Proof. Let $n \in \mathbb {N}$ . Since $a \precsim (b - \delta )_{+}$ , there exists $w_{n} \in A$ such that $\big \| a - w_{n} (b - \delta )_{+}w_{n}^{*} \big \| < \frac {1}{n}$ . By Lemma 2.1, there exists $d_{n} \in A$ such that

$$\begin{align*}\big( a - \tfrac{1}{n} \big)_{+} = d_{n} w_{n} (b - \delta)_{+} w_{n}^{*}d_{n}^{*}. \end{align*}$$

By [Reference Kirchberg and Rørdam34, Lemma 2.4(i)], there exists $v_{n} \in A$ such that

$$\begin{align*}\big( a - \tfrac{1}{n} \big)_{+} = v_{n} b v_{n}^{*} \qquad {\mbox{and}} \qquad \| v_{n} \| \leq \| \big( a - \tfrac{1}{n} \big)_{+} \|^{1/2} \delta^{- 1/2}. \end{align*}$$

Now $v_{n} b v_{n}^{*} \to a$ and $\| v_{n} \| \leq \| a \|^{1/2} \delta ^{- 1/2}$ .

Proof. For the first statement, we have

$$ \begin{align*} \big\| x b x^{*} - x \big( b - \varepsilon / \| x \|^{2} \big)_{+} x^{*} \big\| & \leq \| x \|^{2} \big\| b - \left( b - \varepsilon / \| x \|^{2} \right)_{+} \big\| \\ & \leq \| x \|^{2} \cdot \left( \frac{\varepsilon}{\| x \|^{2}} \right) = \varepsilon. \end{align*} $$

Let $\rho> 0$ . Using [Reference Kirchberg and Rørdam33, Lemma 2.5(i)] at the first step and Lemma 2.1 at the second step, we have

$$\begin{align*}\big( (x b x^{*} - \varepsilon)_{+} - \rho \big)_{+} = (x b x^{*} - \varepsilon - \rho)_{+} \precsim x \big( b - \varepsilon / \| x \|^{2} \big)_{+} x^{*}. \end{align*}$$

Since $\rho> 0$ is arbitrary, it follows from [Reference Kirchberg and Rørdam33, Proposition 2.6] that

$$\begin{align*}(x b x^{*} - \varepsilon)_{+} \precsim x \big( b - \varepsilon / \| x \|^{2} \big)_{+} x^{*}. \end{align*}$$

Clearly, $x \big ( b - \varepsilon / \| x \|^{2} \big )_{+} x^{*} \precsim \big ( b - \varepsilon / \| x \|^{2} \big )_{+}$ .

If $\| x \| \leq 1$ , then $\varepsilon / \| x \|^{2} \geq \varepsilon $ , and so

$$\begin{align*}(x b x^{*} - \varepsilon)_{+} \precsim x \big( b - \varepsilon / \| x \|^{2} \big)_{+} x^{*} \leq x (b - \varepsilon)_{+} x^{*} \precsim (b - \varepsilon)_{+}. \end{align*}$$

This completes the proof.

Proof. First, note that S is norm bounded, by $\sup ( \{ |\lambda | \colon \lambda \in K \} )$ .

Let $(x_{i})_{i \in I}$ be a net in S which tends to $x \in S$ strictly. We have to show that $f (x_{i}) \to f (x)$ strictly. Since $(x_{i})_{i \in I}$ is bounded and the product is jointly continuous in the strict topology on bounded subsets of $M (A)$ , and since $a \mapsto a^*$ is strictly continuous, the statement holds when $f (\zeta )$ is a polynomial in $\zeta $ and ${\overline {\zeta }}$ . Now let $y \in A$ . We need to show that $f (x_{i}) y \to f (x) y$ and $y f (x_{i}) \to y f (x)$ in norm in A. We do only the first; the second is similar. Let $\varepsilon> 0$ . Choose $\delta> 0$ such that $\delta \| y \| < \frac {\varepsilon }{3}$ . There is a polynomial $g (\zeta )$ in $\zeta $ and ${\overline {\zeta }}$ such that $\sup _{\zeta \in K} | g (\zeta ) - f (\zeta ) | < \delta $ . Since $g (x_{i}) y \to g (x) y$ , there is $i_{0} \in I$ such that $\|g (x_{i}) y - g (x) y\| < \frac {\varepsilon }{3}$ for any $i \in I$ with $i \geq i_{0}$ . Thus, for $i \geq i_{0}$ , we have

$$ \begin{align*} & \| f (x_{i}) y - f (x) y \| \\ & \hspace{3em} {\mbox{}} \leq \| f (x_{i}) y - g (x_{i}) y \| + \| g (x_{i}) y - g (x) y \| + \| g (x) y - f (x) y \| \\ & \hspace{3em} {\mbox{}} < \delta \| y \| + \tfrac{\varepsilon}{3} + \delta \| y \| < \varepsilon. \end{align*} $$

So $f (x_{i}) \to f (x)$ strictly.

Since $M (A)_{+}$ is strictly closed in $M (A)$ , the second part of the statement follows from the first.

Proof. Consider $A \subseteq M (A)$ and note that $x^{1/n} \to 1_{M (A)}$ strictly. So $x^{1/n} b x^{1/n} \to b$ strictly. Thus, by Proposition 2.4, $(x^{1/n} b x^{1/n} - \delta )_{+} \to (b - \delta )_{+}$ strictly. Hence, $x \big ( x^{1/n} b x^{1/n} - \delta \big )_{+} x \to x (b - \delta )_{+} x$ in A, and so there is $n_{0} \in \mathbb {N}$ such that for every $n \geq n_{0}$ ,

$$\begin{align*}\big\| x \big( x^{1/n} b x^{1/n} - \delta \big)_{+} x - x (b - \delta)_{+} x \big\| < \varepsilon. \end{align*}$$

Therefore, using Lemma 2.1 at the first step, we have

$$\begin{align*}\big( x (b - \delta)_{+} x - \varepsilon \big)_{+} \precsim x \big( x^{1/n} b x^{1/n} - \delta \big)_{+} x \precsim \big( x^{1/n} b x^{1/n} - \delta \big)_{+}. \end{align*}$$

This completes the proof.

Proof. The forward implication is obvious because $( 1 - \varphi (1) - \varepsilon )_{+} \precsim 1 - \varphi (1)$ .

Now suppose that we are given F, $\varepsilon $ , a, and n as in the statement. We have to find a c.p.c. order zero map $\psi \colon M_{n} \to A$ such that

1. (3) $1 - \psi (1) \precsim a$ .

2. (4) $\| [\psi (z), b] \| < \varepsilon $ for any $z \in M_{n}$ with $\| z \| \leq 1$ and any $b \in F$ .

Choose $\delta> 0$ such that

$$\begin{align*}\delta < \frac{\varepsilon}{ 1 + 2 \max \big( \{ \| x \| \colon x \in F \} \big)}. \end{align*}$$

By assumption, there is a c.p.c. order zero map $\varphi \colon M_{n} \to A$ such that $(1 - \varphi (1) - \delta )_{+} \precsim a$ and $\| [\varphi (z), b] \| < \delta $ for any $z \in M_{n}$ with $\| z \| = 1$ . Define a continuous function $f \colon [0, 1] \to [0, 1]$ by

$$\begin{align*}f (\lambda) = \begin{cases} \frac{\lambda}{1 - \delta} & \hspace{1em} 0 \leq \lambda \leq 1 - \delta \\ 1 & \hspace{1em} 1 - \delta < \lambda \leq 1. \end{cases} \end{align*}$$

Using functional calculus for c.p.c. order zero maps ([Reference Winter and Zacharias55, Corollary 4.2]), set $\psi = f (\varphi )$ . Thus, $\psi \colon M_{n} \to A$ is a c.p.c. order zero map, and, similar to the proof of [Reference Archey, Buck and Phillips3, Lemma 2.8], we have

(3.1) $$ \begin{align} \| \varphi (z) - \psi (z) \| \leq \delta \| z \| \end{align} $$

for any $z \in M_{n}$ with $\| z \| \leq 1$ , and

(3.2) $$ \begin{align} 1 - \psi (1) = \tfrac{1}{1 - \delta} (1 - \varphi (1) - \delta)_{+} \sim (1 - \varphi (1) - \delta)_{+} \precsim a. \end{align} $$

By (3.2), we get (3). To prove (4), let $z \in M_{n}$ with $\| z \| \leq 1$ and let $x \in F$ . By (3.1), we get

$$ \begin{align*} & \|\psi (z) x - x \psi (z) \| \\ & \hspace{3em} {\mbox{}} \leq \| \psi (z) x - \varphi (z) x \| + \|\varphi (z) x - x \varphi (z) \| + \| x \varphi (z)- x \psi (z) \| \\ & \hspace{3em} {\mbox{}} \leq \delta \| x \| + \delta + \delta \| x \| = \delta (1 + 2 \| x \|) < \varepsilon. \end{align*} $$

This completes the proof.

Proof. The forward implication is obvious. To prove the converse, we use Lemma 3.3. Let $ \varepsilon> 0$ , a finite set $ F $ , $ n \in \mathbb {N} $ , and $ a \in A_{+}\setminus \{0\}$ be given. We choose $ 0 < \varepsilon _{0} < \frac {\varepsilon }{2} $ such that $ \varepsilon _{0} < \frac {\varepsilon }{4(1 + \max \{ \| x \| : x \in F \})} $ . By [Reference Kirchberg and Winter35, Proposition 2.5], there is $ \delta < \frac {\varepsilon }{2}$ such that any c.p.c. $ \delta $ -order zero map is $ \varepsilon _{0} $ -close to a c.p.c. order zero map. By assumption, there is a c.p.c. $\delta $ -order zero map $\psi :\mathbb {M}_{n}\to A$ such that

1. (1) $ (1- \psi (1) -\delta )_{+}\precsim a $ ;

2. (2) $\| [\psi (z),b]\| < \delta $ for any normalized element $z\in \mathbb {M}_{n}$ and any $b\in F$ .

Thus, there exists a c.p.c. order zero $ \varphi : \mathbb {M}_{n} \rightarrow A $ such that $ \| \varphi (z) - \psi (z) \| < \varepsilon _{0} $ for any $ z \in \mathbb {M}_{n} $ with $ \| z \| = 1 $ . Hence,

$$\begin{align*}\| x\varphi(z) - \varphi(z)x \| < \|x\|\varepsilon_{0} + \delta + \|x\|\varepsilon_{0} < \delta + \frac{\varepsilon}{2} < \varepsilon. \end{align*}$$

Also, we have

$$\begin{align*} \| (1 - \varphi (1)) - (1 - \psi (1) - \delta )_{+} \| < \delta + \| \phi (1) - \psi (1) \| < \frac {\varepsilon }{2} + \varepsilon _{0}<\varepsilon. \end{align*}$$

Therefore, by Lemma 2.1, $ ( 1 - \varphi (1) - \varepsilon )_{+} \precsim (1 - \psi (1) - \delta )_{+} \precsim a. $

Proof. Let $F \subseteq A$ be a finite subset, let $\varepsilon> 0$ , and let $n \in \mathbb {N}$ . Let $\delta> 0$ satisfy

$$\begin{align*}\delta < \min \left( 1, \, \frac{\varepsilon}{4 (2 \| y \| + 1)} \right). \end{align*}$$

Since $y \in \overline {A x}$ , there is $c \in A \setminus \{ 0 \}$ such that $\| y - c x \| < \delta $ . Set

$$\begin{align*}\eta = \min \left( \varepsilon, \, \frac{\varepsilon}{2 \|c\|^{2}} \right). \end{align*}$$

By assumption, there is a c.p.c. order zero map $\varphi \colon M_{n} \to A$ such that

1. (3) $\big ( x^{2} - x \varphi (1) x - \eta \big )_{+} \precsim a$ .

2. (4) $\| [\varphi (z), b] \| < \eta $ for any $z \in M_{n}$ with $\| z \| \leq 1$ and any $b \in F$ .

By Lemma 2.3, we have

$$ \begin{align*} \left( c x^{2} c^{*} - c x \varphi (1) x c^{*} - \frac{\varepsilon}{2} \right)_{+} & \precsim \left( x^{2} - x \varphi (1) x - \frac{\varepsilon}{2 \|c\|^{2}} \right)_{+} \\ & \precsim \left( x^{2} - x \varphi (1) x - \eta \right)_{+} \precsim a. \end{align*} $$

Also, we have

$$ \begin{align*} & \big\| y^{2} - y \varphi (1) y - \left( c x^{2} c^{*} - c x \varphi (1) x c^{*} - \tfrac{\varepsilon}{2} \right)_{+} \big\| \\ & \hspace{3em} {\mbox{}} \leq \big\| y^{2} - y \varphi (1) y - c x^{2} c^{*} + c x \varphi (1) x c^{*} \big\| + \frac{\varepsilon}{2} \\ & \hspace{3em} {\mbox{}} \leq \| y^{2} - c x y \| + \| c x y - c x^{2} c^{*} \| + \big\| y \varphi (1) y - y \varphi (1) x c^{*} \big\| \\ & \hspace{3em} {\mbox{}} \hspace{3em} {\mbox{}} + \big\| y \varphi (1) x c^{*} - c x \varphi (1) x c^{*} \big\| + \frac{\varepsilon}{2} \\ & \hspace{3em} {\mbox{}} \leq \delta \| y \| + \delta \|c x\| + \delta \| y \| + \delta \|c x\| + \frac{\varepsilon}{2} & \hspace{3em} {\mbox{}} \\ & \hspace{3em} {\mbox{}} < 2 \delta (2 \| y \| + 1 ) + \frac{\varepsilon}{2} < \varepsilon. \end{align*} $$

Thus,

$$\begin{align*}\big( y^{2} - y \varphi (1) y - \varepsilon \big)_{+} \precsim \big(c x^{2} c^{*} - c x \varphi (1) x c^{*} - \tfrac{\varepsilon}{2} \big)_{+} \precsim a. \end{align*}$$

Since $\eta < \varepsilon $ , for any $b \in F$ , we have $\| [\varphi (z), b] \| < \varepsilon $ .

Proof. Suppose $\dim (A) < \infty $ . Choose $n \in \mathbb {N}$ such that $n^{2}> \dim (A)$ . We claim that there is no nonzero c.p.c. order zero map $\varphi \colon M_{n} \to A$ . Indeed, if we had such a map $\varphi $ , then [Reference Winter and Zacharias55, Theorem 3.3] would provide a homomorphism $\pi \colon M_{n} \to M(C)\cap \{\varphi (1)\}'$ such that $\varphi (z) = \pi (z) \varphi (1)$ for all $z \in M_{n}$ , where C denotes the C*-subalgebra of A generated by the range of $\varphi $ and $M(C)$ denotes the multiplier algebra of C. But $\dim (A) < n^{2}$ , so $\pi = 0$ . Thus, $\varphi = 0$ . This proves the claim.

Now it follows from Definition 3.6 that for every $x, a \in A_{+}$ with $a \neq 0$ and every $\varepsilon> 0$ , we have $(x^2 - \varepsilon )_{+} \precsim a$ . So $x \precsim a$ for every $x, a \in A_{+}$ with $a \neq 0$ . Since ${\mathrm {dim}} (A) < \infty $ , this can only happen if $A \cong \mathbb {C}$ , which is excluded in Definition 3.6, or $A = 0$ .

Proof of Lemma 3.13.

We prove the following statement, which clearly implies the conclusion. Let $a \in A_{+} \setminus \{ 0 \}$ , let $F \subseteq A$ be finite, let $\varepsilon> 0$ , and let $n \in \mathbb {N}$ . Then there is a homomorphism $\varphi \colon M_{n} \to A$ such that

1. (1) $1 - \varphi (1) \precsim a$ .

2. (2) $\| [\varphi (z), x] \| < \varepsilon $ for any $z \in M_{n}$ with $\| z \| \leq 1$ and any $x \in F$ .

Since A has real rank zero (by [Reference Lin38, Theorem 3.6.11]) and is unital, infinite dimensional, and simple, there are nonzero orthogonal projections $q_1, q_2 \in {\overline {a A a}}$ . Apply the definition of tracial rank zero to find a projection $p \in A$ , $m, r (1), r (2), \ldots , r (m) \in \mathbb {N}$ , and a unital injective homomorphism $\psi $ from $B = \bigoplus _{l = 1}^m M_{r (l)}$ to $p A p$ such that

1. (3) $\| [p, x] \| < \frac {\varepsilon }{6}$ for all $x \in F$ .

2. (4) ${\operatorname {dist}} (p x p, \, \psi (B)) < \frac {\varepsilon }{6}$ for all $x \in F$ .

3. (5) $1 - p \precsim q_1$ .

Choose $n_0 \in \mathbb {N}$ such that

$$\begin{align*}\frac{1}{n_0} < \inf_{\tau \in T (A)} \tau (q_2). \end{align*}$$

Choose $\nu \in \mathbb {N}$ so large that $\nu n> n_0$ .

For $l = 1, 2, \ldots , m$ , let $\big ( e_{j, k}^{(l)} \big )_{j, k = 1, 2, \ldots , r(l)}$ be the standard system of matrix units for $M_{r (l)}$ . Use [Reference Osaka and Phillips44, Lemma 2.3] to choose projections $f_0^{(l)}, f_1^{(l)}, f_2^{(l)}, \ldots , f_{\nu n}^{(l)} \in \psi \big ( e_{1, 1}^{(l)} \big ) A \psi \big ( e_{1, 1}^{(l)} \big )$ such that

$$\begin{align*}f_1^{(l)} \sim f_2^{(l)} \sim \cdots \sim f_{\nu n}^{(l)}, \qquad f_0^{(l)} \precsim f_1^{(l)}, \qquad {\mbox{and}} \qquad \sum_{j = 0}^{\nu n} f_j^{(l)} = \psi \big( e_{1, 1}^{(l)} \big). \end{align*}$$

There is then a homomorphism $\beta _l \colon M_{\nu n} \to A$ such that the images of the diagonal matrix units in $M_{\nu n}$ are $f_1^{(l)}, f_2^{(l)}, \ldots , f_{\nu n}^{(l)}$ . Define a homomorphism $\beta \colon M_{\nu n} \to A$ by

$$\begin{align*}\beta (z) = \sum_{l = 1}^m \sum_{k = 1}^{r (l)} \psi \big( e_{k, 1}^{(l)} \big) \beta_l (z) \psi \big( e_{1, k}^{(l)} \big) \end{align*}$$

for $z \in M_{\nu n}$ . Clearly, $\beta (z)$ commutes with $\psi (b)$ for all $z \in M_{\nu n}$ and $b \in B$ , and $\beta (1) \leq p$ .

Let $x \in F$ and let $z \in M_{n}$ satisfy $\| z \| \leq 1$ . Apply (4) to choose $b \in B$ such that $\| \psi (b) - p x p \| < \frac {\varepsilon }{6}$ . We have $[ \beta (z), \, (1 - p) x (1 - p) ] = 0$ , and so by (3) at the third step, we get

$$ \begin{align*} \| [\beta (z), x] \| & \leq \| [\beta (z), \, p x p] \| + \| [\beta (z), \, p x (1 - p)] \| + \| [\beta (z), \, (1 - p) x p] \| \\ & \leq 2 \| \beta (z) \| \cdot \| p x p - \psi (b) \| \\ & \hspace{3em} {\mbox{}} + \| \beta (z) \| \cdot \| p x (1 - p) \| + \| \beta (z) \| \cdot \| (1 - p) x p \| \\ & < 2 \left( \frac{\varepsilon}{6} \right) + \left( \frac{\varepsilon}{6} \right) + \left( \frac{\varepsilon}{6} \right) < \varepsilon. \end{align*} $$

This is (2) with $\beta $ in place of $\varphi $ .

For every $\tau \in T (A)$ , for $l = 1, 2, \ldots , m$ , we have

$$\begin{align*}\tau \big( f_1^{(l)} \big) = \tau \big( f_2^{(l)} \big) = \cdots = \tau \big( f_{\nu n}^{(l)} \big), \qquad \tau \big( f_0^{(l)} \big) \leq \tau \big( f_1^{(l)} \big), \end{align*}$$

and

$$\begin{align*}\sum_{j = 0}^{\nu n} \tau \big( f_j^{(l)} \big) = \tau \big( \psi \big( e_{1, 1}^{(l)} \big) \big). \end{align*}$$

Therefore,

$$\begin{align*}\tau \big( f_0^{(l)} \big) \leq \frac{1}{\nu n + 1} \tau (\big( \psi \big( e_{1, 1}^{(l)} \big) \big). \end{align*}$$

Since

$$\begin{align*}p - \beta (1) = \sum_{l = 1}^m \sum_{k = 1}^{r (l)} \psi \big( e_{k, 1}^{(l)}\big) f_0^{(l)} \psi \big( e_{1, k}^{(l)}\big), \end{align*}$$

it follows that

$$ \begin{align*} \tau (p - \beta (1)) & = \sum_{l = 1}^m r (l) \tau \big( f_0^{(l)} \big) \leq \frac{1}{\nu n + 1} \sum_{l = 1}^m r (l) \tau \big( \psi \big( e_{1, 1}^{(l)} \big) \big) \\ & = \frac{\tau (p)}{\nu n + 1} \leq \frac{1}{\nu n + 1} < \frac{1}{n_0} < \tau (q_2). \end{align*} $$

Since A has strict comparison of projections using traces (Remark 3.12), it follows that $p - \beta (1) \precsim q_2$ . We already have $1 - p \precsim q_1$ , $q_1 q_2 = 0$ , and $q_1 + q_2 \in {\overline {a A a}}$ , so $1 - \beta (1) \precsim a$ . This is (1) with $\beta $ in place of $\varphi $ .

To complete the proof, we define $\varphi \colon M_n \to A$ by $\varphi (z) = \beta (1_{M_{\nu }} \otimes z)$ for $z \in M_n$ .

Notation 3.14. If $(A_{i})_{i \in I}$ is a family of C*-algebra, we let $\prod _{i \in I} A_i$ denote that set of all families $(a_{i})_{i \in I}$ in the set theoretic product of the $A_{i}$ such that $\sup _{i \in I} \| a_{i} \| < \infty $ . (This is the product in the category of C*-algebras.)

For a C*-algebra A, we write

$$\begin{align*}A_{\infty} = \prod_{n \in \mathbb{N}} A \Big/ \bigoplus_{n \in \mathbb{N}} A. \end{align*}$$

We identify $a \in A$ with the equivalence class in $A_{\infty }$ of the corresponding constant sequence $(a)_{n \in \mathbb {N}}$ . We write $A_{\infty } \cap A^{\prime }$ for the relative commutant of A in $A_{\infty }$ .

Proof. (1) $\Rightarrow $ (2): Let $\{a_{1}, a_{2}, \ldots \}$ be a dense subset of the closed unit ball of A. Put $F_{m} = \{a_{1}, a_{2}, \ldots , a_{m} \}$ for $m \in \mathbb {N}$ . Let $a, x \in A_{+}$ with $a \neq 0$ and let $n \in \mathbb {N}$ . We may assume that $\| a \| = 1$ . Tracial $\mathcal {Z}$ -absorption of A implies that for every $m \in \mathbb {N}$ , there exists a c.p.c. order zero map $\varphi _{m} \colon M_{n} \to A$ such that

$$\begin{align*}\left( x^{2} - x \varphi_{m} (1) x - \tfrac{1}{m} \right)_{+} \precsim \left(a - \tfrac{1}{2} \right)_{+} \end{align*}$$

and $\| [\varphi _{m} (z), b] \| < \frac {1}{m}$ for every $z \in M_{n}$ with $\| z \| \leq 1$ and every $b \in F_{m}$ . Now let $\pi \colon \prod _{m \in \mathbb {N}} A \to A_{\infty }$ be the quotient map, and define $\psi \colon M_n \to A_{\infty }$ by

$$\begin{align*}\psi (z) = \pi \big( ( \varphi_m (z) )_{m \in \mathbb{N}} \big) \end{align*}$$

for $z \in M_n$ . Then $\psi (M_n) \subseteq A_{\infty } \cap A^{\prime }$ . By Lemma 2.2, for every $m \in \mathbb {N}$ , there exists $v_{m} \in A$ such that $\|v_{m} \| \leq 2 \| x \|^{2}$ and

$$\begin{align*}\big\| \left( x^{2} - x \varphi_{m} (1) x - \tfrac{1}{m} \right)_{+} - v_{m} a v_{m}^{*} \big\| < \tfrac{1}{m}. \end{align*}$$

Hence, $\big \| x^{2} - x \varphi _{m} (1) x - v_{m} a v_{m}^{*} \big \| < \frac {2}{m}$ . Put $v = \pi \big ( ( v_m )_{m \in \mathbb {N}} \big ) \in A_{\infty }$ . Then we have $x^{2} - x \psi (1) x = vav^{*} \precsim a$ in $A_{\infty }$ .

(2) $\Rightarrow $ (1): Suppose that (2) holds. To show that A is tracially $\mathcal {Z}$ -absorbing, we verify the condition in Remark 3.7(3) (Definition 3.6 with order zero replaced by $\varepsilon $ -order zero). So let a, x, $\varepsilon $ , F, and n be as in Definition 3.6. Choose a c.p.c. order zero map $\psi \colon M_{n} \to A_{\infty } \cap A^{\prime }$ such that $x^{2} - x \psi (1) x \precsim a$ in $A_{\infty }$ . Since $M_{n}$ is nuclear, there is a sequence $(\psi _{m})_{m \in \mathbb {N}}$ of c.p.c. maps from $M_{n}$ to A such that, if we define

$$\begin{align*}\widetilde{\psi} = (\psi_m)_{m \in \mathbb{N}} \colon M_{n} \to \prod_{m \in \mathbb{N}} A, \end{align*}$$

then $\psi = \pi \circ \widetilde {\psi }$ . Since F is finite and $M_n$ is finite dimensional, it follows that there is $m \in \mathbb {N}$ such that $\| [\psi _{m} (z), y] \| < \varepsilon $ for every $z \in M_{n}$ with $\| z \| \leq 1$ and every $y \in F$ , such that $\| \psi _{m} (y) \psi _{m} (z) \| < \varepsilon $ for all $y, z \in M_{n}$ with $0 \leq y, z \leq 1$ and $y z = 0$ , and such that there is $w \in A$ with $\big \| x^{2} - x \psi _{m} (1) x - w a w^{*} \big \| < \varepsilon $ . Thus, $\big ( x^{2} - x \psi _{m} (1) x - \varepsilon \big )_{+} \precsim a$ . Hence, $\psi _{m}$ satisfies the conditions of Remark 3.7(3), and so A is tracially $\mathcal {Z}$ -absorbing. Thus, (1) holds.

Suppose that A is unital. The implication (3) $\Rightarrow $ (2) is obvious since $x^{2} - x \psi (1) x \precsim 1_{A_{\infty }} - \psi (1)$ . Also, (2) $\Rightarrow $ (3) follows by taking $x=1_A$ in (2).

Proof. Choose $(b_{n})_{n \in \mathbb {N}}$ in A such that $x b_{n} \to y$ . Then $b_{n}^{*} x c x b_{n} \to y c y$ and for any $n \in \mathbb {N}$ , we have $b_{n}^{*} x c x b_{n} \precsim x cx$ . Taking the limit in the latter gives the result.

Proof. Assume first that A is strongly tracially $\mathcal {Z}$ -absorbing. Let a, F, $\varepsilon $ , and n be as in the hypotheses, and apply Definition 3.16 with these elements and x as given, obtaining a c.p.c. order zero map $\varphi \colon M_{n} \to A$ . This map satisfies (2) by construction. For (1), let $m \in \mathbb {N}$ . Then $x^{1 / m} \in {\overline {x A}}$ , so, using Lemma 3.17 at the second step,

$$\begin{align*}\big( x^{2 / m} - x^{1 / m} \varphi (1) x^{1 / m} - \varepsilon \big)_{+} \precsim x^{2 / m} - x^{1 / m} \varphi (1) x^{1 / m} \precsim x^{2} - x \varphi (1) x. \end{align*}$$

For the converse, let a, F, $\varepsilon $ , and n be as in Definition 3.16. Choose $\delta> 0$ with

$$\begin{align*}\delta < \frac{\varepsilon}{1 + 2 \max \big( \{ \| y \| \colon y \in F \} \big)}. \end{align*}$$

By assumption, there is a c.p.c. order zero map $\varphi \colon M_{n} \to A$ such that (1) and (2) hold for $\delta $ instead of $\varepsilon $ . Define a map $f \colon [0, 1] \to [0, 1]$ by

$$\begin{align*}f (\lambda) = \begin{cases} \tfrac{\lambda}{1 - \delta} & \hspace{1em} 0 \leq \lambda \leq 1 - \delta \\ 1 & \hspace{1em} 1 - \delta < \lambda \leq 1. \end{cases} \end{align*}$$

Using functional calculus for c.p.c. order zero maps ([Reference Winter and Zacharias55, Corollary 4.2]), set $\psi = f (\varphi )$ . As in the proof of Lemma 3.3, in ${A}^{\sim }$ , we have

(3.3) $$ \begin{align} 1 - \psi (1) = \frac{1}{1 - \delta} \big( 1 - \varphi (1) - \delta \big)_{+}. \end{align} $$

We show that (1) in Definition 3.16 holds (i.e., $x^{2} - x \psi (1) x \precsim a$ ). Let $\eta> 0$ . Then by (3.3), we have

$$\begin{align*}\big((1 - \delta) (x^{2} - x \psi (1) x ) - \eta \big)_{+} = \big( x (1 - \varphi (1) - \delta)_{+} x - \eta \big)_{+}. \end{align*}$$

By Lemma 2.5, there is $m \geq 1$ such that

(3.4) $$ \begin{align} \big( x (1 - \varphi (1) - \delta)_{+} x - \eta \big)_{+} \precsim \big( x^{1 / m} (1 - \varphi (1)) x^{1 / m} - \delta \big)_{+}. \end{align} $$

The right-hand side of (3.4) is Cuntz subequivalent to a by (1). Since $\eta> 0$ is arbitrary, and using (3.3) at the first step, we get

$$\begin{align*}x^{2} - x \psi (1) x \sim x (1 - \varphi (1)- \delta)_{+} x \precsim a. \end{align*}$$

The proof of (2) in Definition 3.16 is similar to the last part of the proof of Lemma 3.3. So A is strongly tracially $\mathcal {Z}$ -absorbing.

Proof. Let $x \in B_{+}$ , let $a \in B_{+} \setminus \{ 0 \}$ , let $\varepsilon> 0$ , let $F \subseteq B$ be finite, and let $n \in \mathbb {N}$ . We must find a c.p.c. order zero map $\varphi \colon M_{n} \to B$ such that

1. (1) $\big ( x^{2} - x \varphi (1) x - \varepsilon \big )_{+} \precsim a$ .

2. (2) $\| [\varphi (z), b] \| < \varepsilon $ for any $z \in M_{n}$ with $\| z \| \leq 1$ and any $b \in F$ .

The case $x = 0$ is trivial, so by Lemma 3.5, we may assume that $\| x \| = 1$ . Also, we may assume that $\| b \| \leq 1$ for all $b \in F$ . Let $y_0$ be the sum of x with the positive and negative parts of the real and imaginary parts of all the elements of F, and set $y = \| y_0 \|^{-1} y_0$ . Then $y \in B_{+}$ , $\| y \| = 1$ , and $F \cup \{ x \} \subseteq {\overline {y B y}}$ . Since $\big ( y^{1 / m} \big )_{m \in \mathbb {N}}$ is an approximate identity for ${\overline {y B y}}$ , there is $m \in \mathbb {N}$ such that

(4.1) $$ \begin{align} \big\| [y^{1 / k}, b] \big\| < \tfrac{\varepsilon}{6} \qquad {\mbox{and}} \qquad \big\| y^{1 / k} b - b \big\| < \tfrac{\varepsilon}{6} \end{align} $$

for $k \in \{ m, 2 m, m / 2 \}$ and $b \in F \cup \{ x \}$ . Choose $\delta> 0$ for $\frac {\varepsilon }{6}$ according to [Reference Archey, Buck and Phillips3, Lemma 2.7]. We may assume that $\delta < \frac {\varepsilon }{4}$ . Since A is tracially $\mathcal {Z}$ -absorbing, there exists a c.p.c. order zero map $\psi \colon M_{n} \to A$ such that

1. (3) $\big ( y^{2 / m} - y^{1 / m} \psi (1) y^{1 / m} - \delta \big )_{+} \precsim a$ .

2. (4) $\| [\psi (z) , b] \| < \delta $ for every $z \in M_{n}$ with $\| z \| \leq 1$ and every $b \in F \cup \left \{ x, y^{2 / m}, y^{1 / m} \right \}$ .

Let us show that the conditions of [Reference Archey, Buck and Phillips3, Lemma 2.7] are satisfied for $y^{1 / m}$ , $\psi $ , and B in place of x, $\varphi _0$ , and B. Since $\| y \| \leq 1$ , we have $0 \leq y^{1 / m} \leq 1$ . Also by (4), $\big \| [\psi (z), \ , y^{1 / m}] \big \| < \delta $ for any $z \in M_{n}$ with $\| z \| \leq 1$ . Finally, we claim that ${\operatorname {dist}} \big ( \psi (z) y^{1 / m}, \, B \big ) < \delta $ for any $z \in M_{n}$ with $\| z \| \leq 1$ . In fact, $y^{1 / (2 m)} \psi (z) y^{1 / (2 m)} \in B$ (since B is hereditary), and by (4), we have

$$\begin{align*}\big\| \psi (z) y^{1 / m} - y^{1 / (2 m)} \psi (z) y^{1 / (2 m)} \big\| \leq \big\| \psi (z) y^{1 / (2 m)} - y^{1 / (2 m)} \psi (z) \big\| < \delta, \end{align*}$$

proving the claim. Hence, by the choice of $\delta $ using [Reference Archey, Buck and Phillips3, Lemma 2.7], there is a c.p.c. order zero map $\varphi \colon M_{n} \to B$ such that for every $z \in M_{n}$ with $\| z \| \leq 1$ , we have

(4.2) $$ \begin{align} \big\| \psi (z) y^{1 / m} - \varphi (z) \big\| < \tfrac{\varepsilon}{6}. \end{align} $$

Now we show that $\varphi $ has properties (1) and (2) above. For (2), let $b \in F$ and let $z \in M_{n}$ satisfy $\| z \| \leq 1$ . Using (4.1), (4.2), and (4) at the second step, we have

$$ \begin{align*} \| \varphi (z) b - b \varphi (z) \| & \leq \big\| \varphi (z) - \psi (z) y^{1 / m} \big\|\cdot \| b \| + \| \psi (z) \|\cdot \big\| y^{1 / m} b - b y^{1 / m} \big\| \\ & \hspace{3em} {\mbox{}} + \| \psi (z) b - b \psi (z) \|\cdot \big\| y^{1 / m} \big\| + \| b \|\cdot \big\| \psi (z) y^{1 / m} - \varphi (z) \big\| \\ & < \tfrac{\varepsilon}{6} + \tfrac{\varepsilon}{6} + \delta + \tfrac{\varepsilon}{6} < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{4} < \varepsilon. \end{align*} $$

To prove (1), first we use (4.2) and (4) at the third step to get

$$ \begin{align*} & \big\| \big( y^{2 / m} - y^{1 / m} \psi (1) y^{1 / m} - \delta \big)_{+} - \big( y^{2 / m} - y^{1 / (2 m)} \varphi (1) y^{1 / (2 m)} \big) \big\| \\ & \hspace{3em} {\mbox{}} \leq \delta + \big\| y^{1 / m} \psi (1) y^{1 / m} - y^{1 / (2 m)} \varphi (1) y^{1 / (2 m)} \big\| & \hspace{3em} {\mbox{}} & \hspace{3em} {\mbox{}} \hspace{3em} {\mbox{}} & \hspace{3em} {\mbox{}} \\ & \hspace{3em} {\mbox{}} \leq \delta + \big\| y^{1 / (2 m)} \big\| \cdot \big\| y^{1 / (2 m)} \psi (1) - \psi (1) y^{1 / (2 m)} \big\| \cdot \big\| y^{1 / m} \big\| \\ & \hspace{3em} {\mbox{}} \hspace{3em} {\mbox{}} + \big\| y^{1 / (2 m)} \big\| \cdot \big\| \psi (1) y^{1 / m} - \varphi (1) \big\|\cdot \big\| y^{1 / (2 m)} \big\| \\ & \hspace{3em} {\mbox{}} < \delta + \tfrac{\varepsilon}{6} < \tfrac{\varepsilon}{4} + \tfrac{\varepsilon}{6} < \tfrac{\varepsilon}{2}. \end{align*} $$

Thus, by (3), we have

(4.3) $$ \begin{align} \big( y^{2 / m} - y^{1 / (2 m)} \varphi (1) y^{1 / (2 m)} - \tfrac{\varepsilon}{2} \big)_{+} \precsim \big( y^{2 / m} - y^{1 / m} \psi (1) y^{1 / m} - \delta \big)_{+} \precsim a. \end{align} $$

At the second step using $\big \| x - x y^{1 / (2 m)} \big \| = \big \| x - y^{1 / (2 m)} x \big \|$ and (4.1) several times, we get

$$ \begin{align*} & \big\| \big( x^{2} - x \varphi (1) x \big) - \big( x y^{2 / m} x - x y^{1 / (2 m)} \varphi (1) y^{1 / (2 m)} x - \tfrac{\varepsilon}{2} \big)_{+} \big\| \\ & \hspace{3em} {\mbox{}} \leq \tfrac{\varepsilon}{2} + \| x \| \cdot \big\| x - y^{2 / m} x \big\| + \big\| x - x y^{1 / (2 m)} \big\| \cdot \| \varphi (1) x \| \\ & \hspace{3em} {\mbox{}} \hspace{3em} {\mbox{}} + \big\| x y^{1 / (2 m)} \varphi (1) \big\| \cdot \big\| x - y^{1 / (2 m)} x \big\| \\ & \hspace{3em} {\mbox{}} < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon. \end{align*} $$

Using Lemma 2.3 at the second step and (4.3) at the third step, we now have

$$ \begin{align*} \big( x^{2} - x \varphi (1) x - \varepsilon \big)_{+} & \precsim \big( x y^{2 / m} x - x y^{1 / (2 m)} \varphi (1) y^{1 / (2 m)} x - \tfrac{\varepsilon}{2} \big)_{+} \\ & \precsim \big( y^{2 / m} - y^{1 / (2 m)} \varphi (1) y^{1 / (2 m)} - \tfrac{\varepsilon}{2} \big)_{+} \precsim a. \end{align*} $$

It remains to prove that $B \not \cong \mathbb {C}$ . So suppose $B \cong \mathbb {C}$ . Since B is hereditary and A is simple, it follows that there is a Hilbert space H such that $A \cong K (H)$ . We have $A \not \cong \mathbb {C}$ and $A \neq 0$ , so A has a hereditary subalgebra D isomorphic to $M_2$ . Since $D \not \cong \mathbb {C}$ , by what we have already done, D is tracially $\mathcal {Z}$ -absorbing. This contradicts Lemma 3.9.

Proof. Suppose otherwise that A is type I. Since A is simple, it is elementary, by [Reference Blackadar7, Corollary IV.1.2.6]. Thus, there is a Hilbert space H such that $A\cong {K} (H)$ . In particular, $K(H)$ must be tracially $\mathcal {Z}$ -absorbing. By Lemma 3.9, H is infinite dimensional. Let p a projection in ${K} (H)$ with $\mathrm {rank} (p) = 2$ . Then $p{K} (H) p \cong M_{2}$ would be tracially $\mathcal {Z}$ -absorbing by Theorem 4.1, which is not possible by Lemma 3.9.

Proof. By Theorem 4.1, (1) implies (2). Clearly, (2) is equivalent to (3) since a hereditary C*-subalgebra $B \subseteq A$ is $\sigma $ -unital if and only if there is $x \in A_{+}$ such that $B = \overline {x A x}$ . It remains to show that (3) implies (1). Suppose that (3) holds. Thus, in particular, $A \ncong \mathbb {C}$ . Suppose that we are given $a, x \in A_{+} \setminus \{ 0 \}$ , $\varepsilon> 0$ , a finite set $F \subseteq A$ , and $n \in \mathbb {N}$ . Choose $y \in A_{+}$ such that $F \cup \{ x \} \subseteq \overline {y A y}$ (e.g., let y be the sum of x with the positive and negative parts of the real and imaginary parts of the elements of F). We get $\varphi $ as in Definition 3.6 by applying this definition to $\overline {y A y}$ .

Proof. For $j, k = 1, 2, \ldots , n$ , let $e_{j, k} \in M_n$ be the standard matrix unit. Since a is a positive and nonzero element of ${M}_{n} (A)$ , there exists j such that $(e_{j, j} \otimes 1_{A^{\sim }}) a (e_{j, j} \otimes 1_{A^{\sim }}) \neq 0$ . By replacing a with $(e_{j, j} \otimes 1_{A^{\sim }}) a (e_{j, j} \otimes 1_{A^{\sim }})$ , we may assume that a is of the form $e_{j, j} \otimes c$ for some nonzero positive $c \in A$ . Use [Reference Phillips46, Lemma 2.1] to find $b_1, b_2, \ldots , b_n \in A_{+} \setminus \{ 0 \}$ such that $b_1 \sim b_2 \sim \cdots \sim b_n$ , such that $b_j b_k = 0$ for $j \neq k$ , and such that $b_1 + b_2 + \cdots + b_n \in {\overline {c A c}}$ . Then

$$\begin{align*}1 \otimes b \sim e_{j, j} \otimes ( b_1 + b_2 + \cdots + b_n ) \precsim e_{j, j} \otimes c, \end{align*}$$

as desired.

Proof. Let $x, a \in {M}_{n} (A)_{+}$ with $a \neq 0$ , let $\varepsilon> 0$ , let $m \in \mathbb {N}$ , and let $F \subseteq {M}_{n} (A)$ be finite.

By Lemma 4.4, it is enough to consider the case where $a = 1 \otimes b \in {M}_{n} \otimes A = {M}_{n} (A)$ for some nonzero element $b \in A_{+}$ . Also, there exists $y \in A_{+}$ such that $x \in {\overline {M_n (A) (1 \otimes y) }}$ . Thus, by Lemma 3.5, it is enough to take $x = 1 \otimes y$ with $y \in A_{+}$ . Let E be the set of all matrix entries of elements of F. Put $\delta = \varepsilon / n^{2}$ . Since A is tracially $\mathcal {Z}$ -absorbing, there exists a c.p.c. order zero map $\varphi \colon M_{m} \to A$ such that

1. (1) $\big ( y^{2} - y \varphi (1) y - \delta \big )_{+} \precsim b$ .

2. (2) $\| [\varphi (z), d] \| < \delta $ for any $z \in M_{n}$ with $\| z \| \leq 1$ and any $d \in E$ .

Define a c.p.c. order zero map $\psi \colon M_{m} \to {M}_{n} \otimes A$ by $\psi (z) = 1 \otimes \varphi (z)$ for $z \in {M}_{n}$ . Then

$$ \begin{align*} \big( x^{2} - x \psi (1) x - \varepsilon \big)_{+} & = 1 \otimes \big( y^{2} - y \varphi (1) y - \varepsilon \big)_{+} \\ & \precsim 1 \otimes \big( y^{2} - y \varphi (1) y - \delta \big)_{+} \precsim 1 \otimes b = a. \end{align*} $$

Also, for any $z \in M_{n}$ with $\| z \| \leq 1$ and any $d = (d_{j, k})_{1 \leq j, k \leq n} \in F$ , we have

$$ \begin{align*} \| [\psi (z), d] \| & = \| [1 \otimes \varphi (z), d] \| = \big\| ( [\varphi (z), \, d_{j, k}] )_{1 \leq j, k \leq n} \big\| \\ & \leq \sum_{j, k = 1}^{n} \|[\varphi (z), d_{j, k}] \| < n^{2} \delta = \varepsilon. \end{align*} $$

We have shown that $M_{n} (A)$ is tracially $\mathcal {Z}$ -absorbing.

Proof. It follows from Definition 4.6 that $A \ncong \mathbb {C}$ . Let x, a, F, $\varepsilon $ and n be as in Definition 3.6. Write $F = \{f_{1}, f_{2}, \ldots , f_{m} \}$ . We may assume that $\| a \| = 1$ , that $\| x \| \leq \frac {1}{4}$ , and that $\varepsilon < 1$ . Set $E = F \cup \big \{ x^{1/2}, \, a^{1/2} \big \}$ . Choose $\delta> 0$ such that

$$\begin{align*}\delta < \tfrac{\varepsilon}{4}, \qquad (2 + \delta) \delta < \tfrac{\varepsilon}{12}, \qquad {\mbox{and}} \qquad \big( 2 \|a^{1/2} \| + \delta \big) \delta < \tfrac{\varepsilon}{12}. \end{align*}$$

By assumption, there is a simple tracially $\mathcal {Z}$ -absorbing C*-subalgebra B of A such that $E \subseteq _{\delta } B$ .

In particular, there exists $b \in B$ such that $\| a^{1/2} - b \| < \delta $ . Then

$$ \begin{align*} \| b^{*} b - a \| & \leq \| b^{*} b - b^{*} a^{1/2} \| + \|b^{*} a^{1/2} - a \| \\ & \leq \| b \|\delta + \|a^{1/2} \| \delta \leq \big( \|a^{1/2} \| + \delta + \|a^{1/2} \| \big) \delta < \tfrac{\varepsilon}{12}. \end{align*} $$

Set $d = (b^{*} b - \frac {\varepsilon }{12})_{+}$ . Then the estimate just done implies that

(4.4) $$ \begin{align} d \precsim a. \end{align} $$

Also, $d \neq 0$ since $\| b^{*} b \|> 1 - \frac {\varepsilon }{12}$ and $\varepsilon < 1$ .

Similarly, there exists $y \in B$ such that $\| x^{1/2} - y \| < \delta $ , and the element $w = y^{*} y$ satisfies $\| w - x \| < \tfrac {\varepsilon }{12}$ . Since $\| x \| \leq \frac {1}{4}$ and $\tfrac {\varepsilon }{12} < \tfrac {1}{12}$ , we have $\|w\| < 1$ . Therefore,

(4.5) $$ \begin{align} \| w^2 - x^2 \| < \tfrac{\varepsilon}{6}. \end{align} $$

Choose $e_{1}, e_{2}, \ldots , e_{m} \in B$ such that $\| e_{j} - f_{j} \| < \frac {\varepsilon }{4}$ for $j = 1, 2, \ldots , m $ . Since B is tracially $\mathcal {Z}$ -absorbing, there exists a c.p.c. order zero map $\varphi \colon {M}_{n} \to B$ such that

1. (1) $\big ( w^{2} - w \varphi (1) w - \frac {\varepsilon }{4} \big )_{+} \precsim d$ .

2. (2) $\| [\varphi (z), e_{j}] \| < \frac {\varepsilon }{4}$ for $j \in \{ 1, 2, \ldots , m \}$ and $z \in {M}_{n}$ with $\|z\| \leq 1$ .

Using (4.5), $\| w \| \leq 1$ , and $\| x \| \leq \frac {1}{4}$ at the third step, we then get

$$ \begin{align*} & \big\| \big( w^{2} - w \varphi (1) w - \tfrac{\varepsilon}{4} \big)_{+} - \big( x^{2} - x \varphi (1) x \big) \big\| \\ & \hspace{3em} {\mbox{}} \leq \big\| \big( w^{2} - w \varphi (1) w - \tfrac{\varepsilon}{4} \big)_{+} - w^{2} - w \varphi (1) w \big\| \\ & \hspace{6em} {\mbox{}} + \big\| \big( w^{2} - w \varphi (1) w \big) - \big( x^{2} - x \varphi (1) x \big) \big\| \\ & \hspace{3em} {\mbox{}} \leq \tfrac{\varepsilon}{4} + \|w^{2} - x^{2} \| + \|w \varphi (1) w - w \varphi (1) x \| + \|w \varphi (1) x - x \varphi (1) x \| \\ & \hspace{3em} {\mbox{}} < \tfrac{\varepsilon}{4} + \tfrac{\varepsilon}{6} + \tfrac{\varepsilon}{12} + \tfrac{\varepsilon}{12} < \varepsilon. \end{align*} $$

Hence, using (1) at the second step and (4.4) at the third step, we get

$$\begin{align*}\big( x^{2} - x \varphi (1) x - \varepsilon \big)_{+} \precsim \big( w^{2} - w \varphi (1) w - \tfrac{\varepsilon}{4} \big)_{+} \precsim d \precsim a. \end{align*}$$

Finally, for any $z \in {M}_{n}$ with $\| z \| \leq 1$ and for $j \in \{ 1, 2, \ldots , m \}$ , by (2), we have

$$ \begin{align*} \| [\varphi (z), f_{j}] \| & \leq \|\varphi (z) f_{j} - \varphi (z) e_{j} \| + \|\varphi (z) e_{j} - e_{j} \varphi (z) \| + \| e_{j} \varphi (z) - f_{j} \varphi (z) \| \\ & < \tfrac{\varepsilon}{4} + \tfrac{\varepsilon}{4} + \tfrac{\varepsilon}{4} < \varepsilon. \end{align*} $$

So A is tracially $\mathcal {Z}$ -absorbing.

Proof. This follows immediately from Theorem 4.7.

Proof. The statement follows from Remark 3.7(1), Theorem 4.1 and Theorem 4.7.

Proof. Note that A is not assumed to be unital, and by definition, its minimal unitization is tracially AF. However, by [Reference Forough and Golestani21, Corollary A.22(1)], the algebra A has real rank zero. Therefore, A has an approximate identity $(p_{i})_{i \in I}$ (not necessarily increasing) consisting of projections. (This also follows from [Reference Lin37, Corollary 2.8].) The algebras $p_i A p_i$ have tracial rank zero by [Reference Forough and Golestani21, Corollary A.22(3)]. So they are tracially $\mathcal {Z}$ -absorbing by Lemma 3.13. Now Proposition 4.9 implies that A is tracially $\mathcal {Z}$ -absorbing.

Proof. The forward implication follows from Proposition 4.5 and Corollary 4.8. The converse follows from Theorem 4.1 because A is isomorphic to a hereditary subalgebra of $A \otimes \mathcal {K}$ . In fact, if p is a rank one projection in $ \mathcal {K}$ , then $p \mathcal {K} p =\mathbb {C}p$ , and hence, it follows that $A \otimes \mathbb {C}p$ is a hereditary subalgebra of $A \otimes \mathcal {K}$ isomorphic to A.

Proof. Part (1) follows from Proposition 4.11. For (2), let A and B be $\sigma $ -unital and Morita equivalent. By [Reference Raeburn and Williams47, Theorem 5.55], $A \otimes \mathcal {K} \cong B \otimes \mathcal {K}$ . Now the statement follows from (1).

Proof. Suppose we are given $\varepsilon> 0$ , a finite subset $F \subseteq A \otimes _{\mathrm {min}} B$ , $v, c \in ( A \otimes _{\mathrm {min}} B )_{+}$ with $c \neq 0$ , and $n \in {\mathbb {N}}$ . We have to find a c.p.c. order zero map $\varphi \colon {M}_{n} \to A \otimes _{\mathrm {min}} B$ such that the following hold:

1. (1) $\left ( v^{2} - v \varphi (1) v - \varepsilon \right )_{+} \precsim c$ .

2. (2) $\| [\varphi (z), f] \| < \varepsilon $ for any $z \in {M}_{n}$ with $\| z \| \leq 1$ and any $f \in F$ .

We may assume that there are $m \in {\mathbb {N}}$ and

$$\begin{align*}r_1, r_2, \ldots, r_m \in A \qquad {\mbox{and}} \qquad s_1, s_2, \ldots, s_m \in B \end{align*}$$

such that $\| r_j \| \leq 1$ and $\| s_j \| \leq 1$ for $j = 1, 2, \ldots , m$ and such that $F = \big \{ r_{j} \otimes s_{j} \colon j = 1, 2, \ldots , m \big \}$ . By Kirchberg’s Slice Lemma ([Reference Rørdam, Rørdam and Størmer49, Lemma 4.1.9]), there are nonzero elements $a \in A_{+}$ and $b \in B_{+}$ such that $a \otimes b \precsim c$ in $A \otimes _{\mathrm {min}} B$ . Choose $\delta> 0$ such that $\delta < \varepsilon / 3$ . Since $A \otimes _{\mathrm {min}} B$ has an approximate identity consisting of elementary tensors $x \otimes y$ with x and y positive and of norm at most one, by Remark 3.8, we may assume that $v = x \otimes y$ with $x \in A_{+}$ , $y \in B_{+}$ , and $\| x \|, \, \| y \| \leq 1$ . By [Reference Kirchberg and Rørdam33, Proposition 2.7(v)], there is $k \in \mathbb {N}$ such that

1. (3) $\left ( y^{2} - \delta \right )_{+} \precsim b \otimes 1_{k}$ in ${M}_{\infty } (B)$ .

Since A is simple and tracially $\mathcal {Z}$ -absorbing, Corollary 4.2 implies that A is not type I. Now Lemma 4.4 implies that there is $a_{0} \in A_{+} \setminus \{ 0 \}$ such that

1. (4) $a_{0} \otimes 1_{k} \precsim a$ in ${M}_{\infty } (A)$ .

By assumption, A is tracially $\mathcal {Z}$ -absorbing. Applying Definition 3.6 with $\{ r_1, r_2, \ldots , r_m \}$ in place of F, with $\delta $ in place of $\varepsilon $ , with n as given and with $a_{0}$ in place of a, we obtain a c.p.c. order zero map $\varphi _{0} \colon {M}_{n} \to A$ such that the following hold:

1. (5) $\left ( x^{2} - x \varphi _{0} (1) x - \delta \right )_{+} \precsim a_{0}$ .

2. (6) $\| [\varphi _{0} (z), r_j] \| < \delta $ for $j = 1, 2, \ldots , m$ and any $z \in {M}_{n}$ with $\| z \| \leq 1$ .

Choose $e \in B_{+}$ such that

1. (7) $\| e \| \leq 1$ , $\| y e y - y^{2} \| < \delta $ , and $\| e s_{j} - s_{j} \| < \delta $ and $\| s_{j} e - s_{j} \| < \delta $ for $j = 1, 2, \ldots , m$ .

Now define $\varphi \colon {M}_{n} \to A \otimes _{\mathrm {min}} B$ by $\varphi (z) = \varphi _{0} (z) \otimes e$ for $z \in {M}_{n}$ . Then $\varphi $ is a c.p.c. order zero map.

We show that (1) and (2) hold. To prove (1), first recall that $v = x \otimes y$ at the first step and use (7) at the last step to get

$$ \begin{align*} & \big\| \big( v^{2} - v \varphi (1) v \big) - \big( x^{2} - x \varphi_{0} (1) x - \delta \big)_{+} \otimes \big( y^{2} - \delta \big)_{+} \big\| \\ & \hspace{3em} {\mbox{}} \leq \big\| x^{2} \otimes y^{2} - x \varphi_{0} (1) x \otimes y e y - \big( x^{2} - x \varphi_{0} (1) x \big) \otimes y^{2} \big\| + 2 \delta \\ & \hspace{3em} {\mbox{}} = \big\| x \varphi_{0} (1) x \otimes ( y^{2} - y e y ) \big\| + 2 \delta \\ & \hspace{3em} {\mbox{}} \leq \| y e y - y^{2} \| + 2 \delta < 3 \delta. \end{align*} $$

Then use Lemma 2.1 at the first step, use (3) and (5) at the second step, and use (4) at the fourth step to get

$$ \begin{align*} \big( v^{2} - v \varphi (1) v - 3 \delta \big)_{+} & \precsim \big( x^{2} - x \varphi_{0} (1) x - \delta \big)_{+} \otimes \big( y^{2} - \delta \big)_{+} \\ & \precsim a_{0} \otimes ( b \otimes 1_{k} ) \sim ( a_{0} \otimes 1_{k} ) \otimes b \precsim a \otimes b \precsim c. \end{align*} $$

Thus,

$$\begin{align*}\big( v^{2} - v \varphi (1) v - \varepsilon \big)_{+} \precsim \big( v^{2} - v \varphi (1) v - 3 \delta \big)_{+} \precsim c. \end{align*}$$

For (2), let $f \in F$ . Then there is $j \in \{ 1, 2, \ldots , m \}$ such that $f = r_{j} \otimes s_{j}$ . Let $z \in {M}_{n}$ satisfy $\| z \| \leq 1$ . Using (6) and (7) at the fourth step, we get

$$ \begin{align*} & \| [\varphi (z), f] \| \\ & \hspace{1em} {\mbox{}} = \big\| [\varphi_{0} (z) \otimes e, \, r_{j} \otimes s_{j} ] \big\| \\ & \hspace{1em} {\mbox{}} = \big\| \varphi_{0} (z) r_{j} \otimes e s_{j} - r_{j} \varphi_{0} (z) \otimes s_{j} e \big\| \\ & \hspace{1em} {\mbox{}} \leq \big\| [ \varphi_{0} (z) , r_{j}] \otimes e s_{j} \big\| + \big\| r_{j} \varphi_{0} (z) \otimes (e s_{j} - s_{j}) \big\| + \big\| r_{j} \varphi_{0} (z) \otimes (s_{j} - s_{j} e) \big\| \\ & \hspace{1em} {\mbox{}} < \delta + \delta + \delta < \varepsilon. \end{align*} $$

This completes the proof.

Proof. By definition, $A \cong \mathcal {Z} \otimes A$ . By [Reference Hirshberg and Orovitz28, Proposition 2.2], $\mathcal {Z}$ is tracially $\mathcal {Z}$ -absorbing. Now Theorem 5.1 implies that A is tracially $\mathcal {Z}$ -absorbing.

Proof. The statement follows from Corollaries 5.2 and 4.8 and Remark 5.3.

Proof of Proposition 5.5.

Let D be the dimension drop interval algebra

$$\begin{align*}D = \big\{ g \in C \big( [0, 1], \, M_2 \otimes M_3 \big) \colon {g\ (0) \in M_2 \otimes {\mathbb{C}} \cdot 1 \text{and} \ g (1) \in {\mathbb{C}} \cdot 1 \otimes M_3} \big\}. \end{align*}$$

It suffices to show that there is no approximately central sequence of unital homomorphisms from D to A. To do this, we set

$$\begin{align*}N = 24 \qquad {\mbox{and}} \qquad \varepsilon = \frac{1}{70}. \end{align*}$$

We let $(e_{j, k})_{j, k = 1, 2}$ be the standard system of matrix units for $M_2$ , and we let $(f_{j, k})_{j, k = 1, 2, 3}$ be the standard system of matrix units for $M_3$ . For $n = 1, 2, \ldots , N$ , define $h_n \colon [0, 1] \to [0, 1]$ by

$$\begin{align*}h_n (\lambda) = \begin{cases} 0 & \hspace{1em} 0 \leq \lambda \leq \frac{n - 1}{N} \\ N \lambda - n + 1 & \hspace{1em} \frac{n - 1}{N} < \lambda < \frac{n}{N} \\ 1 & \hspace{1em} \frac{n}{N} \leq \lambda \leq 1. \end{cases} \end{align*}$$

Then define functions $r_{j, n} \in D$ for $j = 1, 2$ and $s_{j, n} \in D$ for $j = 1, 2, 3$ by

$$\begin{align*}r_{j, n} (\lambda) = (1 - h_n (\lambda)) (e_{j, j} \otimes 1) \qquad {\mbox{and}} \qquad s_{j, n} (\lambda) = h_n (\lambda) (1 \otimes f_{j, j}). \end{align*}$$

Next, define a compact set $S \subseteq D$ to consist of the functions defined for $\lambda \in [0, 1]$ by

$$\begin{align*}\lambda \mapsto (u \otimes 1) r_{1, n} (\lambda) (u \otimes 1)^* \end{align*}$$

for $u \in U (M_2)$ and $n = 1, 2, \ldots , N$ , and

$$\begin{align*}\lambda \mapsto (1 \otimes v) s_{1, n} (\lambda) (1 \otimes v)^* \end{align*}$$

for $v \in U (M_3)$ and $n = 1, 2, \ldots , N$ .

If there were an approximately central sequence of unital homomorphisms from D to A, then, in particular, there would be a unital homomorphism $\varphi \colon D \to A$ such that

(5.1) $$ \begin{align} \| [ \varphi (g), a ] \| < \varepsilon, \qquad \| [ \varphi (g), b ] \| < \varepsilon, \qquad {\mbox{and}} \qquad \| [ \varphi (g), c ] \| < \varepsilon \end{align} $$

for all $g \in S$ . We will show that this cannot happen.

Suppose therefore that $\varphi \colon D \to A$ is a unital homomorphism such that (5.1) holds for all $g \in S$ . Regard $C ([0, 1])$ as a unital subalgebra of D in the obvious way. Let $\mu $ be the Borel probability measure on $[0, 1]$ such that $(\omega \circ \varphi ) (g) = \int _{[0, 1]} g \, d \mu $ for all $g \in C ([0, 1])$ . We have

$$\begin{align*}\sum_{n = 1}^N \mu \left( \left[ \frac{n - 1}{N}, \, \frac{n}{N} \right] \right) \leq 2 \mu ([0, 1]) = 2, \end{align*}$$

so there exists $n \in \{ 1, 2, \ldots , N \}$ such that if we set $I = \left [ \frac {n - 1}{N}, \, \frac {n}{N} \right ]$ , then $\mu (I) \leq \frac {2}{N}$ . Set $\alpha = (\omega \circ \varphi ) (h_n)$ .

We have $r_{1, n} + r_{2, n} = 1 - h_n$ , so

$$\begin{align*}(\omega \circ \varphi) (r_{1, n}) + (\omega \circ \varphi) (r_{2, n}) = 1 - \alpha. \end{align*}$$

Therefore, there are $j, k \in {1, 2}$ such that

$$\begin{align*}(\omega \circ \varphi) (r_{j, n}) \geq \frac{1 - \alpha}{2} \qquad {\mbox{and}} \qquad (\omega \circ \varphi) (r_{k, n}) \leq \frac{1 - \alpha}{2}. \end{align*}$$

The set of functions

$$\begin{align*}\lambda \mapsto (u \otimes 1) r_{1, n} (\lambda) (u \otimes 1)^* \end{align*}$$

for $u \in U (M_2)$ , is connected and contains both $r_{j, n}$ and $r_{k, n}$ . Therefore, there is $u \in U (M_2)$ such that the function

$$\begin{align*}r (\lambda) = (u \otimes 1) r_{1, n} (\lambda) (u \otimes 1)^* \end{align*}$$

satisfies $(\omega \circ \varphi ) (r) = \frac {1}{2} (1 - \alpha )$ . A similar argument, starting with

$$\begin{align*}(\omega \circ \varphi) (s_{1, n}) + (\omega \circ \varphi) (s_{2, n}) + (\omega \circ \varphi) (s_{3, n}) = \alpha, \end{align*}$$

produces $v \in U (M_3)$ such that the function

$$\begin{align*}s (\lambda) = (1 \otimes v) s_{1, n} (\lambda) (1 \otimes v)^* \end{align*}$$

satisfies $(\omega \circ \varphi ) (s) = \frac {\alpha }{3}$ . Then $(\omega \circ \varphi ) (r + s) \in \big [ \frac {1}{3}, \frac {1}{2} \big ]$ . It follows that

(5.2) $$ \begin{align} (\omega \circ \varphi) (r + s) - \big[ (\omega \circ \varphi) (r + s) \big]^2 \geq \frac{1}{6}. \end{align} $$

Define $l \in C ([0, 1])$ by

$$\begin{align*}l (\lambda) = \big\| r (\lambda) + s (\lambda) - (r (\lambda) + s (\lambda))^2 \big\| \end{align*}$$

for $\lambda \in [0, 1]$ . We have $0 \leq r + s \leq 1$ , so $0 \leq (r + s)^2 \leq r + s$ , and

$$\begin{align*}0 \leq r + s - (r + s)^2 \leq l \leq 1. \end{align*}$$

Also, $l (\lambda ) = 0$ for $\lambda \not \in I$ . Therefore,

$$ \begin{align*} 0 & \leq (\omega \circ \varphi) (r + s) - (\omega \circ \varphi) \big( (r + s)^2 \big) \\ & \leq (\omega \circ \varphi) (l) = \int_{[0, 1]} l \, d \mu \leq \mu (I) \leq \frac{2}{N} = \frac{1}{12}. \end{align*} $$

Combining this inequality with (5.2) gives

(5.3) $$ \begin{align} (\omega \circ \varphi) \big( (r + s)^2 \big) - \big[ (\omega \circ \varphi) (r + s) \big]^2 \geq \frac{1}{12}. \end{align} $$

Recall that for $x \in P$ , we have $\| x \|_{\omega } = \omega (x^* x)^{1/2}$ . In particular, $\| x \|_{\omega } \leq \| x \|$ . The proof of [Reference Barnett10, Theorem 11] shows that for every $x \in P$ , we have

$$\begin{align*}\| x - \omega (x) \cdot 1 \|_{\omega} \leq 14 \max \big( \| [x, a] \|_{\omega}, \, \| [x, b] \|_{\omega}, \, \| [x, c] \|_{\omega} \big). \end{align*}$$

If $0 \leq x \leq 1$ , a calculation shows that

$$\begin{align*}\| x - \omega (x) \cdot 1 \|_{\omega}^2 = \omega (x^2) - \omega (x)^2. \end{align*}$$

Put $x = r + s$ and use (5.3) at the first step and (5.1) at the third step to get

$$ \begin{align*} \frac{1}{12} & \leq 14^2 \max \big( \| [x, a] \|_{\omega}^2, \, \| [x, b] \|_{\omega}^2, \, \| [x, c] \|_{\omega}^2 \big) \\ & \leq 14^2 \max \big( \| [x, a] \|^2, \, \| [x, b] \|^2, \, \| [x, c] \|^2 \big) < 14^2 \varepsilon^2 = \frac{1}{25}, \end{align*} $$

a contradiction.

Proof of Lemma 6.3.

We may assume that $\| a \| = \| b \| = 1$ . Let $\varepsilon> 0$ ; we prove that $(a - \varepsilon )_{+} \precsim b$ .

The assumption says $a \otimes 1_{n} \precsim b \otimes 1_{n}$ . By [Reference Kirchberg and Rørdam33, Proposition 2.6], there is $\delta> 0$ such that

$$\begin{align*}\big( a \otimes 1_{n} - \tfrac{\varepsilon}{2} \big)_{+} \precsim ( b \otimes 1_{n} - \delta )_{+}. \end{align*}$$

Choose $c_0 \in M_n (A)$ such that

$$\begin{align*}\big\| c_0 \big[ (b \otimes 1_{n} - \delta)_{+} \big] c_0^* - \big( a \otimes 1_{n} - \tfrac{\varepsilon}{2} \big)_{+} \big\| < \tfrac{\varepsilon}{2}. \end{align*}$$

It follows from Lemma 2.1 that there is $c_1 \in M_n (A)$ such that

$$\begin{align*}c_1 c_0 \big[ (b \otimes 1_{n} - \delta)_{+} \big] c_0^* c_1^* = \big( \big( a \otimes 1_{n} - \tfrac{\varepsilon}{2} \big)_{+} - \tfrac{\varepsilon}{2} \big)_{+}. \end{align*}$$

Thus,

$$\begin{align*}c_1 c_0 \big[ (b - \delta)_{+} \otimes 1_{n} \big] (c_1 c_0)^{*} = (a - \varepsilon)_{+} \otimes 1_{n}. \end{align*}$$

Let $f_{1}, f_{2} \in C_{0} ((0 , 1])$ be nonnegative functions such that $f_{1} = 0$ on $\big [ \frac {\delta }{2}, 1 \big ]$ , $f_{1}> 0$ on $\big ( 0, \frac {\delta }{2} \big )$ , and $\| f_{1} \| = 1$ , and such that $f_{2} = 0$ on $\big [ 0, \frac {\delta }{2} \big ]$ and $f_{2} = 1$ on $[\delta , 1]$ . Put $d =f_{1} (b)$ . Since b is weakly purely positive, $d \neq 0$ . Put $c = c_1 c_0 [f_{2} (b) \otimes 1_{n}]$ . Since $f_{2} (\lambda ) (\lambda - \delta )_{+} = (\lambda - \delta )_{+}$ for $\lambda \in [0, 1]$ and $ f_{1} f_{2} = 0$ , we get

$$\begin{align*}c \big[ (b - \delta)_{+} \otimes 1_{n} \big] c^{*} = (a - \varepsilon)_{+} \otimes 1_{n} \qquad {\mbox{and}} \qquad c (d \otimes 1_{n}) = 0. \end{align*}$$

Write $c = (c_{j, k})_{1 \leq j, k \leq n}$ with $c_{j, k} \in A$ for $j, k = 1, 2, \ldots , n$ . Then

(6.1) $$ \begin{align} c_{j, k} d = 0. \end{align} $$

For any $\mu> 0$ , we will find $z \in A$ such that

(6.2) $$ \begin{align} \big\| z \big[ (b - \delta)_{+} + d \big] z^{*} - (a - \varepsilon)_{+} \big\| < \mu. \end{align} $$

This will prove the first step in the calculation

$$\begin{align*}(a - \varepsilon)_{+} \precsim (b - \delta)_{+} + d \precsim b; \end{align*}$$

since the second step is clear, and $\varepsilon> 0$ is arbitrary, we will have finished the proof.

So fix $\mu> 0$ . We have

(6.3) $$ \begin{align} \sum_{l = 1}^{n} c_{j, l} (b - \delta)_{+} c_{k, l}^{*} = \begin{cases} (a - \varepsilon)_{+} & \hspace{1em} j = k \\ 0 & \hspace{1em} j \neq k. \end{cases} \end{align} $$

As in the proof of [Reference Hirshberg and Orovitz28, Lemma 3.1], define $g, h \in C_{0} ((0, 1])$ by

$$\begin{align*}g (\lambda) = \begin{cases} \sqrt{ 13 \lambda / \mu} & \hspace{1em} \lambda < \mu / 13 \\ 1 & \hspace{1em} \lambda \geq \mu / 13 \end{cases} \qquad {\mbox{and}} \qquad h (\lambda) = 1 - \sqrt{1 - \lambda} \end{align*}$$

for $\lambda \in [0, 1]$ . Then

(6.4) $$ \begin{align} | g (\lambda)^{2} \lambda - \lambda | < \tfrac{\mu}{12} \qquad {\mbox{and}} \qquad 1 - h (\lambda) = \sqrt{1 - \lambda} \end{align} $$

for all $\lambda \in [0, 1]$ .

Put

$$\begin{align*}F = \big\{ (a - \varepsilon)_{+} \big\} \cup \big\{ c_{j, k} (b - \delta)_{+} c^{*}_{l, m} \colon j, k, l, m = 1, 2, \ldots, n \big\}. \end{align*}$$

At this point, we need functional calculus for c.p.c. order zero maps for a function $k \in C_0 ((0, 1])$ ([Reference Winter and Zacharias55, Corollary 4.2]). We follow the notation there, writing $k (\varphi )$ (and then $\varphi ^{1/2}$ ), rather than $k [\varphi ]$ (and $\sqrt {[\varphi ]}$ ) as in [Reference Hirshberg and Orovitz28, Notation 2.13]. We briefly recall its construction, slightly simplified since the domain is unital. By [Reference Winter and Zacharias55, Theorem 3.3]), there is a unital homomorphism $\pi $ from $M_n$ to the multiplier algebra of the C*-algebra generated by the range of $\varphi $ such that $\varphi (w) = \varphi (1) \pi (w) = \pi (w) \varphi (1)$ for all $w \in M_n$ . Then $k (\varphi ) (w) = k (\varphi (1)) \pi (w)$ for all $w \in M_n$ . We need two consequences:

(6.5) $$ \begin{align} k (\varphi) (w) \cdot \varphi (1) = \varphi (1) \cdot k (\varphi) (w) \end{align} $$

for any $k \in C_0 ((0, 1])$ and $w \in M_n$ , and

(6.6) $$ \begin{align} k (\varphi) (w_1) \cdot k (\varphi) (w_2) = k (\varphi (1)) \cdot k (\varphi) (w_1 w_2) \end{align} $$

for any $f \in C_0 ((0, 1])$ and $w_1, w_2 \in M_n$ .

For any C*-algebras B and A, any map $\varphi \colon B \to A$ and any subset $F \subseteq A$ , we define

$$\begin{align*}C (\varphi, F) = \sup \big( \big\{ \| [\varphi (w), y] \| \colon {{w \in B \text{satisfies} \| w \| \leq 1 \text{and} y \in F}} \big\} \big). \end{align*}$$

It follows from [Reference Hirshberg and Orovitz28, Lemma 2.8] that there is $\eta _{0}> 0$ such that if $\varphi \colon M_{n} \to A$ is a c.p.c. order zero map with $C (\varphi , F) < \eta _{0}$ , then

(6.7) $$ \begin{align} C ( g (\varphi), F ) < \frac{\mu}{24n^{4}}, \qquad C ( \varphi^{1/2}, F ) < \frac{\mu}{24n^{4}}, \qquad {\mbox{and}} \qquad C ( h (\varphi), F ) < \frac{\mu}{6}. \end{align} $$

Use [Reference Archey and Phillips4, Lemma 2.5] to choose $\eta> 0$ such that $\eta < \min \big ( \eta _{0}, \tfrac {\mu }{30} \big )$ and such that whenever $e, s \in A_{+}$ satisfy

$$\begin{align*}\| e \| \leq 1, \qquad \| s \| \leq 1, \qquad {\mbox{and}} \qquad \| e s - s e \| < 9 \eta, \end{align*}$$

then $\big \| e^{1/2} s - s e^{1/2} \big \| < \mu / 6$ . Choose $x \in A_{+}$ such that

(6.8) $$ \begin{align} \| x \| \leq 1 \qquad {\mbox{and}} \qquad \big\| x (a - \varepsilon)_{+} - (a - \varepsilon)_{+} \big\| < \eta. \end{align} $$

Applying Definition 3.6, with $\eta $ in place of $\varepsilon $ , and with n, F and x as given, we obtain a c.p.c. order zero map $\varphi \colon M_{n} \to A$ such that

(6.9) $$ \begin{align} C ( \varphi, F) < \eta \qquad {\mbox{and}} \qquad \big( x^{2} - x \varphi (1) x - \eta \big)_{+} \precsim d. \end{align} $$

Put

(6.10) $$ \begin{align} r_0 = 1 - \varphi (1) \qquad {\mbox{and}} \qquad r = \big( x^{2} - x \varphi (1) x - \eta \big)_{+} = (x r_0 x - \eta)_{+}. \end{align} $$

Then $r_0 \in A^{\sim }$ and $r \in A$ . Set

$$\begin{align*}a_{1} = \varphi (1) (a - \varepsilon)_{+}, \ \ \ a_{2} = r_0^{1/2} (a - \varepsilon)_{+} r_0^{1/2}, \ \ \ {\mbox{and}} \ \ \ a_{3} = r^{1/2} (a - \varepsilon)_{+} r^{1/2}. \end{align*}$$

Since

(6.11) $$ \begin{align} \big\| \big[ r_0, \, (a - \varepsilon)_{+} \big] \big\| = \big\| \big[ 1 - \varphi (1), \, (a - \varepsilon)_{+} \big] \big\| = \big\| \big[ \varphi (1), \, (a - \varepsilon)_{+} \big] \big\| < \eta, \end{align} $$

the choice of $\eta $ implies

(6.12) $$ \begin{align} \big\| (a - \varepsilon)_{+} - (a_{1} + a_{2}) \big\| < \frac{\mu}{6}. \end{align} $$

For $j, k = 1, 2, \ldots , n$ , let $e_{j, k} \in M_n$ be the standard matrix unit, and set

(6.13) $$ \begin{align} g_{j, k} = g (\varphi) (e_{j, k}) \qquad {\mbox{and}} \qquad \widehat{c}_{j, k} = \varphi^{1/2} (1) g_{j, k} c_{j, k}. \end{align} $$

Then set $\widehat {c} = \sum _{j, k = 1}^{n} \widehat {c}_{j, k}$ .

We claim that

(6.14) $$ \begin{align} \big\| \widehat{c} (b - \delta)_{+} \widehat{c}^{*} - a_{1} \big\| < \frac{\mu}{6}. \end{align} $$

We follow the proof of [Reference Hirshberg and Orovitz28, Lemma 3.1]. One checks, following the steps there and using (6.6) and (6.3), that

(6.15) $$ \begin{align} \sum_{j, k, l, m = 1}^{n} g_{j, k} g_{l, m} c_{j, k} (b - \delta)_{+} c_{m, l}^{*} = g (\varphi) (1)^2 \cdot (a - \varepsilon)_{+}. \end{align} $$

Now, using (6.13) at the second step, using (6.7) at the third step, using (6.7) and (6.5) at the fourth step, using (6.15) at the fifth step, and using (6.4) at the sixth step, we have

$$ \begin{align*} \widehat{c} (b - \delta)_{+} \widehat{c}^{*} & = \sum_{j, k, l, m = 1}^{n} \widehat{c}_{j, k} (b - \delta)_{+} \widehat{c}^{*}_{m, l} \\ & = \varphi^{1/2} (1) \left( \sum_{j, k, l, m = 1}^{n} g_{j, k} c_{j, k} (b - \delta)_{+} c_{m, l}^{*} g_{l, m} \right) \varphi^{1/2} (1) \\ & \approx_{\mu / 24} \varphi^{1/2} (1) \left( \sum_{j, k, l, m = 1}^{n} g_{j, k} g_{l, m} c_{j, k} (b - \delta)_{+} c_{m, l}^{*} \right) \varphi^{1/2} (1) \\ & \approx_{\mu / 24} \varphi (1) \left(\sum_{j, k, l, m = 1}^{n} g_{j, k} g_{l, m} c_{j, k} (b - \delta)_{+} c_{m, l}^{*} \right) \\ & = \varphi (1) \cdot g (\varphi) (1)^2 \cdot (a - \varepsilon)_{+} \\ & \approx_{\mu / 12} \varphi (1) (a - \varepsilon)_{+} \\ & = a_{1}. \end{align*} $$

This proves (6.14).

We claim that

(6.16) $$ \begin{align} \| a_{2} - a_{3} \| < \frac{\mu}{2}. \end{align} $$

First, using (6.10) at the first step, using (6.8) at the second and fourth steps, and using (6.11) and (6.10) at the third and fifth steps, we have

$$ \begin{align*} r (a -\varepsilon)_{+} & \approx_{\eta} x r_0 x (a - \varepsilon)_{+} \approx_{\eta} x r_0 (a - \varepsilon)_{+} \\ & \approx_{\eta} x (a - \varepsilon)_{+} r_0 \approx_{\eta} (a - \varepsilon)_{+} r_0 \approx_{\eta} r_0 (a - \varepsilon)_{+}. \end{align*} $$

Thus,

$$\begin{align*}\big\| r (a - \varepsilon)_{+} - r_0 (a - \varepsilon)_{+} \big\| < 5 \eta \qquad {\mbox{and}} \qquad \big\| r (a - \varepsilon)_{+} - (a - \varepsilon)_{+} r_0 \big\| < 4 \eta. \end{align*}$$

Taking adjoints in the second inequality and combining gives

$$\begin{align*}\big\| r (a - \varepsilon)_{+} - (a - \varepsilon)_{+} r \big\| < 9 \eta. \end{align*}$$

By the choice of $\eta $ , and using (6.11) for the first inequality,

$$\begin{align*}\big\| r_0^{1/2} (a - \varepsilon)_{+} - (a - \varepsilon)_{+} r_0^{1/2} \big\| < \frac{\mu}{6} \quad {\mbox{and}} \quad \big\| r^{1/2} (a - \varepsilon)_{+} - (a - \varepsilon)_{+} r^{1/2} \big\| < \frac{\mu}{6}. \end{align*}$$

Now

$$ \begin{align*} \| a_{2} - a_{3} \| & = \big\| r_0^{1/2} (a - \varepsilon)_{+}r_0^{1/2} - r^{1/2} (a - \varepsilon)_{+} r^{1/2} \big\| \\ & \leq \frac{\mu}{6} + \frac{\mu}{6} + \big\| r_0 (a - \varepsilon)_{+} - r (a - \varepsilon)_{+} \big\| \\ & < \frac{\mu}{3} + 5 \eta < \frac{\mu}{3} + \frac{\mu}{6} = \frac{\mu}{2}. \end{align*} $$

This proves (6.16).

Using (6.9) and (6.10) for the second step, we get $a_{3} \precsim r \precsim d$ . Choose $s \in A$ such that $\| s d s^{*} - a_{3} \| < \frac {\mu }{6}$ . Since $d = f (b)$ , as in the proof of [Reference Hirshberg and Orovitz28, Lemma 3.1], replacing s with $s k (b)$ for a function $k \in C ([0, 1])$ which is $1$ on $\big [ 0, \frac {\delta }{2} \big ]$ and vanishes on $[\delta , 1]$ , we may assume that $s (b - \delta )_{+} = 0$ . Set $z = \widehat {c} + s$ . Using (6.1) and (6.13) at the first step, and using (6.14), (6.16), and (6.12) at the third step, we get

$$ \begin{align*} & \big\| z[(b - \delta)_{+} + d]z^{*} - (a - \varepsilon)_{+} \big\| \\ & \hspace{1em} {\mbox{}} = \big\| \widehat{c} (b - \delta)_{+} \widehat{c}^{*} + s d s^{*} - (a - \varepsilon)_{+} \big\| \\ & \hspace{1em} {\mbox{}} \leq \big\| \widehat{c} (b - \delta)_{+} \widehat{c}^{*} - a_{1} \big\| + \| s d s^{*} - a_{3} \| + \| a_{3} - a_{2} \| + \| a_{1} + a_{2} - (a - \varepsilon)_{+} \| \\ & \hspace{1em} {\mbox{}} < \frac{\mu}{6} + \frac{\mu}{6} + \frac{\mu}{2} + \frac{\mu}{6} = \mu. \end{align*} $$