Nuclear dimension and Jiang–Su stability without elementary subquotients

Proof. The ideal criterion above, applied to the projection 1_D, gives [1_D]\le N[v] for some N. A sufficiently accurate Cuntz approximation, compressed to the 1_D corner, gives x_1,\ldots,x_N\in D with \|1_D-\sum_lx_l^*vx_l\|<1. The positive element z=\sum_lx_l^*vx_l is invertible. Taking y_l=x_lz^{-1/2} gives the identity.□

Proof. Let I=\Ideal_D(v). The map that is zero on I_+ and infinity on D_+\setminus I is a lower semicontinuous extended trace: additivity follows from heredity of I, the trace identity from x^*x\in I\Longleftrightarrow xx^*\in I, and lower semicontinuity from closedness of I. The assumed inequality for this trace forces u\in I.

Work in D\otimes\mathcal K, whose W-semigroup identifies with \Cu(D) and is almost unperforated. Fix a dimension function d on this semigroup with d([v])=1. Its regularization is

By [ERS, Lemma 4.7] and the spectral-cut description of compact containment, \overline d is a functional on \Cu(D). The correspondence in [ERS, Proposition 4.2] realizes it as d_\lambda for a lower semicontinuous extended 2-quasitrace. Haagerup proved that finite-valued 2-quasitraces on exact unital algebras are traces. Blanchard–Kirchberg extended this conclusion to lower semicontinuous 2-quasitraces on exact algebras, allowing a possibly proper finite-domain ideal [ERS, Remarks 4.3 and 4.5]. Thus \lambda is a trace. Restrict \lambda to the original corner D. For each fixed \eta>0 the hypothesis gives

The strict inequality is against d([v]), so it remains valid when \overline d([v])=0. Theorem 2.4 now compares every (u-\eta)_+ with v. Taking the supremum over cuts gives u\precsim v. Finally, comparisons between elements of the original corner can be witnessed in that corner, by compressing each witness on both sides by its multiplier projection.□

Proof. Write e_{ij}=1_A\otimes e_{ij}. Additivity and the trace identity give

Unitary equivalence of the diagonal corners gives \lambda(a\otimes e_{ii})=k^{-1}\tau(a) for a\in A_+. Summing proves the formula. Only finite sums and multiplication by positive scalars occur, so the argument also applies to infinite values.□

Proof. Choose (x_n^{(m)}) solving the first m conditions with ultralimits less than 1/(2m). Choose A_m\in\omega on which the first m errors of this choice are less than 1/m, and set E_m=A_1\cap\cdots\cap A_m. These sets are decreasing and belong to \omega. At coordinate n, use x_n^{(l(n))}, where l(n)=\max\{m\le n:n\in E_m\}, using any element of X_n if the set is empty. Then l(n)\to_\omega\infty, and every fixed error tends to zero.□

Proof. Choose s with \|u-s^*(y-b)_+s\|<\varepsilon and put

where the ratio is zero when its denominator vanishes. This is a continuous function vanishing near zero, and

Consequently

which proves the claim. If the original comparison is formulated in the stabilization, compression to the original corner gives s\in D.□

Proof. Choose an approximate unit (e_l) for D consisting of positive contractive continuous functions of P which vanish at zero. Such elements satisfy d_\tau(e_l)\leq d_\tau(P). For any positive contraction z\in D, the elements e_lze_l converge to z in norm, and

Lower semicontinuity therefore gives \tau(z)\leq d_\tau(P). Monotonicity and additivity of d_\tau on Cuntz classes give the second inequality in (3.4). Thus \tau is bounded on the positive unit ball of D and extends to the asserted bounded positive functional. Its norm bound proves (3.5). This argument also applies when d_\tau(b)=0, in which case the restriction is identically zero.□

Proof. Write A_\infty=\ell^\infty(A)/c_0(A), identifying A with its constant sequences. There are commuting positive contractions u_1,\ldots,u_m\in A_\infty\cap A' such that u_j is represented by positive contractions in I_j and u_jx_j=x_j. Indeed, at coordinate i choose successively an element of a quasicentral approximate unit of each I_j. Require its commutators with the first i elements of a fixed dense subset of A to have norm less than 1/i, require that it act on x_j within 1/i, and require that it commute within 1/i with the elements already chosen at that coordinate. These are finitely many approximate-unit and quasicentrality requirements. They give exactly the stated relations on passing to A_\infty.

Choose N pairwise disjoint closed intervals contained in (0,1). For the rth interval choose continuous functions g_r^-,g_r^+\colon[0,1]\to[0,1] satisfying

whose sum is one outside that interval. We choose g_r^- equal to one below the interval and zero above it, and g_r^+ equal to zero below the interval and one above it. Within the interval the two functions can decrease and increase, respectively, meeting at zero. Set h_r=g_r^-+g_r^+. At every point of [0,1] at most one interval can contribute a loss, so

For t=(t_1,\ldots,t_m)\in\mathcal T define

These are central positive contractions, and for fixed t they are pairwise orthogonal as S varies. Families for different values of t may overlap and are combined by averaging. Joint functional calculus and independent averaging of the coordinates of t give

If j\in S, then P_{S,t} belongs to the image of \ell^\infty(I_j) in A_\infty, since g_{t_j}^+(0)=0. If 1\leq j\leq m and j\notin S, then g_{t_j}^-(u_j)x_j=g_{t_j}^-(1)x_j=0, whence P_{S,t}x_j=0.

We construct the required orthogonal representatives directly, before imposing exact ideal membership. Start with positive contractive representatives of a finite orthogonal family, whose pairwise products tend to zero. To separate a first representative a_i from a later representative b_i, choose r_i>0 tending to zero slowly enough that \|a_ib_i\|/r_i\to0. Let f_i\colon[0,1]\to[0,1] be one on [0,r_i/2] and zero on [r_i,1], with 1-f_i(t)\leq2t/r_i. Replace the first element by (a_i-r_i)_+ and every later one by f_i(a_i)b_if_i(a_i). These replacements are orthogonal to the first cut, and

Choose r_i using the maximum of the finitely many relevant products. Repeating with the later elements preserves orthogonality already obtained and changes no quotient element. This gives, for every t, orthogonal positive contractive lifts R_{S,t,i} of P_{S,t}.

For j\in S, the distance from R_{S,t,i} to I_j tends to zero. Let \varepsilon_i>0 tend to zero and dominate all these distances, over the finite collection of triples (S,t,j) under consideration. The first cuts

belong exactly to every I_j with j\in S: their images in A/I_j vanish by functional calculus. Choose another sequence \sigma_i>0 tending to zero, and put

Because T_{S,t,i}\in\Ideal(b_j), the usual ideal criterion for Cuntz comparison gives a finite integer K_{S,t,i,j} for which (3.6) holds. Here the strict positivity of the second cut is essential: every positive cut of an element in \Ideal(b_j) is Cuntz below a finite direct sum of copies of b_j. Both cuts preserve orthogonality and change the representatives by at most 2\varepsilon_i+\sigma_i\to0. Consequently they preserve all the quotient relations established above, while imposing all the ideal and finite-multiple conditions simultaneously.

Finally, centrality of P_{S,t} gives P_{S,t}^{1/2}aP_{S,t}^{1/2}=P_{S,t}a in A_\infty. Thus (3.10) implies (3.7), and P_{S,t}x_j=0 proves the required vanishing of the omitted compressions. No bound on K_{S,t,i,j} uniform in i has been asserted or used.□

Proof. Fixing the corners. Choose N so large that m/N<\varepsilon/4, and use Lemma 3.3. Write

Asymptotic centrality implies, for (c,d)\in\mathcal O,

Together with (3.7) and the omitted compression estimates, this permits us to fix one coordinate i such that

All these requirements range over finite sets. From now on this coordinate is fixed. For each occurrence indexed by (S,t), suppress i and write

Every subscript P below retains this (S,t) occurrence index. Thus coincident positive elements, including zero pieces, still receive separate corner approximations and outcome labels. In particular, the finitely many constants K_{P,j} have already been fixed before any approximation accuracy within D_P is chosen.

Approximating within the corners. Each D_P is separable and nuclear. The Brown–Carrión–White approximation theorem, Theorem 2.2, supplies finite-dimensional approximations with asymptotically order-zero first maps and return maps which are finite convex combinations of completely positive contractive order-zero maps. For a fixed corner, take a sequence of these first maps \zeta_r\colon D_P\to E_r. The asymptotic product estimate (2.2) gives

Thus sufficiently late first maps have product norm less than \varepsilon on the finitely many pairs (l_P(c),l_P(d)) in (3.17). This uses the approximation sequence for the fixed corner; it requires no uniformity in a varying hereditary subalgebra.

Choose positive numbers \xi_P such that

There are only finitely many constants in these requirements, so such choices are possible. Choose a corresponding approximation in each D_P, denoted

where the \lambda_{P,r} are nonnegative and sum to one, and each \beta_{P,r} is completely positive contractive and order zero. All component return maps for this corner have the same domain E_P and use the same first map \zeta_P. Arrange that

For a zero corner take E_P=0 and the zero maps, with a single return component of weight one.

Individual outcomes and their retained parts. Consider the finite probability space whose outcomes \nu consist of a tuple t\in\mathcal T and, for each subset S, a choice r_S\in\{1,\ldots,h_{P_{S,t,i}}\}. Give t uniform probability and make the return-map choices independently with the probabilities \lambda_{P,r}. Write \lambda_\nu for the resulting probability, and let

The map \alpha_t is completely positive contractive because a finite direct sum has the maximum norm. The hereditary subalgebras D_{P_{S,t,i}} are mutually orthogonal for the single tuple t used by the outcome. Consequently \beta_\nu is completely positive contractive and order zero: each summand has these properties, and products between distinct summand ranges vanish. Moreover, (3.21) bounds the product defect of every \alpha_t by \varepsilon, uniformly over the outcomes.

Independence of the return-map choices gives

For each fixed t, the differences in (3.20) lie in orthogonal hereditary subalgebras, so their sum has norm at most \max_P\xi_P. Equations (3.15), (3.19), and (3.22) therefore imply

For 1\leq j\leq m, retain only the corners indexed by S\ni j:

This is a positive contraction. Contractivity of each component and orthogonality of the omitted corners give, for every individual outcome,

The maximum over an empty family is taken to be zero. In particular, the estimate holds before taking any average.

The uniform relative trace estimate. Fix j and a lower semicontinuous extended trace \tau with d_\tau(b_j)<\infty. In every retained corner, Lemma 3.2 applies with the already fixed constant K_{P,j}. Thus (3.20) implies

All quantities in this inequality are finite. For each fixed t, traciality and \sum_{S\ni j}P_{S,t,i}\leq1 give

Taking expectations in (3.24) and using (3.19) proves

For later use there is also an individual bound

Every coefficient here was chosen independently of \tau. If d_\tau(b_j)=0, then each retained output has trace exactly zero, as does x_j. Hence the same inequalities hold in this case without an absolute additive error. No condition is being placed on traces for which d_\tau(b_j)=\infty.

Rounding the probabilities and forming the matrix model. After all corner maps and the constants M_{\nu,j} have been fixed, choose k=2^s and nonnegative integers n_\nu with \sum_\nu n_\nu=k so that \mu_\nu=n_\nu/k satisfies

Such choices exist for arbitrarily large s: round the finitely many numbers k\lambda_\nu down to integers and distribute the remaining units among the outcomes. The resulting probabilities stay in the probability simplex, and their total absolute error tends to zero as s\to\infty. Contractivity of the T_\nu shows that replacing \lambda_\nu by \mu_\nu changes the expected map on the unit ball by less than \varepsilon/3. Equations (3.27) and (3.28) show that the expected retained trace changes in absolute value by at most (\varepsilon/2)d_\tau(b_j).

List the outcomes with their multiplicities as \nu_1,\ldots,\nu_k, and write t_l for the threshold tuple in \nu_l. Define

The map \alpha is completely positive contractive, and \beta is completely positive contractive and order zero, by the orthogonality of the matrix diagonal positions. We have

Consequently (3.23) and (3.28) give (3.11), and the maximum norm in the direct sum gives (3.12). Finally, put

These are positive contractions, and the estimate for individual outcomes in (3.25) proves (3.13). Since L_k(w_j)=\sum_\nu\mu_\nu W_{\nu,j}, (3.26) and the trace part of the rounding estimate prove (3.14).□

Proof of Lemma 3.1. Fix a countable dense subset (c_l)_{l\geq1} of the unit ball of A. The space

is a separable metric space, so choose a countable dense family (c'_l,d'_l)_{l\geq1} in this space. At stage n, apply Lemma 3.4 to the first n tests, \mathcal F_n=\{c_1,\ldots,c_n\}, \mathcal O_n=\{(c'_1,d'_1),\ldots,(c'_n,d'_n)\}, and \varepsilon=2^{-n}. Denote the resulting maps by \alpha_n and \beta_n, and the retained elements by w_{n,j} for j\leq n.

Contractivity extends (3.11) from the dense family to point-norm convergence on A. Likewise, if c,d are orthogonal positive contractions and (c',d') is one of the chosen orthogonal pairs, then

First taking the limit superior in n, then approximating (c,d) by such pairs, proves asymptotic preservation of orthogonality. Scaling proves the assertion for arbitrary orthogonal positive elements.

For j>n set w_{n,j}=0. Take \delta_{n,j}=2^{-n} for all n,j, and set

Equations (3.2) and (3.3) now hold for every n,j, and both error sequences tend to zero for each fixed j. At each stage the trace inequality holds simultaneously for all the indicated extended traces. Thus the diagonalization involves only the countable lists of elements, tests, and orthogonal pairs; it requires neither a countable family of traces nor a restriction on the ideal lattice of A.□

Proof. Choose a multiple d\geq2 of k with (d+1)^{-1}<\xi, and fix a unital embedding \mu\colon M_k\to M_d. Put p=e_{d+1,d+1}\in M_{d+1}. In M_d\otimes M_{d+1}, the complement of 1\otimes p is the tensor product of two d-dimensional spaces. Let V be the flip on this complement, extended by the identity on the range of 1\otimes p. The unitary group of each finite matrix block is path connected. There is therefore a continuous unitary path U(t), 0\leq t\leq1, from 1 to V, all of whose members commute with 1\otimes p.

Define

This is a c.p.c. order-zero map into C([0,1],M_d\otimes M_{d+1}). Its endpoint values are

Consequently its range lies in the prime dimension-drop algebra

The commutation of U(t) with 1\otimes p gives the exact defect

For 1\leq r\leq d+1, the functions

belong to I_{d,d+1} and satisfy

Thus every tracial state \sigma on I_{d,d+1} satisfies

The prime dimension-drop algebra I_{d,d+1} embeds unitally into \Z by [Rør, Theorem 2.1(i)]. Composing such an embedding with \iota_0 gives the trace estimate. Finally, \|\iota_0(e_{11})(0)\|=1, so isometry of the embedding gives the asserted norm.□

Proof. Take \iota\colon M_m\to\Z from Lemma 4.1, and set

Order-zero functional calculus and the product identity (2.1) give the first identity in (4.1). The element (D_m-1/2)_+ is nonzero, hence full in the simple unital algebra \Z. Lemma 2.3 supplies the second identity.□

Proof. Both ranges \pi(A) and \theta(A) belong to the multiplicative domain of L. For \theta, the two Schwarz inequalities give

The analogous equalities hold with a and a^* interchanged. The assertion for \pi follows in the same way, or directly from its multiplicativity.

Let D_0=C^*(\pi(A),\theta(A))\subset B_1. This is a separable unital algebra, L|_{D_0} is a homomorphism, and

is an ideal containing \theta(a)-\pi(a) for every a\in A. Choose a positive contractive quasicentral approximate unit (f_l) for J in D_0. Then L(f_l)=0, the commutators of f_l with both ranges tend to zero, and

Lift the f_l to positive contractive sequences in A\otimes\Z_1. Apply the countable diagonalization of Lemma 2.7 to the preceding equations on dense countable subsets, together with the coordinate test \|L(f_{l,n})\|. The latter tends to zero along \omega for each fixed l. The resulting positive contraction e\in B_1 satisfies (4.7). Taking a=1 gives qh=q.□

Proof. We first work under the hypothesis d_\tau(b)<\infty. On the hereditary subalgebra \her_A(b)\otimes\Z_1 one has

To justify this formula, every positive contraction in \her_R(b\otimes1) has \lambda-value at most d_\lambda(b\otimes1)=d_\tau(b). The restriction of \lambda to \her_A(b)\otimes\Z_1 therefore extends to a bounded trace; the same holds for \tau on \her_A(b). For a positive s\in\her_A(b), the functional

is a bounded positive trace on \Z_1, of mass \tau(s). Its trace identity follows by applying the trace identity for \lambda to s^{1/2}\otimes z. Uniqueness of the tracial state of \Z gives \lambda(s\otimes v)=\tau(s)\tau_{\Z}(v). Linearity and density prove (4.12). In particular,

There is a separate estimate for w_n, which need not lie in that hereditary subalgebra. Define

Order-zero square-root calculus shows that \rho_n is an extended trace; its lower semicontinuity follows from continuity of T_n. Furthermore,

Lemma 2.6, applied to \rho_n, gives

This identity includes infinite values. It follows from (4.4) that

Put

The common-part identity implies \Delta_{n,j}\to_\omega0: in the quotient, the two expressions are respectively q\theta(x) and q\pi(x). Compressing S'_n\leq S_n+\Delta_{n,j}1 by x^{1/(2r)} gives

Here the compression of S_n has trace at most \lambda(S_n), by traciality and contractivity of x^{1/(2r)}. The compressed S'_n converges in norm to S'_n as r\to\infty. Lower semicontinuity yields

All of \lambda(w_n), \lambda(S_n), \lambda(S'_n), and \tau(x) are finite in the case under consideration. We may therefore subtract these trace values. Since x=S'_n+u_{n,0} and w_n=S_n+u_{n,1}, (4.15) and (4.16) give

Together with (4.13), this proves, for both i=0,1,

Set

The trace-cut estimate (2.3) proves (4.10). In particular, when d_\tau(b)=0 the right side is exactly zero, not an absolute error term.

We now prove (4.11). Suppose \ell_n\leq1/4 and write

Because (q_n-1/2)_+\geq1/2-e_n and t_n\geq z_n-(\gamma/16)1, we have

On the support of b, the element a is at least \gamma/2. Consequently compression by b^{1/(2r)} gives

If d_\tau(b)<\infty, the compression b^{1/(2r)}t_nb^{1/(2r)} belongs to \her_A(b)\otimes\Z_1, so (4.12) and traciality imply

Let r\to\infty and use d_\lambda(t_n)\geq\lambda(t_n), since t_n is a contraction. This proves (4.11) in the finite-rank case, including rank zero.

For the infinite-rank case we use ideals, without appealing to (4.12). First,

To see this, apply the domination theorem, Theorem 2.9, on \Z_1 to v\mapsto\tau_{\Z}(v)1 and the identity map. The source is exact, the maps are nuclear, the identity has order zero, and the required pointwise ideal containment follows from simplicity of \Z. There are therefore completely positive maps of the form

converging pointwise in norm to \tau_{\Z}(v)1. Including 1 in the approximation tests allows us to require \|V_l\|\leq2. Tensoring with \id_A then preserves pointwise norm convergence: it holds first on elementary tensors, and the uniform norm bound extends it to their completion. Each resulting approximation to L(t_n)\otimes1 is a sum of compressions of t_n in R, proving (4.20).

Let I=\Ideal_R(t_n). In R/I, equation (4.18) and (4.20) give

Thus \|a+I\|\leq\gamma/4, and (a-\gamma/4)_+\in I. Taking one more positive cut gives

for some finite positive integer M. If d_\tau(b)=\infty, then d_\lambda(b)=\infty, so this comparison forces d_\lambda(t_n)=\infty. This finishes the proof in all cases.□

Proof. Temporarily fix m\geq2 and j, and keep the abbreviations from the preceding proof. The copy of D_m is now in \Z_2, separate from z_n and t_n=(z_n-\gamma/16)_+. The compression identity in (4.1) gives

For the second inequality, use joint functional calculus and the scalar inequality

Where the left side is positive, the difference between the uncut expression on the right and the left side is s/2+(\gamma/16)t-\gamma/16\geq0; elsewhere the assertion is immediate.

Enumerate the countable set of pairs (j,\eta) with j\geq1 and \eta\in\mathbb Q_{>0}. For each m\geq2 let \Omega_m be an \omega-large set on which \ell_n\leq1/4 and

for all the first m pairs. Such a set exists because the finitely many error coefficients tend to zero along \omega. Define

The sets E_m are decreasing and \omega-large. If n belongs to at least one E_m, let m(n) be the largest such m; otherwise put m(n)=2. The largest value exists since membership in E_m requires m\leq n. This construction gives m(n)\to_\omega\infty and ensures that, for every fixed pair (j,\eta), condition (4.23) with m=m(n) holds on an \omega-large set.

Set P_n=D_{m(n)}. For each fixed (j,\eta), on the preceding \omega-large set and for every lower semicontinuous extended trace \lambda on R, we claim

When d_\tau(b_j)<\infty, combine (4.10), (4.11), (4.22), and (4.23):

This includes d_\tau(b_j)=0. If d_\tau(b_j)=\infty, then (4.11) gives d_\lambda(t_n)=\infty, and (4.22), with finite K_{m(n)}, forces the right side of (4.24) to be infinite. Thus (4.24) also holds in that case. Its coordinate set is independent of \lambda.

The element P commutes with the entire first subsystem B_1. Also q commutes with \pi(A). Hence \Psi is a homomorphism \pi multiplied by commuting positive contractions, and is c.p.c. order zero. In B we have

Since R is exact and \Z-stable, Lemma 2.5 applies to (4.24). Thus, on an \omega-large set,

We explain the passage to the ultrapower, because the norm bound on witnesses is essential. Fix j, a positive rational \eta, i\in\{0,1\}, and an accuracy \epsilon>0, and let u_i=(u_{n,i})_\omega. By Lemma 2.8, choose witnesses r_n\in R on this large set such that

Set r_n=0 elsewhere. This is a bounded sequence, so it defines r\in B. By (4.25),

As \epsilon is arbitrary, each displayed source cut is Cuntz below (\Psi(a_j)-c_j)_+. Letting positive rational \eta decrease to zero proves (4.21).□

Proof. At coordinates where m(n)\geq L, use the elements v_{m(n),l} from Lemma 4.2; at all other coordinates use zero. The representatives have norm at most one. Since m(n)\to_\omega\infty, their relations in (4.1) give (4.26). Their tensor-factor location gives the stated commutation.□

Proof. Consider the order-zero root maps \theta^{1/l}, for l\geq1. They commute with e and the constant third leg. Indeed, each value \theta^{1/l}(a) is a norm limit of sandwiches of \theta(a) by bounded continuous functions of h. One obtains these sandwiches by truncating t^{(1/l-1)/2} from above; the resulting scalar weights converge uniformly to t^{1/l} after multiplication by t. The same observation gives

Here e commutes with the values involved, so the sandwiches also approximate the positive element on the left.

Taking a=1_A in (4.7) gives qh=q. Continuous functional calculus therefore gives qh^{1/l}=q, and the sandwich formula together with (4.7) yields

Moreover,

by the uniform convergence of t^{1+1/l} to t on [0,1].

For every fixed j, combining (5.4) with (4.21) gives

The bounded-witness estimate of Lemma 2.8, with the spare cut c_j, now shows that this positive element can be approximated arbitrarily closely by r_j^*\Psi(a_j)r_j, with

The bound depends on j but, crucially, not on l.

We apply the countable diagonalization principle of Lemma 2.7. To make the boundedness issue explicit, lift each root map to a sequence of c.p.c. maps into R by the Choi–Effros lifting theorem. Keep fixed representatives of e,q,h,\theta,\Psi and \pi. The tests impose the factorization and common-part identities on countable dense subsets of A, commutation with e and a dense subset of the constant third leg, and order zero on a countable dense family of orthogonal pairs of positive contractions. For each j include a witness variable with a fixed representative bound strictly larger than \sqrt{2/c_j}, and impose

Every finite list is approximately solvable: take l sufficiently large for the factorization tests, and then use the witnesses above with sufficiently small errors. Their fixed bounds allow the ultrapower tests to be realized by bounded coordinate variables. The diagonalization produces a c.p.c. map satisfying all the tests. Continuity extends the identities and commutations to all the specified elements; the dense orthogonality tests make the map order zero. This proves (5.1) and (5.2), as well as the witness identities for all j.

It remains to extend ideal containment from the listed cuts in (3.1) to an arbitrary positive contraction a'\in A. Such an a' can be approximated arbitrarily closely by x_j with a_j\in\Ideal_A(a'). To see this, choose d in the fixed dense positive family close to a', and a small positive rational r>\|d-a'\|. The image of d in A/\Ideal_A(a') has norm less than r, so (d-r)_+\in\Ideal_A(a'). This cut belongs to the chosen family; take it as a_j and choose the accompanying rational \gamma_j small. Then x_j=(a_j-\gamma_j)_+ is as close to a' as desired.

The map \Psi is a positive weight times the homomorphism \pi, and the weight commutes with \pi(A). Consequently

apply \Psi to finite sums of products through a' and pass to norm limits. The witness identity places e\widehat\theta(x_j) in this ideal. The comparison (4.21) similarly places e\pi(x_j) there. Taking norm limits, and then scaling, proves (5.3) for every a'\in A_+.□

Proof. Fix m\geq1 and set t=c^{1/2}ec^{1/2}. This positive contraction commutes with \rho_i(A) and satisfies t\rho_i(1_A)=t. The order-zero product identity therefore gives

Apply Theorem 2.9 to

Both maps are completely positive and nuclear, since their domain is nuclear, and \Psi is order zero. For a\in A_+,

by Proposition 5.1. Hence finite sums \sum_{l=1}^L y_l^*\Psi(\,\cdot\,)y_l approximate \Phi pointwise in norm.

Write h_q=(q-1/2)_+. For one such approximation, take mL second-leg elements v_{sl} from (4.26) and set

These second-leg elements commute with q and \pi(A), and satisfy v_{sl}^*v_{s'l'}=\delta_{ss'}\delta_{ll'}P. Expanding the products therefore gives

For f(u)=\min\{1,2u\} on [0,1] one also has f(q)r_s=r_s, since f(q)h_q^{1/2}=h_q^{1/2}. Include 1_A in the approximation set. As \Phi(1_A)=t and \|t\|\leq1, sufficiently accurate approximations in (5.11) have \|r_s\|^2\leq2. Thus, for this fixed m, there is a fixed bound on every packed variable, independent of L and of the approximation accuracy.

Using bounded coordinate lifts and the diagonalization principle of Lemma 2.7 on a countable dense subset of A, we obtain r_1,\ldots,r_m\in B with

These elements are intertwiners. Indeed, using (5.10) and (5.12), the right modulus square of \pi(a)r_s-r_s\rho_i(a) is

Consequently \pi(a)r_s=r_s\rho_i(a) for all a\in A. Taking a=1_A gives r_s\rho_i(1_A)=r_s as well. The matrix R_s with r_s in position (0,i) thus belongs to C, including its support condition.

For the row V=(R_1,\ldots,R_m) over C, the positive matrix V^*V has m diagonal entries equal to t^{(i)} and all other entries zero. Meanwhile VV^*\in\her_C(Q_0): (5.12) gives f(Q_0)VV^*f(Q_0)=VV^*, and f(0)=0. Equality of the Cuntz classes of the two modulus squares now proves (5.8).

Finally,

The second summand, placed in corner i, is Cuntz below Q_i inside C. Apply (5.8) with m=1, the subadditivity of Cuntz classes of positive sums, and (5.7). This proves (5.9).□

Proof. Every positive element of M_2(B) is bounded by twice its diagonal part. For an element of C_+, each diagonal corner also lies in C, because \rho and its support condition are diagonal. After scaling, (5.9) shows that these corners belong to the ideal generated by Q_0. Hence so does the original positive element. Thus Q_0 is full; the same holds for Q_1 by (5.7), and for H because Q_0\leq H. The corner projection H is full in the ambient matrix algebra M_2(B), so the latter also contains C as a full subalgebra.□

Proof. Suppose (N+1)x\leq Ny in W(C), with N\geq1. Choose a separable C^*-subalgebra C_*\subset C containing the matrix entries of representatives of x,y and of countably many comparison witnesses. The inequality then already holds over C_*.

There is a unital copy of \Z in M_2(B) commuting with both C_* and \rho(A). Here is the coordinate argument, which also allows the representatives of C_* to vary with the coordinate. Choose isomorphisms \sigma_l\colon\Z\otimes\Z\to\Z with \sigma_l(d\otimes1)\to d, and put \nu_l(z)=\sigma_l(1\otimes z). For fixed d\in\Z,

Thus the commutators tend to zero uniformly for z in the unit ball. Place these embeddings diagonally on the third tensor leg of M_2(R). Tensor approximation gives the same uniform commutation property with any fixed element of M_2(R). At coordinate n, choose l sufficiently large for the entries of representatives of the first n elements of countable dense commutation data, with tolerance tending to zero. The resulting coordinate embeddings induce a unital embedding \nu\colon\Z\to M_2(B) commuting with C_* and \rho(A). It is isometric because each coordinate embedding is isometric.

Multiplication gives a homomorphism

Commutation gives a homomorphism on the maximal tensor product; nuclearity of \Z identifies it with the spatial tensor product. The range satisfies the support condition because c does and \nu(z) commutes with \rho(1_A). This homomorphism restricts to the original inclusion on C_*\otimes1.

The algebra C_*\otimes\Z is \Z-stable. Its Cuntz semigroup is therefore almost unperforated by Theorem 2.4. Tensoring the original inequality with 1_\Z gives comparison of the images of x and y there. Mapping the comparison back to C yields x\leq y.

Here the reindexed copy \nu(\Z) is used for this inequality alone. The constant third-leg copy retained above is the separate source of the weight k below; no commutation between these two copies is required.□

Proof. Fullness of e_l,e_r in D and fullness of D in C imply that both elements are full in C. Since H is a projection, there is a common finite integer N\geq1 such that

Use the positive contraction D_{N+1} from the order-zero supplies on a constant \Z leg, and put p_0=(D_{N+1})^{(0)}. It belongs to C, since the first-corner map is \pi, and it has norm one. Simplicity of the constant leg gives fullness of p_0 in the corner with unit H; because H is full in C, so is p_0.

The N+1 accompanying elements from Lemma 4.2, placed in this corner, form a row over C whose right modulus square is the diagonal sum of N+1 copies of p_0, while its left modulus square belongs to HCH. Hence

Combining this with (5.13) and Proposition 5.4 gives p_0\precsim e_l,e_r in C. We may therefore choose w_l,w_r\in C so that

are as close to p_0 as prescribed.

We specify the accuracy before choosing these witnesses. Fullness of p_0^2 in the constant simple leg gives fixed z_1,\ldots,z_L\in HCH with

Choose \varepsilon>0 such that

and choose w_l,w_r with \|A_l-p_0\|,\|A_r-p_0\|<\varepsilon. Set

Heredity gives s\in D, since DCD\subset D. Functional calculus applied to the annihilation identities gives e_l^{1/2}y=ye_r^{1/2}=0, so sy=ys=0.

Let J=\Ideal_C(s). It contains

Thus the element b=\sum_j z_j^*A_rA_lz_j\in J satisfies \|H-b\|<1 by (5.14). The image of H in C/J is a projection of norm less than one, and is therefore zero. Since H is full in C, this forces J=C.

It remains to prove that s is full inside D. Let I=\Ideal_D(s). For any ideal I of a hereditary subalgebra D\subset C, one has

Indeed, the ideal in the middle is the closed linear span of products through elements of I. By linear decomposition it is enough to use a positive middle element i\in I_+, and then

the two outer factors lie in D by heredity. Here \Ideal_C(I) contains J=C, so the displayed inclusion and an approximate identity of D give D\subset I. Hence I=D, proving that s is full in D.□

Proof. The case g=0 is immediate. Otherwise rescale so that \|g\|\leq1, and put

All these elements are positive and belong to C. The factors commute because e commutes with h and is represented in the first subsystem. Since qh=q, functional calculus gives T_i=Q_i g(k). The matrix with q^{1/2}g(k)^{1/2} in position (0,1) belongs to C and has modulus squares T_1,T_0. Consequently

The positive element g(k) is nonzero in the simple unital constant third leg. Choose finitely many elements z_1,\ldots,z_K in that leg with

For \delta>0 let

The diagonalized z_l commute with Q_i and \rho(A) and multiply C into itself. In particular z_l b_\delta(Q_i)\in C, and

Taking all cuts proves

Since Q_i is full, this shows that g(h_i) is full for every nonzero g. In particular the two elements in (5.15) are totally full; the same argument also ensures that every such functional calculus is nonzero.

Let d be any dimension function on W(C) normalized at either [G_0] or [G_1], if one exists. Equations (5.9) and (5.17), together with (5.7), imply

Apply (5.8) with c^{(i)}=G_i. Since G_i commutes with e^{(i)}, it follows that m[E_i]\leq[Q_0] for every m, and hence d([E_i])=0. The inequalities

and (5.16) now give

Fix \eta>0. By continuity, g(0)=0, and g\ne0, there is a nonzero continuous function g' with 0\leq g'\leq g and support contained in the open set where 0<g<\eta. For the dimension functions under consideration, (5.17) applied to g' gives d([g'(h_i)])>0. This element is orthogonal to (G_i-\eta)_+, and their sum is bounded by G_i. Therefore (5.18) gives

This holds for every dimension function normalized at [G_{1-i}]. The target G_{1-i} is full, so the ideal condition for the comparison criterion in Theorem 2.4 holds. Proposition 5.4 and that criterion yield (G_i-\eta)_+\precsim G_{1-i}. This use of the criterion also permits an empty family of normalized dimension functions. Taking all positive cuts, and then exchanging i=0,1, proves the required Cuntz equivalence.□

Proof. Apply the relative-commutant unitary-equivalence criterion of Lemma 2.10, namely [BBSTWW, Lemma 5.3], to (5.6). Its coordinate algebras are M_2(R), which are separable, unital and \Z-stable, and its domain is the separable unital algebra A. The map \rho is c.p.c. order zero. Corollary 5.3 verifies fullness of C in the ambient ultrapower; Proposition 5.5 verifies the required condition in every full hereditary subalgebra. Proposition 5.6 supplies total fullness and all the functional-calculus comparisons. Thus there is a unitary in the unitization of C conjugating h_1 to h_0.

Realizing this unitary in the ambient algebra as explained after Lemma 2.10 gives U_0\in M_2(B) commuting with \rho(A). Multiplying U_0h_1U_0^*=h_0 by \rho(a) and using (5.1) gives

The element 1-k also has spectrum [0,1] in the constant third leg. Repeating the same argument with this element, with all the previously constructed maps unchanged, gives U_1 and the other identity in (5.19).□

Proof. Use the notation and the two unitaries of Proposition 5.7. Write k^{[0]}=k and k^{[1]}=1-k. Each unitary U_i\in M_2(R_\omega) has unitary representatives U_{i,n}\in M_2(R). Indeed, a bounded representative is asymptotically unitary on both sides. On an \omega-large set its coordinates are invertible and polar correction gives unitaries with the same quotient class; put the identity on the remaining coordinates.

For the matrix-model maps \alpha_n:A\to F_n and \beta'_n:F_n\to A\otimes\Z_1\subset R, define

Each map is c.p.c. order zero: its weight is a positive contraction in the third tensor factor, commuting with the range of \beta'_n; corner inclusion and unitary conjugation preserve these properties. The conjugacies in Proposition 5.7 give, for every a\in A,

since k^{[0]}+k^{[1]}=1. Choose one coordinate in the intersection of the finitely many \omega-large approximation sets for a\in\mathcal F. This proves the assertion. The first map into F_n\oplus F_n is a\mapsto(\alpha_n(a),\alpha_n(a)); it is contractive because the direct-sum norm is the maximum norm.□

Proof. Fix m\ge1. For a finite set of elements of A\otimes M_m, choose a two-colour approximation of A\to M_2(A) accurate on all their matrix entries. Amplify every map by \id_{M_m}. The intermediate algebra is still the direct sum of two finite-dimensional algebras, namely E_0\otimes M_m and E_1\otimes M_m. The first map remains c.p.c. Each amplified return map remains c.p.c. order zero: if \varphi=h\varphi_{\mathrm{hom}}, then

with a commuting positive contractive weight.

Embed M_2(A)\otimes M_m into A\otimes\mathcal K so that its first diagonal m\times m block is the prescribed first m-corner. The amplified approximation then approaches the inclusion of A\otimes M_m into A\otimes\mathcal K. Entrywise accuracy suffices: an m\times m matrix with entries of norm at most \delta has norm at most m\delta.

Let p_m=1_A\otimes\sum_{r=1}^m e_{rr}. Precompose the amplified first map with the c.p.c. compression x\mapsto p_mxp_m on A\otimes\mathcal K. For an arbitrary prescribed finite set in the stabilization, first choose m so that these compressions approximate every element, then approximate the compressed finite set as above. The resulting composite has the same value on x and p_mxp_m, so the total error is bounded by the compression error plus the finite-corner approximation error. This gives the required two-colour approximation on the stabilization.□

Proof of Theorem 1.3. First let A_0 be nonzero and unital, and set A=A_0\otimes\Z. Strong self-absorption gives isomorphisms

Identify the three extra copies of \Z with one copy by a fixed unital isomorphism. The maps \id_{A_0}\otimes\sigma_l then give isomorphisms

The convergence holds first on elementary tensors and then on all of A by contractivity.

Given a finite subset of A, choose l so that \Gamma_l has the required accuracy on its first-factor images. Apply Proposition 6.1 with a sufficiently small error, and postcompose its return maps with \id_{M_2}\otimes\Gamma_l. This gives two-colour approximations to A\to M_2(A). Lemma 6.3 shows that \nucdim(A\otimes\mathcal K)\le1.

Nuclear dimension decreases on hereditary subalgebras [WZ2, Proposition 2.5]. Applying this to the first corner A\otimes e_{11} gives \nucdim(A_0\otimes\Z)\le1. For nonunital separable nuclear A_0, its unitization \widetilde A_0 is separable and nuclear. The unital case applies to \widetilde A_0\otimes\Z, and A_0\otimes\Z is a closed two-sided ideal, hence a hereditary subalgebra, of that algebra. The same permanence result gives the bound. The zero algebra is immediate.

Finally, if A_0\cong A_0\otimes\Z, transport these finite-dimensional approximations across the isomorphism. The first map is c.p.c., the two return maps are individually c.p.c. order zero, and their sum approximates the identity, exactly as in Definition 2.1.□

Proof. Choose

The first expression is positive. A binary Hamming ball of radius \lfloor n/4\rfloor has cardinality at most

Gilbert’s greedy deletion argument [Gil, Theorem 1] therefore produces, for all sufficiently large n, a set S_n\subset\{0,1\}^n of cardinality N=\lfloor e^{an}\rfloor whose distinct words have Hamming distance at least n/4.

Let the coordinates of G be independent standard real Gaussian variables. Fix an admissible list and set

These variables are jointly centered Gaussian. Their covariance matrix \Sigma has diagonal entries one and off-diagonal entries of absolute value at most r^{n/4}. Consequently its eigenvalues lie in [1-\lambda_n,1+\lambda_n], where

For large n, the density of this Gaussian vector is bounded pointwise by

times the density of N independent centered Gaussians of variance 1+\lambda_n. Indeed, \det\Sigma\geq(1-\lambda_n)^N and \Sigma^{-1}\geq(1+\lambda_n)^{-1}I.

Writing \Phi for the standard normal distribution function gives

for one q<1 and every sufficiently large n. The last inequality is uniform over admissible lists because the bracket tends to \Phi(1)<1. By symmetry the same bound holds for the event X_\alpha\geq-1 for all \alpha.

Set \eta_n=(4n k^{n/2})^{-1}. The compact space of admissible lists has an \eta_n-net, in the maximum Euclidean distance between corresponding vectors, of size at most

For example, mesh the coordinate cubes and select an admissible representative from each occupied box. The logarithm of this bound is O(n^2). A union bound using (7.1) shows that the probability that some net list lacks a value above 1 or below -1 is at most

Also \mathbb E\|G\|^2=k^n, so \Pr(\|G\|\leq2k^{n/2})\geq3/4. For large n, both the successful-net event and this norm bound therefore hold for some G.

If an admissible list is approximated by a net list, telescoping the elementary tensors of unit vectors bounds the change of each evaluation by

The two evaluations exceeding 1 and below -1 on the net list retain values above 1/2 and below -1/2, respectively. This proves the assertion for every admissible list.□

Proof. First, a product of nonzero projections p_j\in M_j is nonzero. Induct on n. By induction q=p_2\cdots p_n is a nonzero projection commuting with M_1. The restriction M_1\to\mathcal B(qH) is a nonzero normal representation. Its kernel is a weak-* closed ideal in a factor and is therefore zero. Thus p_1q\ne0.

For 0<\varepsilon<1, the spectral projections p_j=\mathbf1_{(1-\varepsilon,1]}(e_j) are nonzero. Their product is nonzero, and on its range the positive product \prod_j e_j is at least (1-\varepsilon)^n. Letting \varepsilon\downarrow0 proves the lower bound one; the reverse bound follows from contractivity.□

Proof. Let (\gamma,H,\xi) be the irreducible GNS representation of \rho, with \|\xi\|=1, and put

We first show that \ker\rho=L+L^*. The inclusion from right to left is immediate. If \rho(x)=0, write \eta=\gamma(x)\xi\perp\xi. If \eta=0, then x\in L. Otherwise, on \operatorname{span}\{\xi,\eta\} let T be the rank-one operator with T\xi=\eta and T\eta=0; thus T^*\xi=0. Apply the selfadjoint form of Kadison transitivity [Kad, Theorem 1] separately to the real and imaginary parts of T. This gives y\in E such that

Then x-y\in L and y\in L^*, as desired.

The algebra L\cap L^* is hereditary. Let (g_\lambda) be a positive contractive approximate identity for it. If z\in L, then z^*z\in L\cap L^*, whence

The adjoint assertion holds for elements of L^*. Therefore, with f_\lambda=1-g_\lambda, the equality \ker\rho=L+L^* gives

Each f_\lambda is a positive contraction of norm one because \gamma(f_\lambda)\xi=\xi. One sufficiently large \lambda works for \mathcal F.□

Proof of Theorem 1.4. The closed ideal generated by commutators is all of D; otherwise a nonzero unital commutative quotient would supply a character. Decomposing elements into selfadjoint parts and rescaling, choose selfadjoint contractions s_i,t_i and elements v_i,w_i, for 1\leq i\leq m, such that

Put s_0=1, k=m+1, and M=\sum_i\|v_i\|\|w_i\|>0. Under any nonzero unital representation \sigma, some represented commutator has norm greater than \delta=(2M)^{-1}. Since

when S=S^*, \|T\|\leq1, and c is the midpoint of \spc S, the spectrum of some \sigma(s_i) has diameter at least \delta.

There are pure states \rho^0,\rho^1 of \sigma(D) taking that element to its two spectral endpoints. Indeed, a state attaining an endpoint extends from the unital abelian algebra C^*(1,\sigma(s_i)), and the states attaining that endpoint form a nonempty weak-* compact face of the state space. An extreme point of this face is pure. The associated real expectation vectors

differ by at least \delta in one coordinate. The set of pairs in \big(\{1\}\times[-1,1]^m\big)^2 with \|v^0-v^1\|_\infty\geq\delta is compact. Consequently there is 0<r<1, independent of \sigma, such that

To justify the strict bound, two vectors in this compact set cannot be negatively collinear because both first coordinates are one; positive collinearity forces them to coincide, contrary to their separation.

Choose G,n from Lemma 7.1 for k,r and form the selfadjoint element

Let \pi be any nonzero irreducible representation of this tensor product, with factor restrictions \pi_j:D\to\mathcal B(H). For each j, choose the two pure states of \pi_j(D) described above. By (8.2) and Lemma 7.1, choosing one state \rho_j from each pair makes

strictly positive for one choice and strictly negative for another. Removing the normalizations multiplies each tensor evaluation by a positive number and so preserves its sign.

The von Neumann algebras M_j=\pi_j(D)'' are mutually commuting factors: a central element of any M_j commutes with every factor image and hence with the irreducible joint representation, so is scalar. For one of the choices in (8.4), use Lemma 8.2 to choose positive norm-one elements e_j\in\pi_j(D) with

The product e=\prod_j e_j is positive and has norm one by Lemma 8.1. Commuting distinct factors and telescoping their products in (8.3) yields

All factors in this estimate are contractions.

If p>0 and \pi(h)\leq0, then pe^2-e\pi(h)e\geq pe^2, so the left side of (8.5) is at least p. Choosing \varepsilon sufficiently small is a contradiction. The negative choice of p, with its own excisers, similarly rules out \pi(h)\geq0. Thus

for every nonzero irreducible \pi. Each positive part is full: a proper closed ideal containing it would yield a nonzero quotient and hence an irreducible representation killing it. The two positive parts are orthogonal by functional calculus.□

Proof. A hereditary subalgebra H is strongly Morita equivalent to the ideal I it generates. If H had a matrix quotient, ideal correspondence would give an ideal J\subseteq I for which I/J is strongly Morita equivalent to a matrix algebra, hence to \C. By imprimitivity, I/J would be elementary. Ideals of I are ideals of A, contradicting nowhere scatteredness. This is also the equivalence in [TV, Theorem 3.1].

Apply Theorem 9.1(1) to A, put B=F(A), and retain its maps \varphi_j:E_j\to B, elements c_j, and displayed identity.

If \chi:B\to\C were a character, its compositions with these maps would be order-zero positive functionals. Such a functional on M_r, r\geq2, is zero. To see this, a nonzero positive functional has a positive density matrix with an eigenvector \xi of positive eigenvalue. For a unit eigenvector \eta\perp\xi, the orthogonal rank-one projections onto (\xi+\eta)/\sqrt2 and (\xi-\eta)/\sqrt2 both have strictly positive values, contrary to order zero. Thus \chi\circ\varphi_j=0 for every j, contradicting the displayed identity and \chi(1_B)=1.□

Proof. It suffices to prove that for every separable subalgebra C\subseteq F(A), there is a unital *-homomorphism from D into F(A) with range commuting with C. This is Kirchberg’s reindexing mechanism, in the form recorded in [KR, Proposition 2.2 and Corollary 2.3] for unital A. We include the annihilator details needed for nonunital A.

Choose a countable dense unital *-algebra D_0\subset D over \Q+i\Q. For each x\in D_0, choose a bounded lift (a_l^x)_l in A that represents a central element mapping to x, with

Take the zero lift for zero and the approximate identity for 1. The quotient norm permits these choices: first lift within a bound strictly below 2\|x\|, then truncate coordinate norms without changing the ultrapower class. The additive, scalar, multiplicative, and adjoint relation defects between these lifts annihilate A on both sides along \omega.

Choose a countable dense family in C with bounded lifts (b_s^u)_s in A. Also fix a countable dense subset of A. List the following countably many requirements on an old index l: centrality of each a_l^x against that subset; left and right products of every relation defect with that subset tending to zero; and

for its elements a. All these requirements hold along \omega.

For each output index s, choose l(s) so that the first s requirements hold with error less than 1/s, and also

for the first s pairs (x,u) in an enumeration. For this fixed s, the finitely many b_s^u are fixed elements of A. Original centrality therefore makes all these commutator conditions \omega-large as conditions on l. A finite intersection with the other \omega-large conditions is nonempty, so the choice is possible.

The new sequences (a_{l(s)}^x)_s are central, and all their relation defects annihilate A on both sides: the tests hold in the ordinary limit, and the uniform bounds extend them from the dense subset to all of A. Thus they define a unital rational-complex *-homomorphism on D_0 into F(A). The unit assertion follows from the approximate-identity tests. Bound (9.1) permits continuous extension to a complex unital *-homomorphism on D. Its range commutes with the dense family in C, hence with C. Injectivity is not needed.

Starting with one copy and repeatedly taking C to be generated by the previous images gives n commuting unital images of D. The universal property of the maximal tensor product gives the stated homomorphism.□

Proof of Theorem 1.1. The assertion is immediate for A=0, so assume A\ne0. Assume first that \nucdim(A)<\infty; we prove (1)\Rightarrow(2). By Lemma 9.2, B=F(A) has no characters. Choose in B a finite commutator witness as in (8.1), and let D be the unital C^*-subalgebra generated by all its entries. Then D is separable and has no characters, since applying a character of D to that inequality would give 1<1/2.

By Theorem 1.4, some D^{\otimes_{\max}n} contains full orthogonal positive elements a,b. Lemma 9.3 supplies a unital homomorphism into B. Their images remain orthogonal and full: applying a unital homomorphism to finite ideal combinations approximating the unit proves that fullness passes to the target. Theorem 9.1(2) now gives A\cong A\otimes_{\min}\Z. Conversely, for separable nuclear A, absorption and Theorem 1.3 imply \nucdim(A)\le1. This bound implies finite nuclear dimension, completing all three equivalences in Theorem 1.1.□

Proof of Corollary 1.2. Theorem 1.1 gives the equivalence of its three conditions. The companion [OAI-Cu, Corollaries 1.3 and 1.4] gives that (i) implies absorption and that (ii) is equivalent to absorption, even without the exclusion on elementary ideal subquotients.

To prove that absorption implies (i), the case A=0 is immediate. Put D=A\otimes\Z and j_X(x)=x\otimes1_\Z, so j_A=\iota_A. Strong self-absorption gives an isomorphism \varphi:\Z\to\Z\otimes\Z approximately unitarily equivalent to z\mapsto z\otimes1_\Z. The companion’s nonunital calculation [OAI-Cu, proof of Corollary 1.4] gives \Cu(j_D)=\Cu(\id_A\otimes\varphi), an isomorphism: the implementing unitaries u_n\in\Z\otimes\Z act through 1_{M(A)}\otimes u_n\in M(A\otimes\Z\otimes\Z), and positive cuts and approximate units after stabilization give equality of the completed Cuntz maps.

For an absorption isomorphism \theta:A\to D, both sides of the first identity below send a to \theta(a)\otimes1_\Z. Functoriality then gives

All three factors on the right are isomorphisms, proving (i). Thus (ii) reaches the canonical condition (i) through absorption, not by identifying its abstract isomorphism with \Cu(\iota_A).□