Proof Given $\varepsilon>0$ , choose $\delta <\frac {1}{3}$ small enough so that

(2.1) $$ \begin{align} \frac{2\delta}{\sqrt{1-2\delta}}+\delta<\varepsilon. \end{align} $$

Let $\alpha \colon {\mathbb {T}}\to {\mathrm {Aut}}(A)$ and $u\in {\mathcal {U}}(A)$ be as in the statement. Set $x =\int _{{\mathbb {T}}}\overline { z}\alpha _ z(u)\ d z\in A$ . Then $\|x\|\leq 1$ and $\|x-u\|\leq \delta $ . One checks that $\|x^*x-1\|\leq 2\delta < 1$ , so $x^*x$ is invertible. Moreover,

(2.2) $$ \begin{align} \big\|(x^*x)^{-1}\big\|\leq \frac{1}{1-\|1-x^*x\|}\leq\frac{1}{1-2\delta}. \end{align} $$

Set $u= x(x^*x)^{-\frac {1}{2}}$ . Then u is a unitary in A. Using that $\|x\|\leq 1$ at the first step, and that $0 \leq 1- (x^*x)^{\frac {1}{2}} \leq 1- x^*x$ at the second step, we get

$$ \begin{align*} \|u-x\| &\leq \big\|(x^*x)^{-\frac{1}{2}}-1\big\| \leq \big\|(x^*x)^{-\frac{1}{2}}\big\| \big\|1-(x^*x)^{\frac{1}{2}}\big\|\\ &\stackrel{(2.2)}{\leq} \frac{1}{\sqrt{1-2\delta}} \|1-x^*x\| \leq \frac{2\delta}{\sqrt{1-2\delta}}. \end{align*} $$

Thus

$$\begin{align*}\|u-v\|\leq \|u-x\|+\|x-v\|\leq \frac{2\delta}{\sqrt{1-2\delta}}+\delta\stackrel{(2.1)}{<} \varepsilon.\end{align*}$$

For $ z\in {\mathbb {T}}$ , we have

$$\begin{align*}\alpha_ z(x)=\int_{\mathbb{T}} \overline{\omega}\alpha_{ z\omega}(u)d\omega= \int_{\mathbb{T}} z\overline{\omega}\alpha_{\omega}(u)d\omega= z x. \end{align*}$$

It follows that $\alpha _ z(x^*x)=x^*x$ and hence $\alpha _ z(u)=\alpha _ z\big (x(x^*x)^{-\frac {1}{2}}\big )= z u$ , for all $ z\in {\mathbb {T}}$ , so u satisfies the condition in the statement.

Proof (1). Using Proposition 2.2, let $u\in {\mathcal {U}}(A)$ be a unitary satisfying $\alpha _z (u)= z u$ for all $ z \in {\mathbb {T}}$ . For $a\in A^\alpha $ , we have $\alpha _z (uau^*)=uau^*$ for all $ z \in {\mathbb {T}}$ , and thus conjugation by u determines an automorphism $\theta $ of $A^\alpha $ . Let $v\in A^\alpha \rtimes _\theta {\mathbb {Z}}$ denote the canonical unitary implementing $\theta $ . Since the pair $({\mathrm {id}}_{A^\alpha }, u)$ is a covariant representation of $(A^\alpha ,\theta )$ on A, there is a unique homomorphism $\varphi \colon A^\alpha \rtimes _\theta {\mathbb {Z}}\to A$ satisfying $\varphi (a)=a$ for all $a\in A^\alpha $ and $\varphi (v)=u$ .

We claim that $\varphi $ is an equivariant isomorphism. Equivariance of $\varphi $ is clear, since for all $ z \in {\mathbb {T}}$ we have $\widehat {\theta }_z (a)=a$ for all $a\in A^\alpha $ and $\widehat {\theta }_z (v)= z v$ . Injectivity of $\varphi $ follows from the fact that ${\mathrm {id}}_{A^\alpha }$ is injective (and that ${\mathbb {Z}}$ is amenable). Surjectivity can be deduced using spectral subspaces, as follows. Given $n\in {\mathbb {Z}}$ , we set

$$\begin{align*}A_n=\{a\in A\colon \alpha_z (a)= z ^n a\mbox{ for all } z \in{\mathbb{T}}\}, \end{align*}$$

which is a closed subspace of A. It is well-known that $\sum _{n\in {\mathbb {Z}}}A_n$ is dense in A; see, for example, part (ix) of Theorem 8.1.4 in [Reference Pedersen34]. Note that $A_0=A^\alpha $ and that u belongs to $A_1$ . Moreover, using that u is a unitary, it is easy to see that $A_n=u^nA_0$ for all $n\in {\mathbb {Z}}$ . In particular, A is generated as a $C^*$ -algebra by $A_0$ and u. Since $A_0\cup \{u\}$ is contained in the image of $\varphi $ , we conclude that $\varphi $ is surjective.

(2). Let $\theta '$ be as in the statement, and let $\varphi \colon (A^\alpha \rtimes _{\theta }{\mathbb {Z}},\widehat {\theta })\to (A,\alpha )$ and $\varphi '\colon (A^\alpha \rtimes _{\theta '}{\mathbb {Z}},\widehat {\theta '})\to (A,\alpha )$ be equivariant isomorphisms. Let v be the canonical unitary in $A^\alpha \rtimes _\theta {\mathbb {Z}}$ that implements $\theta $ , and let $v'$ be the canonical unitary in $A^\alpha \rtimes _{\theta '}{\mathbb {Z}}$ that implements $\theta '$ . Set $w=\varphi (v)\varphi '(v')^*$ , which is a unitary in A. We claim that w is fixed by $\alpha $ . For $ z\in {\mathbb {T}}$ , we use equivariance of $\varphi $ and $\varphi '$ to get

$$ \begin{align*}\alpha_ z(w)=\varphi(\widehat{\theta}_ z(v))\varphi'(\widehat{\theta'}_ z(v'))^*)= z\varphi(v)\overline{ z}\varphi'(v')=w.\end{align*} $$

Finally, given $a\in A^\alpha $ , we have

$$ \begin{align*}({\mathrm{Ad}}(w)\circ\theta)(a)&=(\varphi(v)\varphi'(v')^*)(\varphi'(v')a\varphi'(v')^*)(\varphi'(v')\varphi(v)^*)\\ &=\varphi(v)a\varphi(v)^*=\theta'(a).\\[-34pt] \end{align*} $$

Proof This is an immediate from Theorem 2.3 and Takai duality.

Proof (1). Assume that $\alpha $ has the Rokhlin property. Let $F\subseteq A\rtimes _\alpha {\mathbb {T}}$ be a finite subset, and let $\varepsilon>0$ . For $\xi \in L^1({\mathbb {T}})$ and $a\in A$ , write $\xi a$ for the for the function given by $(\xi a)(z)=\xi (z)a$ for all $z\in {\mathbb {T}}$ . Since the linear span of the elements of this form is dense in $L^1({\mathbb {T}},A)$ , and hence, also in $A\rtimes _\alpha {\mathbb {T}}$ , we may assume that there exist finite subsets $F_A\subseteq A$ and $F_{\mathbb {T}}\subseteq L^1({\mathbb {T}})$ such that every element of F has the form $\xi a$ for $a\in F_A$ and $\xi \in F_{\mathbb {T}}$ . Without loss of generality, we may assume that the sets $F_A$ and $F_{\mathbb {T}}$ contain only self-adjoint contractions.

Let $f\colon {\mathbb {T}}\to {\mathbb {C}}$ be a positive, continuous function whose support is a small enough neighborhood of $1\in {\mathbb {T}}$ so that the following conditions are satisfied:

1. (i) $\|(f\ast f)b-b\|<\varepsilon $ for all $b\in F\cup \widehat {\alpha }(F)$ ;

2. (ii) $\|f\|_1=1=\|f\ast f\|_1$ ;

3. (iii) $f(z)=f(\overline {z})$ for all $z\in {\mathbb {T}}$ ;

4. (iv) with $\widetilde {f}(z)=z f(z)$ for all $z\in {\mathbb {T}}$ , we have

   $$\begin{align*}\|f-\widetilde{f}\|_1<\varepsilon \ \ \mbox{ and } \ \ \|f\ast f-\widetilde{f}\ast \widetilde{f}\|_1<\varepsilon; \end{align*}$$

5. (v) given $\xi \in F_{\mathbb {T}}$ and $a\in F_A$ , if $z,\sigma ,\omega \in {\mathbb {T}}$ satisfy $f(\omega )f(\overline {z}\sigma \omega )\neq 0$ , then

   $$\begin{align*}\|\alpha_\omega(a)-a\|<\frac{\varepsilon}{2} \ \ \mbox{ and } \ \ |\xi(\sigma)-\xi(z)|<\frac{\varepsilon}{2}. \end{align*}$$

Using Proposition 2.2, find a unitary $u\in A$ satisfying:

1. (vi) $\alpha _\zeta (u)=\zeta u$ for all $\zeta \in {\mathbb {T}}$ , and

2. (vii) $\|ua-au\|<\varepsilon /2$ for all $a\in \bigcup _{\omega \in {\mathbb {T}}}\alpha _\omega (F_A)$ .

We regard u as a unitary in the multiplier algebra of $A\rtimes _\alpha {\mathbb {T}}$ via the canonical unital embedding $A\hookrightarrow M(A\rtimes _\alpha {\mathbb {T}})$ , and set $v=fu^*$ . Then v is a contraction in $L^1({\mathbb {T}},A)$ , and hence also in $A\rtimes _\alpha {\mathbb {T}}$ . We proceed to check the conditions in Definition 2.6. Let $z\in {\mathbb {T}}$ . Then

$$ \begin{align*} (v^*\ast v)(z) &= \int_{\mathbb{T}} v^*(\omega)\alpha_\omega(v(\overline{\omega}z))\ d\omega = \int_{\mathbb{T}} \omega f(\omega)u\alpha_\omega(f(\overline{\omega}z)u^*)\ d\omega\\ &\stackrel{\mathrm{(vi)}}{=} \int_{\mathbb{T}} \omega f(\omega)uf(\overline{\omega}z)\overline{\omega}u^*\ d\omega = (f\ast f)(z). \end{align*} $$

Thus, $v^*\ast v=f\ast f$ and hence condition (a) in Definition 2.6 follows from condition (ii) above. In order to check (b), we compute as follows for $z\in {\mathbb {T}}$ :

$$ \begin{align*} (v\ast v^*)(z) &= \int_{\mathbb{T}} v(\omega)\alpha_\omega(v^*(\overline{\omega}z))\ d\omega = \int_{\mathbb{T}} f(\omega)u^*\alpha_\omega(\overline{\omega}z f(\overline{\omega}z)u)\ d\omega\\ &\stackrel{\mathrm{(vi)}}{=} \int_{\mathbb{T}} f(\omega)u^*\overline{\omega}z f(\overline{\omega}z)\omega u\ d\omega = \int_{\mathbb{T}} \omega f(\omega)\overline{\omega}z f(\overline{\omega}z)\ d\omega = (\widetilde{f}\ast \widetilde{f})(z). \end{align*} $$

Thus $v\ast v^*=\widetilde {f}\ast \widetilde {f}$ . We deduce that

$$\begin{align*}\|v^*v-vv^*\|\leq \|v^*\ast v- v\ast v^*\|_1=\|f\ast f- \widetilde{f}\ast \widetilde{f}\|_1\stackrel{\mathrm{(iv)}}{<}\varepsilon,\end{align*}$$

as desired. In order to check (c), let $\zeta \in {\mathbb {T}}$ . Then

$$\begin{align*}\widehat{\alpha}(v)(z)=z f(z) u^*=\widetilde{f}(z)u^*,\end{align*}$$

so $\widehat {\alpha }(v)=\widetilde {f}u^*$ . Using this at the second step, we get

$$\begin{align*}\|\widehat{\alpha}(v)-v\|\leq \|\widehat{\alpha}(v)-v\|_1 =\|f-\widetilde{f}\|_1\stackrel{\mathrm{(iv)}}{<}\varepsilon,\end{align*}$$

as desired. Finally, in order to check (d), it suffices to take $\xi \in F_{\mathbb {T}}$ and $a\in F_A$ , and show the desired inequality for $b=\xi a$ . Given $z\in {\mathbb {T}}$ , we have

$$ \begin{align*} (v\ast b \ast v^*)(z) &=\int_{\mathbb{T}} v(\omega)\alpha_\omega((b \ast v^*)(\overline{\omega}z)) \ d\omega \\ &=\int_{\mathbb{T}} f(\omega)u^*\alpha_\omega\Big( \int_{\mathbb{T}} \xi(\sigma)a \alpha_\sigma(v^*(\overline{\sigma\omega}z)) \ d\sigma \Big) d\omega \\ &=\int_{\mathbb{T}}\int_{\mathbb{T}} f(\omega)u^*\xi(\sigma)\alpha_\omega(a) \alpha_{\omega\sigma }\big(\overline{\sigma\omega}z f(\overline{\sigma\omega}z)u \big) d\sigma d\omega \\ &\stackrel{\mathrm{(vi)}}{=} \int_{\mathbb{T}}\int_{\mathbb{T}} f(\omega)u^* \xi(\sigma)\alpha_\omega(a) \overline{\sigma\omega}z f(\overline{\sigma\omega}z) \omega\sigma u \ d\sigma d\omega \\ &= z\int_{\mathbb{T}}\int_{\mathbb{T}} \xi(\sigma)u^*\alpha_\omega(a)u f(\omega) f(\overline{\sigma\omega}z) \ d\sigma d\omega\\ &\stackrel{\mathrm{(vii)}}{\approx}_{\!\!\frac{\varepsilon}{2}} z\int_{\mathbb{T}}\int_{\mathbb{T}} \xi(\sigma)\alpha_\omega(a) f(\omega) f(\overline{\sigma\omega}z) \ d\sigma d\omega. \end{align*} $$

By (iv), if in the above expression, we replace $\xi (\sigma )\alpha _\omega (a)$ by $\xi (z)a$ , we obtain an element in A whose distance to $(v\ast b \ast v^*)(z)$ is at most $\varepsilon /2$ . Hence,

$$ \begin{align*} (v\ast b \ast v^*)(z)&\approx_{\varepsilon} z \xi(z)a\int_{\mathbb{T}}\int_{\mathbb{T}} f(\omega) f(\overline{\sigma\omega}z) \ d\sigma d\omega \\ &= z b(z) \int_{\mathbb{T}} (f\ast f)(\overline{\sigma}z)\ d\sigma = z b(z)\|f\ast f\|_1\stackrel{\mathrm{(ii)}}{=} \widehat{\alpha}(b)(z).\end{align*} $$

We conclude that

$$\begin{align*}\|vbv^*-\widehat{\alpha}(b)\|\leq \|v\ast b \ast v^*-\widehat{\alpha}(b)\|_1<\varepsilon, \end{align*}$$

as desired. This shows that $\widehat {\alpha }$ is approximately representable.

Conversely, assume that $\widehat {\alpha }$ is approximately representable. Denote the left regular representation of G by $\lambda \colon {\mathbb {T}}\to {\mathcal {U}}(L^2({\mathbb {T}}))$ . By Takai duality, there is a canonical equivariant identification

(2.3) $$ \begin{align} (A\rtimes_\alpha{\mathbb{T}}\rtimes_{\widehat{\alpha}}{\mathbb{Z}},\widehat{\widehat{\alpha}})\cong (A\otimes{\mathcal{K}}(L^2({\mathbb{T}})),\alpha\otimes{\mathrm{Ad}}(\lambda)). \end{align} $$

Let $p\in {\mathcal {K}}(L^2({\mathbb {T}}))$ be the projection onto the constant functions, and let $e\in M(A\rtimes _\alpha {\mathbb {T}}\rtimes _{\widehat {\alpha }}{\mathbb {Z}})$ be the projection corresponding to $1_A\otimes p$ under the identification in (2.3). Then e and p are ${\mathbb {T}}$ -invariant, and there is a canonical equivariant isomorphism

(2.4) $$ \begin{align} \left(e(A\rtimes_\alpha{\mathbb{T}}\rtimes_{\widehat{\alpha}}{\mathbb{Z}})e,\widehat{\widehat{\alpha}}\right)\cong (A,\alpha). \end{align} $$

Let $u\in M(A\rtimes _\alpha {\mathbb {T}}\rtimes _{\widehat {\alpha }}{\mathbb {Z}})$ be the canonical unitary implementing $\widehat {\alpha }$ . Let $F\subseteq A$ be a finite subset, and let $\varepsilon>0$ . Let $\delta>0$ such that whenever $x\in A$ satisfies $\|x^*x-1\|<\delta $ and $\|xx^*-1\|<\delta $ , then there is $w\in {\mathcal {U}}(A)$ with $\|w-x\|<\varepsilon /2$ . Set

$$\begin{align*}F"=\{e\}\cup \{a\otimes p\colon a\in F\}\subseteq A\rtimes_\alpha{\mathbb{T}}\rtimes_{\widehat{\alpha}}{\mathbb{Z}}.\end{align*}$$

Let $F'\subseteq A\rtimes _\alpha {\mathbb {T}}$ be a finite subset and let $n\in {\mathbb {N}}$ such that any element in $F"$ is within $\varepsilon /2$ of the span of $\{bu^k\colon b\in F', -n\leq k\leq n\}$ . Using approximate representability of $\widehat {\alpha }$ , let $v\in A\rtimes _\alpha {\mathbb {T}}$ be a contraction satisfying conditions (a), (b), (c), and (d) in Definition 2.6 for $\varepsilon _0=\min \big \{\frac {\varepsilon }{26n^2|F'|},\delta \big \}$ and $F'$ . Set $y=v^*u$ , which is a contraction in $A\rtimes _\alpha {\mathbb {T}}\rtimes _{\widehat {\alpha }}{\mathbb {Z}}$ . Then

(2.5) $$ \begin{align} \widehat{\widehat{\alpha}}_z(y)=v^*\widehat{\widehat{\alpha}}_z(u)=z v^*u=z y \end{align} $$

for all $z\in {\mathbb {T}}$ . Moreover, given $b\in F'$ and $k\in {\mathbb {Z}}$ with $|k|\leq n$ , we have

$$ \begin{align*} ybu^k&=v^*ubu^k=v^*\widehat{\alpha}(b)u^{k+1}\stackrel{\mathrm{(b)}}{\approx}_{\!\varepsilon_0} v^*\widehat{\alpha}(b)v^*vu^{k+1}\\ &\stackrel{\mathrm{(a)}}{\approx}_{\!\varepsilon_0} v^*\widehat{\alpha}(b)vv^*u^{k+1} \stackrel{\mathrm{(d)}}{\approx}_{\!\varepsilon_0} v^*vbv^*vv^*u^{k+1}\\ &\stackrel{\mathrm{(b)}}{\approx}_{\!2\varepsilon_0} bv^*u^{k+1} \stackrel{\mathrm{(c)}}{\approx}_{\!k\varepsilon_0}bu^kv^*u=bu^ky. \end{align*} $$

It follows from the choice of $F'$ , n and $\varepsilon _0$ that

(2.6) $$ \begin{align} \|yc-cy\|<\frac{\varepsilon}{2} \end{align} $$

for every $c\in F"$ . Set $x=eye$ , which we regard as an element in A. By (2.5), we have $\alpha _z(x)=z x$ for all $z\in {\mathbb {T}}$ , since e is ${\mathbb {T}}$ -invariant. For $a\in F,$ we have

$$\begin{align*}\|xax^*-a\|=\|eye(a\otimes p)ey^*e-(a\otimes p)\|\leq \|y(a\otimes p)y^*-(a\otimes p)\| \stackrel{({2.6})}{<} \frac{\varepsilon}{2},\end{align*}$$

since $a\otimes p$ belongs to $F"$ . Moreover, $\|x^*x-1\|=\|ey^*eye-e\|<\varepsilon _0\leq \delta $ , and similarly $\|xx^*-1\|<\delta $ . By the choice of $\delta $ , there exists a unitary $w\in A$ such that $\|w-x\|<\varepsilon /2$ . It is then straightforward to check that $\|wa-aw\|<\varepsilon $ for all $a\in F$ and $\max _{z\in {\mathbb {T}}}\|\alpha _z(w)-z w\|<\varepsilon /2$ . This shows that $\alpha $ has the Rokhlin property.

(2). Assume that $\beta $ is approximately representable. Let F be a finite subset of $A\rtimes _{\beta }{\mathbb {Z}}$ and let $\varepsilon>0$ . Denote by u the canonical unitary in the crossed product. Since A and u generate $A\rtimes _{\beta }{\mathbb {Z}}$ , one can assume that $F=F'\cup \{u\}$ , where $F'$ is a finite subset of A. Using approximate representability for $\beta $ , find $v\in {\mathcal {U}}(A)$ with:

• • $\|\beta (b)-vbv^*\|<\varepsilon $ for all $b\in \beta (F')$ , and

• • $\|\beta (v)-v\|<\varepsilon $ .

Set $w=v^*u$ , which is a unitary in $A\rtimes _\beta {\mathbb {Z}}$ . For $b\in F',$ we have

$$\begin{align*}wb=v^*ub=v^*\beta(b)u\approx_\varepsilon bv^*u=bw.\end{align*}$$

Moreover,

$$\begin{align*}wu=v^*uu=u\beta^{-1}(v)u\approx_\varepsilon uw.\end{align*}$$

It follows that $\|wa-aw\|<\varepsilon $ for all $a\in F=F'\cup \{u\}$ . On the other hand,

$$\begin{align*}\widehat{\beta}_ z (w)=\widehat{\beta}_ z (v^*u)=v^*( z u)= z w\end{align*}$$

for all $ z\in {\mathbb {T}}$ . Thus, w is the desired unitary, and $\widehat {\beta }$ has the Rokhlin property.

Conversely, assume that $\widehat {\beta }$ has the Rokhlin property. We continue to denote by u the canonical unitary in $A\rtimes _\beta {\mathbb {Z}}$ that implements $\beta $ . Let $F\subseteq A$ be a finite subset and set $F'=F\cup \{u\}\subseteq A\rtimes _\beta {\mathbb {Z}}$ . Use Proposition 2.2 to choose a unitary $w\in {\mathcal {U}}(A\rtimes _\beta {\mathbb {Z}})$ such that:

• • $\widehat {\beta }_ z(w)= z w$ for all $ z\in {\mathbb {T}}$ ;

• • $\|wb-bw\|=0$ for all $b\in F'$ .

Set $v=uw^*\in A\rtimes _\beta {\mathbb {Z}}$ . Then the first condition above implies that $\widehat {\beta }_ z(v)=v$ for all $ z\in {\mathbb {T}}$ , and hence the unitary v belongs to the fixed point algebra $(A\rtimes _\beta {\mathbb {Z}})^{\widehat {\beta }}$ , which equals A by Proposition 7.8.9 in [Reference Pedersen34]. For $a\in F$ , we have

$$\begin{align*}\|vav^*-\beta(a)\|=\|uw^*awu^*-uau^*\|=\|w^*aw-a\|<\varepsilon. \end{align*}$$

Moreover, $\|\beta (v)-v\|=\|u(uw^*)u^*-uw^*\|<\varepsilon $ , since $u\in F'$ . It follows that v satisfies the conditions of Definition 2.6, so $\beta $ is approximately representable.

Proof Since $A^\alpha \otimes {\mathcal {K}}(\ell ^2({\mathbb {Z}}))\cong A\rtimes _\alpha {\mathbb {T}}$ by Corollary 2.5, it is enough to show the statement for $A^\alpha $ . We identify A with $A^\alpha \rtimes _{\check {\alpha }}{\mathbb {Z}}$ , where $\check {\alpha }$ is the predual of $\alpha $ ; see Definition 2.4. Let J be an ideal in $A^\alpha $ . Since $\check {\alpha }$ is approximately inner by part (2) of Proposition 2.8, it follows that $\check {\alpha }(J)=J$ . Hence $I=J\rtimes _{\check {\alpha }}{\mathbb {Z}}$ is canonically an ideal in A satisfying $I\cap A^\alpha =J$ , as desired.

Proof Consider the Pimsner–Voiculescu exact sequence associated with the predual automorphism $\check {\alpha }\in {\mathrm {Aut}}(A^\alpha )$ of $\alpha $ (Definition 2.4):

Since $\check {\alpha }$ is approximately inner by part (2) of Proposition 2.8, we get $K_\ast (\check {\alpha })=1$ and thus $K_\ast (\iota )$ is injective.

We turn to the last part of the statement. Let $x,y\in K_0(A^\alpha )$ , and assume that $K_0(\iota )(x)\leq K_0(\iota )(y)$ . Set $z=y-x$ , so that $K_0(\iota )(z)\geq 0$ in $K_0(A)$ . Find projections $p,q\in \bigcup _{m\in {\mathbb {N}}}M_m(A^\alpha )$ with $z=[p]-[q]$ in $K_0(A^\alpha )$ , and let $e\in \bigcup _{m\in {\mathbb {N}}}M_m(A)$ satisfy $K_0(\iota )(z)=[e]$ in $K_0(A)$ . Then $[p]=[e]+[q]$ in $K_0(A)$ . By increasing the matrix sizes, we may assume that there exist $k,n\in {\mathbb {N}}$ such that:

• • p and q belong to $M_n(A^\alpha )$ ;

• • e belongs to $M_n(A)$ ;

• • e is orthogonal to $q\oplus 1_k$ ;

• • $p\oplus 1_k$ is Murray–von Neumann equivalent to $(e+q)\oplus 1_k$ in $M_{n+k}(A)$ .

Note that $z=[p\oplus 1_k]-[q\oplus 1_k]$ . Without loss of generality, upon replacing p and q with $p\oplus 1_k$ and $q\oplus 1_k$ , respectively, and n with $n+k$ , we may moreover assume that:

• • e is orthogonal to q;

• • p is Murray–von Neumann equivalent to $e+q$ in $M_n(A)$ .

Set $\alpha ^{(n)}=\alpha \otimes {\mathrm {id}}_{M_n}$ , which has the Rokhlin property. Let $s\in M_n(A)$ be a partial isometry satisfying $s^*s=p$ and $ss^*=e+q$ . For $\varepsilon =1/12$ , let $\delta>0$ such that whenever B is a $C^*$ -algebra and $a\in B_{\mathrm {sa}}$ satisfies $\|a^2-a\|<\delta $ , then there exists a projection $r\in B$ with $\|r-a\|<\varepsilon $ . Set $\varepsilon _0=\min \{\varepsilon ,\delta /5\}$ , and let $\sigma \colon M_n(A)\to M_n(A^\alpha )$ be a unital completely positive map as in the conclusion of Theorem 2.11 in [Reference Gardella12] for $\varepsilon _0$ , $F_1=\{p,q,e,s,s^*\}$ and $F_2=\{p,q\}$ . Set $t=\sigma (s)$ and $f=\sigma (e)$ . Then

$$\begin{align*}t^*t\approx_{\varepsilon_0} \sigma(s^*s)=\sigma(p)\approx_{\varepsilon_0} p,\end{align*}$$

so $\|t^*t-p\|<2\varepsilon _0$ . In particular, $(1-q)f\approx _{2\varepsilon _0} f$ and hence $\|qf\|<2\varepsilon _0$ . Similarly, $\|fq\|<2\varepsilon _0$ .

Set $a=(1-q)f(1-q)$ . Then $a=a^*$ and $\|a-f\|\leq 4\varepsilon _0$ . Moreover,

$$\begin{align*}a^2=(1-q)f(1-q)f(1-q)\approx_{2\varepsilon_0}(1-q)f^2(1-q)\approx_{2\varepsilon_0}(1-q)f(1-q)=a,\end{align*}$$

so $\|a^2-a\|\leq 4\varepsilon _0<\delta $ . By the choice of $\delta $ applied to $B=(1-q)M_n(A^\alpha )(1-q)$ , there exists a projection $r\in M_n(A^\alpha )$ with $\|r-a\|<\varepsilon $ and $rq=0$ .

Set $x=(q+r)tp\in M_n(A^\alpha )$ . Then

$$ \begin{align*} x^*x&=pt^*(q+r)tp\approx_\varepsilon pt^*(q+a)tp\\ &\approx_{4\varepsilon_0}pt^*(q+f)tp \approx_{2\varepsilon_0}pt^*tt^*tp\\ &\approx_{2\varepsilon_0}p^4=p, \end{align*} $$

so $\|x^*x-p\|\leq 8\varepsilon _0+\varepsilon <1$ . Similarly, we have

$$ \begin{align*} xx^*&=(q+r)tpt^*(q+r)\approx_{2\varepsilon_0} (q+r)tt^*tt^*(q+r)\\ &\approx_{4\varepsilon_0}(q+r)(q+f)(q+f)(q+r) \approx_{4\varepsilon_0+\varepsilon}(q+r)^4=q+r, \end{align*} $$

so $\|xx^*-(q+r)\|\leq 10\varepsilon _0+\varepsilon <1$ . Since $x=(q+r)xp$ , it follows from Lemma 2.5.3 in [Reference Lin25] that there exists a partial isometry $v\in M_n(A^\alpha )$ such that $v^*v=p$ and $vv^*=q+r$ . It follows that $[p]=[q]+[r]$ in $K_0(A^\alpha )$ , and thus $z=[p]-[q]$ is positive in $K_0(A^\alpha )$ , as desired.

Proof (1). Arguing as in the beginning of the proof of Proposition 3.1, we deduce that the Pimsner–Voiculescu exact sequence associated with $\check {\alpha }$ splits into the short exact sequence

$$\begin{align*}0\to K_\ast(A^\alpha)\to K_\ast(A) \to K_{\ast}(SA^\alpha)\to 0.\end{align*}$$

We claim that the extension is pure. By taking considering the tensor product of A with any unital $C^*$ -algebra which is KK-equivalent to $C_0(\mathbb {R})$ , endowed with the trivial action of $\mathbb {T}$ , it follows that it suffices to prove that $K_0(A^\alpha )$ is a pure subgroup of $K_0(A)$ . Let $x\in K_0(A^\alpha )$ , let $k\in {\mathbb {N}}$ , and let $y\in K_0(A)$ , and suppose that $ky=K_0(\iota )(x)$ . Find projections $p_x,q_x\in \bigcup _{m\in {\mathbb {N}}} M_m(A^\alpha )$ and $p_y,q_y\in \bigcup _{m\in {\mathbb {N}}} M_m(A)$ such that $x=[p_x]-[q_x]$ and $y=[p_y]-[q_y]$ . It follows that $k[p_y]+[q_x]=[p_x]+k[q_y]$ in $K_0(A)$ . Without loss of generality, we may assume that there is $n\in {\mathbb {N}}$ such that:

• • $p_x,q_x$ to $M_n(A)^\alpha $ and $p_y,q_y$ belong to $M_n(A)$ ;

• • $p_x$ is orthogonal to $q_y$ and $p_y$ is orthogonal to $q_x$ ;

• • there exists $s\in M_{nk}(A)$ with

  $$ \begin{align*} s^*s=\left( \begin{array}{cccc} q_x+p_y & & & \\ & p_y & & \\ & & \ddots & \\ & & & p_y \\ \end{array} \right) \ \ \mbox{ and } \ \ ss^*=\left( \begin{array}{cccc} p_x+q_y & & & \\ & q_y & & \\ & & \ddots & \\ & & & q_y \\ \end{array} \right).\end{align*} $$

Note that $\alpha ^{(nk)}\colon {\mathbb {T}}\to {\mathrm {Aut}}(M_{nk}(A))$ has the Rokhlin property. For $\varepsilon =1/12$ , let $\delta>0$ such that whenever B is a $C^*$ -algebra and $a\in B$ is a self-adjoint element satisfying $\|a^2-a\|<\delta $ , then there exists a projection $r\in B$ with $\|r-a\|<\varepsilon $ . Set $\varepsilon _0=\min \{\varepsilon ,\delta /10\}$ , and let $\sigma \colon M_{nk}(A)\to M_{nk}(A^\alpha )$ be a unital completely positive map as in the conclusion of Theorem 2.11 in [Reference Gardella12] for $\varepsilon _0$ , $F_1=\{p_x,q_x,p_y,q_y,s,s^*\}$ and $F_2=\{p_x,q_x\}$ . Then $\|\sigma (p_x)-p_x\|<\varepsilon $ and $\|\sigma (q_x)-q_x\|<\varepsilon $ .

Set $e_y=\sigma (p_y)$ and $f_y=\sigma (q_y)$ , which are self-adjoint contractions in $M_{nk}(A^\alpha )$ satisfying $\|e_y^2-e_y\|<2\varepsilon _0$ and $\|f_y^2-f_y\|<2\varepsilon _0.$ Set $t=\sigma (s)$ . Then

$$\begin{align*}t^*t\approx_{2\varepsilon_0}{\mathrm{diag}}(q_x+e_y,e_y,\ldots,e_y) \ \mbox{ and } \ tt^*\approx_{2\varepsilon_0}{\mathrm{diag}}(p_x+f_y,f_y,\ldots,f_y). \end{align*}$$

Set $a=(1-p_x)f_y(1-p_x)$ and $b=(1-q_x)e_y(1-q_x)$ , which are self-adjoint elements in the corners of $M_{nk}(A^\alpha )$ by $1-p_x$ and $1-q_x$ , respectively. Moreover, $\|a-f_y\|\leq 4\varepsilon _0$ and $\|b-e_y\|\leq 4\varepsilon _0$ . On the other hand,

$$ \begin{align*}a^2&=(1-p_x)f_y(1-p_x)f_y(1-p_x)\\ &\approx_{2\varepsilon_0}(1-p_x)f_y^2(1-p_x)\\ &\approx_{2\varepsilon_0}(1-p_x)f_y(1-p_x)=a,\end{align*} $$

so $\|a^2-a\|<4\varepsilon _0<\delta $ . Similarly, we have $\|b^2-b\|<\delta $ . Using the definition of $\delta $ with $B=(1-p_x)M_{nk}(A^\alpha )(1-p_x)$ , there exits a projection $\widetilde {q}_y \in M_{nk}(A^\alpha )$ satisfying $\|\widetilde {q}_y-a\|<\varepsilon $ and $p_x\widetilde {q}_y=0$ . Similarly, there exits a projection $\widetilde {p}_y \in M_{nk}(A^\alpha )$ satisfying $\|\widetilde {p}_y-b\|<\varepsilon $ and $\widetilde {p}_yq_x=0$ . Set

$$\begin{align*}r={\mathrm{diag}}(p_x+\widetilde{q}_y,\widetilde{q}_y,\ldots,\widetilde{q}_y) t {\mathrm{diag}}(q_x+\widetilde{p}_y,\widetilde{p}_y,\ldots,\widetilde{p}_y),\end{align*}$$

which belongs to $M_{nk}(A^\alpha )$ . One checks that $\|r^*r-{\mathrm {diag}}(q_x+\widetilde {p}_y,\widetilde {p}_y,\ldots ,\widetilde {p}_y)\|<1$ , and that $\|r^*r-{\mathrm {diag}}(p_x+\widetilde {q}_y,\widetilde {q}_y,\ldots ,\widetilde {q}_y)\|<1$ . By Lemma 2.5.3 in [Reference Lin25], there exists a partial isometry $w\in M_{nk}(A^\alpha )$ such that

$$\begin{align*}w^*w=\left( \begin{array}{cccc} q_x+\widetilde{p}_y & & & \\ & \widetilde{p}_y & & \\ & & \ddots & \\ & & & \widetilde{p}_y \\ \end{array} \right) \ \mbox{ and } \ ww^*= \left( \begin{array}{cccc} p_y+\widetilde{q}_y & & & \\ & \widetilde{q}_y & & \\ & & \ddots & \\ & & & \widetilde{q}_y \\ \end{array} \right).\end{align*}$$

It follows that $[p_x]+k[\widetilde {q}_y]=k[\widetilde {p}_y]+[p_y]$ in $K_0(A^\alpha )$ . With $z=[\widetilde {q}_x]-[\widetilde {q}_y]\in K_0(A^\alpha )$ , we have $kz=x$ , as desired.

(2). The displayed isomorphisms follow from Proposition 5.4 in [Reference Schochet40]. Since parts of the proposition are left as an exercise, and since we need to show that the isomorphism is compatible with the classes of the unit, we give a brief argument here for the existence of an isomorphism $K_0(A)\cong K_0(A^\alpha )\oplus K_1(A^\alpha )$ sending $[1_A]$ to $([1_{A^\alpha }],0)$ . (The argument for $K_1(A)$ is identical because the assumptions are symmetric.) Abbreviate $K_j(A)$ to $K_j$ , and $K_j(A^\alpha )$ to $K_j^\alpha $ , for $j=0,1$ . It is a standard result in group theory, usually attributed to Kulikov, that subgroups of direct sums of cyclic groups are again direct sums of cyclic groups; see Theorem 18.1 in [Reference Fuchs6]. Since $K_1$ is a direct sum of cyclic groups, we deduce that the same is true for $K_1^\alpha $ .

By part (1) of this theorem, there is a canonical quotient map $\pi \colon K_0\to K_1^\alpha $ whose kernel is $K_0^\alpha $ . Choose a presentation $K_1^\alpha \cong \bigoplus _{s\in S}C_s$ , where each $C_s$ is a cyclic group with generator $x_s$ . In particular, $\{x_s\colon s\in S\}$ generates $K_1^\alpha $ . Let $s\in S$ . If $x_s$ has infinite order in $K_1^\alpha $ , we let $y_s\in K_0$ be any group element (necessarily of infinite order) satisfying $\pi (y_s)=x_s$ . If $x_s$ has order $n<\infty $ , let $z_s\in K_0$ be any lift of $x_s$ , and note that $nz_s$ belongs to $K_0^\alpha $ , which is a pure subgroup of $K_0$ . Hence, there exists $k_s\in K_0^\alpha $ with $nk_s=nz_s$ , and we set $y_s=z_s-k_s$ , which also lifts $x_s$ .

Let L be the subgroup of $K_0$ generated by $\{y_s\colon s\in S\}$ , which is mapped isomorphically onto $K_1^\alpha $ via $\pi $ . In particular, $K_0^\alpha \cap L=\{0\}$ and $K_0^\alpha +L=K_0$ . (Equivalently, L defines a splitting for the quotient map $\pi $ .) We deduce that the extension $0\to K_0^\alpha \to K_0\to K_1^\alpha \to 0$ splits, and thus $K_0\cong K_0^\alpha \oplus K_1^\alpha $ . The isomorphism can be clearly chosen to send $[1_A]\in K_0$ to $[1_{A^\alpha }]\in K_0^\alpha $ .

(3). Assume that $K_1(A)$ is finitely generated (and in particular a direct sum of cyclic groups). Hence, $K_0(A^\alpha )$ and $K_1(A^\alpha )$ are both finitely generated, being a quotient and a subgroup of $K_1(A)$ , respectively. The argument given in the proof of part (2) above shows that there is an isomorphism $K_0(A)\cong K_0(A^\alpha )\oplus K_1(A^\alpha )$ . Thus $K_0(A)$ is also finitely generated, and repeating the same argument again, exchanging the roles of $K_0$ and $K_1$ , shows that $K_1(A)\cong K_0(A^\alpha )\oplus K_1(A^\alpha )$ .

Proof (1). If A is simple, then $A^\alpha $ is simple by Corollary 2.5. We show that $\check {\alpha }$ is aperiodic. Arguing by contradiction, suppose that there exist $n\geq 1$ and a unitary $v\in A^\alpha $ such that $\check {\alpha }^n={\mathrm {Ad}}(v)$ . Set

$$\begin{align*}w=v\check{\alpha}(v)\dots\check{\alpha}^{n-1}(v).\end{align*}$$

Then $\check {\alpha }^{n^2}={\mathrm {Ad}}(w)$ . Using that $\check {\alpha }^n(v)=v$ at the second step, that $vx=\check {\alpha }^n(x)v$ for all $x\in A^\alpha $ at the third, and that $\check {\alpha }^{-n}(v)=v$ at he fifth, we get

$$ \begin{align*} \check{\alpha}(w)&= \check{\alpha}(v)\check{\alpha}^2(v)\dots\check{\alpha}^{n-1}(v) \check{\alpha}^{n}(v)= \check{\alpha}(v)\check{\alpha}^2(v)\dots\check{\alpha}^{n-1}(v) v\\ &=v\check{\alpha}^{-n}\big(\check{\alpha}(v)\check{\alpha}^2(v)\dots\check{\alpha}^{n-1}(v) \big)= v\check{\alpha}\big(\check{\alpha}^{-n}(v)\big)\check{\alpha}^2\big(\check{\alpha}^{-n}(v)\big)\dots\check{\alpha}^{n-1}(\check{\alpha}^{-n}(v)\big)\\ &=v\check{\alpha}(v)\dots\check{\alpha}^{n-1}(v)=w. \end{align*} $$

It follows that w is $\check {\alpha }$ -invariant. With u denoting the canonical unitary in $A^\alpha \rtimes _{\check {\alpha }}{\mathbb {Z}}$ that implements $\check {\alpha }$ , we therefore have $uwu^*=w$ . Set $z=u^{n^2}w^*$ . It is clear that z commutes with u, and for $a\in A^\alpha $ we have

$$ \begin{align*}zaz^*=u^{n^2}w^*aw\big(u^{n^2}\big)^*=u^{n^2}\check{\alpha}^{-n^2}(a)\big(u^{n^2}\big)^*=a,\end{align*} $$

so z belongs to the center of $A^\alpha \rtimes _{\check {\alpha }}{\mathbb {Z}}\cong A$ . Since A is simple, its center is trivial and thus there is $\lambda \in {\mathbb {C}}$ with $u^{n^2}=\lambda w$ . In particular, $u^{n^2}$ belongs to $A^\alpha $ , which is a contradiction. This shows that $\check {\alpha }$ is aperiodic.

Conversely, if $\check {\alpha }$ is aperiodic and $A^\alpha $ is simple, it follows from Theorem 3.1 in [Reference Kishimoto21] that the crossed product $A^\alpha \rtimes _{\check {\alpha }}{\mathbb {Z}}\cong A$ is simple.

(2). Assume that A is purely infinite simple. Then $\check {\alpha }$ is aperiodic by part (1). Let $a\in A^\alpha $ be nonzero, and let $\varepsilon>0$ small enough so that $\varepsilon ^3+3\varepsilon <1$ . Without loss of generality, we assume that $\|a\|=1$ . Find $x,y\in A$ such that $xay=1$ . By Lemma 4.1.7 in [Reference Rørdam and Størmer37], we may assume that $\|x\|<1+\varepsilon $ and $\|y\|<1+\varepsilon $ . Let $\sigma \colon A\to A^\alpha $ a completely positive unital map as in the conclusion of Theorem 2.11 in [Reference Gardella12] for $\varepsilon>0$ , $F_2=\{x,y,xa,a\}$ and $F_1=\{a\}$ . Then

$$ \begin{align*} \sigma(x)a\sigma(y)&\approx_{(1+\varepsilon)^2\varepsilon}\sigma(x)\sigma(a)\sigma(y) \approx_{\varepsilon} \sigma(xa)\sigma(y) \approx_{\varepsilon} \sigma(xay)=1. \end{align*} $$

Hence $\|\sigma (x)a\sigma (y)-1\|\leq \varepsilon ^3+3\varepsilon <1$ . It follows that $\sigma (x)a\sigma (y)$ is invertible. With $z\in A^\alpha $ denoting its inverse, we have $\sigma (x)a\sigma (y)z=1$ , as desired.

Conversely, assume that $\check {\alpha }$ is aperiodic and $A^\alpha $ is purely infinite simple. Then $A\cong A^\alpha \rtimes _{\check {\alpha }}{\mathbb {Z}}$ is purely infinite simple by Corollary 4.6 in [Reference Jeong and Osaka18].

(3). This follows from (1) and (2), since A is nuclear (respectively, separable) if and only if so is $A^\alpha $ .

(4). If A satisfies the UCT, then so does $A^\alpha $ by Theorem 3.13 in [Reference Gardella10]. The converse follows from the fact that the UCT is preserved by ${\mathbb {Z}}$ -crossed products.

Proof Assume that $\alpha $ and $\beta $ are unitally $KK^{\mathbb {T}}$ -equivalent, and fix a unital $KK^{\mathbb {T}}$ -equivalence $\xi \in KK^{\mathbb {T}}\big ((A,\alpha ),(B,\beta )\big )$ . Denote by $\xi \rtimes {\mathbb {T}}$ the $KK^{\mathbb {Z}}$ -equivalence between $(A\rtimes _\alpha {\mathbb {T}},\widehat {\alpha })$ and $(B\rtimes _\beta {\mathbb {T}},\widehat {\beta })$ that $\xi $ induces. Combining Takai duality with Theorem 2.3, it follows that $\xi \rtimes {\mathbb {T}}$ induces a $KK^{\mathbb {Z}}$ -equivalence $\eta $ between $(A^\alpha ,\check {\alpha })$ and $(B^\beta ,\check {\beta })$ . Since $\xi $ is unital, we can choose $\eta $ to be unital as well.

Since $A^\alpha $ and $B^\beta $ are Kirchberg algebras by part (3) of Proposition 4.2, it follows from Theorem 4.2.1 in [Reference Phillips35] that there exists an isomorphism $\phi \colon A^\alpha \to B^\beta $ such that $KK(\phi )=\eta $ . Since $\eta $ is equivariant, it follows that $\phi \circ \check {\alpha }\circ \phi ^{-1}$ and $\check {\beta }$ determine the same class in $KK(B^\beta ,B^\beta )$ . Since A and B are simple, it follows from part (1) of Proposition 4.2 that $\check {\alpha }$ and $\check {\beta }$ are aperiodic. Thus, by the equivalence between (1’) and (3’) in Theorem 9 in [Reference Nakamura30], $\phi \circ \check {\alpha }\circ \phi ^{-1}$ and $\check {\beta }$ , and thus $\check {\alpha }$ and $\check {\beta }$ , are cocycle conjugateFootnote ² . It follows that their dual actions are conjugate, and hence $(A,\alpha )\cong (B,\beta )$ as desired.

We turn to the last part of the statement. Fix a unital $KK^{\mathbb {T}}$ -equivalence $\rho \in KK^{\mathbb {T}}((A,\alpha ),(B,\beta ))$ . Arguing as in the first part using Baaj–Skandalis duality, we obtain a unital $KK^{\mathbb {Z}}$ -equivalence

$$\begin{align*}\eta\in KK^{\mathbb{Z}}\big((A^\alpha,\check{\alpha}),(B^\beta,\check{\beta})\big).\end{align*}$$

Denote by $\mathcal {F}(\eta )\in KK(A^\alpha ,B^\beta )$ the $KK$ -equivalence that $\eta $ induces under the forgetful functor $KK^{\mathbb {Z}}\to KK$ . Similarly, we let $\mathcal {F}(\eta \rtimes {\mathbb {Z}})\in KK(A,A)$ denote the $KK$ -equivalence that $\eta \rtimes {\mathbb {Z}}$ induces, which can be canonically identified with $\mathcal {F}(\rho )$ . Using naturality of the functors involved, we deduce that the diagram

commutes. Since all vertical maps are isomorphisms and $\mathcal {F}(\eta )_0([1_{A^\alpha }])=[1_{B^\beta }]$ , we deduce that $({\mathrm {Ext}}_\ast (\alpha ),[1_{A^\alpha }])\cong ({\mathrm {Ext}}_\ast (\beta ),[1_{B^\beta }])$ .

We now prove the converse, so assume that A and B satisfy the UCT. By part (4) of Proposition 4.2, $A^\alpha $ and $B^\beta $ also satisfy the UCT. Since $\check {\alpha }$ and $\check {\beta }$ are approximately inner, an isomorphism $({\mathrm {Ext}}_\ast (\alpha ),[1_{A^\alpha }])\cong ({\mathrm {Ext}}_\ast (\beta ),[1_{B^\beta }])$ is equivalent to $\check {\alpha }$ and $\check {\beta }$ having isomorphic Pimsner–Voiculescu 6-term exact sequences (in a unit-preserving way); see the comments before Definition 4.5. It thus follows from Theorem 2.12 in [Reference Meyer28] that $(A^\alpha ,\check {\alpha })$ is unitally $KK^{\mathbb {Z}}$ -equivalent to $(B^\beta ,\check {\beta })$ . Again by Baaj–Skandalis duality, it follows that $(A,\alpha )$ is unitally $KK^{\mathbb {T}}$ -equivalent to $(B,\beta )$ . This finishes the proof.

Proof That (a) implies (b) follows from part (1) of Theorem 3.3. Assume that (b) holds, and write $\mathcal {E}_0$ and $\mathcal {E}_1$ explicitly as

For $j=0,1$ , let $(K_j^{(n)})_{n\in {\mathbb {N}}}$ be an increasing sequence of finitely generated subgroups of $K_j$ whose union equals $K_j$ . Without loss of generality, we assume that $k_0$ belongs to $K_0^{(n)}$ for all $n\in {\mathbb {N}}$ . Fix $j=0,1$ . Since $\mathcal {E}_j$ is pure, for every $n\in {\mathbb {N}}$ there exists a (necessarily finitely generated) subgroup $\widetilde {K}_{1-j}^{(n)}$ of $G_j$ such that $\pi _j$ restricts to an isomorphism $\widetilde {K}_{1-j}^{(n)}\cong K_{1-j}^{(n)}$ .

Let $G_j^{(n)}$ be the subgroup of $G_j$ generated by $\iota _j(K_j^{(n)})$ and $\widetilde {K}_{1-j}^{(n)}$ . We denote by $\mathcal {E}_j^{(n)}$ the restricted short exact sequence

where the maps are the restrictions of the ones for $\mathcal {E}_j$ . Fix $n\in {\mathbb {N}}$ . Since $\mathcal {E}_j$ is pure and $G^{(n)}_j$ is finitely generated, it follows that $\mathcal {E}^{(n)}_j$ is isomorphic to the trivial extension. Thus there are isomorphisms

$$\begin{align*}G_0^{(n)}\cong K_0^{(n)}\oplus K_1^{(n)}\cong G_1^{(n)}.\end{align*}$$

Under the above identifications, let

$$\begin{align*}h_j^{(n)}\colon G_j^{(n)}\cong K_0^{(n)}\oplus K_1^{(n)}\hookrightarrow K_0^{(n+1)}\oplus K_1^{(n+1)}\cong G_j^{(n+1)}\end{align*}$$

be the induced injective map.

Given $n\in {\mathbb {N}}$ , let $\widetilde {B}_{n}$ be a UCT Kirchberg algebra satisfying

$$\begin{align*}\big(K_0(\widetilde{B}_n),K_1(\widetilde{B}_n),[1_B]\big)=\big(K_0^{(n)}, K_1^{(n)}, k_0\big).\end{align*}$$

Fix an aperiodic approximately representable automorphism $\Phi $ of ${\mathcal {O}_{\infty }}$ .Footnote ³ Set $B_n=\widetilde {B}_n\otimes {\mathcal {O}_{\infty }}$ and $\varphi _n={\mathrm {id}}_{\widetilde {B}_n}\otimes \Phi \in {\mathrm {Aut}}(B_n)$ . Then $\varphi _n$ is approximately representable and aperiodic. Set $A_n=B_n\rtimes _{\varphi _n}{\mathbb {Z}}$ and let $\alpha ^{(n)}\colon {\mathbb {T}}\to {\mathrm {Aut}}(A_n)$ denote the dual action of $\varphi _n$ . Then $A_n$ is a Kirchberg algebra satisfying the UCT by part (4) of Proposition 4.2, and $\alpha ^{(n)}$ has the Rokhlin property by Proposition 2.8. Since the K-theory of $A_n$ is finitely generated, it follows from part (3) of Theorem 3.3 that ${\mathrm {Ext}}_\ast (\alpha ^{(n)})$ is isomorphic to the trivial extension, and hence $({\mathrm {Ext}}_\ast (\alpha ^{(n)}),k_0)\cong (\mathcal {E}_0^{(n)},\mathcal {E}_1^{(n)},k_0)$ .

Use Theorem 4.1.1 in [Reference Phillips35] to find a unital homomorphism $\widetilde {\theta }_n\colon \widetilde {B}_n \to \widetilde {B}_{n+1}$ inducing the canonical inclusion

$$\begin{align*}(K_0^{(n)}, K_1^{(n)}, k_0) \hookrightarrow \big(K_0^{(n+1)}, K_1^{(n+1)}, k_0\big) \end{align*}$$

at the level of K-theory. Set $\theta _n=\widetilde {\theta }_n\otimes {\mathrm {id}}_{{\mathcal {O}_{\infty }}}\colon B_n \to B_{n+1}$ , and note that $\theta _n$ is equivariant with respect to the automorphisms $\varphi _n$ and $\varphi _{n+1}$ . Let $\rho _n\colon (A_n,\alpha ^{(n)})\to (A_{n+1},\alpha ^{(n+1)})$ be the unital equivariant homomorphism induced by $\theta _n$ using duality.

Finally, we set $(A,\alpha )=\varinjlim (A_n,\rho _n,\alpha ^{(n)})$ . Then A is a Kirchberg algebra satisfying the UCT, and $\alpha $ has the Rokhlin property by part (4) of Theorem 2.5 in [Reference Gardella12]. Since $K_j(\rho _n)$ is identified with $h_j^{(n)}\colon G_j^{(n)}\to G_j^{(n+1)}$ , it follows that $\rho _n$ induces an embedding $\mathcal {E}_j^{(n)}\hookrightarrow \mathcal {E}_j^{(n+1)}$ . Since the direct limit of $\big (\mathcal {E}_j^{(n)},\rho _n\big )_{n\in {\mathbb {N}}}$ is isomorphic to $\mathcal {E}_j$ , we conclude that $({\mathrm {Ext}}_\ast (\alpha ),[1_{A^\alpha }])=(\mathcal {E}_0,\mathcal {E}_1,k_0)$ , as desired.

Notation 4.10 For every $n\geq 1$ , let $u_n\in M_n\cong {\mathcal {B}}(\ell ^2(\{0,\ldots ,n-1\}))$ be the unitary given by $u_n(\delta _j)=\delta _{j+1}$ for $j=0,\ldots ,n-1$ , where the subscripts are taken modulo n. (In particular, $u_1=1\in {\mathbb {C}}$ .) Let $\rho _n\colon M_n\to {\mathcal {O}_2}$ be a unital homomorphism. Fix an isomorphism $\kappa \colon \bigotimes _{n=1}^\infty {\mathcal {O}_2}\to {\mathcal {O}_2}$ , and let $\Psi \in {\mathrm {Aut}}({\mathcal {O}_2})$ be the automorphism satisfying $\Psi \circ \kappa =\kappa \circ \bigotimes _{n=1}^\infty {\mathrm {Ad}}(\rho _n(u_n))$ . One can check with elementary methods that $\Psi $ is aperiodic.

Proof Fix a unital embedding $\theta \colon B\to {\mathcal {O}_2}$ . Denote by $\pi \colon \ell ^\infty ({\mathcal {O}_2})\to ({\mathcal {O}_2})_\infty $ the canonical quotient map. Let $(u_n)_{n\in {\mathbb {N}}}$ be any sequence of unitaries in B satisfying $\varphi (b)=\lim \limits _{n\to \infty } {\mathrm {Ad}}(u_n)(b)$ for all $b\in B$ , and set $z=\pi \big ((\theta (u_n))_{n\in {\mathbb {N}}}\big )\in ({\mathcal {O}_2})_\infty $ . Then the unitary z satisfies $\varphi (b)=z\theta (b)z^*$ for all $b\in B$ . By the universal property of the crossed product, there exists a unital embedding $\Theta \colon B\rtimes _{\varphi }{\mathbb {Z}}\to ({\mathcal {O}_2})_\infty $ .

We will show that $\Theta $ lifts to a unital completely positive map $B\rtimes _\varphi {\mathbb {Z}}\to \ell ^\infty ({\mathcal {O}_2})$ which agrees with $\theta $ on the canonical copy of B. Denote by $u\in B\rtimes _{\varphi }{\mathbb {Z}}$ the canonical unitary implementing $\varphi $ For $n\in {\mathbb {N}}$ , let $\phi _n\colon B\rtimes _\varphi {\mathbb {Z}}\to M_n(B)$ and $\psi _n\colon M_n(B)\to B\rtimes _\varphi {\mathbb {Z}}$ be the unital completely positive maps given as follows:

$$\begin{align*}\phi_n(bu^j)= \begin{cases} \sum\limits_{k=j}^{n-1}\varphi^{-k}(b)e_{k,k-j}, & \text{ if }0\leq j<n \\ \sum\limits_{k=0}^{j+n-1}\varphi^{-k}(b)e_{k,k-j}, & \text{ if } -n< j\leq 0\\ 0, & otherwise, \end{cases} \ \text{ and } \ \ \psi_n(b\otimes e_{j,k})=\frac{1}{n}\varphi^k(a)u^{k-j}. \end{align*}$$

One readily checks that $\psi _n(\phi _n(bv^j))=\frac {n-j}{n}bv^j$ if $|j|< n$ , and $0$ otherwise. In particular, $\psi _n\circ \phi _n$ converges pointwise in norm to ${\mathrm {id}}_{B\rtimes _\varphi {\mathbb {Z}}}$ , and thus $\Theta \circ \psi _n\circ \phi _n$ converges pointwise in norm to $\Theta $ . Since $B\rtimes _\varphi {\mathbb {Z}}$ is separable, it follows from Theorem 6 in [Reference Arveson2] that the set of maps $B\rtimes _\varphi {\mathbb {Z}}\to ({\mathcal {O}_2})_\infty $ which have unital completely positive lifts into $\ell ^\infty ({\mathcal {O}_2})$ is closed in the point-norm topology. In particular, it suffices to show that for every $n\in {\mathbb {N}}$ , the map $\Theta \circ \psi _n\circ \phi _n$ has a unital completely positive lift which extends  $\theta $ .

Using nuclearity of $M_n$ , let $\widetilde {\rho }_n\colon M_n\to \ell ^\infty ({\mathcal {O}_2})$ be any unital completely positive lift of the composition

Let $\rho _n\colon M_n(B) \to \ell ^\infty ({\mathcal {O}_2})$ be the linear map determined by

$$\begin{align*}\rho_n(b\otimes e_{j,k})=\widetilde{\rho}_n(e_{j,j})^{1/2}\theta(b)\widetilde{\rho}_n(e_{k,k})^{1/2}\end{align*}$$

for all $b\in B$ and $j,k=0,\ldots ,n-1$ . Then $\rho _n$ is unital. We claim that $\rho _n$ is positive. Since a positive element in $M_n(B)$ is the sum of n elements of the form $\sum _{j,k=0}^{n-1} b_j^*b_k\otimes e_{j,k}$ for some $b_0,\ldots ,b_{n-1}\in B$ , it suffices to show that $\rho _n$ preserves positivity of such elements. Given $b_0,\ldots ,b_{n-1}\in B$ , set

$$\begin{align*}b=\sum_{j,k=0}^{n-1} b_j^*b_k\otimes e_{j,k}\in M_n(B) \ \ \mbox{ and } \ \ x=\sum_{j=0}^{n-1}\theta(b_j)\widetilde{\rho}_n(e_{j,j})^{1/2}\in \ell^\infty({\mathcal{O}_2}).\end{align*}$$

Then $\rho _n(b)=x^*x\geq 0$ , as desired. It follows that $\rho _n$ is positive, and one shows in a similar way that it is completely positive.

Note that $\pi \circ \rho _n=\Theta \circ \psi _n$ , and this $\rho _n\circ \phi _n$ is a lift for $\Theta \circ \psi _n\circ \phi _n$ which agrees with $\theta $ on the canonical copy of B. Thus there exists a unital completely positive lift for $\Theta $ which extends $\theta $ .

Using Lemma 2.2 in [Reference Kirchberg and Phillips20], find a unital embedding $\sigma \colon B\rtimes _{\varphi }{\mathbb {Z}}\to {\mathcal {O}_2}$ satisfying $\sigma (b)=\theta (b)$ for all $b\in B$ , and set $w=\sigma (u)$ . Consider the following embeddings, where the first one is the restriction of $\sigma $ to B:

$$\begin{align*}(B,\varphi)\hookrightarrow ({\mathcal{O}_2},{\mathrm{Ad}}(w))\hookrightarrow ({\mathcal{O}_2}\otimes{\mathcal{O}_2},{\mathrm{Ad}}(w)\otimes\Psi).\end{align*}$$

Note that ${\mathrm {Ad}}(w)$ is unitarily equivalent to ${\mathrm {id}}_{{\mathcal {O}_2}}$ , and that ${\mathrm {id}}_{{\mathcal {O}_2}}\otimes \Psi $ is conjugate to $\Psi $ . It follows that ${\mathrm {Ad}}(w)\otimes \Psi $ is unitarily equivalent to ${\mathrm {id}}_{\mathcal {O}_2}\otimes \Psi $ of ${\mathcal {O}_2}\cong {\mathcal {O}_2}\otimes {\mathcal {O}_2}$ . Since ${\mathrm {id}}_{\mathcal {O}_2}\otimes \Psi $ is conjugate to $\Psi $ , and thus there is a unital, equivariant embedding $(B,\varphi )\to ({\mathcal {O}_2},{\mathrm {Ad}}(v)\circ \Psi )$ for some unitary $v\in {\mathcal {O}_2}$ .

Since $({\mathcal {O}_2},{\mathrm {id}}_{{\mathcal {O}_2}})$ embeds unitally into $({\mathcal {O}_2}, \Psi )$ , it follows that is also embeds unitally into $({\mathcal {O}_2}\otimes {\mathcal {O}_2},{\mathrm {Ad}}(w)\otimes \Psi )$ . Thus the last part of the statement also follows from the construction above.

Proof Note that $A^\alpha $ is unital, separable and exact, and that the predual automorphism $\check {\alpha }\in {\mathrm {Aut}}(A^\alpha )$ is approximately inner by Proposition 2.8. Use Proposition 4.11 to find a unitary $v\in {\mathcal {O}_2}$ and a unital embedding

(4.7) $$ \begin{align} \iota\colon (A^\alpha,\check{\alpha})\to ({\mathcal{O}_2},{\mathrm{Ad}}(v)\circ\Psi). \end{align} $$

Set $\Phi ={\mathrm {Ad}}(u)\circ \Psi $ . Since $\Psi $ is aperiodic, the same is true for $\Phi $ . It follows from parts (3) and (4) of Proposition 4.2 that ${\mathcal {O}_2}\rtimes _\Phi {\mathbb {Z}}$ is a Kirchberg algebra satisfying the UCT, and it has trivial K-theory by the Pimsner–Voiculescu exact sequence. Thus ${\mathcal {O}_2}\rtimes _\Phi {\mathbb {Z}}$ is isomorphic to ${\mathcal {O}_2}$ . Moreover, since $\Psi $ is approximately representable by construction, the same is true for $\Phi $ . By Proposition 2.8, it follows that the dual action $\widehat {\Phi }$ has the Rokhlin property, and thus $\widehat {\Phi }$ is conjugate to $\gamma $ by Corollary 4.9.

Applying crossed products to (4.7), and identifying $(A^\alpha \rtimes _{\check {\alpha }}{\mathbb {Z}},\widehat {\check {\alpha }})$ with $(A,\alpha )$ (see Theorem 2.3), and identifying $({\mathcal {O}_2}\rtimes _\Phi {\mathbb {Z}},\widehat {\Phi })$ with $({\mathcal {O}_2},\gamma )$ as in the paragraph above, we obtain a unital homomorphism $\widetilde {\iota }\colon (A,\alpha )\to ({\mathcal {O}_2},\gamma )$ . Finally, $\widetilde {\iota }$ is injective because so is $\iota $ and ${\mathbb {Z}}$ is amenable.

Proof Let $v\in {\mathcal {O}_2}$ be a unitary as in the conclusion of Corollary 4.12, and set $\Phi ={\mathrm {Ad}}(v)\circ \Psi $ . Fix unital embeddings

$$\begin{align*}\tau\colon ({\mathcal{O}_2},{\mathrm{id}}_{\mathcal{O}_2})\hookrightarrow ({\mathcal{O}_2},\Phi) \ \ \mbox{ and } \ \ \rho\colon (A^\alpha,\check{\alpha})\to ({\mathcal{O}_2},\Phi).\end{align*}$$

We prove the theorem in two steps. First, assume that there is a unital equivariant embedding $\sigma \colon ({\mathcal {O}_2},\Phi )\to (A^\alpha ,\check {\alpha })$ . Let $s_1,s_2\in \mathcal {O}_2$ be the canonical generating isometries, and note that $\tau (s_1)$ and $\tau (s_2)$ are $\Phi $ -invariant. Set $t_j=\sigma (\tau (s_j))$ for $j=1,2$ . Then $t_j$ is an isometry, $\check {\alpha }(t_j)=t_j$ , and $t_1t_1^*+t_2t_2^*=1$ . Define a unital equivariant map $\pi \colon (A^\alpha ,\check {\alpha })\to (A^\alpha ,\check {\alpha })$ by

$$\begin{align*}\pi(a)=t_1\sigma(\rho(a))t_1^*+t_2at_2^*\end{align*}$$

for all $a\in A^\alpha $ . Then $\pi $ is full. Indeed, if $a\in A^\alpha \setminus \{0\}$ , then $\rho (a)\neq 0$ and thus pure infiniteness of ${\mathcal {O}_2}$ implies that there exist $x,y\in {\mathcal {O}_2}$ with $x\rho (a)y=1_{\mathcal {O}_2}$ . Then

$$\begin{align*}\sigma(x)t_1^*\pi(a)t_1\sigma(y)=\sigma(x\rho(a)y)=1,\end{align*}$$

and thus $\pi (a)$ generates $A^\alpha $ as an ideal, as desired.

Claim 1 We have $KK^{\mathbb {Z}}(\pi )=KK^{\mathbb {Z}}({\mathrm {id}}_{A^\alpha })$ . By Theorem F in the introduction of [Reference Gardella11] (see also Corollary 3.10 there), the action $\gamma \colon {\mathbb {T}}\to {\mathrm {Aut}}({\mathcal {O}_2})$ has the continuous Rokhlin property, and thus it is unitally $KK^{\mathbb {T}}$ -equivalent to $\big (C({\mathbb {T}})\otimes {\mathcal {O}_2},\texttt {Lt}\otimes {\mathrm {id}}_{{\mathcal {O}_2}}\big )$ by Theorem C of [Reference Gardella11]. By Baaj–Skandalis duality, and since the dual of $\gamma $ is $KK^{\mathbb {Z}}$ -equivalent to $\Phi $ , we deduce that $({\mathcal {O}_2},\Phi )$ is $KK^{\mathbb {Z}}$ -equivalent to $({\mathcal {O}_2},{\mathrm {id}}_{{\mathcal {O}_2}})$ . Thus

$$\begin{align*}KK^{\mathbb{Z}}\big((A^\alpha,\check{\alpha}),({\mathcal{O}_2},\Phi)\big)=0.\end{align*}$$

Since $\sigma \circ \rho $ factors through $({\mathcal {O}_2},\Phi )$ , it must be $KK^{\mathbb {Z}}$ -trivial. We conclude that $KK^{\mathbb {Z}}(\pi )=KK^{\mathbb {Z}}({\mathrm {id}}_{A^\alpha })$ .

We deduce from the claim above that $\pi $ is a $KK^{\mathbb {Z}}$ -equivalence, and that $\widehat {\pi }$ is a $KK^{\mathbb {T}}$ -equivalence. Set $(C_0,\theta _0)=\varinjlim \big ((A^\alpha ,\check {\alpha }),\pi \big )$ . Then $C_0$ is simple because $\pi $ is full (see Remark 4.14), and it is nuclear, unital and separable because so is $A^\alpha $ . Denote by $\Pi \colon (A^\alpha , \check {\alpha })\to (C_0,\theta _0)$ the canonical equivariant map into the limit.

Set $(C,\theta )=(C\otimes {\mathcal {O}_{\infty }},\theta _0\otimes {\mathrm {id}}_{{\mathcal {O}_{\infty }}})$ . Then C is a unital Kirchberg algebra, and $(C,\theta )\sim _{KK^{\mathbb {Z}}}(A^\alpha ,\check {\alpha })$ unitally, since $({\mathcal {O}_{\infty }},{\mathrm {id}}_{{\mathcal {O}_{\infty }}})\sim _{KK^{\mathbb {Z}}}({\mathbb {C}},{\mathrm {id}}_{\mathbb {C}})$ unitally.

Set $D=C\rtimes _\theta {\mathbb {Z}}$ and $\delta =\widehat {\theta }$ . Since C is a Kirchberg algebra and $\theta $ is aperiodic, it follows from part (3) of Proposition 4.2 that D is a unital Kirchberg algebra. Moreover, $\delta $ has the Rokhlin property by Proposition 2.8, since $\theta $ is approximately representable by Claim 3. Finally, by Claim 2, the equivariant map $\Pi \otimes {\mathrm {id}}_{{\mathcal {O}_{\infty }}}$ is a unital $KK^{\mathbb {Z}}$ -equivalent $(A^\alpha ,\check {\alpha })\sim _{KK^{\mathbb {Z}}} (C,\theta )$ . By taking crossed products, we obtain a unital $KK^{\mathbb {T}}$ -equivalence $(A,\alpha )\sim _{KK^{\mathbb {T}}}(D,\delta )$ . This proves the theorem in the case that there is a unital equivariant embedding $\sigma \colon ({\mathcal {O}_2},\Phi )\to (A^\alpha ,\check {\alpha })$ .

We turn to the general case. Let $\mathcal {O}_\infty ^{(0)}$ be the unique unital UCT Kirchberg algebra which is $KK$ -equivalent to ${\mathcal {O}_{\infty }}$ and whose class of the unit is zero.