for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 7.71)

The lemma then follows (with sufficiently large $m$ depending on $n$).

∎

######  Corollary 7.8.

Let $A$ be a non-elementary, $\sigma$-unital and algebraically simple
$C^{*}$-algebra with $QT(A)\not=\emptyset$ and T-tracial approximate
oscillation zero. Let $e_{k}\in A_{+}^{\bf 1}\setminus\\{0\\}$
($k\in\mathbb{N}$) such that $\\{e_{k}\\}\not\in I_{\overline{QT(A)}^{w}}$ and
$\\{e_{k}\\}$ is a permanent projection lifting of $p=\Pi(\\{e_{k}\\}),$ and
let $n>1$ be an integer and $\varepsilon>0.$

Then, there is $k_{0}\in\mathbb{N}$ such that, for any $k\geq k_{0}$ and for
any $1\leq i\leq n,$ there exists $g_{k,i}\in A_{+}$ such that

$\displaystyle\sup\\{|{i\over{n}}d_{\tau}(e_{k})-d_{\tau}(g_{k,i})|:\tau\in\overline{QT(A)}^{w}\\}<\varepsilon.$
(e 7.72)

###### Proof.

By the proof of Lemma 7.7, for $k\geq k_{0},$

$\displaystyle\sup\\{|{1\over{n}}d_{\tau}(e_{k})-d_{\tau}(f_{k,1})|:\tau\in\overline{QT(A)}^{w}\\}<\varepsilon/n.$
(e 7.73)

So, for any $1\leq i\leq n,$ choose $g_{k,i}=\sum_{j=1}^{i}f_{k,j}.$ Then

$\displaystyle\sup\\{|{i\over{n}}d_{\tau}(e_{k})-d_{\tau}(g_{k,i})|:\tau\in\overline{QT(A)}^{w}\\}<\varepsilon.$
(e 7.74)

∎

######  Definition 7.9.

Let $A$ be a $C^{*}$-algebra with
${\widetilde{QT}}(A)\setminus\\{0\\}\not=\emptyset.$ Define

$\displaystyle R_{\tau,f}(A)=\\{\hat{a}:a\in{\rm Ped}(A\otimes{\cal
K})_{+}\\}\subset\operatorname{Aff}_{+}({\widetilde{QT}}(A)).$ (e 7.75)

######  Lemma 7.10.

Let $A$ be a non-elementary, $\sigma$-unital and simple $C^{*}$-algebra with
${\widetilde{QT}}(A)\setminus\\{0\\}\not=\emptyset.$ Suppose that $A$ has
$T$-tracial oscillation zero. Then the image of the canonical map $\Gamma$
(see 2.13) is dense in $R_{\tau,f}(A).$

###### Proof.

Fix an element $0\leq e\leq 1$ in ${\rm Ped}(A)_{+}.$ Let
$A_{1}=\mathrm{Her}(e).$ Then $A_{1}={\rm Ped}(A_{1}).$ By Brown’s stable
isomorphism theorem ([4]), $A_{1}\otimes{\cal K}\cong A\otimes{\cal K}.$ It
suffices to show that the image of the map
$\Gamma_{1}:\mathrm{Cu}(A_{1})=\mathrm{Cu}(A)\to{\rm
LAff}_{+}(\overline{QT(A_{1})}^{w})$ is dense in

${{R_{\tau,f}(A_{1})=\\{\hat{a}:a\in{\rm Ped}(A_{1}\otimes{\cal
K})_{+}\\}\subset\operatorname{Aff}_{+}(\overline{QT(A_{1})}^{w})}}$

(see 2.13).

Fix $a\in{\rm Ped}(A_{1}\otimes{\cal K})_{+}.$ Let $\varepsilon>0.$ It
suffices to show that there is $f\in\mathrm{Cu}(A_{1})$ such that

$\displaystyle\sup\\{|\tau(a)-\widehat{f}(\tau)|:\tau\in\overline{QT(A_{1})}^{w}\\}<\varepsilon.$
(e 7.76)

Without loss of generality, we may assume that $0\leq a\leq 1.$ Since
$a\in{\rm Ped}(A_{1}\otimes{\cal K})_{+},$ there exists $r\geq 1$ such that
$r[f_{\delta}(e)]\geq[a]$ for some $\delta\in(0,1/4).$ Therefore we may assume
that $a\in M_{r}(A_{1})$ for some integer $r\geq 1.$

Put $B:=l^{\infty}(A_{1})/I_{{}_{\overline{QT(A_{1})}^{w}}}.$ Then, by Theorem
6.4, $B$ has real rank zero.

Therefore, for any $\varepsilon>0,$ there are mutually orthogonal projections
$p_{1},p_{2},...,p_{m}\in M_{r}(B)$ and
$\lambda_{1},\lambda_{2},...,\lambda_{m}\in(0,1)$ such that

$\displaystyle\|\Pi(\iota(a))-\sum_{i=1}^{m}\lambda_{i}p_{i}\|<\varepsilon/16.$
(e 7.77)

We may assume that $\lambda_{i}\in(0,1)\cap\mathbb{Q}.$ There are mutually
orthogonal elements $\\{e_{n,i}\\}\in l^{\infty}(M_{r}(A_{1}))$
($i=1,2,...,m$) such that $\Pi(\\{e_{n,i}\\})=p_{i},$ $i=1,2,...,m.$ By
Proposition 6.2, we may assume that $\\{e_{n,i}\\}$ is a permanent projection
lifting. By (2) of Proposition 6.2, $\lim_{n\to\infty}\omega(e_{n,i})=0.$
Without loss of generality, we may assume that, for all $n\in\mathbb{N},$

$\displaystyle
d_{\tau}(e_{n,i})-\tau(e_{n,i})<\varepsilon/16(m+1)^{{2}}\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in{{\overline{QT(A_{1})}^{w},}}\,\,i=1,2,...,m.$ (e
7.78)

Applying Corollary 7.8, without loss of generality, we may also assume that,
there are permanent projection liftings $\\{f_{n,i}\\}$ such that

$\displaystyle\sup\\{|\lambda_{i}d_{\tau}(e_{n,i})-d_{\tau}(f_{n,i})|:\tau\in{{\overline{QT(A_{1})}^{w}}}\\}<\varepsilon/16(m+1)^{{2}},\,\,i=1,2,...,m.$
(e 7.79)

By (e 7.77), there exists $\\{c_{n}\\}\in I_{{}_{\overline{QT(A_{1})}^{w}}}$
and $n_{1}\in\mathbb{N}$ such that, for all $n\geq n_{1},$

$\displaystyle\|a-\sum_{i=1}^{m}\lambda_{i}e_{i,n}+c_{n}\|<\varepsilon/8.$ (e
7.80)

Then, for $n\geq n_{1},$

$\displaystyle\sup\\{|\tau(a-\sum_{i=1}^{m}\lambda_{i}e_{n,i}+c_{n})|:\tau\in\overline{QT(A_{1})}^{w}\\}<\varepsilon/8.$
(e 7.81)

Since $\\{c_{n}\\}\in I_{{}_{\overline{QT(A_{1})}^{w}}},$ we have
$\\{|c_{n}|^{1/2}\\}\in I_{{{}_{\overline{QT(A_{1})}^{w}}}}.$ It follows that

$\lim_{n\to\infty}\sup\\{|\tau(c_{n})|:\tau\in{{\overline{QT(A_{1})}^{w}}}\\}=0.$

Therefore there exists $n_{2}\geq n_{1}$ such that

$\displaystyle|\tau(a-\sum_{i=1}^{m}\lambda_{i}e_{n,i})|<|\tau(c_{n})|+\varepsilon/8<\varepsilon/4\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A_{1})}^{w}\,\,{\rm and}\,\,\,{\rm
for\,\,\,all}\,\,\,n\geq n_{2}.$ (e 7.82)

Thus, by (e 7.78) and (e 7.79), for $n\geq n_{2},$

$\displaystyle|\tau(a)-\sum_{i=1}^{m}d_{\tau}(f_{n,i})|<\varepsilon/2\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A_{1})}^{w}.$ (e 7.83)

Put $e={\rm diag}(f_{n,1},f_{n,2},...,f_{n,m}).$ Then

$\displaystyle|\tau(a)-d_{\tau}(e)|<\varepsilon\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A_{1})}^{w}.$ (e 7.84)

This completes the proof. ∎

######  Theorem 7.11.

Let $A$ be a non-elementary and $\sigma$-unital simple $C^{*}$-algebra with
$\widetilde{QT}(A)\setminus\\{0\\}\not=\emptyset$ and strict comparison.
Suppose that $A$ has T-tracial approximate oscillation zero. Then $\Gamma$ is
surjective (see 2.13).

###### Proof.

We keep the same setting as in the proof of Theorem 7.10.

Let $b\in{\rm Ped}(A_{1}\otimes{\cal K})_{+}$ with $0\leq b\leq 1.$ Choose
$b_{n}=(1-1/(n+1))b.$ Then $h_{n}:=b_{n+1}-b_{n}\in{\rm Ped}(A_{1}\otimes{\cal
K})_{+}\setminus\\{0\\}.$ Put

$\displaystyle\sigma_{n}:=\inf\\{\tau(h_{n}):\tau\in\overline{QT(A_{1})}^{w}\\}>0.$
(e 7.85)

Applying Theorem 7.10, for each $n\in\mathbb{N},$ we obtain $f_{n}\in{\rm
Ped}(A_{1}\otimes{\cal K})_{+}$ such that

$\displaystyle\eta_{n}:=\sup\\{|\tau(b_{n})-d_{\tau}(f_{n})|:\tau\in\overline{QT(A_{1})}^{w}\\}<{\min\\{\sigma_{j}:1\leq
j\leq n+1\\}\over{2^{n+2}}}.$ (e 7.86)

In particular, for all $n\in\mathbb{N},$

$\displaystyle d_{\tau}(f_{n})<\eta_{n}+{{\tau}}(b_{n})<\tau(b)\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in{{\overline{QT(A_{1})}^{w}.}}$ (e 7.87)

Then, for all $\tau\in\overline{QT(A_{1})}^{w}$ (note that
$b_{n}b_{n+1}=b_{n+1}b_{n}$),

$\displaystyle d_{\tau}(f_{n+1})-d_{\tau}(f_{n})$ $\displaystyle>$
$\displaystyle(\tau(b_{n+1})-\eta_{n+1})-(\tau(b_{n})+\eta_{n})$ (e 7.88)
$\displaystyle=$
$\displaystyle{{\tau}}(h_{n})-\eta_{n+1}-\eta_{n}>\tau(h_{n})-\sigma_{n}/2>0.$
(e 7.89)

It follows from the strict comparison that $[f_{n}]$ is an increasing sequence
in $\mathrm{Cu}(A).$ Let $f$ be the supremum of $\\{[f_{n}]\\}$ in
$\mathrm{Cu}(A).$ On the other hand, for all
$\tau\in\overline{QT(A_{1})}^{w},$

$\displaystyle d_{\tau}(f_{n+1})-{{\tau}}(b_{n})$ $\displaystyle>$
$\displaystyle{{\tau}}(b_{n+1})-\eta_{n+1}-{{\tau}}(b_{n})$ (e 7.90)
$\displaystyle\geq$ $\displaystyle{{\tau}}(h_{n})-\sigma_{n}/2^{n+1}>0.$ (e
7.91)

It follows that $\widehat{f}(\tau)\geq\tau(b_{n})$ for all
$\tau\in\overline{QT(A_{1})}^{w}$ for each $n\in\mathbb{N}.$ Hence
$\widehat{f}(\tau)\geq\tau(b)$ for all $\tau\in\overline{QT(A_{1})}^{w}.$

Let $\varepsilon>0.$ By Theorem 7.10, there is
$c_{\varepsilon}\in\mathrm{Cu}(A_{1})$ such that

$\displaystyle\sup\\{|\tau(b+(\varepsilon/2)b)-\Gamma_{1}(c_{\varepsilon})(\tau)|:\tau\in\overline{QT(A_{1})}^{w}\\}<\varepsilon\sigma_{1}/8.$
(e 7.92)

Then, for all $n\in\mathbb{N}$ (see also (e 7.87)),

$\displaystyle\Gamma_{1}(c_{\varepsilon})(\tau)>\tau(b+\varepsilon/2b)-\varepsilon\sigma_{1}/8>\tau(b)>d_{\tau}(f_{n})\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A_{1})}^{w}.$ (e 7.93)

Hence $[f]\leq[c_{\varepsilon}].$ It follows that

$\displaystyle\widehat{f}(\tau)\leq\widehat{c_{\varepsilon}}(\tau)<\tau(b)+\varepsilon\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in{{\overline{QT(A_{1})}^{w}.}}$ (e 7.94)

Let $\varepsilon\to 0.$ We conclude that $\widehat{f}(\tau)=\tau(b)$ for all
$\tau\in\overline{QT(A_{1})}^{w}.$

So far we have shown that, for any $b\in{\rm Ped}(A\otimes{\cal K})_{+},$
there is $f\in\mathrm{Cu}(A)$ such that $\widehat{f}(\tau)=\tau(b)$ for all
$\tau\in\overline{QT(A)}^{w}.$ Note that, for any $a\in(A\otimes{\cal
K})_{+},$ $(a-\|a\|/n)_{+}\in{\rm Ped}(A\otimes{\cal K}).$ Thus, there are
$f_{n}\in\mathrm{Cu}(A)$ such that
$\widehat{f_{n}}(\tau)=\tau((a-\|a\|/n)_{+})$ for all
$\tau\in\overline{QT(A)}^{w}.$ Since $(a-\|a\|/n)_{+}\nearrow a,$ we conclude,
using the similar argument used above, that there is $f\in\mathrm{Cu}(A)$ such
that $\widehat{f}(\tau)=\tau(a)$ for all $\tau\in\overline{QT(A)}^{w}.$
Applying Theorem 5.7 of [13] and repeating the argument above, we conclude
that $\Gamma$ is surjective. ∎

##  8 Tracially matricial property

######  Definition 8.1.

Let $A$ be a $C^{*}$-algebra and $S\subset\widetilde{QT}(A)\setminus\\{0\\}$
be a non-empty compact subset. $C^{*}$-algebra $A$ is said to have property
(TM) relative to $S,$ if for any $a\in{\rm Ped}(A\otimes{\cal K})_{+},$ any
$\varepsilon>0,$ any $n\in\mathbb{N},$ there is a c.p.c. order zero map
$\varphi:M_{n}\to\mathrm{Her}(a)$ such that
$\|a-\varphi(1_{n})a\|_{{}_{2,S}}<\varepsilon.$

A $\sigma$-unital simple $C^{*}$-algebra $A$ with
$\widetilde{QT}(A)\setminus\\{0\\}\not=\emptyset$ is said to have property
(TM), if for some $e\in{\rm Ped}(A)_{+}^{\bf 1}\setminus\\{0\\},$
$\mathrm{Her}(e)$ has property (TM) relative to $\overline{QT(A)}^{w}.$

From the definition, it is clear that, if $A$ is a $\sigma$-unital simple
$C^{*}$-algebra which has property (TM), then $A\otimes{\cal K}$ has property
(TM), and by Brown’s stable isomorphism theorem [4], every $\sigma$-unital
hereditary $C^{*}$-subalgebra has property (TM). Since we only need to
consider $\mathrm{Her}(a)$ for each $a\in{\rm Ped}(A\otimes{\cal K})_{+}^{\bf
1},$ it follows that every hereditary $C^{*}$-subalgebra has property (TM).

In the absence of strict comparison, one may also define the following:

A $C^{*}$-algebra $A$ is said to have property (CM), if, for any
$n\in\mathbb{N}$ and any $a\in{\rm Ped}(A\otimes{\cal K})_{+}^{\bf 1},$ there
is a c.p.c. order zero map $\varphi:M_{n}\to l^{\infty}(C)/N_{cu}(C)$ such
that

$\displaystyle\iota(b)\varphi(1_{n})=\iota(b)\,\,\,{\rm
for\,\,\,all}\,\,\,b\in C,$ (e 8.1)

where $C=\overline{a(A\otimes{\cal K})a}.$

######  Remark 8.2.

In the definition 8.1, let $\psi:C_{0}((0,1])\otimes M_{n}\to\mathrm{Her}(a)$
be the homomorphism induced by $\varphi,$ i.e., $\psi(\iota\otimes
e_{i,j})=\varphi(e_{i,j})$ ($1\leq i,j\leq n$). Then $\mathrm{Her}(a)$ has
some “tracially large” matricial structure (see Proposition 8.3).

Let $x\in{\rm Ped}(A\otimes{\cal K}).$ Then $a_{0}=x^{*}x+xx^{*}\in{\rm
Ped}(A\otimes{\cal K})_{+}.$ Note $x\in\mathrm{Her}(a_{0}).$ Let ${\cal
F\subset}\mathrm{Her}(a_{0})^{\bf 1}$ be a finite set. For any
$1>\varepsilon>0,$ there is $a\in\mathrm{Her}(a_{0})_{+}^{\bf 1}$ such that

$\displaystyle\|ay-y\|<(\varepsilon/4)^{3}\,\,\,{\rm
and}\,\,\,\|ya-y\|<(\varepsilon/4)^{3}\,\,\,{\rm for\,\,\,all}\,\,\,y\in{\cal
F}.$ (e 8.2)

If $A$ has property (TM) relative to $S$ such that $\|\tau\|\leq 1$ for all
$\tau\in S,$ then, there is a c.p.c. order zero map
$\varphi:M_{n}\to\mathrm{Her}(a_{0})$ such that

$\displaystyle\|a-\varphi(1_{n})a\|_{{}_{2,S}}<(\varepsilon/2)^{3}.$ (e 8.3)

Then, for all $y\in{\cal F},$

$\displaystyle\|y-\varphi(1_{n})y\|_{{}_{2,S}}^{2/3}$ $\displaystyle\leq$
$\displaystyle\|y-ya\|^{2/3}+\|ya-ya\varphi(1_{n})\|_{{}_{2,S}}^{2/3}$ (e 8.4)
$\displaystyle<$
$\displaystyle(\varepsilon/2)^{2}+\|y\|\|a-a\varphi(1_{n})\|_{{}_{2,S}}^{2/3}<\varepsilon.$
(e 8.5)

Similarly,

$\displaystyle\|y-\varphi(1_{n})y\|_{{}_{2,S}}^{2/3}<\varepsilon.$ (e 8.6)

The following fact is well known. For completeness, we include a proof here.

######  Proposition 8.3.

Let $A$ be a $C^{*}$-algebra, $n\in\mathbb{N},$ and $\varphi:M_{n}\to A$ be a
c.p.c. order zero map. Then
$\mathrm{Her}(\varphi(1_{n}))\cong\mathrm{Her}(\varphi(e_{1,1}))\otimes
M_{n}.$

###### Proof.

Let $\psi:C_{0}((0,1])\otimes M_{n}\to A$ be the homomorphism defined by
$\psi(\imath\otimes e_{ij})=\varphi(e_{i,j}),$ where $\imath$ is the identity
function on $(0,1],$ $i,j\leq n.$ In particular, $\psi(\imath\otimes
1_{n})=\varphi(1_{n}).$ Write $\psi(\imath\otimes e_{i,j})=u_{i,j}r_{{j}}$ as
a polar decomposition of $\varphi(\imath\otimes e_{i,j})$ in $A^{**}.$ Hence
$r_{{j}}=|{{\psi}}(\imath\otimes e_{i,j})|$ and $u_{i,j}$ is a partial
isometry in $A^{**}.$ Note that $au_{i,j}b\in A$ for all $a\in
p_{i}A^{**}p_{i}\cap A$ and $b\in p_{j}A^{**}p_{j}\cap A,$ where $p_{i}$ is
the open projection of $r_{i},$ $i,j=1,2,...,n.$ Since $\psi$ is a
homomorphism, we compute that $p_{i}=u_{i,i},$ $i=1,2,...,n,$ and
$\\{u_{i,j}\\}_{1\leq i,j\leq n}$ forms a system of matrix units for $M_{n}.$

Define $\Phi:M_{n}({\rm Her}({{\varphi}}(e_{1,1})))\cong{\rm
Her}(\varphi(e_{1,1}))\otimes M_{n}\to{\rm Her}(\varphi(1_{n}))$ by defining
$\Phi(a\otimes e_{i,j})=u_{i,1}au_{1,j}$ for all $a\in{\rm
Her}({{\varphi}}(e_{1,1})),$ $i,j=1,2,...,n.$ Then $\Phi$ is a homomorphism.
Since $\Phi$ is the identity map on ${\rm Her}{{(\varphi(e_{1,1}))}}$ and
$M_{n}$ is simple, the map $\Phi$ is injective. Put $B=\Phi(M_{n}({\rm
Her}({{\varphi}}(e_{1,1})))).$ To see $\Phi$ is surjective, let $x\in A,$
then, for any $i,j=1,2,...,n,$ $b_{i,j}=u_{1,i}r_{i}xr_{j}u_{j,1}\in{\rm
Her}(\varphi(e_{1,1}){{)}}.$ Therefore

$\displaystyle\varphi(1_{n})x\varphi(1_{n})$ $\displaystyle=$
$\displaystyle\sum_{{1\leq i,j\leq n}}u_{i,i}r_{i}xr_{j}u_{j,j}$ (e 8.7)
$\displaystyle=$ $\displaystyle\sum_{1\leq i,j\leq
n}u_{i,1}(u_{1,i}xr_{j}u_{j,1})u_{1,j}=\sum_{1\leq i,j\leq
n}u_{i,1}b_{i{{,}}j}u_{1,j}\in B.$ (e 8.8)

It follows that ${\rm Her}(\varphi(1_{n}))=B.$ The lemma follows. ∎

######  Lemma 8.4.

Let $A$ be a simple $C^{*}$-algebra with
${\widetilde{QT}}(A)\setminus\\{0\\}\not=\emptyset.$ Suppose that $A={\rm
Ped}(A),$ $A$ has strict comparison and $\Gamma$ is surjective. Suppose that
$b\in{\rm Ped}(A\otimes{\cal K})_{+}^{\bf 1}.$ Then, for any $\varepsilon>0$
and any integer $n\geq 1,$ there is a c.p.c. order zero map
$\varphi:M_{n}\to{\rm Her}(b)$ such that

$\displaystyle\|b-\varphi(1_{n})b\|_{{}_{2,\overline{QT(A)}^{w}}}<\|b\|\sqrt{\omega(b)+\varepsilon}.$
(e 8.9)

Moreover, if $QT(A)=T(A),$ then

$\displaystyle\|b-\varphi(1_{n})b\|_{{}_{2,\overline{T(A)}^{w}}}<\|b\|_{{{}_{2,\overline{T(A)}^{w}}}}\sqrt{\omega(b)+\varepsilon}.$
(e 8.10)

###### Proof.

Fix $\varepsilon\in(0,1/2)$ and $n\in\mathbb{N}.$ Since $\Gamma:{\rm
Cu}(A)\to{\rm LAff}_{+}(\overline{QT(A)}^{w})$ is surjective, there is
$b_{1}\in(A\otimes{\cal K})_{+}^{\bf 1}$ such that
$d_{\tau}(b_{1})=(1/n)d_{\tau}(b)$ for all $\tau\in\overline{QT(A)}^{w}.$ We
compute that $\omega(b_{1})=(1/n)\omega(b).$ Choose $\delta>0$ such that

$\displaystyle\sup\\{d_{\tau}(b_{1})-\tau(f_{2\delta}(b_{1})):\tau\in{{\overline{QT(A)}^{w}}}\\}<\omega(b_{1})+\varepsilon/4n.$
(e 8.11)

Put $b_{2}=f_{\delta}(b_{1})$ and

$\displaystyle d_{1}={\rm diag}(\overbrace{b_{1},b_{1},...,b_{1}}^{n})\in
A\otimes{\cal K}.$ (e 8.12)

Then

$\displaystyle f_{\delta}(d_{1})={\rm
diag}(\overbrace{{{b_{2},b_{2},....,b_{2}}}}^{n})\in A\otimes{\cal K}.$ (e
8.13)

Since $A$ has strict comparison, by Proposition 2.2 and 2.3 of [42], there is
$x\in A\otimes{\cal K}$ such that

$\displaystyle x^{*}x=f_{\delta}(d_{1})\,\,\,{\rm and}\,\,\,xx^{*}\in{\rm
Her}(b).$ (e 8.14)

Then one obtains an isomorphism $\psi:{\rm Her}(x^{*}x)\to{\rm
Her}(xx^{*})\subset{\rm Her}(b)$ such that $\psi(f(x^{*}x))=f(xx^{*})$ for all
$f\in C_{0}((0,\|x\|^{2}]).$ It induces a homomorphism $\varphi_{c}:C_{0}({\rm
sp}(f_{\delta}({{b_{1}}})))\otimes M_{n}\to{\rm Her}(b)$ such that
$\varphi_{c}(\imath\otimes 1_{n})=xx^{*},$ where $\imath\in
C_{0}(f_{\delta}({{b_{1}}}))$ is the identity function on ${\rm
sp}({{f_{\delta}(b_{1})}}).$ Define a c.p.c. order zero map
$\varphi:M_{n}\to\mathrm{Her}(b)$ by
$\varphi(e_{i,j})=\varphi_{c}(\imath\otimes{{e_{i,j}}})$ ($1\leq i,j\leq 1$).

Let $p$ be the open projection in $A^{**}$ corresponding to $b$ which may be
identified with the identity of $\widetilde{\mathrm{Her}(b)}.$ We extend each
$\tau\in\overline{QT(A)}^{w}$ to a 2-quasitrace on
$\widetilde{\mathrm{Her}(b)}$ (see II.2.5 of [2]) such that
$\tau(p)=\|\tau|_{\mathrm{Her}(b)}\|=d_{\tau}(b).$ Then, we have, for all
$\tau\in\overline{QT(A)}^{w},$

$\displaystyle\tau((p-\varphi(1_{n}))^{2})\leq\tau(p-\varphi(1_{n}))=\tau(p)-\tau(\varphi(1_{n}))$
(e 8.15)
$\displaystyle=d_{\tau}(b)-\tau(f_{\delta}(d_{1}))<\omega(b)+\varepsilon.$ (e
8.16)

It follows that

$\displaystyle\|b-\varphi(1_{n})b\|_{{}_{2,\overline{QT(A)}^{w}}}\leq\|b\|\|p-\varphi(1_{n})\|_{{}_{2,\overline{QT(A)}^{w}}}<\|b\|\sqrt{\omega(b)+\varepsilon}.$
(e 8.17)

In the case that $QT(A)=T(A),$ one obtains (e 8.10). ∎

######  Theorem 8.5.

Let $A$ be a $\sigma$-unital simple $C^{*}$-algebra with
$QT(A)\setminus\\{0\\}\not=\emptyset.$ Suppose that $A$ has strict comparison
and T-tracial approximate oscillation zero. Then $A$ has property (TM).

###### Proof.

Choose $e\in{\rm Ped}(A)_{+}\setminus\\{0\\}$ and define
$A_{1}=\mathrm{Her}(e).$ By Brown’s stable isomorphism theorem ([4]),
$A_{1}\otimes{\cal K}\cong A\otimes{\cal K}.$ To show that $A$ has property
(TM), it suffices to show that $A_{1}$ has property (TM). To simplify
notation, without loss of generality, we may assume that $A={\rm Ped}(A).$

Fix $a\in{\rm Ped}(A\otimes{\cal K})$ such that $0\leq a\leq 1.$ Since $A$ has
T-tracial approximate oscillation zero, $\Omega^{T}(a)=0.$ It follows that
there is a sequence $c_{k}\in{\rm Her}(a)$ with $0\leq c_{k}\leq 1$ such that

$\displaystyle\Pi(\iota(a))=\Pi(c)\,\,\,{\rm
and}\,\,\,\lim_{k\to\infty}\omega(c_{k})=0,$ (e 8.18)

where $c=\\{c_{k}\\}$ and $\Pi:l^{\infty}(A)\to
l^{\infty}(A)/I_{{}_{\overline{QT(A)}^{w}}}$ is the quotient map. By Theorem
7.11, $\Gamma$ is surjective. Then, applying Lemma 8.4, we obtain a c.p.c.
order zero map $\varphi_{k}:M_{n}\to{\rm Her}(c_{k})\subset{\rm Her}(a)$ such
that

$\displaystyle\|c_{k}-\varphi_{k}(1_{n})c_{k}\|_{{}_{2,\overline{QT(A)}^{w}}}\leq\sqrt{\omega(c_{k})+1/k^{2}},\,\,k=1,2,....$
(e 8.19)

Fix $1>\varepsilon>0.$ Choose $k_{0}\geq 1$ such that

$\displaystyle\sqrt{\omega(c_{k})+1/k^{2}}<\min\\{(\varepsilon/4)^{3},1/4\\}\,\,\,{\rm
for\,\,\,all}\,\,\,k\geq k_{0}.$ (e 8.20)

Since $\Pi(\iota(a))=\Pi(c),$ there exists $k_{1}\geq k_{0}$ such that

$\displaystyle\|a-c_{k}\|_{{{}_{2,\overline{QT(A)}^{w}}}}<(\varepsilon/3)^{3}\,\,\,{\rm
for\,\,\,all}\,\,\,k\geq k_{1}.$ (e 8.21)

Choose $b=c_{k_{1}+1}.$ Then

$\displaystyle\|a-\varphi_{k}(1_{n})a\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}$
$\displaystyle\leq$
$\displaystyle\|a-b\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}+\|b-\varphi_{k}(1_{n})b\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}$
(e 8.23)
$\displaystyle+\|\varphi_{k}(1_{n})(b-a)\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}$
$\displaystyle<$
$\displaystyle\varepsilon/3+\varepsilon/4+\varepsilon/3<\varepsilon.$ (e 8.24)

The lemma follows. ∎

######  Lemma 8.6.

Let $A$ be a $\sigma$-unital algebraically simple $C^{*}$-algebra with
$QT(A)\not=\emptyset.$ Suppose that $A$ has property (TM) and $a\in{\rm
Ped}(A\otimes{\cal K})_{+}^{\bf 1}\setminus\\{0\\}.$ Then, for any integer
$n,r\in\mathbb{N},$ any $k\in\mathbb{N}$ ($k\geq 2$) and $\varepsilon>0,$
there exist an integer $m_{k}\geq r\in\mathbb{N}$ and mutually orthogonal
elements $b_{k,1},b_{k,2},...,b_{k,n}\in\mathrm{Her}(a)_{+}^{\bf 1}$ such that
$[b_{k,i}]=[b_{k,1}],$ $i=1,2,...,n,$

$\displaystyle d_{\tau}(f_{1/k}(a^{1/m_{k}}))\leq
nd_{\tau}(f_{1/k}(b_{k,1}))+\varepsilon\,\,\,\text{for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}$
(e 8.25)

and, an integer $l(k)\in\mathbb{N}$ such that

$\displaystyle nd_{\tau}(f_{1/k}(b_{k,1}))\leq
d_{\tau}(f_{1/l(k)}(a))\,\,\,\text{for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$
(e 8.26)

###### Proof.

Fix $n\in\mathbb{N}.$ Since $A$ has property (TM), for each $m\in\mathbb{N},$
there exists a c.p.c. order zero map $\varphi_{m}:M_{n}\to{\rm Her}(a)$ such
that

$\displaystyle\|a^{1/m}-a^{1/m}\varphi_{m}(1_{n})\|_{{}_{2,\overline{QT(A)}^{w}}}<1/2^{m}.$
(e 8.27)

Define $c:=\Pi(\\{a^{1/m}\\}{{{}_{m\in\mathbb{N}}}})$ and $\varphi:M_{n}\to
l^{\infty}(A)/I_{{}_{\overline{QT(A)}^{w}}}$ such that
$\varphi(f)=\Pi(\\{\varphi_{m}(f)\\})$ for all $f\in M_{n}.$ Then

$\displaystyle c=c\varphi(1_{n})=\varphi(1_{n})c=c.$ (e 8.28)

It follows that $c=cf_{1/2}(\varphi(1_{n}))=f_{1/2}(\varphi(1_{n}))c=c.$ Thus
$c\leq f_{1/2}(\varphi(1_{n})).$

Let $\varepsilon>0.$ By (1) of Lemma 7.2, for each integer $k\geq 2,$ there
exists $m_{k}{{\geq r}}\in\mathbb{N}$ such that, for all $m\geq m_{k},$

$\displaystyle[f_{1/k}(a^{1/m})]$ $\displaystyle\leq$
$\displaystyle[f_{1/2}(\varphi_{m}(1_{n}))]+[d_{m}]$ (e 8.29)
$\displaystyle\leq$
$\displaystyle[f_{1/k}(\varphi_{m}(e_{1,1}))+f_{1/k}(\varphi_{m}(e_{2,2}))+\cdots
f_{1/k}(\varphi_{m}(e_{n,n}))]+[d_{m}],$ (e 8.30)

where $\sup\\{d_{\tau}(d_{m}):\tau\in\overline{QT(A)}^{w}\\}<\varepsilon/2.$
Put $b_{k,i}=\varphi_{m_{k}}(e_{i,i})$ $1\leq i\leq n$ and $k\in\mathbb{N}.$
On the other hand, since $\varphi_{m_{k}}(1_{n})\in\mathrm{Her}(a),$ for each
$k,$ there is $l(k)\in\mathbb{N}$ such that

$\displaystyle\|f_{1/l(k)}(a)\varphi_{m_{k}}(1_{n})f_{1/l(k)}(a)-\varphi_{k}(1_{n})\|<1/2k.$
(e 8.31)

It follows that (see Proposition 2.2 of [42]), for $k\geq 2,$

$\displaystyle f_{1/k}(\varphi_{{m_{k}}}(1_{n}))\lesssim
f_{{1/l(k)}}(a)\,\,\,{\rm
and}\,\,\,n[f_{1/k}(b_{k,1})]=[f_{1/k}(\varphi_{{m_{k}}}(1_{n}))].$ (e 8.32)

The lemma then follows. ∎

######  Theorem 8.7.

Let $A$ be a $\sigma$-unital simple $C^{*}$-algebra with $QT(A)\not=\\{0\\}.$
If $A$ has strict comparison and property (TM), then $\Gamma$ is surjective
(see 2.13).

###### Proof.

It suffices to prove the proposition for the case that $A={\rm Ped}(A).$

We claim that, for any $a\in{\rm Ped}(A\otimes{\cal K})_{+}^{\bf
1}\setminus\\{0\\}$ and any integer $n\in\mathbb{N},$ there exists
$b\in(A\otimes{\cal K})_{+}$ such that

$\displaystyle n\widehat{[b]}\leq\widehat{[a]}\leq(n+1)\widehat{[b]}.$ (e
8.33)

Case (1): $0$ is an isolated point of ${\rm sp}(a).$ In this case we may
assume that $a=p$ for some projection $p.$ Choose

$\displaystyle\eta:=({1\over{(n+1)^{2}}})\inf\\{\tau(p):\tau\in\overline{QT(A)}^{w}\\}>0.$
(e 8.34)

Note that $f_{1/k}(p)=p$ for all $k>1.$ Applying Lemma 8.6, we obtain
$b\in\mathrm{Her}(a)_{+}$ such that

$\displaystyle nd_{\tau}(b)\leq d_{\tau}(p)<nd_{\tau}(b)+\eta\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.35)

Then we compute that

$\displaystyle n\widehat{[b]}\leq\widehat{[a]}\leq(n+1)\widehat{[b]}.$ (e
8.36)

Case (2): $0$ is not an isolated point of ${\rm sp}(a).$ We will use Lemma 8.6
for an induction argument. Put

$\displaystyle\sigma_{0}:=\inf\\{\tau(a):\tau\in\overline{QT(A)}^{w}\\}>0.$ (e
8.37)

Fix $n\in N.$ Since $0$ is a not an isolated point of ${\rm sp}(a),$ for each
integer $k,$ there is a smallest integer ${{J}}(k)>k$ such that

$\displaystyle f_{1/J(k)}(a)-f_{1/k}(a)\not=0.$ (e 8.38)

Define, for each $k,$

$\displaystyle\sigma_{k}$ $\displaystyle:=$
$\displaystyle\inf\\{d_{\tau}(f_{1/J(k)}(a))-d_{\tau}(f_{1/k}(a)):\tau\in\overline{QT(A)}^{w}\\}>0\,\,\,{\rm
and}\,\,\,$ (e 8.39) $\displaystyle\eta_{k}$ $\displaystyle:=$
$\displaystyle\min\\{\sigma_{j}:1\leq j\leq k+1\\}/2^{k+1}(n+1).$ (e 8.40)

Applying Lemma 8.6, there are mutually orthogonal elements
$b_{1,1},b_{1,2},...,b_{1,n}\in\mathrm{Her}(a)_{+}^{\bf 1}$ such that
$[b_{1,i}]=[b_{1,1}],$ $i=1,2,...,n,$ and, for some $m_{1}\in\mathbb{N},$

$\displaystyle{{d_{\tau}}}(f_{1/2}(a^{1/m_{1}}))\leq
nd_{\tau}(f_{1/2}(b_{1,1}))+\eta_{1}\,\,\,\text{for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w},$
(e 8.41)

and, an integer $l(1)\in\mathbb{N}$ such that

$\displaystyle
nd_{\tau}(f_{1/2}(b_{1,1}))\leq{{d_{\tau}}}(f_{1/l(1)}(a))\,\,\,\text{for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$
(e 8.42)

Put $c_{1}:=f_{1/2}(b_{1,1}).$ Then, for all $\tau\in\overline{QT(A)}^{w},$

$\displaystyle\widehat{[f_{1/2}(a^{1/m_{1}})]}(\tau)\leq
n\widehat{[c_{1}]}(\tau)+\eta_{1}\,\,\,{\rm
and}\,\,\,n\widehat{[c_{1}]}(\tau)\leq\widehat{[f_{1/l(1)}(a)]}(\tau).$ (e
8.43)

Choose $k_{2}>l(1)$ such that $k_{2}\geq J(l(1)).$ Applying Lemma 8.6, we
obtain $m_{2}\geq m_{1}$ and mutually orthogonal
$b_{2,1},b_{2,2},...,b_{2,n}\in\mathrm{Her}(a)_{+}^{\bf 1}$ with $b_{2,j}\sim
b_{2,1}$ ($1\leq j\leq n$) and $l(k_{2})\in\mathbb{N}$ such that

$\displaystyle
d_{\tau}(f_{1/k_{2}}(a^{1/m_{2}}){{)}}<nd_{\tau}(f_{1/k_{2}}(b_{2,1}))+\eta_{l(1)}\,\,\,{\rm
and}\,\,\,$ (e 8.44) $\displaystyle nd_{\tau}(b_{2,1})\leq
d_{\tau}(f_{1/l(2)}(a))\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.45)

Put $c_{2}=f_{1/k_{2}}(b_{2,1}).$ Then, for all $\tau\in\overline{QT(A)}^{w},$

$\displaystyle[f_{1/k_{2}}(a^{1/m_{2}}){{\widehat{]}}}(\tau)\leq
n\widehat{[c_{2}]}(\tau)+\eta_{l(1)}\,\,\,{\rm
and}\,\,\,n\widehat{[c_{2}]}(\tau)\leq[f_{1/l(2)}(a){{\widehat{]}}}(\tau).$ (e
8.46)

We compute that, for all $\tau\in\overline{QT(A)}^{w}$ (recall that
$k_{2}>J(l(2)),$ and also (e 8.44) and (e 8.42)),

$\displaystyle nd_{\tau}(f_{1/k_{2}}(b_{2,1}))$ $\displaystyle>$
$\displaystyle d_{\tau}(f_{1/k_{2}}(a^{1/m_{2}}))-\eta_{l(1)}$ (e 8.47)
$\displaystyle=$ $\displaystyle
d_{\tau}(f_{1/l(1)}(a))+(d_{\tau}(f_{1/k_{2}}(a^{1/m_{2}}))-d_{\tau}(f_{1/l(1)}(a)))-\eta_{l(1)}$
(e 8.48) $\displaystyle\geq$ $\displaystyle
nd_{\tau}(f_{1/2}(b_{1,1}))+(d_{\tau}(f_{1/k_{2}}(a^{{1/m_{2}}}))-d_{\tau}(f_{1/l(1)}(a)))-\eta_{l(1)}$
(e 8.49) $\displaystyle>$ $\displaystyle
nd_{\tau}(f_{1/2}(b_{1,1}))+\sigma_{l(1)}-\eta_{l(1)}>nd_{\tau}(f_{1/2}(b_{1,1})).$
(e 8.50)

Since $A$ has strict comparison, we obtain

$\displaystyle[c_{1}]\leq[c_{2}].$ (e 8.51)

Suppose that we have constructed integers $k_{i},m_{i},l(i)\in\mathbb{N}$ and
$b_{i,1}\in\mathrm{Her}(a)_{+}^{\bf 1},$ $1\leq i\leq I$ such that, for all
$2\leq i\leq I$ and $\tau\in\overline{QT(A)}^{w},$

$\displaystyle k_{i}>J(l(i-1))>l(i-1),$ (e 8.52)
$\displaystyle[f_{1/k_{i}}(a){{\widehat{]}}}(\tau)<n[f_{1/k_{i}}(b_{i,1}){{\widehat{]}}}(\tau)+\eta_{l(i-1)}\,\,\,{\rm
and}\,\,\,$ (e 8.53) $\displaystyle
n[f_{1/k_{i}}(b_{i,1}){{\widehat{]}}}(\tau)\leq[f_{1/l(i)}(a){\widehat{]}}(\tau),$
(e 8.54)

and verified that $[f_{1/k_{i}}(b_{i,1})]\leq[f_{1/k_{i+1}}(b_{i+1,1})]$
$1\leq i\leq I-1.$

Define $c_{i}=f_{1/k_{i}}(b_{i,1}),$ $1\leq i\leq I.$ By applying Lemma 8.6,
there is $k_{I+1}>J(l(i))>l(i),$ $l(I+1)\geq k_{I+1},$ $m_{I+1}\geq m_{I},$
and $b_{I+1,1}\in\mathrm{Her}(a)_{+}^{\bf 1}$ such that, for all
$\tau\in\overline{QT(A)}^{w},$

$\displaystyle[f_{1/k_{I+1}}(a^{1/m(I+1)}){\widehat{]}}(\tau)<n[f_{1/k_{I+1}}(b_{I+1,1}){\widehat{]}}(\tau)+\eta_{l(I)}\,\,\,{\rm
and}\,\,\,$ (e 8.55) $\displaystyle
n[f_{1/k_{I+1}}(b_{I+1,1}{\widehat{]}}(\tau)\leq[f_{1/l(I+1)}(a){\widehat{]}}(\tau).$
(e 8.56)

Then, for all $\tau\in\overline{QT(A)}^{w},$

$\displaystyle nd_{\tau}(f_{1/k_{I+1}}(b_{I+1,1}))$ $\displaystyle>$
$\displaystyle d_{\tau}(f_{1/k_{I+1}}(a^{1/m_{I+1}}))-\eta_{l(I)}$ (e 8.57)
$\displaystyle=$ $\displaystyle
d_{\tau}(f_{1/l(I)}(a))+(d_{\tau}(f_{1/k_{I+1}}(a^{1/m_{I+1}}))-d_{\tau}(f_{1/l(I)}(a)))-\eta_{l(I)}$
$\displaystyle\geq$ $\displaystyle
nd_{\tau}(f_{1/k_{I}}(b_{I,1}))+(d_{\tau}(f_{1/k_{I+1}}(a^{1/m_{I+1}}))-d_{\tau}(f_{1/l(I)}(a)))-\eta_{l(I)}$
$\displaystyle>$ $\displaystyle
nd_{\tau}(f_{1/k_{I}}(b_{I,1}))+\sigma_{l(I)}-\eta_{l(I)}>nd_{\tau}(f_{1/k_{I}}(b_{I,1})).$

Put $c_{I+1}=f_{1/k_{I+1}}(b_{I+1,1}).$ Then, by the strict comparison,
estimates above imply that

$\displaystyle[c_{I}]\leq[c_{I+1}].$ (e 8.58)

Thus, by induction, we obtain an increasing sequence
$c_{i}\in\mathrm{Her}(a)_{+}^{\bf 1}$ such that, for all
$\tau\in\overline{QT(A)}^{w},$

$\displaystyle[f_{1/k_{i}}(a){\widehat{]}}(\tau)<n\widehat{[c_{i}]}(\tau)+\eta_{l(i-1)}\,\,\,{\rm
and}\,\,\,$ (e 8.59) $\displaystyle
n\widehat{[c_{i}]}(\tau)\leq[f_{1/l(i)}(a){\widehat{]}}(\tau),\,\,\,i\geq 2.$
(e 8.60)

Let $c\in(A\otimes{\cal K})_{+}$ be such that $[c]$ is the supremum of
$\\{[c_{i}]\\}.$ Then, by (e 8.59), for all $i,$

$\displaystyle d_{\tau}(f_{1/k_{i}}(a))\leq
n\widehat{[c]}(\tau)+\eta_{l(i)}\leq
n\widehat{[c]}(\tau)+\sigma_{0}/2^{i+1}(n+1)\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.61)

Thus (since $A$ is simple), for all sufficiently large $i,$

$\displaystyle d_{\tau}(f_{1/k_{i}}(a))\leq(n+1)\widehat{[c]}(\tau)\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.62)

It follows that (let $i\to\infty$)

$\displaystyle d_{\tau}(a)\leq(n+1)\widehat{[c]}(\tau)\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.63)

On the other hand, by (e 8.60)

$\displaystyle n\widehat{[c_{i}]}\leq\widehat{[a]}\,\,\,{\rm
for\,\,\,all}\,\,\,i\in\mathbb{N}.$ (e 8.64)

For any $\varepsilon>0,$ choose a non-zero element $e\in A_{+}$ such that
$d_{\tau}(e)<\varepsilon/2$ for all $\tau\in\overline{QT(A)}^{w}.$ Then, by
the strict comparison,

$\displaystyle n[c_{i}]\leq[a]+[e]\,\,\,{\rm
for\,\,\,all}\,\,\,i\in\mathbb{N}.$ (e 8.65)

If follows that

$\displaystyle n[c]\leq[a]+[e].$ (e 8.66)

Hence

$\displaystyle n\widehat{[c]}<d_{\tau}(a)+\varepsilon\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.67)

Let $\varepsilon\to 0.$ We also obtain

$\displaystyle n\widehat{[c]}\leq\widehat{[a]}\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.68)

Combining (e 8.68) and (e 8.63), the claim also holds for Case (2).

We now show that the proved claim implies that $\mathrm{Cu}(A)$ has the
property D stated in the proof of Proposition 6.21 of [39]. Let $x^{\prime}\ll
x,$ where $x=[a]$ for some $a\in(A\otimes{\cal K})_{+}.$ Since
$x=\sup\\{[(a-\|a\|/k)_{+}]:k\in\mathbb{N}\\},$
$x^{\prime}\leq[(a-\|a\|/k)_{+}]$ for some $k\in\mathbb{N}.$ Note
$(a-\|a\|/k)_{+}\in{\rm Ped}(A\otimes{\cal K})_{+}.$ Then the claim implies
that there is $y\in\mathrm{Cu}(A)$ such that

$\displaystyle\widehat{x^{\prime}}\leq[(a-\|a\|/k)_{+}{\widehat{]}}\leq(n+1)\widehat{y}\,\,\,{\rm
and}\,\,\,n\widehat{y}\leq[(a-\|a\|/k)_{+}{\widehat{]}}\leq[a]=x.$ (e 8.69)

Therefore, as observed by L. Robert (see Property D in the proof of
Proposition 6.21 of [39]) following Corollary 5.8 of [13], $\Gamma$ is
surjective. ∎

######  Lemma 8.8.

Let $A$ be a $C^{*}$-algebra and $a,b\in A_{+}^{\bf 1}.$ Suppose that there is
$x\in A$ such that

$\displaystyle x^{*}x=a\,\,\,{\rm and}\,\,\,xx^{*}\in{\rm Her}(b).$ (e 8.70)

Then, for any $\varepsilon>0,$ there exists a unitary $U\in
M_{2}({{\widetilde{A}}})$ such that

$\displaystyle U^{*}{\rm diag}(f_{\varepsilon}(a),0)U\in{\rm Her}({\bar{b}}),$
(e 8.71)

where ${\bar{b}}={\rm diag}(b,0).$

###### Proof.

First we claim that, for any $y\in A,$ ${\rm dist}({\rm
diag}(y,0),GL(M_{2}({{\widetilde{A}}})))=0.$ To see this, let $\varepsilon>0.$
Choose $V=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ which is a unitary in
$M_{2}(\mathbb{C}).$ Then

$Y:=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}\begin{pmatrix}y&0\\\
0&0\end{pmatrix}=\begin{pmatrix}0&0\\\ y&0\end{pmatrix}$

which is a nilpotent. Therefore $Y+\varepsilon\cdot 1_{2}\in
GL(M_{2}({{\widetilde{A}}})).$ Then ${\rm
diag}(y,0)\approx_{\varepsilon}V^{*}(Y+\varepsilon)\in
GL(M_{2}({{\widetilde{A}}})).$ This proves the claim.

To prove the lemma, we will combine the claim with an argument of Rørdam. By
Proposition 2.4 of [42], for any $\varepsilon>0,$ there exist $\delta>0$ and
$r\in A$ such that

$\displaystyle f_{\varepsilon/2}(a)=rf_{\delta}(b)r^{*}.$ (e 8.72)

Put $z=rf_{\delta}(b)^{1/2}$ and $Z={\rm diag}(z,0).$ By the claim and Theorem
5 of [35], there is a unitary $U\in M_{2}({{\widetilde{A}}})$ such that

$\displaystyle U^{*}f_{1/2}(ZZ^{*})U=U^{*}f_{1/2}({\rm diag}(zz^{*},0))U={\rm
diag}(f_{1/2}(z^{*}z),0)=f_{1/2}(Z^{*}Z).$ (e 8.73)

Note that $Z^{*}Z\in{\rm Her}({\bar{b}}),$ where ${\bar{b}}={\rm diag}(b,0).$
Moreover (with ${\bar{a}}={\rm diag}(a,0)$)

$\displaystyle U^{*}f_{\varepsilon}({\bar{a}})U\leq
U^{*}f_{1/2}(f_{\varepsilon/2}({\bar{a}})U=U^{*}f_{1/2}(ZZ^{*})U=f_{1/2}(Z^{*}Z).$
(e 8.74)

∎

######  Lemma 8.9.

Let $A$ be an algebraically simple $C^{*}$-algebra which has strict
comparison. Suppose that the canonical map $\Gamma$ is surjective.

Suppose that $a,a^{\prime}\in{\rm Ped}(A)^{\bf 1}\setminus\\{0\\}$ with
$a\in\mathrm{Her}(a^{\prime}).$ Then, there exists $1/2>\varepsilon_{0}$
satisfying the following: for any $0<\eta<\varepsilon<\varepsilon_{0},$ any
$\sigma>0,$ there exist $c\in{\rm Her}(f_{\eta}(a))_{+}^{\bf 1}$ with
$\|c\|\leq\|a\|$ and unitary $U\in M_{2}({\rm Her}(a^{\prime})^{\sim})$ such
that (with $b=U^{*}{\rm diag}(c,0)U$)

$\displaystyle(1)\hskip 14.45377pt{\rm diag}(f_{\varepsilon}(a),0)\leq b,$ (e
8.75) $\displaystyle(2)\hskip 14.45377ptd_{\tau}(f_{\varepsilon}(a))\leq
d_{\tau}(b)\leq
d_{\tau}(f_{\eta}(a))\,\,\,\text{for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w},$
(e 8.76)

and, for some $1>\delta>0,$

$\displaystyle(3)$
$\displaystyle\,\,\,|d_{\tau}(b)-\tau(f_{\delta}(b))|<\sigma\,\,\,\text{for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$
(e 8.77)

Moreover,

$\displaystyle(4)$ $\displaystyle\hskip 14.45377ptU^{*}{\rm
diag}(g_{\eta/2}(a),a^{\prime})U\in B,$ (e 8.78)

where $B:={\rm Her}(b)^{\perp}\cap{\rm Her}({\rm diag}(a,a^{\prime}))$ and
$g_{\eta}(t)\in C_{0}((0,1])$ is defined as in 2.5.

Consequently, if $e$ is a strictly positive element in $({\rm
Her}(b)^{\perp}\cap{\rm Her}({\rm diag}(a,a^{\prime}))),$ then

$\displaystyle
d_{\tau}(e)>d_{\tau}(a^{\prime})+d_{\tau}(g_{\eta}(a))\,\,\,\text{for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$
(e 8.79)

###### Proof.

Without loss of generality, we may assume that $\|a\|\leq 1.$ Let us first
assume that $[0,\varepsilon_{0})\subset{\rm sp}(a)$ for some
$\varepsilon_{0}>0.$

Fix $0<\varepsilon<\varepsilon_{0}.$ Note that, without loss of generality, we
may assume that

$\displaystyle
d_{\tau}(f_{\varepsilon}(a))<\tau(f_{\delta_{1}}(a))<d_{\tau}(f_{\eta_{1}}(a))<\tau(f_{\delta_{2}}(a))<d_{\tau}(f_{\eta}(a))\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w},$ (e 8.80)

where
$\varepsilon/2>\delta_{1},\delta_{1}/2>\eta_{1},\eta_{1}/2>\delta_{2},\delta_{2}/2>\eta.$
Put $h_{i}(\tau)=\tau(f_{\delta_{i}}(a))\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w},$ $i=1,2.$ Then
$h_{i}\in\operatorname{Aff}_{+}(\overline{QT(A)}^{w}),$ $i=1,2.$

Since $\Gamma$ is surjective, there is $c_{0}\in(A\otimes{\cal K})_{+}$ such
that $d_{\tau}(c_{0})=h_{2}(\tau)$ for all $\tau\in{{\overline{QT(A)}^{w}.}}$
Choose $\delta_{0}>0$ such that (as $h_{2}$ is continuous)

$\displaystyle
d_{\tau}(c_{0})-\tau(f_{\delta_{0}}(c_{0}))<\sigma/2\,\,\,\text{for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$
(e 8.81)

Since $h_{1}<h_{2}$ are continuous, we may also assume that

$\displaystyle
d_{\tau}(f_{\delta_{0}}(c_{0}))>d_{\tau}(f_{\varepsilon}(a))\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.82)

Since $A$ has strict comparison, by (e 8.80), there is $x\in A\otimes{\cal K}$
such that

$\displaystyle x^{*}x=f_{\delta_{0}/4}(c_{0})\,\,\,{\rm
and}\,\,\,xx^{*}\in{\rm Her}(f_{\eta}(a)).$ (e 8.83)

Choose $c=xx^{*}.$ Then $0\leq c\leq 1$ and
$d_{\tau}(c)=d_{\tau}(f_{\delta_{0}/4}(c_{0}))$ for all
$\overline{QT(A)}^{w}.$ Let $C={\rm Her}(f_{\eta}(a)).$ By (e 8.82), the
strict comparison and Lemma 8.8, we obtain a unitary $U\in
M_{2}(\widetilde{C})$ such that

$\displaystyle U{\rm diag}(f_{\eta_{1}}(a),0)U^{*}\in{\rm Her}({\bar{c}}),$ (e
8.84)

where ${\bar{c}}={\rm diag}(c,0).$ Let $b=U^{*}{\bar{c}}U.$ Then

$\displaystyle{\rm diag}(f_{\eta_{1}}(a),0)\leq b.$ (e 8.85)

(So (1) holds). Hence $d_{\tau}(b)=d_{\tau}(c)$ for all
$\tau\in\overline{T(A)}^{w}.$ Consequently

$\displaystyle d_{\tau}(f_{\eta_{1}}(a))\leq d_{\tau}(b)\leq
d_{\tau}(f_{\eta}(a))\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}$ (e 8.86)

(so (2) holds). Moreover, there is $1>\delta>0$ such that

$\displaystyle d_{\tau}(f_{\delta}(b))\geq\tau(f_{\delta_{0}}(c))\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.87)

It then follows from (e 8.81) that

$\displaystyle|d_{\tau}(b)-\tau(f_{\delta}(b))|=|d_{\tau}(f_{\delta_{0}/4}(c_{0}))-\tau(f_{\delta}(b))|<\sigma\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}$ (e 8.88)

(so (3) holds). To show the “ Moreover” part, put $B={\rm
Her}(b)^{\perp}\cap{\rm Her}({\rm diag}(a,a^{\prime}))$ and consider
$e^{\prime}=U^{*}{\rm diag}(g_{\eta/2}(a),a^{\prime})U.$ Since
$g_{\eta/2}(a)\perp f_{\eta}(a),$ we have ${\rm
diag}(g_{\eta/2}(a),a^{\prime})\perp{\rm diag}(f_{\eta}(a),0)$ and

$\displaystyle e^{\prime}\perp b.$ (e 8.89)

It follows that

$\displaystyle U^{*}{\rm diag}(g_{\eta/2}(a),a^{\prime})U\in B.$ (e 8.90)

If $e$ is a strictly positive element in ${\rm Her}(b)^{\perp}\cap{\rm
Her}({\rm diag}(a,a^{\prime})),$ then

$\displaystyle d_{\tau}(e)\geq
d_{\tau}(e^{\prime})=d_{\tau}(a^{\prime})+d_{\tau}(g_{\eta/2}(a))\,\,\,{\rm
for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}.$ (e 8.91)

This proves the case that $[0,\varepsilon_{0}]\subset{\rm sp}(a).$

If there exists $r_{n}\in(0,1]$ with $r_{n}>r_{n+1}$ and
$\lim_{n\to\infty}r_{n}=0$ such that $r_{n}\not\in{\rm sp}(a),$ then
$b_{n}=f_{2r_{n}}(a)$ has the property that $\omega(b_{n})=0.$ Then the lemma
easily follows by choosing $U={\rm diag}(1,1)$ and $b=b_{n}$ for some
sufficiently large $n.$

∎

######  Lemma 8.10.

Let $A$ be a $\sigma$-unital algebraically simple $C^{*}$-algebra with
$QT(A)=\emptyset.$ Suppose that $A$ has strict comparison and $\Gamma$ is
surjective. Suppose that $a={\rm diag}(0,a_{1},a_{2},...,a_{n})$ in
$M_{n+1}(A){{{}^{\bf 1}_{+}}}$ for some integer $n\geq 1.$ Then, for any
$1/2>\varepsilon>0$ and $1/2>\sigma>0,$ there exists $d\in M_{n+1}(A){{{}^{\bf
1}_{+}}}$ such that

$\displaystyle f_{\varepsilon}(a)\leq
d\,\,\,\text{and}\,\,\,\omega(d)<\sigma.$ (e 8.92)

###### Proof.

For $n=1,$ this follows immediately from Lemma 8.9.

Assume that the lemma holds for $n\geq 1.$

Let $0\leq e_{A}\leq 1$ be a strictly positive element of $A.$ Fix
$1/2>\varepsilon>0.$ Choose $\eta=\varepsilon/4(n+2)$ and
$\sigma_{0}:=\sigma/2(n+2).$ We will apply Lemma 8.9 with
$a_{j}\in\mathrm{Her}(e_{A})$ ($1\leq j\leq n+1$), $\eta$ as above, and
$\sigma_{0}$ (in place of $\sigma$).

By Lemma 8.9, there is $c_{1}\in{\rm Her}(f_{\eta}(a_{1}))_{+}^{\bf 1},$ a
unitary $U_{1}\in M_{2}({{\widetilde{A}}})$ and
${{b_{1}^{\prime}}}=U_{1}^{*}{\rm diag}(0,c_{1})U_{1}$ such that

$\displaystyle{\rm diag}(0,f_{\varepsilon}(a_{1}))\leq b_{1}^{\prime},$ (e
8.93) $\displaystyle d_{\tau}(f_{\varepsilon}(a_{1}))\leq
d_{\tau}(b_{1}^{\prime})\leq
d_{\tau}(f_{\eta}(a_{1}))\,\,\,\text{for\,\,\,all}\,\,\,\tau\in\overline{QT(A)}^{w}{\color[rgb]{0,0,1}{,}}$
(e 8.94) $\displaystyle\omega(b_{1})<\sigma_{0},$ (e 8.95) $\displaystyle
U_{1}^{*}{\rm diag}(e_{A},g_{\eta/2}(a_{1}))U_{1}\in B_{1},$ (e 8.96)

where $B_{1}:=({\rm Her}({{b_{1}^{\prime}}})^{\perp}\cap{\rm Her}({\rm
diag}(a_{1},e_{A})))_{+}.$ Put

$\displaystyle V_{1}={\rm
diag}(U_{1},\overbrace{1_{{{\widetilde{A}}}},...,1_{{{\widetilde{A}}}}}^{n-1}),\,\,{\alpha}_{2}=V_{1}^{*}{\rm
diag}(0,0,a_{2},...,a_{n+1})V_{1}\,\,\,{\rm and}\,\,\,$ (e 8.97)
$\displaystyle{{b_{1}}}=V_{1}^{*}(0,c_{1},0,...,0)V_{1}.$ (e 8.98)

Define $C_{1}={\rm Her}(V_{1}^{*}(e_{A},0,e_{A},...,e_{A})V_{1}).$ Then

$\displaystyle b_{1}\in C_{1}^{\perp}\,\,\,{\rm and}\,\,\,C_{1}\cong
M_{n+1}(A).$ (e 8.99)

In $C_{1},$ we may write ${\alpha}_{2}={\rm diag}(0,a_{2},a_{3},...,a_{n+1})$
(the number of possible nonzero elements is now reduced to $n$).

By the induction assumption, there is $b_{2}\in C_{1}$ with $0\leq b_{2}\leq
1$ such that

$\displaystyle f_{\varepsilon}({\alpha}_{2})\leq b_{2}\,\,\,{\rm
and}\,\,\,\omega(b_{2})<\sigma_{0}.$ (e 8.100)

Define $d:=b_{1}+b_{2}.$ Note that $b_{1}\perp b_{2}.$ Then

$\displaystyle f_{\varepsilon}(a)={\rm
diag}(0,f_{\varepsilon}(a_{1}),f_{\varepsilon}(a_{2}),...,f_{\varepsilon}(a_{n+1}))\leq
b_{1}+b_{2}.$ (e 8.101)

and by (2) of Lemma 4.4,

$\displaystyle\omega(d)\leq\omega(b_{1})+\omega(b_{2})<2\sigma_{0}<\sigma.$ (e
8.102)

This completes the induction. The lemma follows.

∎

######  Theorem 8.11.

Let $A$ be a $\sigma$-unital simple $C^{*}$-algebra with
$\widetilde{QT}(A)\setminus\\{0\\}\not=\emptyset.$ Suppose that $A$ has strict
comparison and property (TM). Then $A$ has $T$-tracial approximate oscillation
zero.

###### Proof.

Choose $e\in{\rm Ped}(A)_{+}\setminus\\{0\\}$ and $A_{1}=\mathrm{Her}(e).$
Then ${\rm Ped}(A_{1})=A_{1}.$ To prove the theorem, without loss of
generality, we may assume that $A={\rm Ped}(A).$ By Theorem 8.7, $\Gamma$ is
surjective.

Fix $a\in{\rm Ped}(A\otimes{\cal K})_{+}^{\bf 1}$ and $1/2>\varepsilon>0.$
Choose $n\in\mathbb{N}$ such that $1/n<(\varepsilon/8)^{3}.$ Put
$B=\mathrm{Her}(a).$ Then ${\rm Ped}(B)=B$ and $B\otimes{\cal K}\cong
A\otimes{\cal K}.$ Therefore, to simplify notation, without loss of
generality, we may assume that $a\in A.$

We claim that ${\rm Her}(a)$ has a T-tracial approximate identity
$\\{d_{n}\\}$ such that $\lim_{n\to\infty}\omega(d_{n})=0.$ Since $A$ has
property (TM), there is a c.p.c. order zero map $\varphi:M_{n+1}\to{\rm
Her}(a)$ such that

$\displaystyle\|a(1-\varphi(1_{n+1})a\|_{{}_{2,\overline{QT(A)}^{w}}}<(\varepsilon/8)^{3}.$
(e 8.103)

By Proposition 8.3, let $B={\rm Her}(\varphi(1_{n+1}))\cong M_{n+1}(({\rm
Her}(\varphi(e_{1,1}))).$ Write

$\displaystyle\varphi(1_{n+1})={\rm diag}(c,c,...,c)\in M_{n+1}({\rm
Her}(\varphi(e_{1,1}))).$ (e 8.104)

Choose $1/2>\eta>0$ such that

$\displaystyle\|f_{\eta}(\varphi(1_{n+1}){{)}}\varphi(1_{n+1})-\varphi(1_{n+1})\|<\varepsilon/8.$
(e 8.105)

Put $c_{n}={\rm diag}(0,c,c,...,c).$ It follows from Lemma 8.10 that there
exists $d\in B_{+}^{\bf 1}$ such that

$\displaystyle f_{\eta}(c_{n})\leq d\,\,\,{\rm and}\,\,\,\omega(d)<1/2^{n}.$
(e 8.106)

Note that

$\displaystyle\|\varphi(1_{n+1})-c_{n}\|_{{}_{2,\overline{QT(A)}^{w}}}<{1\over{n+1}}{\color[rgb]{0,0,1}{.}}$
(e 8.107)

Then

$\displaystyle\|(d-\varphi(1_{n+1}))a\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}<\|(d-c_{n})a\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}+({1\over{n+1}})^{2/3}.$
(e 8.108)

Next, by II.2.5 of [2], we extend each $\tau\in\overline{QT(A)}^{w}$ to a
2-quasitrace $\widetilde{A}.$ Then, by (e 8.107) and (e 8.103), we estimate
that

$\displaystyle\|(d-c_{n})a\|_{{}_{2,\overline{QT(A)}^{w}}}^{4/3}$
$\displaystyle=$
$\displaystyle(\sup_{{}_{\tau\in\overline{QT(A)}^{w}}}\\{\tau(a(d-c_{n})^{2}a)\\})^{2/3}\leq\sup_{{}_{\tau\in\overline{QT(A)}^{w}}}\\{\tau(a(1-c_{n})a)\\}^{2/3}$
(e 8.109) $\displaystyle\leq$
$\displaystyle\sup_{{}_{\tau\in\overline{QT(A)}^{w}}}\\{\tau((a(1-c_{n})a)^{2})^{1/3}\\}\sup_{{}_{\tau\in\overline{QT(A)}^{w}}}\\{\tau(1_{\widetilde{A}})^{1/3}\\}$
$\displaystyle\leq$
$\displaystyle\|a(1-c_{n})a\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}$
$\displaystyle\leq$
$\displaystyle\|a(1-\varphi(1_{n+1}))a\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}+({1\over{n+1}})^{2/3}$
(e 8.110) $\displaystyle<$
$\displaystyle(\varepsilon/8)^{2}+({1\over{n+1}})^{2/3}<(\varepsilon/4)^{2}.$
(e 8.111)

Therefore (from (e 8.108))

$\displaystyle\|(d-\varphi(1_{n+1}))a\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}<\varepsilon/4+({1\over{n+1}})^{2/3}<3\varepsilon/8.$
(e 8.112)

Finally

$\displaystyle\|a-ad\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}$
$\displaystyle\leq$
$\displaystyle\|a-a\varphi(1_{n+1})\|_{{}_{2,\overline{QT(A)}^{w}}}^{2/3}+\|a\varphi(1_{n+1})-ad\|^{2/3}_{{}_{2,\overline{QT(A)}^{w}}}$
(e 8.113) $\displaystyle<$
$\displaystyle\varepsilon/8+3\varepsilon/8<\varepsilon.$ (e 8.114)

This proves the claim. The theorem then follows from the claim and Lemma 5.7.

∎

##  9 Stable rank one

Let $A$ be a $C^{*}$-algebra and $n\in\mathbb{N}.$ Recall that we view
$M_{n}(A)$ as a $C^{*}$-subalgebra of $M_{n+1}(A)$ in the canonical way, i.e.,
$M_{n}(A)$ is the upper left block of $M_{n+1}(A).$

Recall an element $a=(a_{i,j})_{n\times n}$ in $M_{n}(A)$ is called upper
(resp. lower) triangular, if $a_{i,j}=0$ whenever $i<j$ (resp. $i>j$), and,
$a$ is called strictly upper (resp. lower) triangular, if $a_{i,j}=0$ whenever
$i\leq j$ (resp. $i\geq j$).

The following proposition is a generalization of an elementary fact (and
proof) in linear algebra.

######  Proposition 9.1.

Let $A$ be a $C^{*}$-algebra such that
$A\subset\overline{GL({{\widetilde{A}}})}.$ Then, for any $n\in\mathbb{N},$
any $a\in M_{n}(A),$ any $\varepsilon>0,$ there is an upper triangular matrix
$x\in M_{n}(A)$ and a lower triangular matrix $y\in M_{n}(A)$ such that
$a\approx_{\varepsilon}xy.$

###### Proof.

We prove this by induction on $n.$ For $n=1,$ let $a\in M_{n}(A)=A$ and
$\varepsilon>0.$ By the existence of approximate identity, there is $e\in
A_{+}$ such that $a\approx_{\varepsilon}ae.$ Note that $a$ and $e$ are
triangular matrices in $M_{n}(A).$ Thus the proposition holds for $n=1.$

Assume the proposition holds for $n\geq 1.$ Let
$a=\sum_{i,j=1}^{n+1}a_{i,j}\otimes e_{i,j}\in A\otimes M_{n+1},$ where
$a_{i,j}\in A$ and $\\{e_{i,j}\\}$ is the matrix units of $M_{n+1},$
$i,j=1,...,n+1.$ Let $\varepsilon>0.$ Since
$A\subset\overline{GL({{\widetilde{A}}})},$ there is $\tilde{a}\in
GL(\widetilde{A})$ such that $a_{n+1,n+1}\approx_{\varepsilon/2}\tilde{a}.$ In
what follows in this proof, $1$ is the identity of ${{\widetilde{A}}}$ and
$1_{n+1}$ is the identity of $M_{n+1}({{\widetilde{A}}}).$

Let $b^{(0)}:=\sum_{i=1}^{n}a_{i,n+1}\tilde{a}^{-1}\otimes e_{i,n+1}$ and
$c^{(0)}:=\sum_{j=1}^{n}\tilde{a}^{-1}a_{n+1,j}\otimes e_{n+1,j}.$ Then
$b^{(0)}$ and $c^{(0)}$ are nilpotents.

$\displaystyle a^{\prime}:=a+(\tilde{a}-a_{n+1,n+1})\otimes e_{n+1,n+1},$ (e
9.1) $\displaystyle b:=1_{n+1}-b^{(0)}\quad\text{and}\quad
c:=1_{n+1}-c^{(0)}.$ (e 9.2)

Let $s:=ba^{\prime}c.$ Note that $a^{\prime},b,c,s$ are in
$M_{n+1}(\widetilde{A}).$ Let $a^{\prime}_{i,j}$ (resp.
$b_{i,j},c_{i,j},s_{i,j}$) be the $(i,j)$-th entry of $a^{\prime}$ (resp.
$b,c,s$), $1\leq i,j\leq n+1.$ Note that

$\displaystyle
s_{i,j}=\sum_{m=1}^{n+1}\left(\sum_{k=1}^{n+1}b_{i,k}a^{\prime}_{k,m}c_{m,j}\right)\quad(1\leq
i,j\leq n+1).$ (e 9.3)

Denote by $\pi:M_{n+1}({{\widetilde{A}}})\to M_{n+1}$ the quotient map. Then
$\pi(b)=\pi(1_{n+1})=\pi(c)$ and $\pi(a)=\pi(\tilde{a}\otimes e_{n+1,n+1}).$
It follows that $\pi(s)=\pi(\tilde{a}\otimes e_{n+1,n+1}).$ Thus

$\displaystyle s_{i,j}\in A\quad(1\leq i,j\leq n).$ (e 9.4)

Note that $b_{n+1,k}=0$ for $1\leq k\leq n$ and $c_{m,j}=0$ for
$m\notin\\{j,n+1\\},$ if $1\leq j\leq n.$ By (e 9.3), we have, for $1\leq
j\leq n,$

$\displaystyle s_{n+1,j}$ $\displaystyle=$ $\displaystyle
b_{n+1,n+1}a^{\prime}_{n+1,j}c_{j,j}+b_{n+1,n+1}a^{\prime}_{n+1,n+1}c_{n+1,j}$
(e 9.5) $\displaystyle=$ $\displaystyle
a_{n+1,j}+\tilde{a}\cdot(-\tilde{a}^{-1}a_{n+1,j})=0.$ (e 9.6)

If $1\leq i\leq n,$ then $b_{i,k}=0$ for $k\notin\\{i,n+1\\}$ and
$c_{m,n+1}=0$ for $1\leq m\leq n.$ By (e 9.3), we compute

$\displaystyle s_{i,n+1}$ $\displaystyle=$ $\displaystyle
b_{i,i}a^{\prime}_{i,n+1}c_{n+1,n+1}+b_{i,n+1}a^{\prime}_{n+1,n+1}c_{n+1,n+1}$
(e 9.7) $\displaystyle=$ $\displaystyle
a_{i,n+1}+(-a_{i,n+1}\tilde{a}^{-1})\cdot\tilde{a}=0.$ (e 9.8)

We also have

$\displaystyle
s_{n+1,n+1}=\sum_{m=1}^{n+1}\left(\sum_{k=1}^{n+1}b_{n+1,k}a^{\prime}_{k,m}c_{m,n+1}\right)=b_{n+1,n+1}a^{\prime}_{n+1,n+1}c_{n+1,n+1}=\tilde{a}.$
(e 9.9)

Therefore (e 9.4), (e 9.6), (e 9.8), and (e 9.9) show that
$ba^{\prime}c=d+\tilde{a}\otimes e_{n+1,n+1},$ where $d\in M_{n}(A).$ Note
that $b$ and $c$ are invertible in $M_{n+1}(\widetilde{A}),$ as both $b^{(0)}$
and $c^{(0)}$ are nilpotents. Let
$\varepsilon_{1}=\frac{\varepsilon}{4(1+\|b^{-1}\|\cdot\|c^{-1}\|)}.$ By our
assumption, there is an upper triangular matrix $x_{1}$ and a lower triangular
matrix $y_{1}$ in $M_{n}(A)$ such that

$\displaystyle d\approx_{\varepsilon_{1}}x_{1}y_{1}.$ (e 9.10)

Let $e\in M_{n+1}(A)_{+}^{\bf 1}$ be a diagonal matrix such that
$a\approx_{\varepsilon/4}eae.$

Note that $b^{-1}=1_{n+1}+b^{(0)}$ and $x{{:}}=eb^{-1}(x_{1}+\tilde{a}\otimes
e_{n+1,n+1})$ are upper triangular matrix in $M_{n+1}(A).$ Similarly,
$c^{-1}{{=1_{n+1}+c^{(0)}}}$ is a lower triangular matrix in
$M_{n+1}(\widetilde{A}),$ and $y:=(y_{1}+{{1}}\otimes 1_{n+1,n+1})c^{-1}e$ is
a lower triangular matrix in $M_{n+1}(A).$ Then

$\displaystyle a\approx_{\varepsilon/4}eae$
$\displaystyle\approx_{\varepsilon/2}$ $\displaystyle
ea^{\prime}e=eb^{-1}ba^{\prime}cc^{-1}e$ (e 9.11) $\displaystyle=$
$\displaystyle eb^{-1}(d+\tilde{a}\otimes e_{n+1,n+1})c^{-1}e$ (e 9.12)
$\displaystyle\approx_{\varepsilon/4}$ $\displaystyle
eb^{-1}(x_{1}y_{1}+\tilde{a}\otimes e_{n+1,n+1})c^{-1}e$ (e 9.13)
$\displaystyle=$ $\displaystyle eb^{-1}(x_{1}+\tilde{a}\otimes
e_{n+1,n+1})\cdot(y_{1}+1\otimes e_{n+1,n+1})c^{-1}e$ (e 9.14)
$\displaystyle=$ $\displaystyle xy.$ (e 9.15)

Thus the proposition holds for $n+1.$ By induction, the proposition holds. ∎

######  Proposition 9.2.

Let $A$ be a $C^{*}$-algebra such that
$A\subset\overline{GL({{\widetilde{A}}})}$ and let $n\in\mathbb{N}.$ Then for
any $a\in M_{n}(A)$ and any $\varepsilon>0,$ there is a strictly upper
triangular matrix $x\in M_{n+1}(A)$ and a strictly lower triangular matrix
$y\in M_{n+1}(A)$ such that $a\approx_{\varepsilon}xy.$

In particular, any element in $M_{n}(A)$ can be approximated in norm by
product of two nilpotent elements in $M_{n+1}(A).$

###### Proof.

By Proposition 9.1, there is an upper triangular matrix $x_{1}\in M_{n}(A)$
and a lower triangular matrix $y_{1}\in M_{n}(A)$ such that
$a\approx_{\varepsilon}x_{1}y_{1}.$ Let
$v=\sum_{i=1}^{n}1_{\widetilde{A}}\otimes e_{i,i+1}\in
M_{n+1}(\widetilde{A}).$ Then $x=x_{1}v\in M_{n+1}(A)$ is a strict upper
triangular matrix and $y=v^{*}y_{1}\in M_{n+1}(A)$ is a strict lower
triangular matrix, and $xy=x_{1}vv^{*}y_{1}=x_{1}y_{1}\approx_{\varepsilon}a$
(Recall that we identify $M_{n}(A)$ with the upper left $n\times n$ corner of
$M_{n+1}(A)$).

The last part of the proposition follows from the fact that strictly
triangular matrices are nilpotents. ∎

######  Lemma 9.3.

Let $A$ be a $\sigma$-unital algebraically simple $C^{*}$-algebra with
$QT(A)\not=\emptyset$ which has strict comparison. Suppose that $A$ also has
the property (TM). Let $a\in A.$ If there are $b_{1},b_{2}\in
A_{+}\setminus\\{0\\}$ such that $a^{*}a+aa^{*},b_{1},b_{2}$ are mutually
orthogonal, then, for any $\varepsilon>0,$ there are two nilpotents $x,y\in A$
such that $\|a-xy\|<\varepsilon.$

###### Proof.

Let $B=l^{\infty}(A)/I_{{}_{\overline{QT(A)}^{w}}}.$ Recall that
$\Pi:l^{\infty}(A)\to B$ is the quotient map and $\iota:A\to l^{\infty}(A)$ is
the canonical embedding. Denote ${\bar{\iota}}:=\Pi\circ\iota.$ Fix $a\in A.$
Without loss of generality, we may assume $\|a\|\leq 1.$ Put
$a_{0}=a^{*}a+aa^{*}.$ Assume that there are $b_{1},b_{2}\in A_{+}$ such that
$0=b_{1}b_{2}=ab_{1}=b_{1}a=ab_{2}=b_{2}a{{}}.$ Let $\varepsilon>0.$ Since $A$
is simple, one can choose $n\in\mathbb{N}$ such that
$1/n<\inf\\{d_{\tau}(b_{i}):\tau\in\overline{QT(A)}^{w}\\}$ $(i=1,2).$

Since $A$ has property (TM), by Theorem 8.11, $A$ has T-tracial approximate
oscillation zero. Then, by Theorem 6.6, $B$ has stable rank one. Also, by the
last part of Remark 8.2, for each $m\in\mathbb{N},$ there is a c.p.c. order
zero map $\varphi_{m}:M_{n}\to\mathrm{Her}({{a_{0}}})$ such that

$\displaystyle\|a-\varphi_{m}(1_{n})a\|_{{}_{2,\overline{QT(A)}^{w}}}<1/m.$ (e
9.16)

Let $\varphi:M_{n}\to l^{\infty}(A)$ be the map induced by
$\\{\varphi_{m}\\}_{m\in\mathbb{N}}$ and ${\bar{\varphi}}:=\Pi\circ\varphi.$
Then (e 9.16) shows that

$\displaystyle{{{\bar{\varphi}}(1_{n}){\bar{\iota}}(a)={\bar{\iota}}(a).}}$ (e
9.17)

Denote by $\\{e_{i,j}:1\leq i,j\leq n\\}$ a system of matrix units for $M_{n}$
and $\\{e_{i,j}:1\leq\i,j\leq n+1\\}$ an expanded system of matrix units for
$M_{n+1}.$ In particular we view $M_{n}$ generated by $\\{e_{i,j}:1\leq
i,j\leq 1\\}$ as a $C^{*}$-subalgebra of $M_{n+1}$ generated by
$\\{e_{i,j}:1\leq i,j\leq n+1\\}.$

Since $A$ has strict comparison, and for all $m\in\mathbb{N},$

$\displaystyle\sup\\{d_{\tau}(\varphi_{m}(e_{1,1})):\tau\in\overline{QT(A)}^{w}\\}\leq
1/n<\inf\\{d_{\tau}(b_{2}):\tau\in\overline{QT(A)}^{w}\\},$ (e 9.18)

we have $\varphi_{m}(e_{1,1})\lesssim b_{2}$ for all $m\in\mathbb{N}.$ By [42,
Proposition 2.4 (iv)], there are $v_{m}\in A$ such that

$\displaystyle v_{m}^{*}v_{m}=(\varphi_{m}(e_{1,1})-1/m)_{+}\quad\text{ and
}\quad v_{m}v_{m}^{*}\in\mathrm{Her}_{A}(b_{2})\quad(m\in\mathbb{N}).$ (e
9.19)

Note that
$\|v_{m}\|^{2}=\|v_{m}^{*}v_{m}\|=\|(\varphi_{m}(e_{1,1})-1/m)_{+}\|\leq 1.$
Let $v=\\{v_{1},v_{2},...\\}\in l^{\infty}(A).$ Since
$\|(\varphi_{m}(e_{1,1})-1/m)_{+}-\varphi_{m}(e_{1,1})\|\leq 1/m$
($m\in\mathbb{N}$), we have

$\displaystyle{{\Pi}}(v^{*}v)={\bar{\varphi}}(e_{1,1}).$ (e 9.20)

The facts that $\varphi_{m}(1_{n})\in\mathrm{Her}({{a_{0}}}),$
$\mathrm{Her}({{a_{0}}})\bot\mathrm{Her}(b_{2}),$ and
$v_{m}v_{m}^{*}\in\mathrm{Her}(b_{2})$ show that $\varphi_{m}(1_{n})\bot
v_{m}v_{m}^{*}$ for all $m\in\mathbb{N}.$ Hence

$\displaystyle\Pi(vv^{*}){\bar{\varphi}}(1_{n})=0.$ (e 9.21)

Let $h:C_{0}((0,1])\otimes M_{n}\to B$ the homomorphism defined by
$h(\imath\otimes e_{i,j})={\bar{\varphi}}(e_{i,j})$ ($1\leq i,j\leq n$).
Extend ${\widetilde{h}}:C_{0}((0,1])\otimes M_{n+1}\to B$ by
$\widetilde{h}(\imath\otimes e_{i,j})=h({{\imath\otimes e_{i,j}}})$ and
$\widetilde{h}(\imath\otimes e_{1,n+1})=v^{*}.$ By (e 9.20) and (e 9.21),
$\widetilde{h}$ is indeed a homomorphism. Define
$\widetilde{\varphi}(e_{i,j})=\widetilde{h}({{\imath\otimes e_{i,j}}})$ for
$1\leq i,j\leq n+1.$

As we view $M_{n}$ as $C^{*}$-subalgebra of $M_{n+1},$ $\widetilde{\varphi}$
is an extension of ${\bar{\varphi}},$ i.e.,
$\widetilde{\varphi}|_{M_{n}}={\bar{\varphi}}.$ By Proposition 8.3,

$\displaystyle\mathrm{Her}_{B}(\widetilde{\varphi}(1_{n+1}))\cong\mathrm{Her}_{B}(\widetilde{\varphi}(e_{1,1}))\otimes
M_{n+1}=\mathrm{Her}_{B}({\bar{\varphi}}(e_{1,1}))\otimes M_{n+1},$ (e 9.22)

and

$\displaystyle\mathrm{Her}_{B}({\bar{\varphi}}(1_{n}))\cong\mathrm{Her}_{B}({\bar{\varphi}}(e_{1,1}))\otimes
M_{n}.$ (e 9.23)

Moreover, as $\\{e_{i,j}:1\leq i,j\leq n\\}\subset\\{e_{i,j}:1\leq i,j\leq
n+1\\},$ we also write

$\mathrm{Her}_{B}({\bar{\varphi}}(1_{n}))\subset\mathrm{Her}_{B}(\widetilde{\varphi}(1_{n+1})).$

Since $B$ has stable rank one, by [6, Corollary 3.6],
$\mathrm{Her}_{B}(\widetilde{\varphi}(1_{n+1}))$ also has stable rank one.
Note, by (e 9.17),
${\bar{\iota}}(a)\in\mathrm{Her}_{B}({\bar{\varphi}}(1_{n}))\cong\mathrm{Her}_{B}({\bar{\varphi}}(e_{1,1}))\otimes
M_{n},$ Then by Proposition 9.2, there are nilpotents
$x_{1},y_{1}\in\mathrm{Her}_{B}(\widetilde{\varphi}(1_{n+1})){{\cong\mathrm{Her}_{B}({\bar{\varphi}}(e_{1,1}))\otimes
M_{n+1}}}$ such that $\|{\bar{\iota}}(a)-x_{1}y_{1}\|<\varepsilon/8.$

Recall that $\Pi(vv^{*}+\varphi(1_{n}))=\widetilde{\varphi}(1_{n+1}).$ Thus

$\displaystyle\mathrm{Her}_{B}(\widetilde{\varphi}(1_{n+1}))=\Pi(\mathrm{Her}_{l^{\infty}(A)}(vv^{*}+\varphi(1_{n}))).$
(e 9.24)

Also note that $(vv^{*}+\varphi(1_{n}))\bot\iota(b_{1}).$ Thus

$\displaystyle\mathrm{Her}_{l^{\infty}(A)}(vv^{*}+\varphi(1_{n}))\subset\\{\iota(b_{1})\\}^{\bot}.$
(e 9.25)

By (e 9.24) and the fact that nilpotents can be lifted (see [31, Theorem
6.7]), there are nilpotents
$x_{2},y_{2}\in\mathrm{Her}_{l^{\infty}(A)}(vv^{*}+\varphi(1_{n}))$ such that

$\displaystyle\Pi(x_{2})=x_{1},\quad\Pi(y_{2})=y_{1}.$ (e 9.26)

It follows from (e 9.25) that we also have

$\displaystyle x_{2}\bot\iota(b_{1}),\quad\text{ and }\quad
y_{2}\bot\iota(b_{1}).$ (e 9.27)

Since $\|{\bar{\iota}}(a)-x_{1}y_{1}\|<\varepsilon/8,$ there is $\bar{z}\in
I_{\overline{QT(A)}^{w}}$ such that
$\iota(a)-x_{2}y_{2}\approx_{\varepsilon/8}{{{\bar{z}}.}}$ Note that
$\iota(a)-x_{2}y_{2}\in\\{\iota(b_{1})\\}^{\bot}.$ Hence there is
$d\in\\{\iota(b_{1})\\}^{\bot}$ such that

$\iota(a)-x_{2}y_{2}\approx_{\varepsilon/8}d(\iota(a)-x_{2}y_{2})d\approx_{\varepsilon/8}d\bar{z}d.$

Let

$\displaystyle
z:=d\bar{z}d\in\\{\iota(b_{1})\\}^{\bot}\cap{{I_{{}_{\overline{QT(A)}^{w}}}}}.$
(e 9.28)

Then $\|\iota(a)-(x_{2}y_{2}+z)\|<\varepsilon/4.$ Choose $\delta>0$ such that
$\|zf_{\delta}(|z|)-z\|<\varepsilon/8.$ Then

$\displaystyle\|\iota(a)-(x_{2}y_{2}+zf_{\delta}(|z|))\|<\varepsilon/2.$ (e
9.29)

Write $z=\\{z_{1},z_{2},...\\}$ with $z_{i}\perp b_{1}$ $(i\in\mathbb{N}).$
Since $z\in{{I_{{}_{\overline{QT(A)}^{w}}}}},$ there is $i\in\mathbb{N}$ such
that (note, the first inequality of the following always holds):

$\displaystyle\sup_{\tau\in{{\overline{QT(A)}^{w}}}}\\{d_{\tau}(f_{\delta/2}(|z_{i}|))\\}\leq\frac{{4}}{\delta}\sup_{\tau\in{{\overline{QT(A)}^{w}}}}\tau(|z_{i}|)<\inf_{\tau\in
T(A)}\\{d_{\tau}(b_{1})\\}.$ (e 9.30)

Since $A$ has strict comparison, $f_{\delta/2}(|z_{i}|)\lesssim b_{1}.$ By
[42, Proposition 2.4 (iv)], there is $r\in A$ such that

$\displaystyle r^{*}r=f_{\delta}(|z_{i}|)\quad\text{and}\quad
rr^{*}\in\mathrm{Her}(b_{1}).$ (e 9.31)

Write $x_{2}=\\{x_{2,j}\\}_{j\in\mathbb{N}}$ and
$y_{2}=\\{y_{2,j}\\}_{j\in\mathbb{N}}.$ By (e 9.27), $x_{2,i}\bot b_{1}$ and
$y_{2,i}\bot b_{1}.$ Together with (e 9.31), we have

$\displaystyle x_{2,i}r=r^{*}x_{2,i}=y_{2,i}r=r^{*}y_{2,i}=0.$ (e 9.32)

Thus

$\displaystyle
x_{2,i}y_{2,i}+z_{i}f_{\delta}(|z_{i}|)=x_{2,i}y_{2,i}+z_{i}r^{*}r=(x_{2,i}+z_{i}r^{*})(y_{2,i}+r).$
(e 9.33)

Since $x_{2}$ and $y_{2}$ are nilpotents, so are $x_{2,i},y_{2,i},$ By (e
9.28) and (e 9.31), $r^{*}z_{i}=0.$ Hence
$(z_{i}r^{*})^{2}=z_{i}r^{*}z_{i}r^{*}=0.$ By (e 9.31), $r^{2}=0.$ By (e
9.32), $(z_{i}r^{*})x_{2,i}=0$ and $y_{2,i}r=0.$

Let ${\alpha}_{1}:=x_{2,i},$ ${\alpha}_{2}:=z_{i}r^{*},$
${\beta}_{1}:=y_{2,i},$ and ${\beta}_{2}:=r.$ Then the last paragraph shows
that ${\alpha}_{1},{\alpha}_{2},{\beta}_{1},{\beta}_{2}$ are all nilpotents,
and ${\alpha}_{2}{\alpha}_{1}={\beta}_{1}{\beta}_{2}=0.$ Then it is standard
to conclude that $x:={\alpha}_{1}+{\alpha}_{2}$ and
${y:=}{\beta}_{1}+{\beta}_{2}$ are nilpotents (see the proof of Claim 1 in the
proof of [16, Lemma 5.6]).

By (e 9.29) and (e 9.33),

$\displaystyle
a\approx_{\varepsilon/2}x_{2,i}y_{2,i}+z_{i}f_{\delta}(|z_{i}|)=(x_{2,i}+z_{i}r^{*})(y_{2,i}+r){{=xy}}.$
(e 9.34)

The lemma follows. ∎

######  Theorem 9.4.

Let $A$ be a $\sigma$-unital simple $C^{*}$-algebra with continuous scale and
with $QT(A)\neq\emptyset.$ If $A$ has strict comparison and has $T$-tracial
approximate oscillation zero, then $A$ has stable rank one.

###### Proof.

We may assume that $A$ is non-elementary. There are two cases.

Case 1. $A$ has a nonzero projection $p$.

Set $A_{1}:=pAp.$ Then $A_{1}$ is unital, simple, has nonempty
${{QT(A_{1})}},$ and has strict comparison as well as T-tracial approximate
oscillation zero. Hence $l^{\infty}(A_{1})/I_{{}_{\overline{QT(A_{1})}}}$ has
stable rank one (see Theorem 6.6). Let $a\in A_{1}$ be a non-invertible
element and let $\varepsilon>0.$ Since $A_{1}$ is simple and finite, by [41,
Proposition 3.2, Lemma 3.5], there is a unitary $u\in U(A_{1}),$ an element
$\bar{a}\in A_{1},$ and a positive element $b\in(A_{1})_{+}\backslash\\{0\\}$
such that $\|a-\bar{a}\|<\varepsilon/4$ and $b(u\bar{a})=(u\bar{a})b=0.$ Note
that $\mathrm{Her}(b)$ is also infinite dimensional. Hence there are two
nonzero orthogonal positive elements $b_{1},b_{2}\in\mathrm{Her}(b).$

By Theorem 8.5, $A$ has property (TM). We then apply Lemma 9.3 to obtain two
nilpotent elements $x,y$ such that $u\bar{a}\approx_{\varepsilon/4}xy.$ Let
$\delta>0$ be such that $xy\approx_{\varepsilon/4}(x+\delta)(y+\delta).$ Note
that $x+\delta$ and $y+\delta$ are invertible since $x,y$ are nilpotents. Then
that
$a\approx_{\varepsilon/4}u^{*}u\bar{a}\approx_{\varepsilon/2}u^{*}(x+\delta)(y+\delta)$
shows that $a$ can be approximated by an invertible element
$u^{*}(x+\delta)(y+\delta)$ up to tolerance $\varepsilon.$ Hence $A_{1}$ has
stable rank one. It follows that $A$ also has stable rank one.

Case 2. $A$ has no nonzero projections.

Let $A_{1}$ be a $\sigma$-unital hereditary $C^{*}$-subalgebra of $A.$ Let
$a\in A_{1}^{\bf 1}$ and let $\varepsilon>0.$ Let $a_{1}:=a^{*}a+aa^{*}.$ Then
there is $\delta>0$ such that
$\|a-f_{\delta}({{a_{1}}})af_{\delta}({{a_{1}}})\|<\varepsilon/2.$ Let
$\bar{a}:=f_{\delta}({{a_{1}}})af_{\delta}({{a_{1}}}).$ Since $A_{1}$ has no
nonzero projections, we may assume that $[0,\delta]\subset{\rm
sp}({{a_{1}}}).$ Let $g\in C_{0}((0,1])_{+}$ with ${\rm
supp}(g)\subset[\delta/4,\delta/2],$ then $b{{:}}=g({{a_{1}}})\neq 0$ and
$b\bar{a}=\bar{a}b=0.$

Let us consider $A_{2}=\mathrm{Her}_{A_{1}}(f_{\delta/8}({{a_{1}}})).$ Note
that $A_{2}$ is simple, $A_{2}={\rm Ped}(A_{2})$ and $QT(A_{2})\not=\emptyset$
and has strict comparison. Moreover, by Proposition 5.4, $A_{2}$ has T-tracial
approximate oscillation zero. Hence, by Theorem 6.6,
$l^{\infty}(A_{2})/I_{{}_{{\overline{QT(A_{2})}^{w}}}}$ has stable rank one.
Note that $\bar{a},b\in A_{2}.$ Note also that, since $A$ is non-elementary,
$\mathrm{Her}_{A_{2}}(b)$ is infinite dimensional. It follows that there are
$b_{1},b_{2}\in{{\mathrm{Her}_{A_{2}}(b)_{+}\setminus\\{0\\}}}$ such that
$b_{1}\bot b_{2}.$ Since $\bar{a}^{*}\bar{a}+\bar{a}\bar{a}^{*},b_{1},b_{2}$
are mutually orthogonal, applying Lemma 9.3, we get two nilpotents $x,y\in
A_{2}\subset A_{1}$ such that $\|\bar{a}-xy\|<\varepsilon/2.$ It follows that
$\|a-xy\|<\varepsilon.$

Therefore, for any $\sigma$-unital hereditary $C^{*}$-subalgebra $A_{1}\subset
A,$ any $a\in A_{1},$ any $\varepsilon>0,$ there are nilpotents $x,y\in A_{1}$
such that $\|a-xy\|<\varepsilon.$ Together with the facts that $A$ is
projectionless and has continuous scale, applying [14, Theorem 6.4], we
conclude that $A$ has stable rank one. ∎

The proof of Theorem 1.1

###### Proof.

For (1) $\Rightarrow$ (2), applying Theorem 7.11, we know that $\Gamma$ is
surjective. Choose $e\in{\rm Ped}(A)_{+}^{\bf 1}\setminus\\{0\\}$ such that
$\widehat{[a]}$ is continuous on ${\widetilde{QT}}(A).$ Then, by Theorem 2.19,
$\mathrm{Her}(e)$ has continuous scale. It follows from Theorem 9.4 that
$\mathrm{Her}(e)$ has stable rank one. By Brown’s stable isomorphism theorem
([4]), $A\otimes{\cal K}\cong\mathrm{Her}(e)\otimes{\cal K}.$ Hence $A$ also
has stable rank one (see [38, Theorem 6.4]).

Both (2) $\Rightarrow$ (3) and (2) $\Rightarrow$ (4) are obvious. That (3)
$\Rightarrow$ (2) follows from [1].

For (4) $\Rightarrow$ (1), we apply Theorem 5.10. That (1) $\Leftrightarrow$
(5) follows from Theorem 8.5 and Theorem 8.11. This ends the proof of Theorem
1.1. ∎

We learned that the following is also obtained by S. Geffen and W. Winter.

######  Corollary 9.5.

Let $A$ be a separable stably finite simple $C^{*}$-algebra of real rank zero
which has strict comparison. Then $\Gamma$ is surjective and $A$ has stable
rank one.

###### Proof.

Let $p\in A$ be a non-zero projection and $B=pAp.$ It suffices to show the
statement holds for $B.$ Note that $B$ is unital and stably finite. By the
paragraph right after the proof of Theorem 3.3 of [3], $B$ has a 2-quasitrace
(see also III. 1.4 of [2]). By Proposition 5.8, $A$ has tracial approximate
oscillation zero. Then the corollary follows from Theorem 1.1. ∎

Let $A$ be a separable simple $C^{*}$-algebra with
${\widetilde{QT}}(A)\setminus\\{0\\}\not=\emptyset.$ Let $e\in{\rm
Ped}(A)_{+}\setminus\\{0\\}.$ Denote
$T_{e}=\\{\tau\in{\widetilde{QT}}(A):\tau(e)=1\\}.$ It is a compact convex set
which is also a basis for the cone $\widetilde{QT}(A).$

######  Corollary 9.6.

Let $A$ be a separable simple $C^{*}$-algebra with $QT(A)\not=\emptyset$ which
has strict comparison. Suppose that, for some $e\in{\rm Ped}(A)_{+}^{\bf
1}\setminus\\{0\\}$ and $\partial_{e}(T_{e})$ has countably many points. Then
$\Gamma$ is surjective, $A$ has stable rank one, property (TM) and T-tracial
approximate oscillation zero.

###### Proof.

It follows from Theorem 5.9 that $A$ has norm approximate oscillation zero.
Thus corollary follows from Theorem 1.1 immediately. ∎

The following is perhaps known but we are not able to locate it in the
literature.

######  Proposition 9.7.

Let $A$ be a separable $C^{*}$-algebra which has local finite nuclear
dimension. Then every hereditary $C^{*}$-subalgebra $B\subset A$ also has
local finite nuclear dimension.

###### Proof.

Let $B$ be a hereditary $C^{*}$-subalgebra of $A$. Let $\varepsilon>0$ and
${\cal F}\subset B$ be a finite subset. To simplify notation, without loss of
generality, we may assume that ${\cal F}\subset B^{\bf 1}$ and there is
$e_{B}\in B_{+}^{\bf 1}$ such that $e_{B}x=xe_{B}=x$ for all $x\in{\cal F}.$

Choose $\delta>0$ as in Lemma 3.3 of [10] associated with $\varepsilon/4$ (in
place of $\varepsilon$) and $\sigma=\varepsilon/4.$ We may assume that
$\delta<\varepsilon/4.$

Since $A$ has local finite nuclear dimension, there is a $C^{*}$-subalgebra
$C\subset A$ with finite nuclear dimension, say $k$
($k\in\mathbb{N}{{\cup\\{0\\}}}$), such that

$\displaystyle x\in_{\delta/2}C\,\,\,{\rm for\,\,\,all}\,\,\,x\in{\cal
F}\cup\\{e_{B}\\}.$ (e 9.35)

Choose $d\in C_{+}^{\bf 1}$ such that $\|e_{B}-d\|<\delta.$ Then, by Lemma 3.3
of [10], there is a partial isometry $w\in A^{**}$ such that

$\displaystyle
ww^{*}f_{\varepsilon/4}(d)=f_{\varepsilon/4}(d)ww^{*}=f_{\varepsilon/4}(d),\,\,w^{*}cw\in\mathrm{Her}(e_{B})\subset
B\,\,\,{\rm and}\,\,\,$ (e 9.36)
$\displaystyle\|w^{*}cw-c\|<(\varepsilon/4)\|c\|\,\,\,{\rm
for\,\,\,all}\,\,\,c\in\overline{f_{\varepsilon/4}(d)Af_{\varepsilon/4}(d)}.$
(e 9.37)

Set $C_{1}=w^{*}\overline{f_{\varepsilon/4}(d)Cf_{\varepsilon/4}(d)}w\subset
B.$ By Proposition 2.5 of [50],
$\overline{f_{\varepsilon/4}(d)Cf_{\varepsilon/4}(d)}$ has nuclear dimension
$k.$ Since $C_{1}\cong\overline{f_{\varepsilon/4}(d)Cf_{\varepsilon/4}(d)},$
$C_{1}$ has nuclear dimension $k.$ We then estimate that

$\displaystyle x\in_{\varepsilon}C_{1}\,\,\,{\rm for\,\,\,all}\,\,\,x\in{\cal
F}.$ (e 9.38)

Thus $B$ has local finite nuclear dimension. ∎

As in [44], we have the following (note that, by [17], since $A$ is exact,
$\widetilde{T}(A)=\widetilde{QT}(A)$):

######  Corollary 9.8.

Let $A$ be a separable exact simple $C^{*}$-algebra with
${\widetilde{T}}(A){{\setminus\\{0\\}}}\not=\emptyset.$ Suppose that $A$ has
strict comparison, T-tracial approximate oscillation zero and has local finite
nuclear dimension. Then $A\otimes{\cal Z}\cong A.$

###### Proof.

Choose $e\in{\rm Ped}(A)_{+}^{\bf 1}\setminus\\{0\\}$ and $B=\mathrm{Her}(e),$
Then ${\rm Ped}(B)=B.$ It suffices to show (see Cor. 3.1 of [47]) that $B$ is
${\cal Z}$-stable. Note that $B$ has strict comparison, and by Proposition
9.7, has local finite nuclear dimension. Since $B$ also has T-tracial
approximate oscillation zero (see 5.4), by Theorem 7.11, $\Gamma$ is
surjective. It follows that $B$ has $m$-almost divisibility for some $m$ (in
fact $m$ can be zero). By [45, Theorem 8.5 (iii)], $B$ is ${\cal Z}$-stable. ∎

######  Remark 9.9.

At least in the unital case, the condition that $A$ has local finite nuclear
dimension in Corollary 9.8 can be further weakened to that $A$ is amenable and
has weak tracial finite nuclear dimension (see Definition 8.1 and Theorem 8.3
of [27]).

######  Remark 9.10.

(1) Note that, in Theorem 1.1 and Corollary 9.6, we do not assume that $A$ is
amenable or even exact.

(2) Usually, the condition that $A$ has strict comparison implies that $A$ has
at least one non-zero 2-quasitrace. However, one may insist that the condition
that $A$ has strict comparison means that, if $A$ has no non-zero
2-quasitraces, $A$ is purely infinite. In that case the assumption in Theorem
1.1 (part of the assumption of Corollary 9.8) may be replaced by that $A$ is
finite and has strict comparison.

(3) On the other hand, if one assumes that $\mathrm{Cu}(A)$ is almost
unperforated and $A$ is not purely infinite, then, by [43], $A$ has strict
comparison (in the usual sense) (see also Remark 2.5 and Proposition 4.9 of
[16]). Conversely, if $A$ has strict comparison (in usual sense),
$\mathrm{Cu}(A)$ is almost unperforated. Therefore, if one prefers not to
mention 2-quasitraces in each statement, one could use the condition that $A$
is finite and $\mathrm{Cu}(A)$ is almost unperforated.

(4) If $A$ is a unital stably finite simple $C^{*}$-algebra, then, by [42,
Theorem 6.1]), see aslo [8, Corollary 4.7] and [2, Theorem II.2.2], $A$ has at
least one non-trivial 2-quasitrace. So, in the unital case, we may assume that
$A$ is stably finite instead assume that $A$ has a non-trivial 2-quasitrace.
This also works for the case that $A$ is not unital but
$K_{0}(A)_{+}\not=\\{0\\}.$ However, when $A$ is stably projectionless, the
situation is somewhat different. Nevertheless, we may proceed this as follows:

We assume that $A$ is a separable simple $C^{*}$-algebra. Recall that an
element $a\in{\rm Ped}(A)_{+}$ is infinite, if there are nonzero elements
${{b,\ }}c\in{\rm Ped}(A)_{+}$ such that $bc=0,$ $b+c\lesssim c$ and
$c\lesssim a.$ $A$ is said to be finite, if there are no infinite elements in
${\rm Ped}(A)_{+}.$ $A$ is said to be stably finite, if $M_{n}(A)$ is finite
for each $n$ (see Definition 1.1 of [29] and Definition 4.7 of [16], for
example).

Choose $e\in{\rm Ped}(A)_{+}\setminus\\{0\\}$ and consider
$B=\mathrm{Her}(e).$ Without loss of generality, we may well assume that $A=B$
for convenience. Define $K_{0}^{*}(A)$ (using $W(A)$ not $\mathrm{Cu}(A)$)
exactly the same way as in section 4 of [8]. Note that Lemma 4.1 of [8] holds
automatically with the definition above. The same definition of order there
(before Proposition 4.2 of [8]) also works in this case. In other words, so
defined $K_{0}^{*}(A)$ is a (directed) ordered group and the stably finiteness
ensures that $K_{0}^{*}(A)$ is not zero. Since $A$ is simple, Proposition 4.2
of [8] still holds. We now return to the paragraph right after the proof
Theorem 3.3 of [3]. Note that $(K_{0}^{*}(A),K_{0}^{*}(A)_{+},[e])$ is a
scaled ordered group which has a state, and which gives a dimension function.
By II. 2.2 of [2], the dimension function just mentioned gives a 2-quasitrace
on $A.$ Therefore, we may replace the condition that
${\widetilde{QT}}(A)\setminus\\{0\\}\not=\emptyset$ by the condition that $A$
is stably finite (recall that we assume that $A$ is simple) in Theorem 1.1
(see also [19] for the case that $A$ is exact).

(5) One may notice that the condition that $A$ has strict comparison and
$\Gamma$ is surjective implies that there is an isomorphism
$\Gamma^{\sim}:\mathrm{Cu}(A)\to V(A)\sqcup({\rm
LAff}_{+}(\widetilde{QT}(A))\setminus\\{0\\})$ (see also 2.13).

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Xuanlong Fu

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Huaxin Lin

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