Proof To prove that $\psi $ is surjective, let $c\in C$ . Since $C=\overline {\bigcup _{i\in I}\gamma _i(C_i)}$ and the image of $\psi $ is closed, it suffices to prove that $c \in \text {Im}(\psi )$ , when c is of the form $c=\gamma _i(c_i)$ for some i. Since $\psi _i$ is surjective, there exists $b_i\in B_i$ such that $\psi _i(b_i) = c_i$ . The definition of $\psi $ gives us

$$\begin{align*}\psi(\beta_i(b_i)) = \gamma_i(\psi_i(b_i))=\gamma_i(c_i)=c. \end{align*}$$

Hence, $c \in \text {Im}(\psi )$ . This proves that $\psi $ is surjective.

As $\psi _i\circ \varphi _i =0$ for all $i\in I$ , it is clear that $\psi \circ \varphi =0$ , and hence $\text {Im}(\varphi ) \subset \text {Ker}(\psi )$ . To show the other inclusion, let $b \in B$ be such that $\psi (b) =0$ . Let $\epsilon>0$ be given. Choose i and $b_i \in B_i$ such that $\|b-\beta _i(b_i)\|<\epsilon $ . Since $\psi (b)=0$ ,

$$\begin{align*}\|\gamma_i(\psi_i(b_i))\|=\|\psi(b)-\psi(\beta_i(b_i))\|\leq \|b-\beta_i(b_i)\|<\epsilon. \end{align*}$$

Hence, for large j, $\|\gamma _{j,i}(\psi _i(b_i))\|<\epsilon $ . Choose one such j and let $b_j=\beta _{j,i}(b_i)$ . Then,

$$\begin{align*}\|\psi_j(b_j)\|=\|\psi_j(\beta_{j,i}(b_i))\|=\|\gamma_{j,i}(\psi_i(b_i))\|<\epsilon. \end{align*}$$

Since $\text {Ker}(\psi _j)\kern1.4pt{=}\kern1.4pt\text {Im}(\varphi _j)$ , it follows that there exists $a_j \kern1.4pt{\in}\kern1.4pt A_j$ such that $ {\|b_j-\varphi _j(a_j)\|\kern1.4pt{<}\kern1.4pt\epsilon }$ . Let $a=\alpha _j(a_j)$ . Then,

$$ \begin{align*} \|b-\varphi(a)\| &\leq \|b-\beta_j(b_j)\|+\|\beta_j(b_j)-\beta_j(\varphi_j(a_j))\|\\& \leq \|b-\beta_i(b_i)\|+\|b_j-\varphi_j(a_j)\|\\ & < 2\epsilon. \end{align*} $$

As $\text {Im}(\varphi )$ is closed, it follows that $b\in \text {Im}(\varphi )$ . Hence, $\text {Ker}(\psi ) \subset \text {Im}(\varphi )$ . Consequently, $\text {Im}(\varphi )=\text {Ker}(\psi )$ .

The argument required to prove that $\varphi $ is injective is similar to the one used to prove $\text {Im}(\varphi )=\text {Ker}(\psi )$ . So, we omit the proof.

Proof Let $a \in A$ be such that $\varphi (a) =0$ . Then, $\varphi _z(a(z)) =0$ for all $z\in Z$ . Since each $\varphi _z$ is injective, $a(z)=0$ for all $z\in Z$ , hence $a=0$ . Thus, $\varphi $ is injective.

To show that $\psi $ is onto, note that $\psi (B)$ is a $C_0(Z)$ -subalgebra of C as $\psi $ is a $C_0(Z)$ -linear map. For $z\in Z$ ,

$$\begin{align*}\overline{\text{span}\{\psi(b)(z) : b\in B\}} = \overline{\text{span}\{\psi_z(b(z)) : b\in B\}} =C(z), \end{align*}$$

because $\psi _z$ is onto. Then, by [Reference Muhly and Williams10, Lemma A.4], we conclude that $\psi $ is onto.

As $\psi _z\circ \varphi _z =0$ for all $z\in Z$ , we have $\psi \circ \varphi = 0$ . Hence, $\text {Im}(\varphi ) \subseteq \text {Ker}(\psi )$ .

Let $b \in B$ be such that $\psi (b)=0$ . First, consider the case when b is compactly supported. Let $K \subset Z$ be compact such that b vanishes outside K. Let $\epsilon>0$ . Let $z \in Z$ be given. Since $\psi $ is $C_0(Z)$ -linear, $\psi _z(b(z))=0$ . Since $\text {Im}(\varphi _z)=\text {Ker}(\psi _z)$ , there exists $\widetilde {a}_z \in A(z)$ such that $b(z)=\varphi _z(\widetilde {a}_z)$ . Choose $a \in A$ such that $a(z)=\widetilde {a}_z$ . Then, there exists a neighborhood of z, say $U_z$ , such that $\|b(y)-\varphi _y(a(y))\|<\epsilon $ for $y \in U_z$ . A partition of unity argument implies that there exists a compactly supported section $a \in A$ such that $\|b(y)-\varphi _y(a(y))\|<\epsilon $ for all $y \in Z$ . Then, $\|b-\varphi (a)\| <\epsilon $ . Hence, $b \in \overline {\text {Im}{(\varphi )}}=\text {Im}{(\varphi )}$ .

Now, let $b \in B$ be such that $\psi (b)=0$ . Let $(f_n)_n$ be an approximate identity in $C_0(Z)$ such that $f_n$ is compactly supported for each n. Since $C_0(Z)B$ is total in B, it follows that $(f_nb)_n \to b$ in norm. Since $\psi $ is a $C_0(Z)$ -homomorphism, it follows that $\psi (f_nb)=0$ for all n. From the earlier case, we deduce that $f_nb \in \text {Im}{(\varphi )}$ . As $\text {Im}{(\varphi )}$ is closed and $(f_nb)_n \to b$ in norm, it follows that $b \in \text {Im}{(\varphi )}$ .

Proof Let $(\pi _A,V)$ be the regular representation of $(A,P,\alpha )$ on $\ell ^2(P) \otimes A$ and $(\pi _{I \otimes A},\widetilde {V})$ be the regular representation of $(I\otimes A,P,1 \otimes \alpha )$ on $\ell ^2(P)\otimes (I \otimes A)$ . Note that $I\otimes (A\rtimes _{\mathrm {red}}P) \subseteq \mathcal {L} (I\otimes \mathcal {E})$ , where $\mathcal {E} =\ell ^2(P)\otimes A$ . Define a canonical unitary $U\colon I\otimes (\ell ^2(P)\otimes A) \to \ell ^2(P)\otimes (I\otimes A)$ by

$$\begin{align*}U(x\otimes (\delta_t\otimes y)) = \delta_t\otimes (x\otimes y). \end{align*}$$

Now, for $x\in I, a\in A$ , we have

$$ \begin{align*} &U(x\otimes \pi_A(a))U^*(\delta_t\otimes (z\otimes y)) = U(x\otimes \pi_A(a)) (z\otimes(\delta_t\otimes y))\\ &\quad =U(xz\otimes(\delta_t \otimes\alpha_t(a)y)) = \delta_t \otimes (xz\otimes \alpha_t(a)y)\\ &\quad =\pi_{I\otimes A}(x\otimes a) (\delta_t\otimes (z\otimes y)). \end{align*} $$

Thus, $U(x\otimes \pi _A(a)) U^* = \pi _{I\otimes A}(x\otimes a)$ for $x\in I$ and $a\in A$ . For $s\in P$ , we have

$$ \begin{align*} U(1\otimes V_s)U^*(\delta_t\otimes (z\otimes y)) &=U(1\otimes V_s) (z\otimes (\delta_t\otimes y)) =U(z\otimes (\delta_{ts}\otimes y))\\ &=\delta_{ts}\otimes (z\otimes y) = \widetilde{V}_s(\delta_t\otimes (z\otimes y)). \end{align*} $$

Hence, for $s \kern1.4pt{\in}\kern1.4pt P$ , $U(1 \otimes V_s)U^*\kern1.4pt{=}\kern1.4pt\widetilde {V}_s$ . Thus, $\mathrm {Ad}(U)$ maps the generators of ${I \kern1.4pt{\otimes}\kern1.4pt (A \rtimes _{\mathrm {red}} P)}$ onto the generators of ${(I \otimes A) \rtimes _{\mathrm {red}} P}$ . Hence, $\mathrm {Ad}(U)$ is an isomorphism from ${I \otimes (A \rtimes _{\mathrm {red}} P)}$ onto $(I \otimes A)\rtimes _{\mathrm {red}}P$ .

Proof Let $(\mathcal {D}^{I},\mathcal G)$ , $(\mathcal {D}^{A},\mathcal G),$ and $(\mathcal {D}^{B},\mathcal G)$ be the groupoid dynamical system associated with $(I,P,\alpha )$ , $(A,P,\beta ),$ and $(B,P,\gamma )$ , respectively. We also consider the equivalent ones $(\widetilde {\mathcal {D}}^I, \widetilde {\Omega }\rtimes G), (\widetilde {\mathcal {D}}^A, \widetilde {\Omega } \rtimes G),$ and $(\widetilde {\mathcal {D}}^B, \widetilde {\Omega }\rtimes G)$ , where the acting groupoid is the transformation groupoid $\widetilde {\Omega }\rtimes G$ . The section algebra of $\widetilde {\mathcal {D}}^{\bullet }$ is denoted by $\widetilde {\mathrm {D}}^{\bullet }$ , and that of $\mathcal {D}^{\bullet }$ is denoted by $\mathrm {D}^{\bullet }$ for $\bullet \in \{I, A, B\}$ .

Let $F\in \widetilde {\Omega }$ . By assumption, F is directed. Consider the inductive systems of $C^{*}$ -algebras

$$\begin{align*}I_s =I, \quad A_s =A, \quad B_s=B \end{align*}$$

with connecting maps

$$\begin{align*}\varphi_{t,s} = \alpha_{st^{-1}}, \quad \beta_{st^{-1}}, \quad \gamma_{st^{-1}} \quad \text{for } t \leq s\in F, \end{align*}$$

respectively. The natural map from $I_s \to \varinjlim _{s \in F}I_s$ will be denoted by $\lambda _s$ . We use the same letter $\lambda _s$ to denote the natural maps $A_s \to \varinjlim _{s \in F}A_s$ and $B_s \to \varinjlim _{s \in F}B_s$ .

We claim that the fiber $\widetilde {\mathcal {D}}^I_{F}$ of the bundle $\widetilde {\mathcal {D}}^I$ is isomorphic to the inductive limit $\varinjlim _{s\in F} I_s $ . The analogous statement (i.e., the unital version) for $\widetilde {\mathcal {D}}^{A}_F$ and $\widetilde {\mathcal {D}}^{B}_F$ is [Reference Hossain and Sundar3, Proposition 6.2].

Fix $F \in \widetilde {\Omega }$ . We embed $\ell ^{\infty }(G,I^{+})$ inside $\ell ^{\infty }(G,A^{+})$ and abuse notation. As A is unital,

$$ \begin{align*} \widetilde{\mathrm{D}}^A &= C^*(\{j_g(x): x\in A, g\in G\}), \quad \text{and}\\ \widetilde{\mathrm{D}}^I &= C^*(\{\varphi j_g(x): x\in I, g\in G, \varphi \in C_0(\widetilde{\Omega})\}). \end{align*} $$

Since $\widetilde {\mathrm {D}}^I$ is an ideal of $\widetilde {\mathrm {D}}^A$ , we have the following commutative diagram:

In the above diagram, q stands for the quotient map. It was shown in [Reference Hossain and Sundar3, Proposition 6.2] that there exists a unique homomorphism $\lambda _1:\widetilde {\mathrm {D}}^{A} \to \varinjlim _{s \in F}A_s$ such that, for ${a \in A}$ ,

$$\begin{align*}\lambda_1(j_s(a)) = \begin{cases} \lambda_s(a) & \text{ if } s\in F,\\ 0 & \text{ if } s\notin F. \end{cases} \end{align*}$$

Moreover, it was also shown that $\lambda _1$ factors through to give an isomorphism, denoted $\widetilde {\lambda }_1$ , between $\widetilde {\mathcal {D}}^A_F$ and the inductive limit $ \varinjlim _{s \in F}A_s$ . Since, for $x \in A$ and $g,g_1,g_2,\ldots ,g_n \in G$ ,

$$\begin{align*}1_{\Omega g_1}1_{\Omega g_2}\dots 1_{\Omega g_n} j_g(x) =j_{g_1}(1)j_{g_2}(1)\dots j_{g_n}(1)j_g(x), \end{align*}$$

we have

$$ \begin{align*} \lambda_1( 1_{\Omega g_1}1_{\Omega g_2}\dots 1_{\Omega g_n} j_g(x)) &= \lambda_1(j_{g_1}(1))\lambda_1(j_{g_2}(1))\dots \lambda_1(j_{g_n}(1)) \lambda_1(j_{g}(x))\\ &= 1_{\Omega g_1}(F)1_{\Omega g_2}(F)\dots 1_{\Omega g_n}(F) \lambda_1(j_{g}(x)). \end{align*} $$

Since $C_0(\widetilde {\Omega })$ is the closure of $\text {span}\{1_{\Omega g_1}1_{\Omega g_2}\dots 1_{\Omega g_n}: g_1, g_2,\ldots , g_n\in G\}$ , the above computation gives us

(3.2) $$ \begin{align} \lambda_1(\varphi j_g(x)) = \varphi(F)\lambda_1(j_g(x)) \end{align} $$

for $x \in A$ . Hence,

$$ \begin{align*} \lambda_1(\varphi d)=\varphi(F)\lambda_1(d) \end{align*} $$

for $ \varphi \in C_{0}(\widetilde {\Omega })$ and $d \in \widetilde {\mathrm {D}}^{A}$ . In particular, the above equation together with Equation (3.2) imply that $\lambda _1$ factors through $\widetilde {\mathcal {D}}^{I}_F$ and maps $\widetilde {\mathcal {D}}^{I}_F$ onto the $C^{*}$ -algebra $\varinjlim _{s \in F}I_s \subset \varinjlim _{s \in F}A_s$ . The resulting map thus obtained from $\widetilde {\mathcal {D}}^{I}_F \to \varinjlim _{s \in F}I_s$ is denoted by $\widehat {\lambda }_1$ .

Since $\widetilde {\lambda }_1\colon \widetilde {\mathcal {D}}_F^{A} \to \varinjlim _{s \in F}A_s$ is an isomorphism and $\widetilde {\mathcal {D}}^I_F \hookrightarrow \widetilde {\mathcal {D}}^A_F$ is an embedding, it follows that $\widehat {\lambda }_1$ is an isomorphism. This proves the claim.

Applying Lemma 3.1 to Equation (3.1), we obtain

$$ \begin{align*} 0 \longrightarrow \widetilde{\mathcal{D}}_F^{I} \longrightarrow \widetilde{\mathcal{D}}_F^{A} \longrightarrow \widetilde{\mathcal{D}}_F^{B}\longrightarrow 0. \end{align*} $$

Recall that $\mathcal {D}^{\bullet }$ is the restriction of $\widetilde {\mathcal {D}}^{\bullet }$ onto $\Omega $ . Thus, for $F \in \Omega $ , we have a short exact sequence

$$ \begin{align*} 0 \longrightarrow \mathcal{D}_F^{I} \longrightarrow \mathcal{D}_F^{A} \longrightarrow \mathcal{D}_F^{B}\longrightarrow 0. \end{align*} $$

Applying Lemma 3.2, we obtain the following short exact sequence:

$$\begin{align*}0 \longrightarrow C(\Omega, \mathcal{D}^{I}) \longrightarrow C(\Omega, \mathcal{D}^{A}) \longrightarrow C(\Omega,\mathcal{D}^{B}) \longrightarrow 0. \end{align*}$$

Since, the full groupoid crossed product functor is exact (see [Reference Anantharaman-Delaroche and Renault2, Lemma 6.3.2] or [Reference Ionescu and Williams4, Theorem 3.7]), we have the following short exact sequence:

$$\begin{align*}0 \longrightarrow C(\Omega, \mathcal{D}^{I})\rtimes \mathcal{G} \longrightarrow C(\Omega,{\mathcal{D}}^{A})\rtimes \mathcal{G} \longrightarrow C(\Omega, \mathcal{D}^B)\rtimes \mathcal{G} \longrightarrow 0. \end{align*}$$

Since $\mathcal {G}$ is amenable, we have the following short exact sequence:

$$\begin{align*}0 \longrightarrow C(\Omega, \mathcal{D}^{I})\rtimes_{\mathrm{red}} \mathcal{G} \longrightarrow C(\Omega,{\mathcal{D}}^{A})\rtimes_{\mathrm{red}} \mathcal{G} \longrightarrow C(\Omega, \mathcal{D}^B)\rtimes_{\mathrm{red}} \mathcal{G} \longrightarrow 0. \end{align*}$$

Since $(P,G)$ satisfies the Toeplitz condition, [Reference Sundar13, Theorem 4.3] allows us to rewrite the last equation as

$$\begin{align*}0 \longrightarrow I\rtimes_{\mathrm{red}} P \longrightarrow A\rtimes_{\mathrm{red}} P \longrightarrow B \rtimes_{\mathrm{red}} P\longrightarrow 0. \end{align*}$$

This completes the proof.

Proof Let

$$\begin{align*}0\longrightarrow I \overset{\iota}{\longrightarrow} B \overset{\pi}{\longrightarrow} C \longrightarrow0 \end{align*}$$

be an arbitrary short exact sequence of separable $C^*$ -algebras. After unitizations, we obtain the short exact sequence

$$\begin{align*}0\longrightarrow I \longrightarrow B^{+} \overset{\pi^{+}}{\longrightarrow} C^{+} \longrightarrow0, \end{align*}$$

where $\pi ^{+}(b,\lambda )=(\pi (b),\lambda )$ for $(b,\lambda ) \in B^+$ . Also, we have a commutative diagram with exact columns

Tensoring with A gives the following commutative diagram:

By hypothesis, the third row, the third column, and the fourth column are exact. The other columns are exact is tautological. By diagram chasing, we can conclude that the fourth row is also exact. Hence, A is separably exact.

Proof The “only if” part follows from the fact that A is a subalgebra of $A \rtimes _{\mathrm {red}} P$ and exactness passes to subalgebras. For the “if part,” suppose that A is exact. Let $A^{+}$ be the unitization of A. Then, $A^{+}$ is exact. For $s \in P$ , let $\alpha _{s}^{+}:A^{+} \to A^{+}$ be defined by $\alpha _{s}(x,\lambda )=(\alpha _s(x),\lambda )$ for $(x,\lambda ) \in A^{+}=A \oplus \mathbb {C}$ . It follows from Corollary 2.7 of [Reference Hossain and Sundar3] that $A \rtimes _{\mathrm {red}} P$ is an ideal of $A^{+}\rtimes _{\mathrm {red}} P$ . As exactness passes to ideals and subalgebras, it suffices to prove that $A^{+}\rtimes _{\mathrm {red}} P$ is exact. In other words, to prove the “if part,” we can assume that A is unital and $\alpha _s$ is unital, that is, $\alpha _s(1)=1$ for each $s \in P$ .

Thus, assume that A is exact and unital and $\alpha _s$ is unital for every $s \in P$ . By [Reference LaLonde5, Proposition 6.13], it suffices to prove that $A \rtimes _{\mathrm {red}} P$ is separably exact. Let

$$ \begin{align*} 0 \longrightarrow I \longrightarrow B \longrightarrow C \longrightarrow 0 \end{align*} $$

be a short exact sequence of separable $C^*$ -algebras. Thanks to Lemma 3.7, we may further assume that B and C are unital. Since A is exact, we have the following short exact sequence:

$$\begin{align*}0 \longrightarrow I\otimes A \longrightarrow B\otimes A \longrightarrow C\otimes A \longrightarrow 0. \end{align*}$$

Applying Proposition 3.4 to the above sequence (with the trivial action of P on $I, B, C$ and the given action on A), we obtain the following short exact sequence:

$$\begin{align*}0 \longrightarrow (I\otimes A)\rtimes_{\mathrm{red}}P \longrightarrow (B\otimes A)\rtimes_{\mathrm{red}}P \longrightarrow (C\otimes A)\rtimes_{\mathrm{red}}P \longrightarrow 0. \end{align*}$$

Using Lemma 3.3, we can rewrite the last equation as

$$ \begin{align*} 0 \longrightarrow I\otimes (A\rtimes_{\mathrm{red}}P) \longrightarrow B\otimes (A\rtimes_{\mathrm{red}}P) \longrightarrow C\otimes (A\rtimes_{\mathrm{red}}P) \longrightarrow 0. \end{align*} $$

Therefore, $A\rtimes _{\mathrm {red}}P$ is separably exact. Hence, by [Reference LaLonde5, Proposition 6.13], $A \rtimes _{\mathrm {red}} P$ is exact.

Proof Observe that right translation preserves the pre-order $\leq $ . Thus, a right translate of a directed set is directed. Hence, every element of $\widetilde {\Omega }$ is directed.

We claim that $(P,G)$ satisfies the Toeplitz condition. Let $\rho $ be the right regular representation of G on $\ell ^2(G)$ . For $g \in G$ , let $w_g$ be the compression of $\rho (g)$ to $\ell ^2(P)$ . For $s \in P$ , let $v_s:=w_s$ . Let $\{\delta _s\}_{s \in P}$ be the standard orthonormal basis for $\ell ^2(P)$ . Then, for $s \in P$ and $g \in G$ ,

(3.3) $$ \begin{align} v_s(\delta_t)=\delta_{ts}; \quad w_g(\delta_t):= \begin{cases} \delta_{tg}, & \text{ if } tg \in P,\\ 0, & \text{ otherwise.} \end{cases} \end{align} $$

Let $g \in G$ be given. Since $G=\bigcup _{n=1}^{\infty }Pa^{-n}$ , there exist $n \geq 1$ and $x \in P$ such that ${g=xa^{-n}}$ . A direct check with Equation (3.3) shows that $w_g=v_{a^n}^{*}v_x$ . Hence, $(P,G)$ satisfies the Toeplitz condition.

Since P has an order unit, [Reference Hossain and Sundar3, Theorem 1.2] ensures the existence of a countably generated Hilbert B-module $\mathcal {E}$ and an $E_0$ -semigroup $\alpha $ over P on $\mathcal {L}_B(\mathcal {E})$ such that X is isomorphic to the product system associated with $\alpha $ . Therefore, Theorem 3.8 ensures that $\mathcal {K}_B(\mathcal {E})\rtimes _{\mathrm {red}}P$ is exact if and only if $\mathcal {K}_B(\mathcal {E})$ is exact. On the other hand, by [Reference Hossain and Sundar3, Theorem 1.1], the reduced $C^*$ -algebra $C^*_{\mathrm {red}}(X)$ and $\mathcal {K}_B(\mathcal {E})\rtimes _{\mathrm {red}} P$ are Morita equivalent. Since exactness is preserved under Morita equivalence, $C^*_{\mathrm {red}}(X)$ is exact if and only if $\mathcal {K}_B(\mathcal {E})$ is exact. Since $\mathcal {E}$ is full, $\mathcal {K}_B(\mathcal {E})$ and B are Morita equivalent. Thus, $\mathcal {K}_B(\mathcal {E})$ is exact if and only if B is exact. Consequently, $C^*_{\mathrm {red}}(X)$ is exact if and only if B is exact.

Proof (2). $\implies $ (1). Let $(b_i,e_i)\in \mathcal B*\mathcal E$ be a net such that $(b_i,e_i) \to (b,e)$ . Let $\gamma _i:=p(b_i), \eta _i:= q(e_i), \gamma :=p(b), \textrm {~and~}\eta :=q(e)$ . Then, $\gamma _i\to \gamma $ and $\eta _i\to \eta $ in $\mathcal G$ . Choose open bisections U and V of $\mathcal G$ containing $\gamma $ and $\eta $ , respectively, such that eventually $\gamma _i \in U$ and $\eta _i \in V$ . Since the bundles $\mathcal B$ and $\mathcal E$ have enough sections, there exist sections $f\in \Gamma _{\mathcal B}$ and $g\in \Gamma _{\mathcal E}$ with $\text {supp}(f)\subset U$ and $\text {supp}(g)\subset V$ such that $f(\gamma ) = b$ and $g(\eta )=e$ . Condition (2) ensures that $f*g \in \Gamma _{\mathcal E} $ . Since both f and g are bisection supported, we have

$$\begin{align*}f*g(\gamma_i\eta_i) = f(\gamma_i)g(\eta_i) \quad \text{and} \quad f*g(\gamma \eta)=f(\gamma)g(\eta).\end{align*}$$

Since $f*g \in \Gamma _{\mathcal E}$ (by assumption), $f*g(\gamma _i\eta _i) \to f*g(\gamma \eta )=be$ . Hence,

(4.1) $$ \begin{align} \lim_{i}f(\gamma_i)g(\eta_i)=be. \end{align} $$

Note that

$$\begin{align*}\ \|f(\gamma_i)g(\eta_i)-b_ie_i\| \leq \|f(\gamma_i)-b_i\|\|g(\eta_i)\| + \|b_i\|\|g(\eta_i)-e_i\|. \end{align*}$$

Since $b_i \to b$ , $f(\gamma _i) \to b$ , $e_i \to e,$ and $g(\eta _i) \to e$ , we can conclude from the above inequality that

(4.2) $$ \begin{align} \lim_{i}\|f(\gamma_i)g(\eta_i)-b_ie_i\|= 0. \end{align} $$

Equations (4.1) and (4.2) imply that $b_ie_i\to be$ (see [Reference Williams14, Proposition C.20]Footnote ¹ ). Hence, the map $ (b,e) \mapsto be$ is continuous.

(1). $\implies $ (2). This is standard, and hence omitted.

Proof Let $\alpha>0$ , and let $R= \{F\in \Omega : \varphi (F)<\alpha \}$ . Fix $F_0\in R$ .

Case (I): $p_0\notin F_0$ . Then, $F_0\in \{F\in \Omega :p_0\notin F\} \subseteq R$ . As $\{F \in \Omega : p_0 \notin F\}$ is open, it follows that $F_0$ is an interior point of R.

Case (II): $p_0\in F_0$ . Since $\varphi (F_0) = \inf \{f(p): p\in F_0 \text { and } p\geq p_0\} <\alpha $ , there exists $p_1\in F_0$ such that $f(p_1)<\alpha $ . Consider the basic open set $U_{p_1} = \{F \in \Omega : p_1\in F\}$ . For any $F\in U_{p_1}$ , we have $\varphi (F)\leq f(p_1)<\alpha $ . Hence, $F_0\in U_{p_1}\subseteq R$ , and in this case too, $F_0$ is an interior point of R.

Proof Note that $\text {span}\{1_{\Omega p} \otimes e_g:g \in G, p \in P\}$ is dense in $C_{c}(\mathcal G)$ with the inductive limit topology. To see this, first observe that as P is quasi-lattice ordered, $\Omega p \cap \Omega q$ is empty if $p \vee q=\infty $ and $\Omega p \cap \Omega q=\Omega (p \vee q)$ if $p \vee q<\infty $ . Also, $\{1_{\Omega p}: p \in P\}$ separates points of $\Omega $ . Hence, $\text {span}\{1_{\Omega p}: p \in P\}$ is an algebra and is dense in $C(\Omega )$ . Consequently, $\{1_{\Omega p} \otimes e_g: p \in P, g \in G\}$ is total in $C_{c}(\mathcal G)$ .

Let $p,q \in P$ , $T \in \mathcal K_{B}(E,X_p)$ and $g_1,g_2 \in G$ be given. Suppose p and q have an upper bound. Let $s=(p \vee q)p^{-1}$ . Then,

(4.4) $$ \begin{align} (1_{\Omega q} \otimes e_{g_1})f_{p,g_2, T} = \begin{cases} f_{p \vee q, g_1, (T\otimes 1_{s})\sigma_{s}^{-1}}, & \text{ if } g_1=g_2,\\ 0, & \text{ if }g_1 \neq g_2. \end{cases} \end{align} $$

If p and q do not have an upper bound, then

(4.5) $$ \begin{align} (1_{\Omega q}\otimes e_{g_1})f_{p,g_2,T}=0. \end{align} $$

As $\{1_{\Omega p} \otimes e_g: p\in P, g \in G\}$ is total in $C_{c}(\mathcal G)$ , it follows from Equations (4.4) and (4.5) that for $\varphi \in C_{c}(\mathcal G)$ and $f \in \Gamma $ , $\varphi f \in \Gamma $ . This proves $(1)$ .

Next, we prove $(2)$ . It suffices to prove that $\mathcal G \ni (F,g) \mapsto \|f(F,g)\| \in \mathbb {R}$ is upper semicontinuous when f is of the form

$$\begin{align*}f=\sum_{i=1}^{n}f_{p_i,g_i,T_i}. \end{align*}$$

Let $ f:=\sum _{i=1}^{n}f_{p_i,g_i,T_i}$ be one such section. It suffices to prove that for every $j \in \{1,2,\ldots ,n\}$ , the map

$$\begin{align*}\Omega \ni F \mapsto \|f(F,g_j)\|=\Big\|\sum_{g_i=g_j, p_i \geq g_{i}^{-1}}f_{p_i,g_i,T_i}(F,g_j)\Big\| \in \mathbb{R} \end{align*}$$

is upper semicontinuous. Hence, we can, without loss of generality, assume that $g_i=g_j$ for all $i,j$ , and set $g:=g_i$ . And we can assume that $p_i \geq g^{-1}$ for all i. Moreover, $f_{p,g,T}+f_{p,g,S}=f_{p,g,T+S}$ , and hence we can further assume that $p_i$ ’s are distinct.

Let $\varphi :\Omega \to \mathbb {C}$ be defined by $\varphi (F):=\|f(F,g)\|$ . We need to show that $\varphi $ is upper semicontinuous. Let $J \subset \{1,2,\ldots ,n\}$ , and let

$$\begin{align*}U_{J}:=\{F \in \Omega: p_i \in F \text{~~for }i \in J\text{ and } p_i \notin F \text{~for }i \notin J\}.\end{align*}$$

Suppose $J \neq \emptyset $ and $U_J \neq \emptyset $ . As every element of $\Omega $ is directed and hereditary, $p_J:=\bigvee _{j \in J}p_j<\infty $ . Moreover, if $F\in U_J$ , then $p_j\in F$ for every $j\in J$ . As $F\in \Omega $ is directed, it follows that $p_J\in F$ . For $j \in J$ , set $s_j:=p_{J}p_j^{-1}$ . Set $S:=\sum _{j \in J}(T_j \otimes 1_{s_j})\sigma _{s_j}^{-1} \in \mathcal K_B(E,X_{p_J})$ . Define a function $\chi _0:(p_J,P] \to [0,\infty )$ by

$$\begin{align*}\chi_0(p):=\|(S \otimes 1_{pp_{J}^{-1}})\sigma_{pp_J^{-1}}^{-1}\|. \end{align*}$$

Then, $\chi _0$ is decreasing. Define $\chi :\Omega \to \mathbb {C}$ by

$$\begin{align*}\chi(F) = \begin{cases} 0, & \text{ if } p_J\notin F,\\ \displaystyle \inf_{p\in F, p \geq p_J}\|(S \otimes 1_{pp_J^{-1}})\sigma_{pp_J^{-1}}^{-1}\|, & \text{ if } p_J \in F. \end{cases} \end{align*}$$

It follows from Lemma 4.2 that $\chi $ is upper semicontinuous. Note that for $F \in U_J$ ,

$$\begin{align*}\varphi(F)=\Big\|\sum_{j \in J}i_{p_j}^{(F,g)}(T_j)\Big\|=\Big\|\sum_{j \in J}i_{p_J}^{(F,g)}((T_j \otimes 1_{s_j})\sigma_{s_j}^{-1})\Big\|=\|i_{p_J}^{(F,g)}(S)\|=\chi(F). \end{align*}$$

Hence, $\varphi |_{U_J}$ is upper semicontinuous. If J is empty, then $\varphi |_{U_J}=0$ . Since $\{U_J: J \subset \{1,2,\ldots ,n\}\}$ is an open cover of $\Omega $ , it follows that $\varphi $ is upper semicontinuous. The proof of $(2)$ is now over.

By the definition of $\mathcal E_{(F,g)}$ , the set $\{f_{p,g,T}(F,g): p \in P, T \in \mathcal K_B(E,X_p)\}$ which coincides with $\big \{i_p^{(F,g)}(T): p \in d_{(F,g)}, T \in \mathcal K_B(E,X_p)\big \}$ is total in $\mathcal E_{(F,g)}$ . Hence, $(3)$ follows. The existence of the topology follows by applying [Reference Williams14, Theorem C.25]. (The footnote on page 13 is applicable here as well.)

Proof For $(F,g) \in \mathcal G$ and for $p \in d_{(F,g)}$ , we denote the natural map $\mathcal K_B(X_{pg},X_p) \to \varinjlim _{q \in d_{(F,g)}}\mathcal K_{B}(X_{qg},X_q)$ by $j_{p}^{(F,g)}$ . For $p,q\in P$ , $T\in \mathcal K_B(X_q,X_p)$ , define a section $\varphi _{p,q,T}$ of the bundle $\mathcal A$ by

$$\begin{align*}\varphi_{p,q, T}(F,h) = \begin{cases} j_{p}^{(F,g)}(T), & \text{ if } p\in d_{(F,g)} ~~\text{and }g=p^{-1}q,\\ 0, & \text{ otherwise. } \end{cases} \end{align*}$$

Let

$$ \begin{align*} \Gamma_0&:=\text{span}\{\varphi_{p,q,T}:p,q \in P, T \in \mathcal K_B(X_q,X_p)\}, \textrm{~and~}\\ \Gamma_1&:=\text{span} \{f_{p,g,S}: p \in P, g \in G, S \in \mathcal K_B(E,X_p)\}. \end{align*} $$

Then, by construction, $\Gamma _0$ is dense in $\Gamma _{\mathcal A}$ and $\Gamma _1$ is dense in $\Gamma _{\mathcal E}$ . We now use Lemma 4.1 to prove that the left action of $\mathcal A$ on $\mathcal E$ is continuous. Thus, it suffices to show that $\Gamma _0*\Gamma _1 \subset \Gamma _{\mathcal E}$ .

Fix $p,q,r \in P$ , $g \in G$ , $S \in \mathcal K_B(X_q,X_p),$ and $T \in \mathcal K_B(E,X_r)$ . Note that $\text {supp}(\varphi _{p,q,S}) \subset \Omega \times \{p^{-1}q\}$ and $\text {supp}(f_{r,g,T}) \subset \Omega \times \{g\}$ . Hence, the convolution is supported in $\Omega \times \{p^{-1}q g\}$ . Set $g_0:=p^{-1}qg$ .

Now, for $(F,g_0) \in \mathcal G$ ,

$$\begin{align*}\varphi_{p,q,S}*f_{r,g,T}(F,g_0)=1_{F}(q^{-1}p)\varphi_{p,q,S}(F,p^{-1}q)f_{r,g,T}(Fp^{-1}q,q^{-1}pg_0). \end{align*}$$

The RHS vanishes unless $p \in F$ , $r \geq g^{-1}$ , and $r \in Fp^{-1}q$ . Let $s_0,s_1 \in P$ be such that $r\vee q=s_0q$ and $r\vee q=s_1r$ . Set $R:=(S \otimes 1_{s_0})\cdot ((T \otimes 1_{s_1})\sigma _{s_1}^{-1}) \in \mathcal K_B(E,X_{s_0p})$ . It can now be checked from the definition of the left action (see Equation (4.3)) that

$$\begin{align*}\varphi_{p,q,S}*f_{r,g,T}=f_{s_0p, g_0, R}. \end{align*}$$

Hence, $\Gamma _0*\Gamma _1 \subset \Gamma _{\mathcal E}$ . The conclusion follows from Lemma 4.1.