mathscr{B}} \mathscr{K}^{\mathrm{op}}) & \cong \big(\mathscr{M} \otimes_{\mathscr{B}}(\mathscr{N}^{\mathrm{op}}\otimes_{\mathscr{A}} \mathscr{N})\big)\otimes_{\mathscr{B}} \mathscr{K}^{\mathrm{op}}. \end{align*} $$

By Proposition 5.1 (applied to $\mathscr {N}^{\mathrm {op}}$ ), the Fell bundle $\mathbb {K}(\mathscr {N}^{\mathrm {op}}_{\mathscr {A}})=\mathscr {N}^{\mathrm {op}}\otimes _{\mathscr {A}} \mathscr {N}$ is isomorphic to $\mathscr {B}$ . Since the balanced tensor product of imprimitivity bimodule absorbs the coefficient algebra (meaning that $\mathbf {X}\otimes _{A} A\cong \mathbf {X}$ ), it is easy to see that $\otimes _{\mathscr {B}}$ absorbs the coefficient Fell bundle. Thus,

$$ \begin{align*} (\mathscr{M} \otimes_{\mathscr{B}}\mathscr{N}^{\mathrm{op}})\otimes_{\mathscr{A}} (\mathscr{N}\otimes_{\mathscr{B}} \mathscr{K}^{\mathrm{op}}) & \cong (\mathscr{M} \otimes_{\mathscr{B}} \mathscr{B} )\otimes_{\mathscr{B}} \mathscr{K}^{\mathrm{op}} \cong \mathscr{M}\otimes_{\mathscr{B}}\mathscr{K}^{\mathrm{op}}, \end{align*} $$

as claimed.

Proof Take $X=\mathcal {H}$ , with the source map of $\mathcal {H}$ . The existence of a Haar system implies that the source map of $\mathcal {H}$ (and hence the anchor map of X) is open [Reference Williams28, Proposition 1.23], so we are in good shape to use our main theorem, provided we can find the right $\mathscr {B}$ -demi-equivalence. Let $\mathscr {M}$ to be the bundle over X with fibres for each $h \in \mathcal {H}$ . We define

and

where we choose the inner product on to be conjugate linear in the first coordinate to match up with our definitions of Hilbert $\mathrm {C}^{*}$ -modules. Then, using Corollary 6.1, it is easy to see that

where $\vec {e}_{i}$ is the standard basis vector of and $E_{i,j}$ is the matrix unit with a $1$ in the $i^{\text {th}}$ row and $j^{\text {th}}$ column and $0$ s everywhere else. By Theorem 1.6, the bundles $\mathscr {B}$ and are equivalent. By [Reference Muhly and Williams22, Theorem 6.4], it follows that $\mathrm {C}^{*}(\mathscr {B})$ is strongly Morita equivalent to .