Proof. All but the statement that $\Theta := \bigoplus _{v\in E^0}(\psi _v\times \pi _v)$ is faithful follows from essentially replicating the computations in the proof of [Reference Hawkins16, Lemma 4.1.7], hence we will only deal with the question of faithfulness.

Let $U_v$ be the canonical inclusion of $\ell ^2(E^\ast v)$ into $\ell ^2(E^\ast )$. This $U_v$ is an isometry with adjoint $U_v^\ast :\ell ^2(E^\ast )\to \ell ^2(E^\ast v)$ the corresponding orthogonal projection. We define $U:= \bigoplus _{v\in E^0} U_v :\bigoplus _{v\in E^0}\ell ^2(E^\ast v)\to \ell ^2(E^\ast )$. Then $U$ is unitary, and

\[ U^\ast h= (h|_{E^\ast v})_{v\in E^0}\quad \text{for all }h\in \ell^2(E^\ast). \]

Let $\{\delta _\mu : \mu \in E^\ast \}$ be the canonical basis for $\ell ^2(E^\ast )$. By [Reference Hawkins16, Lemma 4.1.7], there exist $\lambda ^0: C_0(E^0) \to \mathcal {B}(\ell ^2(E^\ast ))$ and $\lambda ^1 : X(E) \to \mathcal {B}(\ell ^2(E^\ast ))$ such that for all $\xi \in X(E),\, \alpha \in C_0(E^0)$ and $\mu \in E^\ast$,

\[ \lambda^1(\xi)\delta_\mu=\sum_{f\in E^1 r(\mu)}\xi(f)\delta_{f\mu}\quad \text{and}\quad \lambda^0(\alpha)\delta_\mu=\alpha(r(\mu))\delta_\mu. \]

Furthermore, $(\lambda ^1,\,\lambda ^0)$ is a Toeplitz representation of $X(E)$ on $\ell ^2(E^\ast )$ and $\lambda ^1\times \lambda ^0$ is faithful. We claim that, for $a\in \mathcal {T}C ^\ast (E)$,

(3.4)$$\begin{align} \Theta(a)=U^\ast\left(\lambda^0\times \lambda^1(a)\right)U. \end{align}$$

To show this, we first prove that $U_v (\psi _v \times \pi _v(a)) = (\lambda ^1 \times \lambda ^0(a)) U_v$ for all $v \in E^0$. By linearity and continuity, it suffices to fix $x=x_1\otimes x_2\otimes \ldots \otimes x_m\in X(E)^{\otimes m}$, $y=y_1\otimes y_2\otimes \ldots \otimes y_n\in X(E)^{\otimes n}$ (with the convention that if $m = 0$, we mean $x = b \in A$, and similarly if $m = 0$ for $y$), and then consider $a = \iota _{X(E)}^{\otimes m}(x)\iota _{X(E)}^{\otimes n}(y)^\ast$. With the convention that $\prod ^0_{i=1} \psi _v(x_i)$ means $\pi _v(b)$ when $x = b \in A$, and similarly for $y$, we have

\[ U_v \Big( \psi_v\times\pi_v(a)\Big)=U_v\Big(\prod_{i=1}^m\psi_v(x_i)\Big)\Big(\prod_{i=0}^{n-1}\psi_v(y_{n-i})^\ast\Big). \]

Direct calculation with basis vectors shows that $U_v\psi _v(x_i)=\lambda ^1(x_i) U_v$ and that $U_v\psi _v(y_j)^\ast =\lambda ^1(y_j)^\ast U_v$. Therefore

\[ U_v \Big(\psi_v\times\pi_v(a)\Big) =\Big(\prod_{i=1}^m\lambda^{1}(x_i)\Big)\Big(\prod_{i=0}^{n-1}\lambda^1(y_{n-i})^\ast\Big) U_v = \big(\lambda^1\times\lambda^0(a)\big)U_v \]

as claimed.

Since $U_v$ is an isometry, we deduce that $\psi _v\times \pi _v(a)=U_v^\ast (\lambda ^1\times \lambda ^0(a))U_v$. Since each $\ell ^2(E^\ast v) \subseteq \ell ^2(E^*)$ is invariant for $\lambda ^1\times \lambda ^0(a)$,

$$\begin{align*} \Theta(a)& =\bigoplus_{v\in E^0}\Big(\psi_v\times \pi_v(a)\Big)=\bigoplus_{v\in E^0} U_v^\ast\Big(\lambda^1\times \lambda^0(a)\Big)U_v\\ & =\Big(\bigoplus_{v\in E^0}U_v\Big)^\ast \Big(\lambda^1\times \lambda^0(a)\Big)\Big(\bigoplus_{v\in E^0}U_v\Big) =U^\ast \left(\lambda^1\times \lambda^0(a)\right)U, \end{align*}$$

as in equation (3.4). We conclude that $\Theta$ is unitarily equivalent to $\lambda ^1\times \lambda ^0$ and hence faithful by [Reference Hawkins16, Lemma 4.1.7].

Proof. We show that $e_v$ is a cyclic vector for $\psi _v\times \pi _v$. It is clear that $\psi _v\times \pi _v(1)e_v=e_v$. Fix $f \in E^1 v$. Since $|E^1 v|<\infty$ there exists a function $x\in X(E)$ such that $x(f)=1$ and $x(g)=0$ for $g\in E^1v\setminus {\{f\}}$, hence

\[ \psi_v\times \pi_v(\iota_{X(E)}(x))e_v = \sum_{g\in E^1 v}x(g) e_g =e_f. \]

Thus, $\{e_f : f\in E^1v\} \subset \{\psi _v\times \pi _v(a) e_v: a \in \mathcal {T}C ^\ast (E)\}$. An induction on $n$, using a similar argument, shows that the set $\{e_\mu : \mu \in E^nv\} \subset \{\psi _v\times \pi _v(a)e_v: a \in \mathcal {T}C ^\ast (E)\}$ for each $n \in \mathbb {N}$. Therefore, $e_v$ is cyclic for $\psi _v\times \pi _v$ and this proves the lemma by the uniqueness of cyclic representations [Reference Murphy28, Theorem 5.1.4].

Proof. Let $x=x_1\otimes x_2\otimes \cdots x_m\in X(E)^{\otimes m}$ and $y=y_1\otimes y_2\otimes \cdots y_n\in X(E)^{\otimes n}$. First, suppose $n>0$. Then,

$$\begin{align*} \varphi_v(\iota^{{\otimes} m}(x)\iota^{{\otimes} n}(y)^*)& =\left\langle \psi_v^{{\otimes} m}(x)\psi_v^{{\otimes} n}(y)^\ast e_v, e_v \right\rangle\\ & =\left\langle \psi_v^{{\otimes} m}(x)\psi_v(y_n)^\ast\cdots\psi_v(y_1)^\ast e_v, e_v \right\rangle = 0, \end{align*}$$

by remark 3.6. Now, suppose $n=0$ and $m>0$. Then,

\[ \varphi_v(\iota^{{\otimes} m}(x)\iota_{C_0(E^0)}(y))=\left\langle \psi_v^{{\otimes} m}(x)\pi_v(y)e_v, e_v \right\rangle=\left\langle \psi_v(x_1)\cdots \psi_v(x_m\cdot y)e_v, e_v \right\rangle. \]

By proposition 3.4,

\[ \left\langle \psi_v(x_1)\cdots \psi_v(x_m\cdot y)e_v, e_v \right\rangle=\sum_{\mu = \mu_1\cdots\mu_m\in E^m v}\left\langle x_1(\mu_1)\cdots x_m(\mu_m)y(v)e_\mu, e_v \right\rangle=0, \]

which concludes the proof.

Proof. First suppose that $v_n\to v$ in $E^0$. We want to show that $\varphi _{v_n}(a) \to \varphi _v (a)$ for every $a\in \mathcal {T}C ^\ast (E)$. By linearity and continuity it is enough to prove this for $a=\iota ^{\otimes m}(x)\iota ^{\otimes n}(y)^\ast$ where $x\in X(E)^{\otimes m},\, y\in X(E)^{\otimes n}$ are elementary tensors. If $n>0$ or $m>0$, then $\varphi _{v_n}(a)=0 = \varphi _v(a)$ by lemma 3.7, so we just need to check the case when $a=\iota _{C(E^0)}(\alpha )$ for $\alpha \in C(E^0)$. Indeed,

\[ \varphi_{v_n}(\iota_{C(E^0)}(\alpha)) = \left\langle \pi_{v_n}(\alpha) e_{v_n}, e_{v_n} \right\rangle = \alpha(v_n) \to \alpha(v) = \varphi_{v}(\iota_{C(E^0)}(\alpha)). \]

This proves the first direction. Now suppose that $v_n$ does not converge in $E^0$. Since $E^0$ is compact, $(v_n)$ has at least two distinct accumulation points $v,\,v'\in E^0$. Take $\alpha \in C(E^0)$ such that $\alpha (v)\neq \alpha (v')$. Let $(v_{n_\ell })_{\ell }$ and $(v_{n_k})_k$ be subsequences of $v_n$ that converge to $v$ and $v'$, respectively. If $\varphi _n$ weak$^\ast$ converges to some $\phi$, then by the previous paragraph,

\[ \alpha(v) = \lim_\ell \varphi_{v_{n_\ell}}(\iota_{C(E^0)}(\alpha)) = \phi(\iota_{C(E^0)}(a)) = \lim_k \varphi_{v_{n_k}}(\iota_{C(E^0)}(\alpha)) = \alpha(v'), \]

a contradiction.

Proof. First, assume that $(\beta _n)$ and $(\phi _n)$ are sequences as in the statement of the proposition such that $\phi _n$ weak$^\ast$ converges to some $\phi$. We need to show that there exists $v\in E^0$ such that $\phi = \varphi _v$. Since $\beta _n\to \infty$, we may assume that $\beta _n>\log (\rho (A_E))$ for all $n$. By remark 3.3, for each $n$ there is a vertex $v_n\in E^0$ such that $\phi _n=\phi ^{\beta _n}_{v_n}$, the extremal KMS$_{\beta _n}$-state described just before theorem 3.1. By equation (3.3), for every $a\in \mathcal {T}C ^\ast (E)$,

$$\begin{align*} & \phi^{\beta_n}_{v_n}(a)=(N_{\phi^{\beta_n}_{v_n}})^{{-}1}\sum_{E^\ast v_n}e^{-\beta_n |\mu|}\left\langle \psi_{v_n}\times \pi_{v_n}(a) e_\mu, e_\mu \right\rangle\\ & \quad=(N_{\phi^{\beta_n}_{v_n}})^{{-}1}\Big( \left\langle \psi_{v_n}\times \pi_{v_n}(a) e_{v_n}, e_{v_n} \right\rangle+ \sum_{E^\ast v_n\setminus{ \{v_n\}}}e^{-\beta_n |\mu|}\left\langle \psi_{v_n}\times \pi_{v_n}(a) e_\mu, e_\mu \right\rangle \Big). \end{align*}$$

We claim that $\lim _{n \to \infty } N_{\phi ^{\beta _n}_{v_n}} = 1$ and that

\[ \lim_{n \to \infty} \sum_{E^\ast v_n\setminus{ \{v_n\}}}e^{-\beta_n |\mu|}\left\langle \psi_{v_n}\times \pi_{v_n}(a) e_\mu, e_\mu \right\rangle = 0. \]

For the first of these equalities, note that by remark 2.2, we have for $n\in \mathbb {N}$,

\[ 1\leq N_{\phi^{\beta_n}_{v_n}}\leq 1+\sum_{i=1}^\infty e^{-\beta_n i}\max_{v\in E^0} |E^iv|. \]

Hence $\lim _{n \to \infty } \sum _{i=1}^\infty e^{-\beta _n i}\max _{v\in E^0} |E^iv| = 0$ by the dominated convergence theorem. Thus $N_{\phi ^{\beta _n}_{v_n}}\to 1$.

The second equality follows similarly once we observe that

\[ \sum_{E^\ast v_n\setminus{ \{v_n\}}}e^{-\beta_n |\mu|}\left\langle \psi_{v_n}\times \pi_{v_n}(a) e_\mu, e_\mu \right\rangle\leq \|a\| \sum_{i=1}^\infty e^{-\beta_n i}\max_{v\in E^0} |E^iv|. \]

The claim implies that $\phi _{v_n}^{\beta _n} - \varphi _{v_n}$ weak$^\ast$ converges to $0$. In particular, $\varphi _{v_n}$ weak$^\ast$ converges to the same weak$^\ast$ limit as $\phi _n = \phi _{v_n}^{\beta _n}$, namely $\phi$. So lemma 3.8 implies that $v_n$ converges to some $v\in E^0$ and $\phi = \varphi _v$.

Now, assume that $v\in E^0$. We have to show that there exist sequences $(\beta _n)$ and $(\phi _n)$ as in the statement of the proposition such that $\phi _n$ weak$^\ast$ converges to $\varphi _v$. Pick any sequence $(\beta _n)$ of real numbers such that $\beta _n > \log (\rho (A_E))$ and $\beta _n \to \infty$. Take the constant sequence $v_n = v$. By remark 3.3, $\phi _n := \phi _{v_n}^{\beta _n} = \phi _v^{\beta _n}$ is an extremal KMS$_{\beta _n}$-state for all $n$. The same argument as before shows that $\phi _{v_n}^{\beta _n} - \varphi _{v_n} = \phi _n - \varphi _v$ weak$^\ast$ converges to $0$ and we are done.

Proof. We begin with existence. As $E^1$ is compact and the source map is a local homeomorphism, there exists a finite open cover $(U_i)_{i=1}^n$ of $E^1$ by s-sections. Let $\{\xi _i \}_{i=1}^n$ be a partition of unity subordinate to the $U_i$ with each $\xi _i\in C_0(U_i,\,[0,\,1])$. Writing $\sqrt {\xi _i}$ for the pointwise square-root of each $\xi _i:E^1\to [0,\,1]$, we define an element $p_E\in \mathcal {T}C ^\ast (E)$ by

(3.6)$$\begin{align} p_E:=1-\sum_{i=1}^n\iota_{X(E)}\big(\sqrt{\xi_i}\big)\iota_{X(E)}\big(\sqrt{\xi_i}\big)^\ast. \end{align}$$

Let $v\in E^0$ be the vertex such that $\varphi = \varphi _v$, and fix $\mu \in E^\ast v\setminus {\{v\}}$. Write $\mu = f\nu$ with $f\in E^1$. By remark 3.6, since each $\xi _i$ is supported in an s-section,

\[ \psi_v\times\pi_v(p_E)e_\mu = e_\mu-\sum_{i=1}^n\sum_{g\in E^1 r(\nu)}\sqrt{\xi_i(f)}\sqrt{\xi_i(g)}e_{g\nu} = e_\mu - \left(\sum_{i=1}^n\xi_i(f) \right)e_\mu = 0. \]

On the other hand, again by remark 3.6, $\psi _v\times \pi _v(p_E)e_v=e_v$, so $\psi _v\times \pi _v(p_E)$ is the projection onto the basis vector $e_v$, thus by lemma 3.5, $p_E$ satisfies (i) and (ii).

Now we prove uniqueness. Fix $a\in \mathcal {T}C ^\ast (E)$ such that $\pi _\varphi (a)$ is a minimal projection in $\pi _\varphi (\mathcal {T}C ^\ast (E))$ and $\varphi (a)=1$ for every $\varphi \in S^\infty$. Fix $v\in E^0$. By lemma 3.5, $\psi _v\times \pi _v(a)$ is a minimal projection in $\psi _v\times \pi _v(\mathcal {T}C ^\ast (E))$. Since $\psi _v\times \pi _v(p_E) \in \psi _v\times \pi _v(\mathcal {T}C ^\ast (E))\cap \mathcal {K}(\ell ^2(E^\ast v))$, we have $\mathcal {K}(\ell ^2(E^\ast v))\subset \pi _{\varphi _v}(\mathcal {T}C ^\ast (E))$ [Reference Murphy28, Theorem 2.4.9]. Hence, $\psi _v\times \pi _v(a)$ is the rank-one projection $\xi \otimes \xi ^\ast$, for a unit vector $\xi \in \ell ^2(E^\ast v)$ and

\[ 1= \varphi_v(a)=\left\langle \psi_v\times \pi_v(a) e_v, e_v \right\rangle=\left\langle \xi\otimes \xi^\ast (e_v), e_v \right\rangle=\vert \left\langle \xi, e_v \right\rangle\vert^2. \]

Since $\xi$ is a unit vector, $\xi =\lambda e_v$, for some $\lambda \in \mathbb {T}$. This implies that $\psi _v\times \pi _v(a)=\xi \otimes \xi ^\ast =\lambda e_v\otimes (\lambda e_v)^\ast = e_v\otimes e_v^\ast$. Hence $\psi _v\times \pi _v(a)=\psi _v\times \pi _v(p_E)$ for every $v\in E^0$. Proposition 3.4 implies that $\bigoplus _{v\in E^0}\psi _v\times \pi _v$ is faithful, so $a=p_E$.

Proof. We start by showing that $\psi _E(X(E))=\mathcal {T}C ^\ast (E)_1p_E$. For this, recall that $\mathcal {T}C ^\ast (E)=\overline {\operatorname {span}}\{\iota _X^{\otimes m}(x)\iota _X^{\otimes n}(y)^\ast : n,\,m\in \mathbb {N},\,x\in X(E)^{\otimes m},\,y\in X(E)^{\otimes n}\}$. Note that $\gamma _z(\iota _X^{\otimes m}(x)\iota _X^{\otimes n}(y)^\ast )=z^{m-n}\iota _X^{\otimes m}(x)\iota _X^{\otimes n}(y)^\ast$, so

\[ \text{$\mathcal{T}C ^\ast(E)$}_1=\overline{\operatorname{span}}\{\iota_X^{{\otimes} n+1}(x)\iota_X^{{\otimes} n}(y)^\ast: n\geq 0, x\in X(E)^{{\otimes} n+1},y\in X(E)^{{\otimes} n}\}. \]

Hence,

\[ \text{$\mathcal{T}C ^\ast(E)$}_1p_E=\overline{\operatorname{span}}\{\iota_X^{{\otimes} n+1}(x)\iota_X^{{\otimes} n}(y)^\ast p_E: n\geq 0, x\in X(E)^{{\otimes} n+1},y\in X(E)^{{\otimes} n}\}. \]

Suppose that $x=x_1\otimes x_2\otimes...\otimes x_{n+1}$ and that $y=y_1\otimes y_2\otimes...\otimes y_n$ if $n>0$ or $y=a\in {_A}{A}{_A}$ if $n=0$. For $\mu \in E^\ast$, the representation $(\lambda ^0,\,\lambda ^1)$ as in remark 3.11 satisfies

\[ \lambda^0\times\lambda^1(\iota_X^{{\otimes} n+1}(x)\iota_X^{{\otimes} n}(y)^\ast p_E)e_\mu=\begin{cases}\lambda^0\times\lambda^1(\iota_X^{{\otimes} n+1}(x)\iota_X^{{\otimes} n}(y)^\ast)e_\mu,\text{ if $|\mu|=0$},\\0, \text{ if $|\mu|>0$.} \end{cases} \]

If $|\mu |=0$, then by the proof of [Reference Hawkins16, Lemma 4.1.7], if $n>0$, we have $\lambda ^1(y_1)^\ast e_\mu =0$. Hence, for $|\mu |=0$,

\[ \lambda^0\times\lambda^1(\iota_X^{{\otimes} n+1}(x)\iota_X^{{\otimes} n}(y)^\ast)e_\mu=\begin{cases}\lambda^1(x\cdot a^\ast)e_\mu, \text{ if $n=0$}, \\ 0, \text{ if $n>0$}. \end{cases} \]

Since $\lambda ^1\times \lambda ^0$ is faithful representation, we conclude that $\iota _X^{\otimes n+1}(x)\iota _X^{\otimes n}(y) p_E=0$ if $n>0$ and

$$\begin{align*} \text{$\mathcal{T}C ^\ast(E)$}_1p_E& =\overline{\operatorname{span}}\{\iota_X(x)\iota_{C(E^0)}(a)^\ast p_E: x\in X(E),a\in C(E^0)\}\\ & =\{\psi_E(\xi): \xi\in X(E)\}=\psi_E(X(E)). \end{align*}$$

Now we show that $(\psi,\,\iota _{C(E^0)})$ is a bimodule morphism. Indeed, for $a\in C(E^0)$ and $\xi \in X(E)$,

\[ \psi_E(a\cdot\xi)=\iota_X(a\cdot \xi)p_E=\iota_{C(E^0)}(a)\iota_X(\xi)p_E=\iota_{C(E^0)}(a)\psi_E(\xi). \]

Since $p_E$ commutes with $M$ as in remark 3.11, it follows from a similar computation that $\psi$ preserves the right actions of $C(E^0)$. So $(\psi _E,\,\iota _{C(E^0)})$ preserves the bimodule structure. It remains to show that $(\psi _E,\,\iota _{C(E^0)})$ is an isomorphism of Hilbert modules. A simple calculation shows that it respects the $M$-valued inner-product of (3.7):

$$\begin{align*} \left\langle \psi_E(\xi), \psi_E(\eta) \right\rangle_M & = \rho^{{-}1}_E(\psi_E(\xi)^\ast\psi_E(\eta))\\ & =\rho^{{-}1}_E((\iota_X(\xi)a)^\ast\iota_X(\eta)p_E) =\iota_X(\xi)^\ast\iota_X(\eta) =\iota_{C(E^0)}(\left\langle \xi, \eta \right\rangle_{C(E^0)}). \end{align*}$$

Hence $X(E)\cong \mathcal {T}C ^\ast (E)_1p_E$ as $C^\ast$-correspondences.

Proof. We first claim that $\theta (\mathcal {T}C ^\ast (E)_1)=\mathcal {T}C ^\ast (F)_1$ and $\theta (p_E)=p_F$. The first is because $\theta$ interwines the gauge actions $\gamma _E$, $\gamma _F$. For the second we show that $\theta (p_E)\in \mathcal {T}C ^\ast (F)$ satisfies properties (i) and (ii) of lemma 3.10. Let $\varphi \in S^\infty _F$ (the subscript $F$ is to make it explicit that we are referring to the states of $\mathcal {T}C ^\ast (F)$). Then writing $\xi$ for the corresponding cyclic vector in the GNS-space $\mathcal {H}_\varphi$, for any $b\in \mathcal {T}C ^\ast (E)$,

\[ \varphi(\theta(b))=\left\langle \pi_\varphi(\theta(b))\xi, \xi \right\rangle. \]

Since $\theta$ is a gauge invariant isomorphism, $\phi \to \phi \circ \theta$ is an isomorphism of the KMS$_\beta$-simplex of $(\mathcal {T}C ^\ast (F),\,\gamma _F)$ to that of $(\mathcal {T}C ^\ast (E),\,\gamma _E)$ for each $\beta$. Thus, given an extremal KMS$_\beta$-state $\phi$ of $\mathcal {T}C ^\ast (F)$, the composition $\phi \circ \theta$ is an extremal KMS$_\beta$-state of $\mathcal {T}C ^\ast (E)$. Since $p_E$ satisfies condition (ii) of lemma 3.10 in $\mathcal {T}C^*(E)$, we have $\phi (\theta (p_E)) = \phi \circ \theta (p_E) = 1$ for each extremal $\phi$, and so $\theta (p_E)$ satisfies condition (ii) of lemma 3.10 in $\mathcal {T}C^*(F)$. So, writing $\eta$ for the cyclic vector in the GNS-space $\mathcal {H}_{\varphi \circ \theta }$, for $b\in \mathcal {T}C ^\ast (E)$ we have

\[ \varphi\circ\theta(b)=\left\langle \pi_{\varphi\circ \theta}(b)\eta, \eta \right\rangle. \]

Hence, by uniqueness of the GNS construction, $\pi _{\varphi }\circ \theta$ is unitarily equivalent to $\pi _{\varphi \circ \theta }$. Since $\pi _\varphi (\theta (b))$ is a minimal projection if and only if $\pi _{\varphi \circ \theta }(b)$ is a minimal projection, $\theta (p_E)$ satisfies condition (i) of lemma 3.10. Thus by uniqueness, $\theta (p_E)=p_F$.

Considering the diagram

(⋆)

we show that the map $\theta _X$ that makes the diagram $\star$ commute satisfies (3.8). For $\xi \in X(E)$,

(3.9)$$\begin{align} \iota_{X(F)}(\theta_X(\xi))p_F& =\iota_{X(F)}(\psi_F^{{-}1}\circ\theta\circ\psi_E(\xi))p_F \\ & =\iota_{X(F)}(\psi_F^{{-}1}\circ\theta(\iota_{X(E)}(\xi)p_E))p_F. \nonumber \end{align}$$

We already saw that $\theta (\mathcal {T}C ^\ast (E)_1)=\mathcal {T}C ^\ast (F)_1$ and $\theta (p_E)=\theta (p_F)$. Consequently, $\theta (\iota _{X(E)}(\xi )p_E)=\theta (\iota _{X(E)}(\xi ))p_F\in \mathcal {T}C ^\ast (F)_1p_F$, so there exists $\xi '\in X(F)$ such that

\[ \theta(\iota_{X(E)}(\xi)p_E)=\iota_{X(F)}(\xi')p_F=\psi_F(\xi'). \]

Hence, we obtain

\[ \iota_{X(F)}(\theta_X(\xi)p_F)\overset{(3.9)}=\iota_{X(F)}(\xi')p_F=\theta(\iota_{X(E)}(\xi)p_E), \]

that is (3.8). Since the maps $\psi _E$ and $\psi _F$ in the diagram are isomorphisms, we deduce that $(\theta _X,\,\theta _M)$ is an isomorphism of $C^\ast$-correspondences.

Proof. Let $A$ be the $k\times k$ matrix with $i$th column $x_i$, and let $S_k$ be the symmetric group. Since $\{x_1,\,\ldots,\,x_k\}$ are linearly independent, $A$ is invertible. Therefore,

\[ \det(A)=\sum_{\sigma\in S_k} \Big(\operatorname{sgn}(\sigma)\prod_{i=1}^k A_{i,\sigma(j)}\Big)\neq 0. \]

Hence there exists $\sigma \in S_k$ such that $\prod _{i=1}^k A_{i,\sigma (i)}\neq 0$, and hence

\[ \left\langle x_i, e_{\sigma(i)} \right\rangle = A_{i,\sigma(i)}\neq 0 \]

for all $i=1,\,\ldots,\,k$.

Proof. Fix $v\in \operatorname {supp}^\circ (h)$. For $i\le k$, regard $g_i|_{E^1v}$ as an element of $\ell ^2(E^1v)$. Then for $i,\,j \le k$,

\[ \left\langle g_i|_{E^1 v}, g_j|_{E^1v} \right\rangle = \sum_{e\in E^1v}\overline{g_i|_{E^1v}(e)}g_j|_{E^1v}(e)= \left\langle g_i, g_j \right\rangle(v) = \delta_{i,j} h(v). \]

Since $h(v)\not = 0$ it follows that $\{g_i|_{E^1v} : i \le k\}$ is linearly independent. We claim that $\operatorname {span}\{g_i|_{E^1v}: i \le k\} = \ell ^2(E^1v)$. For this, note that $\operatorname {res} : \xi \mapsto \xi |_{E^1v}$ is a norm-decreasing linear map from $X(E) \cdot h$ to $\ell ^2(E^1v)$, and is surjective since $h(v)\not = 0$. So condition (2) gives

$$\begin{align*} \ell^2(E^1v) & \subseteq \operatorname{res}\big(\overline{\operatorname{span}}\{g_j\cdot a: j\leq k, a\in C(E^0)\}\big) \\ & \subseteq \overline{\operatorname{span}}\{a(v)g_i|_{E^1 v} : a \in C_0(E^0), i \le k\} = \operatorname{span}\{g_i|_{E^1 v} : i \le k\}. \end{align*}$$

Hence $|E^1 v|=k$, and so lemma 4.2 yields a bijection $\sigma :\{1,\,\ldots,\,k\}\to E^1 v$ such that $g_i(\sigma (i))\neq 0$ for all $i=1,\,\ldots,\,k$.

We claim that whenever $g_i(e) \not = 0$, the function $\alpha _i$ of (3) satisfies $\alpha _i(s(e))=r(e)$. To see this, we prove the contrapositive. So suppose that $\alpha _i(s(e)) \not = r(e)$. By Tietze's theorem there exists $a \in C(E^0)$ such that $a(r(e))=1$ and $a(\alpha _i(s(e)))=0$. The function $a \circ \alpha _i|_{\operatorname {supp} {h}}$ is a continuous function on the compact set $\operatorname {supp} (h)$, so by Tietze's theorem there is a function $\tilde {a} \in C(E^0)$ that extends $a \circ \alpha _i|_{\operatorname {supp} {h}}$. Since $s(e) \in \operatorname {supp} (h)$ we have $\tilde {a}(s(e)) = a(\alpha _i(s(e)))$, and so condition (3) gives

\[ 0 = g_i(e) a(\alpha_i(s(e))) = (g_i \cdot \tilde{a})(e) = (a\cdot g_i)(e) = a(r(e))g_i(e) = g_i(e), \]

proving the claim.

Fix a neighbourhood $W$ of $v$ and s-sections $Z_e$, $e\in E^1 v$ as in lemma 2.1. Since the $g_i$ are continuous and each $g_i(\sigma (i))\neq 0$, for each $i$ there is a neighbourhood $\sigma (i) \in E^1$ on which $g_i$ is everywhere nonzero. Shrinking the $Z_{\sigma (i)}$ and $W$ appropriately, we may therefore assume that $Z_{\sigma (i)}\subseteq \operatorname {supp}g_i$ for each $i$ and $W \subseteq \operatorname {supp} (h)$. Fix $w\in W$. Then $(s|_{Z_{\sigma (i)}})^{-1}(w) \in \operatorname {supp}g_i$ and by the claim above $\alpha _i(w) = \alpha _i(s\circ (s|_{Z_{\sigma (i)}})^{-1}(w)) = r((s|_{Z_{\sigma (i)}})^{-1}(w))$ as needed.

Proof. Let $\theta =(\theta ^0,\,\theta ^1)$ be a bimodule isomorphism from $X(F)$ to $X(E)$, so $\theta ^1:X(F)\to X(E)$ and $\theta ^0:C(F^0)\to C(E^0)$ preserve the bimodule structure. Let $\phi ^0=\widehat {\theta ^0}:E^0\to F^0$ be the Gelfand transform of $\theta ^0$ which is a homeomorphism between $E^0$ and $F^0$. Fix $v\in E^0$. Applying lemma 2.1 to $\phi ^0(v)$ we obtain an open neighbourhood $W'\subseteq F^0$ of $\phi ^0(v)$ and s-sections $Z_e'$ for $e \in F^1\phi ^0(v)$ such that $F^1W' = \bigsqcup _{e\in F^1\phi ^0(v)}Z_e'$. Since $F^0$ is normal, Urysohn's lemma gives a function $h'\in C_c(W',\,[0,\,1])$ (in particular, the compact set $\operatorname {supp} (h)$ is contained in $W'$) such that $h'(\phi ^0(v))=1$. For each $e\in F^1\phi ^0(v)$, define $g_e'\in C_0(Z_e')\subseteq X(F)$ by $g_e'=\sqrt {h'\circ s|_{Z'_e}}$. Define $\alpha _e':\operatorname {supp} (h') \to F^0$ by $\alpha _e'=r_F\circ (s_F|_{Z_e'})^{-1}$. Direct computation shows that $h'$, the $g_e'$ and the $\alpha _e'$ satisfy condition (1) of lemma 4.3. To see that they also satisfy condition (2), fix $\xi \in X(F) = C(F^1)$ and $e \in F^1\phi ^0(v)$. If $f\in Z_e'$. Then

$$\begin{align*} (\xi\cdot h')(f) & = \xi((s_F|_{Z_e'})^{{-}1}(s_F(f)))\sqrt{h'(s_F(f))}\sqrt{h'(s_F(f))} \\ & = \Big(g'_e(f)\cdot\big(\xi\circ(s_F|_{Z'_e})^{{-}1} \sqrt{h'}\big)\Big)(f). \end{align*}$$

For each $e \in F^1\phi ^0(v)$, let $a_e := (\xi \circ (s_F|_{Z_e'})^{-1} \sqrt {h'}) \in C_0(W') \subseteq C(F^0)$. Since the $Z_e'$ are disjoint and $\operatorname {supp}(\xi \cdot h') \subseteq F^1W' = \bigsqcup _{e\in F^1\phi ^0(v)}Z_e'$, we can write $\xi \cdot h' = \sum _{e\in F^1\phi ^0(v)} g_e' \cdot a_e$. This gives condition (2). For condition (3), fix $a' \in C(F^0)$ and suppose that $\tilde {a}' \in C(F^0)$ extends $a' \circ \alpha '_e$. We must show that $(a' \cdot g'_e)(f) = (g'_e \cdot \tilde {a}')(f)$ for all $f \in F^1$. We consider two cases. First suppose that $f \not \in \operatorname {supp} (g'_e)$, then $(a' \cdot g'_e)(f) = a'(r(f))g'_e(f) = 0 = g'_e(f) \tilde {a}'(s(f)) = (g'_e \cdot \tilde {a}')(f)$. Now suppose that $f \in \operatorname {supp} (g'_e)$. Then $f \in Z'_e$, which implies that $\alpha '_e(s(f)) = r(f)$; and $s(f) \in \operatorname {supp} (h')$, giving $\tilde {a}'(s(f)) = a'(\alpha '_e(s(f))) = a'(r(f))$. Hence $(a' \cdot g'_e)(f) = a'(r(f))g'_e(f) = \tilde {a}'(s(f))g'_e(f) = (g'_e \cdot \tilde {a}')(f)$.

Let $h:=\theta ^0(h')=h'\circ \phi ^0$ and for each $e\in F^1 \phi ^0(v)$, let $g_e:=\theta ^1(g_e')$ and $\alpha _e=(\phi ^0)^{-1}\circ \alpha '_e\circ \phi ^0$, defined on $(\phi ^0)^{-1}(\operatorname {supp} (h')) = \operatorname {supp} (h) \subseteq E^0$. Now we claim that $h$, the $g_e$ and the $\alpha _e$ also satisfy (1)–(3) of lemma 4.3. Since $(\theta ^0,\,\theta ^1)$ is a bimodule isomorphism, conditions (1) and (2) follow from straightforward calculations using that $h'$, the $g'_e$ and the $\alpha '_e$ satisfy (1) and (2). We show that condition (3) holds. For this, fix $a \in C(E^0)$ and let $a' := (\theta ^0)^{-1}(a) \in C(F^0)$. Suppose that $\tilde {a} \in C(E^0)$ extends $a \circ \alpha _e$. We claim that $\tilde {a}' := (\theta ^0)^{-1}(\tilde {a})$ extends $a' \circ \alpha '_e|_{\operatorname {supp} (h')}$. To see this, note first that, by definition of $h$, we have $\operatorname {supp} (h) = (\phi ^0)^{-1}(\operatorname {supp} (h'))$. Also, by definition of $\alpha _e$, we have $\alpha _e \circ (\phi ^0)^{-1} = (\phi ^0)^{-1} \circ \alpha '_e$. So for $w \in \operatorname {supp} (h')$,

$$\begin{align*} (\theta^0)^{{-}1}(\tilde{a})(w) & = \tilde{a}((\phi^0)^{{-}1}(w)) = (a \circ \alpha_e)((\phi^0)^{{-}1}(w))\\ & = (a \circ (\phi^0)^{{-}1}) \circ \alpha'_e(w) = a' \circ \alpha'_e(w). \end{align*}$$

Consequently, condition (3) for $h'$, $g'_e$ and $\alpha '_e$ gives $a' \cdot g'_e = g'_e \cdot \tilde {a}'$. Hence

\[ a\cdot g_e = \theta^1(a'\cdot g_e') = \theta^1(g_e' \cdot \tilde{a}') = \theta^1(g'_e)\cdot \theta^0(\tilde{a}') = g_e \cdot \tilde{a}. \]

Thus condition (3) holds for $h$, the $g_e$ and the $\alpha _e$.

As $h(v) = h'(\phi ^0(v)) = 1$, we have $v\in \operatorname {supp} ^\circ (h)$, so by lemma 4.3, there exists an open neighbourhood $W \subseteq (\phi ^0)^{-1}(W')$ of $v$ and s-sections $Z_f$, $f\in E^1v$ as in lemma 2.1, and a bijection $\sigma : F^1\phi ^0(v) \to E^1v$ such that $\alpha _e = r_E\circ (s_E|_{Z_{\sigma (e)}})^{-1}$ on $W$ for each $e\in F^1\phi ^0(v)$.

Define $\phi _W^1: E^1 W\to F^1 \phi ^0(W)$ by $\phi ^1_W|_{Z_{\sigma (e)}}=(s_F|_{Z'_e})^{-1}\circ \phi ^0\circ s_E$ for each $e\in F^1\phi ^0(v)$. Since each $\phi ^1_W|_{Z_{\sigma (e)}}$ is a homeomorphism and their images are disjoint, $\phi ^1_W$ is a homeomorphism by the pasting lemma. We claim that $\phi ^1_W$ satisfies the conditions necessary for local conjugacy, that is, $r_F\circ \phi _W^1=\phi ^0\circ r_E|_{E^1 W}$ and $s_F\circ \phi _W^1=\phi ^0\circ s_E|_{E^1 W}$. That $s_F\circ \phi _W^1=\phi ^0\circ s_E|_{E^1 W}$ is by definition of $\phi ^1_W$. Now, fix $\bar {f}\in E^1 W$. There exists a unique $f\in E^1 v$ such that $\bar {f}\in Z_{f}$. Since $\sigma$ is a bijection there is a unique $e\in F^1\phi ^0(v)$ such that $\sigma (e)=f$, hence $\bar {f}\in Z_{\sigma (e)}$. Thus, we have

$$\begin{align*} \phi^0\circ r_E(\bar{f})& =\phi^0(r_E((s|_{Z_{\sigma(e)}})^{{-}1}(s_E(\bar{f}))))=\phi^0(\alpha_e(s_E(\bar{f})))\\ & = \phi^0((\phi^0)^{{-}1}\circ \alpha'_e\circ \phi^0(s_E(\bar{f})))=\alpha_e'(\phi^0(s_E(\bar{f})))\\ & =r_F((s_F|_{Z_e'})^{{-}1}(\phi^0(s_E(\bar{f}))))=r_F((s_F|_{Z_e'})^{{-}1}(s_F(\phi^1_W(\bar{f})))). \end{align*}$$

Since $\bar {f}\in Z_{\sigma (e)}$, we have $\phi ^1_W(\bar {f})=(s_F|_{Z_e'})^{-1}(\phi ^0\circ s_E(\bar {f}))\in Z_e'$, therefore $\phi ^0\circ r_E(\bar {f})=r_F(\phi ^1_W(\bar {f}))$. Hence $E$ and $F$ are locally conjugate.

Proof. Theorems 3.2 and 4.5 imply that $E$ and $F$ are locally conjugate. Let $\phi ^0 : E^0\to F^0$ be the map on vertices implementing the local conjugacy. Then for each $v\in E^0$ there exists a neighbourhood $U_v$ of $v$ and a homeomorphism $\phi ^1_{U_v}:E^1U_v\to F^1\phi ^0(U_v)$ such that $r_F\circ \phi _{U_v}^1=\phi ^0\circ r_E|_{E^1U_v}$ and $s_F\circ \phi _{U_v}^1=\phi ^0\circ s_E|_{E^1U_v}$. Since $E^0$ is totally disconnected, we may suppose the $(U_v)$ are compact open. The $(U_v)_v$ cover $E^0$, so admit a finite subcover $(U_i)_{i=1}^n$. Put $V_1=U_1$ and let $V_i=U_i\setminus {\bigcup _{j=1}^{i-1} U_j}$ for $i\geq 2$. Then the $V_i$ are mutually disjoint compact open sets that cover $E^0$ and we can define the homeomorphisms $\phi ^1_{V_i}:E^1 V_i\to F^1 \phi ^0(V_i)$ by the restriction of $\phi ^1_{U_i}$ on $E^1V_i$. Note that $(E^1V_i)_i$ are mutually disjoint open sets that cover $E^1$ and, similarly, $(F^1\phi ^0(V_i))_i$ are mutually disjoint open sets that cover $F^1$. We define $\phi ^1:E^1\to F^1$ by $\phi ^1|_{E^1 V_i}=\phi ^1_{V_i}$. By the pasting lemma $\phi ^1$ is continuous and bijective, thus a homeomorphism. It only remains to verify that $r_F\circ \phi ^1=\phi ^0\circ r_E$ and $s_F\circ \phi ^1=\phi ^0\circ s_E$. We show this only for the range map, since it is analogous for the source map. If $f\in E^1$, then $f\in E^1V_i$ for some $i=1,\,\ldots,\,n$ and hence $(r_F\circ \phi ^1) (f)=r_F(\phi ^1_{V_i}(f))=(\phi ^0\circ r_E)(f)$.

Proof. By theorem 4.5, $E\cong _{\operatorname {loc}} F$. Hence there is a homeomorphism $\phi ^0:E^0\to F^0$ as in definition 4.1. Fix $v\in E^0$. There is a neighbourhood $U$ of $v$ and a homeomorphism $\phi ^1_U : E^1 U \to F^1 \phi ^0(U)$ satisfying equations (4.1). Hence, for $w\in E^0$ we have $\phi ^1_U(wE^1v)=\phi ^0(w)F^1\phi ^0(v)$. Since $\phi _U^1$ is a bijection, $|wE^1v|=|\phi ^0(w)F^1\phi ^0(v)|$. Since $v,\,w$ were arbitrary, it follows that there is a range-and-source-preserving bijection $\phi ^1 : E^1 \to F^1$ such that $(\phi ^0,\, \phi ^1)$ is an isomorphism of discrete graphs.

Proof. By lemma 2.1 there exists a finite open cover $\{U_i\}_{i\in F}$, of the vertex space $E^0$, such that, for each $i\in F$ there exists $k(i) \in \mathbb {N}$ and $s$-sections $Z_j^{(i)},\, j \le k(i)$ such that the source map restricts to a homeomorphism $Z_j^{(i)} \cong U_i$ for each $j$, and $E^1 U_i = \bigsqcup _{j=1}^{k(i)}Z_j^{(i)}$.

Fix $i\in F$ and for each $j \le k(i)$, define $\phi _i^j:Z_j^{(i)}\to U_i\times \{1,\,\ldots,\, k(i)\}$ by $\phi _i^j(e)=(s|_{Z_j^{(i)}}(e),\,j)$. Then each $\phi _i^j$ is a homeomorphism of $Z_j^{(i)}$ onto $U_i\times \{j\}$. Hence, by the pasting lemma, $\phi _i=: \bigsqcup _{j=1}^{k(i)}\phi _i^j : E^1 U_i \to U_i\times \{1,\,\ldots,\, k(i)\}$ is a homeomorphism.

If $U_i\cap U_j\neq \emptyset$, then we may consider the composition $\phi _j\circ \phi _i^{-1}$ with domain $(U_i\cap U_j)\times \{1,\,\ldots,\,k(i)\}$. Let $(x,\,\ell )\in (U_i\cap U_j) \times \{1,\,\ldots,\,k(i)\}$. Then $\phi _i^{-1}(x,\,\ell ) = (s|_{Z_\ell ^{(i)}})^{-1}(x)$, and there exists a $m\in \{1,\,\ldots,\, k(j)\}$ such that $(s|_{Z_\ell ^{(i)}})^{-1}(x)\in Z_m^{(j)}$, since otherwise $(s|_{Z_\ell ^{(i)}})^{-1}(x)\notin E^1 U_j$, contradicting $x\in U_j$. This $m$ is unique because the $Z_n^{(j)}$, $n\in \{1,\,\ldots,\, k(j)\}$ are disjoint. Hence

\[ \phi_j\circ \phi_i^{{-}1}(x,\ell) = \phi_j((s|_{Z_\ell^{(i)}})^{{-}1}(x)) = (x,m). \]

So, there is a function $\sigma _x^{j,i} : \{1,\, \dots,\, k(i)\} \to \{1,\, \dots,\, k(j)\}$ such that $(s|_{Z_\ell ^{(i)}})^{-1}(x)\in Z^{(j)}_{\sigma _x^{j,i}(\ell )}$ for all $\ell \in \{1,\,\ldots,\, k(i)\}$.

We claim that $\sigma _x^{j,i}$ is a bijection. Indeed, if $\sigma _x^{j,i}(\ell )=\sigma _x^{j,i}(\ell ')$, then $\phi _j\circ \phi _i^{-1}(x,\,\ell ) = (x,\,\sigma _x^{j,i}(\ell )) = (x,\,\sigma _x^{j,i}(\ell ')) = \phi _j\circ \phi _i^{-1}(x,\,\ell ')$. Since $\phi _j$ and $\phi _i$ are injective, $\ell =\ell '$, thus $\sigma _x^{j,i}$ is injective.

We now claim that $\sigma ^{j,i}_x$ is surjective. For this, fix $m\in \{1,\,\ldots,\,k(j)\}$. Then $\phi _j^{-1}(x,\,m) = (s|_{Z_m^{(j)}})^{-1}(x) \in E^1(U_i\cap U_j) \subseteq E^1 U_i$. Thus, there is a unique $m'\in \{1,\,\ldots,\, k(i)\}$ such that $\phi _j^{-1}(x,\,m)\in Z_{m'}^{(i)}$, so $\phi _j^{-1}(x,\,m) = (s|_{Z_{m'}^{(i)}})^{-1}(x)$. Hence

\[ \phi_j\circ \phi_i^{{-}1}(x,m') =\phi_j((s|_{Z_{m'}^{(i)}})^{{-}1}(x)) =\phi_j\big(\phi_j^{{-}1}(x,m)\big) =(x,m). \]

This implies that $\sigma ^{j,i}_x(m')=m$ and $\sigma ^{j,i}_x$ is a bijection. In particular, $k(i)=k(j)$.

Recall the canonical Hilbert bundle $\mathcal {E}$ associated with the bimodule $X(E)$ discussed in the preliminaries. For each $i$ the map $\phi _i$ induces a right-Hilbert $C_0(U_i)$-module isomorphism $\phi _i^* : C_0(U_i\times \{1,\,\ldots,\, k(i)\}) \cong C_0(U_i,\, \mathbb {C}^{k(i)}) \to X(E) \cdot C_0(U_i)$ by $\phi _i^*(\xi ) = \xi \circ \phi _i$. This induces a vector-bundle isomorphism $\psi _i : U_i \times \mathbb {C}^{k(i)} \cong \mathcal {E}|_{U_i}$ satisfying $\psi _i(x,\, e_m) = \phi _i^{-1}(x,\, m)$ for all $x \in U_i$ and $m \le k(i)$.

The maps $\psi _i,\, i \in F$ are a local trivialization of $\mathcal {E}$, and for $x \in U_i \cap U_j$ and $\ell \le k(i)$, the transition function $\psi _j^{-1} \circ \psi _i$ satisfies $\psi _j^{-1} \circ \psi _i(x,\, e_\ell ) = \psi _j^{-1}(\phi _i^{-1}(x,\,\ell )) = \psi _j^{-1}(\phi _j^{-1}(\sigma _x^{j,i}(\ell ))) = (x,\, e_{\sigma _x^{j,i}(\ell )})$. That is, the matrix implementing $\psi _j^{-1} \circ \psi _i$ in the fibre over $x$ is precisely the permutation matrix corresponding to $\sigma _x^{j,i}$.

Proof. That (1) and (2) are equivalent follows from [Reference Kaliszewski, Patani and Quigg19, Theorem 6.4]. Proposition 6.1 gives (1)$\implies$(3). So it suffices to show that if $\mathcal {E}$ has a local trivialization whose transition functions take values in permutation matrices, then it admits a global choice of orthonormal bases. For this, fix such a local trivialization, say $\{U_i\}_{i \in F}$ is an open cover of $K$ and for each $i$, we have a bundle isomorphism $\psi _i : U_i \times \mathbb {C}^{k(i)} \to \mathcal {E}|_{U_i}$ so that the transition functions $\psi _i^{-1} \circ \psi _j$ are permutation-matrix valued. Fix $x \in K$. If $i,\,j \in F$ satisfy $x \in U_i \cap U_j$ then since $\psi _i^{-1} \circ \psi _j$ takes values in permutation matrices, there is a permutation $\sigma$ such that $\psi _i^{-1} \circ \psi _j(x,\, e_\ell ) = (x,\, e_{\sigma (\ell )})$ for all $\ell \le k(i)$. Hence $\{\psi _i(x,\, e_1),\, \dots,\, \psi _i(x,\, e_{k(i)})\} = \{\psi _j(x,\, e_{\sigma (1)}),\, \dots,\, \psi _j(x,\, e_{\sigma ({k(i)})})\} = \{\psi _j(x,\, e_1),\, \dots,\, \psi _j(x,\, e_{k(i)})\}$. So there is a well-defined set-valued map $\psi : K \to \mathcal {P}(\mathcal {E})$ such that, for each $i$, the restriction $\psi |_{U_i}$ is the map $x \mapsto \{\psi _i(x,\, e_1),\, \dots,\, \psi _i(x,\, e_{k(i)})\}$. The set $B := \bigcup _{x \in K} \psi (x)$ is a subset of $\mathcal {E}$. Each $B \cap \mathcal {E}_x = \psi (x)$ is an orthonormal basis for $\mathcal {E}_x$. For each $i$, the restriction $\pi _i$ of the bundle map to $B \cap \mathcal {E}|_{U_i}$ satisfies $\pi _i(\psi _i(x,\, e_\ell )) = x$ and for each $x,\,\ell$ the set $\psi _i(U_i \times \{e_\ell \}) \cong U_i \times \{e_\ell \}$ is an open neighbourhood of $(x,\, e_\ell )$ on which $\pi _i$ restricts to a homeomorphism. So the bundle map restricts to a local homeomorphism $B \to K$. That is $B$ is a continuous choice of basis.