This paper introduces a new way to represent elements of E-theory for C*-algebras using pairs of maps from the algebra itself, broadening the understanding of asymptotic homomorphisms.
Contribution
It provides a novel representation of E-theory elements via pairs of maps from the algebra, extending the classical asymptotic homomorphism approach.
Findings
01
E(A,B) can be represented by pairs of maps with the same deficiency from being homomorphisms
02
Pairs of maps can be viewed as asymptotic homomorphisms from a surjecting C*-algebra C
03
Examples of full surjections CβA show all classes in E(A,B) can be obtained from such pairs
Abstract
Let A, B be separable C*-algebras, B stable. Elements of the E-theory group E(A,B) are represented by asymptotic homomorphisms from the second suspension of A to B. Our aim is to represent these elements by (families of) maps from A itself to B. We have to pay for that by allowing these maps to be even further from β-homomorphisms. We prove that E(A,B) can be represented by pairs (Ο+,Οβ) of maps from A to B that are not necessarily asymptotic homomorphisms, but have the same deficiency from being ones. Not surprisingly, such pairs of maps can be viewed as pairs of asymptotic homomorphisms from some C*-algebra C that surjects onto A, and the two maps in a pair should agree on the kernel of this surjection. We give examples of full surjections CβA, i.e. those, for which all classes in E(A,B) can be obtained from pairs of asymptoticβ¦
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Full text
A KK-like picture for E-theory of Cβ-algebras
V. Manuilov
Moscow State University,
Leninskie Gory 1, Moscow,
119991, Russia
Let A, B be separable Cβ-algebras, B stable. Elements of the E-theory group E(A,B) are represented by asymptotic homomorphisms from the second suspension of A to B. Our aim is to represent these elements by (families of) maps from A itself to B. We have to pay for that by allowing these maps to be even further from β-homomorphisms. We prove that E(A,B) can be represented by pairs (Ο+,Οβ) of maps from A to B, which are not necessarily asymptotic homomorphisms, but have the same deficiency from being ones. Not surprisingly, such pairs of maps can be viewed as pairs of asymptotic homomorphisms from some Cβ-algebra C that surjects onto A, and the two maps in a pair should agree on the kernel of this surjection. We give examples of full surjections CβA, i.e. those, for which all classes in E(A,B) can be obtained from pairs of asymptotic homomorphisms from C.
Introduction
There are several pictures for Kasparovβs KK-theory of Cβ-algebras [7], [4], [2], [11], [13] each emphasizing one of the faces of KK-cycles, i.e. of representatives of KK-classes. One of the reasons behind this diversity is lack of homomorphisms between Cβ-algebras. Any β-homomorphism Ο:AβB represents an element in KK(A,B), but if one wants to represent arbitrary element of KK(A,B) then one has to use generalized morphisms. For example, the Cuntz picture of KK-theory uses the pairs (Ο+,Οβ) of β-homomorphisms from A to the multiplier algebra MB of B such that Ο+(a)βΟβ(a)βB for any aβA [4].
E-theory of Connes and Higson [3] shares the same problem: one has to use asymptotic homomorphisms in place of β-homomorphisms. Moreover, the asymptotic homomorphisms that represent elements in E(A,B) have domain not in A but in the suspension SA over A (or, even in the second suspension), and take values (if considered as β-homomorphisms) in Cbβ(R+β;B)/C0β(R+β;B), where R+β=[0,β), and Cbβ(R+β;B) denotes the algebra of bounded B-valued functions on R+β.
Our aim is to represent elements of E(A,B) by pairs of maps from A itself to Cbβ(R+β;B)/C0β(R+β;B) similarly to the Cuntz picture for KK-theory. We have to pay for that by allowing these maps to be even further from β-homomorphisms. We shall work with pairs of maps, which should be continuous, but may be not linear or not asymptotically linear, and we require that the two maps in a pair should have the same deficiency from being an asymptotic homomorphism. Namely, we show that elements of E(A,B) can be represented by pairs (Ο+,Οβ) of maps from A to Cbβ(R+β;B)/C0β(R+β;B) that satisfy
[TABLE]
for any n=0,1,2,β¦ and any a,b,c0β,c1β,β¦,cnββA, where MΒ±(a,b) stands for either ΟΒ±(ab)βΟΒ±(a)ΟΒ±(b) or ΟΒ±(a+b)βΟΒ±(a)βΟΒ±(b). Isomorphism between E(A,B) and the group of homotopy classes of such pairs is proved using the explicit form of the Bott isomorphism obtained in [8].
Not surprisingly, such pairs of maps can be viewed as pairs of asymptotic homomorphisms from some Cβ-algebra C that surjects onto A, and the two maps in a pair should agree on the kernel of this surjection. We give examples of full surjections CβA, i.e. those, for which all classes in E(A,B) can be obtained from pairs of asymptotic homomorphisms from C.
Note that in topological applications, A is often a group Cβ-algebra of a discrete group G, or its augmentation ideal.
Our interest in representing E-theory classes by maps from A instead of suspensions over A is caused by our hope that group actions may give more direct constructions of these classes.
Special cases A=C or B=K (the Cβ-algebra of compact operators) were considered earlier in [9] and in [10] respectively.
Similar approach can be used in KK-theory. We shall describe it in a separate paper.
1. Asymptotic KK-cycles
Let A, B be Cβ-algebras, A separable, B stable and Ο-unital, MB the multiplier algebra for B. Let Cbβ(R+β;MB) (resp. Cb,sβ(R+β;MB)) denote the algebras of MB-valued functions on R+β=[0,β) continuous with respect to the norm (resp. to the strict) topology. For tβR+β, evtβ denotes the evaluation at t.
Let C be a Cβ-algebra (usually C is either Cb,sβ(R+β;MB) or Cbβ(R+β;B)).
Let (Ξ¦+,Ξ¦β) be a pair of (not necessarily linear) continuous maps Φ±:AβC. We denote by E=E(Ξ¦+,Ξ¦β)=Cβ(Ξ¦+(A),Ξ¦β(A))βC the Cβ-algebra generated by all Φ±(a), aβA, and by D=D(Ξ¦+,Ξ¦β)βE the ideal, in E, generated by Ξ¦+(a)βΞ¦β(a), aβA.
For x,yβCb,sβ(R+β;MB) we write xβΌy if xβyβC0β(R+β;MB), i.e. if limtβββxtββytβ=0.
Definition 1.1**.**
A pair (Ξ¦+,Ξ¦β) of continuous maps Φ±:AβCbβ(R+β;B) is a compact asymptotic KK-cycle from A to B if
(a1)
Φ± are C-homogeneous and asymptotically involutive, i.e. Φ±(Ξ»a)=λΦ±(a) for any Ξ»βC and for any aβA, and Φ±(aβ)βΦ±(a)ββC0β(R+β;B) for any aβA;
(a2)
(Φ±(a+b)βΦ±(a)βΦ±(b))xβC0β(R+β;B) and (Φ±(ab)βΦ±(a)Φ±(b))xβC0β(R+β;B)
for any a,bβA and for any xβD.
Let MΒ±(a,b) denote either Φ±(a+b)βΦ±(a)βΦ±(b) or Φ±(ab)βΦ±(a)Φ±(b). Note that (a2) is equivalent to
[TABLE]
or, equivalently,
[TABLE]
for any a,b,c1β,β¦,cnβ,dβA and any n=0,1,2,β¦.
Definition 1.2**.**
A pair (Ξ¦+,Ξ¦β) of continuous maps Φ±:AβCbβ(R+β;B)has the same deficiency from being a homomorphism if
(a1)
Φ± are C-homogeneous and asymptotically involutive;
We write xβΌy if xβyβC0β(R+β;B).
It follows from
[TABLE]
and
[TABLE]
that MΒ±(a,b)(Ξ¦+(c)βΞ¦β(c))βΌ0.
Similarly,
[TABLE]
and
[TABLE]
imply MΒ±(a,b)Φ±(c1β)β―Φ±(cnβ)(Ξ¦+(cn+1β)βΞ¦β(cn+1β))βΌ0. Any xβD can be approximated by finite sums of products Ξ¦+(c1β)β―Ξ¦+(cnβ)(Ξ¦+(cn+1β)βΞ¦β(cn+1β))e, with eβE, hence (a2) follows from (a3).
To prove the opposite, note that
[TABLE]
hence M+(a,b)βMβ(a,b)βD. Passing to adjoints, M+(a,b)ββMβ(a,b)ββD. For shortnessβ sake set mΒ±β=MΒ±(a,b). Since mΒ±βxβΌ0 for x=m+βββmββββD, we have (m+ββmββ)(m+ββmββ)ββΌ0. Then, by the Cβ-property of the norm, we conclude that m+ββΌmββ, i.e. that
M+(a,b)βΌMβ(a,b). Now, it is easy to check that (a3) holds for any n.
β
Remark 1.4*.*
It suffices to check (a3) in Definition 1.2 only for n=0 and n=1:
[TABLE]
[TABLE]
Definition 1.5**.**
A pair (Ξ¦+,Ξ¦β) of continuous maps Φ±:AβCb,sβ(R+β;MB) is an asymptotic KK-cycle from A to B if
(a1)
Φ± are C-homogeneous and asymptotically involutive;
(b2)
DβCbβ(R+β;B);
(b3)
(Ξ¦(a+b)βΞ¦(a)βΞ¦(b))xβC0β(R+β;B) and (Ξ¦(ab)βΞ¦(a)Ξ¦(b))xβC0β(R+β;B)
for any a,bβA and for any xβD.
Two (compact) asymptotic KK-cycles (Ξ¦0+β,Ξ¦0ββ) and (Ξ¦1+β,Ξ¦1ββ) from A to B are homotopic if there is a (compact) asymptotic KK-cycle (Ξ¨+,Ξ¨β) from A to IB=C([0,1];B) such that the obvious evaluation maps at 0 and at 1 give (Ξ¦0+β,Ξ¦0ββ) and (Ξ¦1+β,Ξ¦1ββ).
Lemma 1.6**.**
Let Ξ¨t,ΟΒ±β:AβMB, (t,Ο)βR+βΓ[0,1], be a family of continuous maps.
The following properties are equivalent:
(1)
The pair of maps Ψ±:AΓR+βΓ[0,1]βMB is an asymptotic KK-cycle from A to IB;
2. (2)
The maps Ξ¨t,ΟΒ±β satisfy:
(h1)
Ξ¨t,ΟΒ±β* are homogeneous and Ξ¨t,ΟΒ±β(aβ)βΞ¨t,ΟΒ±β(a)βC0β(R+β;IB)=C0β(R+βΓ[0,1];B) for any aβA;*
(h2)
Ξ¨t,ΟΒ±β(a)βCb,sβ(R+β;M(IB))* for any aβA;*
(h3)
Ξ¨t,Ο+β(a)βΞ¨t,Οββ(a)βCbβ(R+β;IB)* for any aβA.*
(h4)
Mt,ΟΒ±β(a,b)Ξ¦t,Ο+β(c1β)β―Ξ¦t,Ο+β(cnβ)(Ξ¦t,Ο+β(d)βΞ¦t,Οββ(d))βC0β(R+β;IB)=C0β(R+βΓ[0,1];B)* for any a,b,c1β,β¦,cnβ,dβA;*
β
Denote the homotopy class of a (compact) asymptotic KK-cycle (Ξ¦+,Ξ¦β) by [(Ξ¦+,Ξ¦β)], and
denote the set of all homotopy classes of (compact) asymptotic KK-cycles from A to B by EL(A,B) (resp. by EM(A,B)).
Stability of B allows to define addition on EL(A,B) and on EM(A,B): let s1β,s2ββMB be isometries with s1βs1ββ+s2βs2ββ=1, and let s~1β,s~2ββCb,sβ(R+β;MB) be the constant functions with values s1β and s2β respectively. Let (Ξ¦+,Ξ¦β), (Ξ¨+,Ξ¨β) be two (compact) asymptotic KK-cycles. Set
[TABLE]
It is easy to see that the pair in the right-hand side is an asymptotic KK-cycle, and that the addition is well defined.
Lemma 1.7**.**
An asymptotic KK-cycle (Ξ¦+,Ξ¦β) with Ξ¦+=Ξ¦β represents the zero element in EL(A,B).
Proof.
Let Ξ¦=Ξ¦+=Ξ¦β.
Note that the ideal D generated by Ξ¦+(a)βΞ¦β(a), aβA, is zero, hence (ΟΞ¦,ΟΞ¦) is an asymptotic KK-cycle for any Οβ[0,1], so [(Ξ¦,Ξ¦)]=[(0,0)].
Let (Ξ¨+,Ξ¨β) be an asymptotic KK-cycle. There are strictly continuous paths s1β(Ο), s2β(Ο), Οβ[0,1], such that, for Οβ(0,1], s1β(Ο) and s2β(Ο) are isometries with s1β(Ο)s1β(Ο)β+s2β(Ο)s2β(Ο)β=1, and s1β(0)=1, s2β(0)=0.
Set
[TABLE]
This is an asymptotic KK-cycle from A to IB ((h1)-(h4) trivially hold), hence
[(Ξ¨+,Ξ¨β)]=[(Ξ¨+,Ξ¨β)]+[(0,0)].
β
The standard rotation argument shows that
[TABLE]
and that
[TABLE]
in EL(A,B), hence EL(A,B) is an abelian group.
Similar results hold for compact asymptotic KK-cycles.
2. Making asymptotic KK-cycles compact
There is a canonical map j:EM(A,B)βEL(A,B) induced by the inclusion BβMB. Here we construct a map in the opposite direction and show that these two maps are inverse to each other.
Let (Ξ¦+,Ξ¦β) be an asymptotic KK-cycle, [(Ξ¦+,Ξ¦β)]βEL(A,B). Recall that DβCbβ(R+β;B) denotes the ideal generated by Ξ¦+(a)βΞ¦β(a), aβA, in the Cβ-algebra E generated by Φ±(A). Let (vrβ)rβR+ββ be an approximate unit in D, quasicentral in E.
Lemma 2.1**.**
Let Ο:AβE be a continuous map. There exists an order-preserving homeomorphism r of [0,β) (a reparametrization) such that
(r1)
limtβββ[vr(t)β,Οtβ(a)]=0* for any aβA;*
(r2)
limtβββvr(t)βxtββxtβ=0* for any xβD.*
Proof.
As vrβ is an approximate unit for D, we have
[TABLE]
for any xβD. Therefore, any function r=r(t) satisfies (r2) if limtβββr(t)=β.
Consider a dense sequence a1β,a2β,β¦βA and positive numbers Ξ΅1β,Ξ΅2β,β¦ such that limnβββΞ΅nβ=0. Inductively, for r=nβN find tnβ such that tnββ₯tnβ1β+1 and β₯[vnβ(t),Οtβ(aiβ)]β₯<Ξ΅nβ for i=1,2,β¦,n when tβ₯tnβ. Extend the map nβ¦tnβ to a homeomorphism of [0,β) and take its inverse as r.
β
As a bonus, we have [v,Ο(a)]βC0β(R+β;B) (i.e. [v,Ο(a)] is norm-continuous) for any aβA.
Set v(t)=vr(t)β(t), vβCbβ(R+β;B), and Ψ±(a)=vΦ±(a)βCbβ(R+β;B). Note that (a1) holds, i.e. that (vΦ±(a))ββvΦ±(aβ)βΌ0.
Then
[TABLE]
[TABLE]
[TABLE]
Properties of v imply that the both terms here lie in C0β(R+β;B).
Similarly, one can show that
[TABLE]
for any a,b,c,dβA, where MΒ±(a,b) is either Ξ¨+(a+b)βΨ±(a)βΨ±(b) or Ξ¨+(ab)βΨ±(a)Ψ±(b), hence (Ξ¨+,Ξ¨β) is a compact asymptotic KK-cycle.
Two different choices of an approximate unit v and of a reparametrization give rise to two homotopic compact asymptotic KK-cycles. Also, if (Ξ¦+,Ξ¦β) and (Ξ+,Ξβ) are two homotopic asymptotic KK-cycles then one can find an approximate unit v and a reparametrization related to this homotopy, so that v0=ev0βv, v1=ev1βv are approximate units for the given two asymptotic KK-cycles, and (v0Ξ¦+,v0Ξ¦β) and (v1Ξ+,v1Ξβ) are homotopic. Thus this construction defines a map
[TABLE]
Theorem 2.2**.**
The map p is an isomorphism.
Claim 1.pβj is the identity map on EM(A,B).
Let (Ξ¨+,Ξ¨β) be a compact asymptotic KK-cycle. Consider it as an asymptotic KK-cycle, and find an appropriate approximate unit v such that [(vΞ¨+,vΞ¨β)]=p([(Ξ¨+,Ξ¨β)]). Since Ψ±(A)βCbβ(R+β;B), there exists wβCbβ(R+β;B) such that 0β€wβ€1 and (1βw)Ψ±(a)βΌ0 for any aβA. Then [(wΞ¨+,wΞ¨β)]=[(Ξ¨+,Ξ¨β)]. Connecting v and w by the linear homotopy, we conclude that p([(Ξ¨+,Ξ¨β)])=[(Ξ¨+,Ξ¨β)]. β
Claim 2.jβp is the identity map on EL(A,B).
Let (Ξ¦+,Ξ¦β) be an asymptotic KK-cycle.
Consider the Cβ-algebra E generated by Φ±(A) and by constant B-valued functions on R+β, and let DβE be the ideal generated by Ξ¦+(a)βΞ¦β(a), aβA, and by constant B-valued functions on R+β. Let (wrβ)rβR+ββ be an approximate unit in D, quasicentral in E. Set w(t)=wr(t)β(t), tβR+β, wβCbβ(R+β;B), where r satisfies [wr(t)β,Ξ¦tΒ±β(a)]βΌ0 for any aβA. Then the linear homotopy connecting v and w shows that the asymptotic KK-cycles (vΞ¦+,vΞ¦β) and (wΞ¦+,wΞ¦β) are homotopic. Note that wΦ±(a)βΌw1/2Φ±(a)w1/2.
Lemma 2.3**.**
Let (Ξ¨t+β,Ξ¨tββ) be an asymptotic KK-cycle from A to B, UtββM(B) a norm-continuous family of unitaries, tβR+β. Then (UtβΞ¨t+βUtββ,UtβΞ¨tββUtββ) is an asymptotic KK-cocycle homotopic to (Ξ¨t+β,Ξ¨tββ).
Proof.
Note that (UΟtβΞ¨t+βUΟtββ,UΟtβΞ¨tββUΟtββ) is an asymptotic KK-cycle from A to IB as UΟtββCbβ(R+β;M(IB)). Then use contractibility of the unitary group of M(B).
β
Set
[TABLE]
V is an isometry, and there is a norm-continuous family Utβ of unitaries such that VΦ±Vββ0=U(Φ±β0)Uβ, hence [(Ξ¦+,Ξ¦β)]=[(VΞ¦+Vβ,VΞ¦βVβ)]. Write VΦ±Vβ as a matrix:
[TABLE]
Note that
[TABLE]
as wxβΌx for any xβD.
For Οβ[0,1] set AΒ±(Ο)=(10β0Οβ)VΦ±(a)Vβ(10β0Οβ).
Lemma 2.4**.**
This is a homotopy connecting (VΞ¦+Vβ,VΞ¦βVβ) with (w1/2Ξ¦+w1/2,w1/2Ξ¦βw1/2).
Proof.
Let a,bβA, c=ab, and let BΒ±=VΦ±(b)Vβ, CΒ±=VΦ±(c)Vβ. As (Ξ¦+,Ξ¦β) is an asymptotic KK-cycle, we have C+βA+B+βΌCββAβBβ, which is equivalent to C+βCβ=(A+βAβ)B++Aβ(B+βBβ). Looking at the upper left corner and taking into account (2.1), we obtain c11+ββc11βββΌ(a11+ββa11ββ)b11+β+a11ββ(b11+ββb11ββ), whence
[TABLE]
Replacing here A,B,C by A(Ο),B(Ο),C(Ο), we see that (2.2) does not depend on Ο.
Let dβA, DΒ±=VΦ±(d)Vβ. We know that (C+(Ο)βA+(Ο)B+(Ο))(D+(Ο)βDβ(Ο))βΌ0 when Ο=1.
Note that D+(Ο)βDβ(Ο)βΌ(d11+ββd11ββ0β00β),
[TABLE]
As (1βw)1/2(d11+ββd11ββ)βΌ0, so aij+βbkl+β(d11+ββd11ββ)βΌ0 for all (ijkl)ξ =(1111), therefore, (D+(Ο)βDβ(Ο))(C+(Ο)βA+(Ο)B+(Ο))βΌ0 for any Ο, and uniformly in Ο. This proves (h4). (h2) and (h3) are obvious.
β
Remark 2.5*.*
We did not use in Lemma 2.4 that wβCbβ(R+β;B). It suffices to require that wβCbβ(R+β;MB)), wΦ±(a)βCbβ(R+β;B), [w,Φ±(a)]βΌ0 and w(Ξ¦+(a)βΞ¦β(a))βΌΞ¦+(a)βΞ¦β(a) for any aβA.
3. A map EL(A,B)βE(A,B)
Let eijβ, i,jβZ, denote the standard matrix units for M(BβK). For an asymptotic KK-cycle (Ξ¦+,Ξ¦β) set
[TABLE]
Let E be the Cβ-algebra generated by {Ξ¦(a):aβA} and by T, D=DβK, where D is the ideal in E generated by Ξ¦+(a)βΞ¦β(a), aβA, and E is the Cβ-algebra generated by Φ±(A). Then DβEβCb,sβ(R+β;MB), and D is an ideal in E.
Let (vrβ)rβR+ββ be an approximate unit in D, quasicentral in E.
Let Ξ¦tβ=evtββΞ¦, vr,tβ=evtβ(vrβ), Mtβ(a,b)=evtβ(M(a,b)), where M(a,b) is either Ξ¦(a+b)βΞ¦(a)βΞ¦(b) or Ξ¦(ab)βΞ¦(a)Ξ¦(b).
Lemma 3.1**.**
There exists a reparametrization r such that
(v1)
vβCbβ(R+β;BβK);
(v2)
f(v)xβx,xf(v)βxβC0β(R+β;BβK)* for any xβD and for any fβC0β(0,1];*
(v3)
f(v)yβyf(v)βC0β(R+β;BβK)* for any yβE and for any fβC[0,1];*
(v4)
f(v)M(a,b)βC0β(R+β;BβK)* for any a,bβA and any fβC0β(0,1].*
Proof.
Items (v1)-(v2) hold for any reparametrization. We have already shown how to satisfy (v3) in Lemma 2.1, where a sequence t1β,t2β,β¦ was constructed. To satisfy (v4), note that M(a,b)vrββΌ0 for any a,bβA and any rβR+β. As A is separable, fix a dense sequence (cnβ)nβNββA. For each nβN find TnββR+β such that
Tnββ₯tnβ and
[TABLE]
for a,bβ{c1β,β¦,cnβ} and for any rβ[0,n].
Then assigning Tnβ to n, we get a reparametrization r=r(t) such that r(Tnβ)=n.
β
For aβA, fβC0β(0,1), hnβ(s)=e2Οins, set
[TABLE]
It follows from (v2)-(v4) that
[TABLE]
for any f,gβC0β(0,1), any a,bβA and any n,mβN, hence (3.1) determines an asymptotic homomorphism C0β(0,1)βC(S1)βAβCbβ(R+β;BβK), which can be restricted to C0β((0,1)2)βA. We denote this asymptotic homomorphism by Ξ±(Ξ¦+,Ξ¦β), Ξ±(Ξ¦+,Ξ¦β)tβ:S2AβCbβ(R+β;BβK).
It is clear that the map Ξ±:EL(A,B)βE(A,BβK) doesnβt depend on the choices of the approximate unit and of the approximation, as any two such choices can be connected by a linear homotopy.
4. A map E(A,B)βEL(A,B)
Let Ο:S2AβCbβ(R+β;B) be a continuous map such that
(Οtβ)tβR+ββ:S2AβB is an asymptotic homomorphism, where Οtβ=evβΟ. We identify here S2A with C0β(D;A), where D is the open unit ball on the complex plane.
Let (Ξ±iβ)i=0,1,2,β¦β be a family of functions in C0β[0,1) such that
(Ξ±1)
suppΞ±0β=[0,21β] and Ξ±0β(0)=1;
(Ξ±2)
suppΞ±iβ=[iiβ1β,i+2i+1β], i=1,2,β¦;
(Ξ±3)
Ξ±iββ₯0 for any i=0,1,2,β¦;
(Ξ±4)
βi=0ββΞ±i2β=1 (convergence here is pointwise).
Let f:[0,1]β[0,1] satisfy f(1βt)=1βf(t) and f(t)=0 for tβ[0,101β], and let g=fβf2β. Set
[TABLE]
where (r,Ο) are the polar coordinates on D.
Note that pΒ±βΞ±iββM2β(C0β(D)) for any i=0,1,2,β¦.
Set
[TABLE]
where Ο΅ijβ, i,j=0,1,2,β¦, are the matrix units in M(BβK) (later on, we shall use two copies of K, so we have to use two different sets of matrix units β eijβ and Ο΅ijβ).
Then the maps Ξ¦tΒ±β are well-defined (the sum is strictly convergent), but are not necessarily asymptotic homomorphisms (because algebraic properties of (Οtβ) hold asymptotically in a non-uniform way).
Lemma 4.1**.**
(Ξ¦+,Ξ¦β)* is an asymptotic KK-cycle.*
Proof.
(a1) is trivial. As p+βΞ±iβ=pββΞ±iβ for iβ₯10, so (b2) holds. To check (b3), let Pnβ=βi=0nβeiiβ. Then, as pΒ±β are projections, so, for any nβN and for any a,bβA, one has
[TABLE]
where MtΒ±β(a,b) is either Ξ¦tΒ±β(a+b)βΞ¦tΒ±β(a)βΞ¦tΒ±β(b) or Ξ¦tΒ±β(ab)βΞ¦tΒ±β(a)Ξ¦tΒ±β(b). As p+β(r,Ο)=pββ(r,Ο) for rβ₯109β, so
[TABLE]
for any aβA.
Note that Ξ±iβΞ±jβ=0 when β£iβjβ£β₯2, so Φ±(a) (4.1) has tri-diagonal form, therefore, combining
(4.2) and (4.3), we get
[TABLE]
for any a,b,c1β,β¦,cnβ,dβA.
β
This construction doesnβt depend, up to homotopy, on the choice of the functions Ξ±iβ, i=0,1,2,β¦, f, and on the choice of Ο within its class [Ο]βE(A,B), so this gives us a map Ξ²(Ο)=(Ξ¦+,Ξ¦β), Ξ²:E(A,B)βEL(A,BβK).
5. Isomorphism EL(A,B)β E(A,B)
Theorem 5.1**.**
The maps Ξ± and Ξ² are isomorphisms between E(A,B) and EL(A,B).
Proof.
Claim 1.Ξ²βΞ±:EL(A,B)βEL(A,BβKβKβM2β) coincides with the isomorphism induced by an isomorphism Bβ BβKβKβM2β.
Set
[TABLE]
First step of homotopy.
Let wr(t)β=βi=0ββΞ»iβ(r(t))Ο΅iiβ, where r is a reparametrization (i.e. an order-preserving homeomorphism of [0,β)), and Ξ»iβ, i=0,1,β¦, are continuous functions such that
(Ξ»1)
0β€β¦β€Ξ»iβ(t)β€Ξ»i+1β(t)β€β¦β€1 for any tβ[0,β) and any i=0,1,β¦;
(Ξ»2)
for any tβ[0,β) there is NβN such that Ξ»iβ(t)=0 for any iβ₯N;
(Ξ»3)
for any i=0,1,β¦limtβββΞ»iβ(t)=1;
(Ξ»4)
limtβββΞ»i+1β(t)βΞ»iβ(t)=0 for any i=0,1,β¦;
Note that tri-diagonal structure of ΞΒ± and (Ξ»4) imply that
[TABLE]
for any aβA for an appropriately chosen reparametrization. In fact, any sufficiently slow reparametrization would do. We shall fix it later.
By (Ξ»2), wrβ(t)ΞtΒ±β(a) is norm-continuous in t (ΞΒ± is only strictly continuous).
Tri-diagonality of ΞΒ± and (Ξ»4) imply that
[TABLE]
for any aβA,
and (Ξ»3) implies that
[TABLE]
for any aβA.
All these hold for any reparametrization r. Therefore, there exists a homotopy connecting (Ξ+,Ξβ) with (wΞ+,wΞβ) (and with (wΞ+w,wΞβw)) as in Theorem 2.2 (see Lemma 2.4 and Remark after it).
Set u=1βv, u=(utβ)tβR+ββ, where an approximate unit (vrβ)rβR+ββ and a reparametrization r are as in Lemma 3.1;
set
[TABLE]
and set
[TABLE]
Note that
[TABLE]
for any i,jβN and for any aβA, and
[TABLE]
if β£iβjβ£>1, so we have
[TABLE]
and the asymptotic KK-cycles (Ξ+,Ξβ) and (Ξ¨+,Ξ¨β) are homotopic.
Second step of homotopy.
Set Ξ³(Ο)=(1βΟ)+Οf(u)β, Ξ΄(Ο)=1βΞ³2(Ο)β, Οβ[0,1]; s=Ξ³(1), c=Ξ΄(1). Note that Ξ³(0)=1, Ξ΄(0)=0. Set
[TABLE]
[TABLE]
Now, let us fix a reparametrization.
Lemma 5.2**.**
Let Ο:AβCb,sβ(R+β;MB) be a continuous map, let vβCbβ(R+β;MB) satisfy 0β€vβ€1 and
[TABLE]
*for any aβA. Let Ξ²iβ, i=0,1,β¦, be the functions defined by Ξ²iβ(x)=Ξ±iβ(1βx), xβ[0,1].
There exists a reparametrization r such that*
[TABLE]
for any aβA, uniformly in Ο;
2) For any reparametrization r and for any i=0,1,β¦ one has
[TABLE]
Proof.
Let a1β,a2β,β¦ be a dense sequence in A, Ξ΅1β>Ξ΅2β>β¦>0, limjβββΞ΅jβ=0. Set gΟβ(x)=βi=0ββΞ²i2β(x)Ξ»i2β(Ο), xβ[0,1], Οβ[0,β). We may think of Ο(a) and the function v~ given by Οβ¦v(Ο)=Οv+(1βΟ)1 as elements of Cbβ(R+β;MB)βC[0,1]. Let q:Cbβ(R+β;MB)βC[0,1]βCbβ(R+β;MB)/C0β(R+β;MB)βC[0,1] be the quotient map. Then (5.4) implies that
q(Ο(a)) and q(v~) commute. Then q(Ο(a)) commutes with q(g(v~))=g(q(v~)) for any gβC[0,1], hence
limtβββ[Οtβ(a),g(vtβ(Ο))]=0 for any aβA uniformly in Ο.
Let k1β,k2β,β¦ be a sequence of numbers, kiββ₯1, iβN.
Start with Ο=1 and find t1ββ[0,β) such that [g1β(vtβ(Ο)),Οtβ(a1β)]<Ξ΅1β for any tβ₯t1β and any Οβ[0,1]. Then take Ο=2 and find t2ββ[0,β) such that t2ββ₯t1β+k1β and [g2β(vtβ(Ο)),Οtβ(aiβ)]<Ξ΅2β for i=1,2, for any tβ₯t2β and for any Οβ[0,1]. Proceeding infinitely, we obtain a map Οβ¦tΟβ when ΟβN. Extend this map to a homeomorphism on [0,β), and take its inverse as r.
Ξ²iβ(vtβ)βi=0ββΞ²i2β(vtβ)Ξ»i2β(r(t))=Ξ²iβ(vtβ)gr(t)β(vtβ), so the claim follows from the fact that grβ is an approximate unit in C0β(0,1].
β
When finding a reparametrization, we should keep in mind that it should be sufficiently slow to satisfy also (5.2). This can be done by adjusting the numbers kiβ, iβN.
Note that
[TABLE]
and
[TABLE]
uniformly in Ο, as [T,Ξ³(Ο)]βΌ0, [T,Ξ΄(Ο)]βΌ0 uniformly in Οβ[0,1] by (v3).
We have to check (h1)-(h4) from Lemma 1.6.
(h1) is trivial. (h2) and (h3) follow from (5.5). To check (h4), set MΒ±Οβ(a,b)=Ψ±(ab)(Ο)βΨ±(a)(Ο)Ψ±(b)(Ο), MΒ±β(a,b)=Φ±β(ab)βΦ±β(a)Φ±β(b), a,bβA. Let cβA. By (5.6) and Lemma 5.2, using that limtβββ[Ξ±iβ(utβ),UΒ±β(Ο)]=0 uniformly in Ο for any i=0,1,β¦, limtβββ[Ξ±iβ(utβ),Ξ¦tΒ±β(a)]=0, and limtβββ[pΒ±β(utβ,T),Ξ¦tβ(a)]=0 we have
Then it would suffice to show that M+β(a,b)(p+β(u,T)βpββ(u,T)), M+β(a,b)pββ(u,T)Ξ¦(c)(U+ββ(Ο)βUβββ(Ο)) and
M+β(a,b)U+ββ(Ο)(U+β(Ο)βUββ(Ο)) vanish as tββ uniformly in Ο.
Note that p+β(u,T)βpββ(u,T)=(0(Tββ1)g(u)βg(u)(Tβ1)0β), where g(0)=g(1)=0, hence, by (v4) M+β(a,b)g(u)βC0β(R+β;B).
To deal with the two remaining terms, note that
[TABLE]
and that Ξ΄(Ο)=1βΞ³2(Ο)β=Ο(1βf(u))β=Οf(v)β, so M+β(a,b)Ξ΄(Ο)βΌ0 by (v4) uniformly in Ο.
As
[TABLE]
[TABLE]
uniformly in Ο by (v3), so, by (v4) M+β(a,b)pββ(u,T)(U+ββ(Ο)βUβββ(Ο))βΌ0 uniformly in Ο.
By (v4),
[TABLE]
uniformly in Ο, so
[TABLE]
uniformly in Ο.
The last summand, M+β(a,b)U+ββ(Ο)(U+β(Ο)βUββ(Ο)), vanishes as tββ uniformly in Ο by similar argument. So, M+Οβ(a,b)(Ξ¨+(c)βΞ¨β(c))βΌ0 uniformly in Ο. Similarly one can show that the same holds also for MβΟβ(a,b)(Ξ¨+(c)βΞ¨β(c)) and for MΒ±Οβ(a,b)Ψ±(c)(Ξ¨+(d)βΞ¨β(d)).
β
Thus, we have a homotopy connecting (Ξ¨+(0),Ξ¨β(0)) with (Ξ¨+(1),Ξ¨β(1)). As UΒ±β(0)=(10β01β), so Ψ±(0)=Ψ±. For Ο=1, Ξ³(1)=s=f(u)β, Ξ΄(1)=c=1βf(u)β, and both s and c asymptotically commute with Ξ¦(a), i.e. [s,Ξ¦(a)]βΌ0, [c,Ξ¦(a)]βΌ0. Then
[TABLE]
As s[T,Ξ¦(a)]βΌ0, we can skip the off-diagonal elements. Similarly,
[TABLE]
hence the first summand also vanishes as tββ, so
Similarly, we have Ξ¨tββ(a)(1)=wr(t)βββ£iβjβ£β€1βΞ±iβ(utβ)(Ξ¦tβ(a)0β00β)Ξ±jβ(utβ)Ο΅ijβΒ wr(t)β.
Third step of homotopy.
Set u(Ο)=1β((1βΟ)v+Ο1). This is the linear homotopy connecting u=u(0)=1βv with [math].
Set
[TABLE]
[TABLE]
Norm-continuity of Ξ±iβ(utβ(Ο)) in (t,Ο) for each i=0,1,β¦ implies norm-continuity of Ψ±(a) in Ο for each tβ[0,β), and strict continuity in t uniformly in Ο, so (h2) holds for Ψ±(a). (h3)
follows from limiβββsupΟβ[0,1]ββ₯Ξ±iβ(u(Ο)(TΞ¦(a)TββΞ¦(a))β₯=0. Finally, (h4) follows from
Lemma 5.2, so (Ξ¨+(Ο),Ξ¨β(Ο)) is an asymptotic KK-cycle from A to IBβK2.
When Ο=1, utβ(1)=0, hence Ξ±0β(utβ(1))=1, Ξ±iβ(utβ(1))=0 for any iβ₯1. As Ξ»0β(t)=1, we get
[TABLE]
We can write (Ξ+,Ξβ)=(TΞ¦Tββ0,Ξ¦β0)=((Ξ¦+0β0Ξ¦β²β),(Ξ¦β0β0Ξ¦β²β)), where Ξ¦β² is the infinite direct sum of copies of Ξ¦+, Ξ¦β and [math]. By Lemma 2.1, [(Ξ+,Ξβ)]=[(Ξ¦+,Ξ¦β)].
β
Claim 2.Ξ±βΞ²:E(A,B)βE(A,BβKβKβM2β) coincides with the isomorphism induced by an isomorphism Bβ BβKβKβM2β.
Let Ο:S2AβCbβ(R;B) represent an element of E(A,B), and let Φ± be defined as in (4.1). Denote Ξ±βΞ²(Ο) by Ο:S2AβCbβ(R+β;B). Then
[TABLE]
where Ξ¦=βiβ₯0βΞ¦+eiiβ+βi<0βΞ¦βeiiβ, fβC0β(0,1), h(x)=e2Οix, and vβDβKβCbβ(R+β;B) satisfies
β’
limtβββf(v)d=0 for any dβDβK;
β’
limtβββf(v)M(a,b)=0 for any a,bβA, where M(a,b) is either Ξ¦(ab)βΞ¦(a)Ξ¦(b)
or Ξ¦(a+b)βΞ¦(a)βΞ¦(b).
Let ΟΒ±(gβa)=βi,jβ₯0βΟ(Ξ±iβΞ±jβgβa)Ο΅ijβ, where (Ο΅ijβ)i,j=0,1,2β¦β is a set of matrix units for M(BβK) and gβaβΞ£2A=C(S2;A), and let ΟβΒ±β(gβa)=βi=0ββΟΒ±(gβa)eiiβ. (eijβ are the matrix units in M(BβKβK) with respect to the second copy of K.) Set W=βi=0ββei,i+1β.
Let E be the Cβ-subalgebra of Cb,sβ(R+β;M(BβK2)) generated by ΟβΒ±β(Ξ£2A) and by W, and let DβE be the ideal generated by Οβ+β(gβa)βΟβββ(gβa), gβaβΞ£2A, and by Ο+(Ξ£2A)e00β. Let (wrβ)rβ[0,β)β be a quasicentral aproximate unit in D. Without loss of generality, we may assume that wrβ is diagonal, wrβ=βi=0ββwr;iβeiiβ. As before, we may find a reparametrization r(t) such that, for any fβC0β(0,1) one has
(w1)
limtβββf(wr(t),tβ)d(t)=0 for any dβD;
(w2)
limtβββ[f(wr(t),tβ),e(t)]=0 for any eβE,
where wr(t),tβ=wr(t)β(t).
Set utβ²β=1ββiβZβwr(t),t;β£iβ£βeiiβ, uβ²=βiβZβuiβ²βeiiβ, and
[TABLE]
It is easy to check that Οβ² defines an element of [[S2A,B]], and the linear homotopy Οu+(1βΟ)uβ² shows that Οβ² is homotopic to Ο.
Let us write Οβ² as a matrix with respect to the decomposition 1=βi=1ββeiiβ+e00β+βi=β1βββeiiβ,
This follows from the definition of uβ² (see (w2)).
β
Set Οβ²β²(fβhnββa)=f(uβ²)0βT0nβΞ¦(a)0β. Then Οβ²β² is homotopic to Ο, and can be written as a direct sum, Οβ²β²=Ο1ββΟ2ββ0, where
[TABLE]
[TABLE]
u~=βi=1ββuiβ²βeiiβ. Therefore, [Ο]=[Ο1β]+[Ο2β]=[Ο1β]β[Ο0β], where
[TABLE]
Let us recall the construction of the Bott isomorphism in E-theory. Let u0ββCbβ(R+β;M(BβK) be defined by u0β=βi=1ββaiβ(t)eiiβ, where tβR+β and aiβ(t)=min{itβ,1}. For g,gβ²βC0β(D2), and for Ο:S2AβCbβ(R+β;B), [Ο]β[[S2A,B]], set
[TABLE]
Consider the commuting diagram
[TABLE]
where iβ is induced by the canonical inclusion S2AβΞ£2A, B(Ο)(gβa)=Ο(gβp+ββa)βΟ(gββpβββa), where gβS2C, gβ(reΟ)=g(reβΟ) in polar coordinates, pΒ±ββΞ£2C, aβA.
Note that Οkβ=B(Οkβ), k=0,1, where
[TABLE]
[TABLE]
where gβ²βC(D2).
Thus, [Ο]=[B(Ο~β)],
where
[TABLE]
The linear homotopy connecting u~ with u0β shows that [Ο]=[\mboxBott(Ο^β)], where
[TABLE]
Finally, [Ο^β]=[Ο] by a homotopy similar to the homotopy via the functions Ξ±Ο as in (Ξ±1β)-(Ξ±4β).
So, we have [Ο]=[B(Ο~β)] and [iβ(Ο~β)]=[\mboxBott(Ο)]. Since Bott is an isomorphism, we conclude from (5.8) that [Ο]=[Ο].
β
6. Surjections and asymptotic KK-cycles
Let A be a separable Cβ-algebra. Consider all surjective β-homomorphisms p:CβA, where C is a separable Cβ-algebra, with a partial order given by (C,p)β€(Cβ²,pβ²) if there exists a surjective β-homomorphism Ξ»:Cβ²βC such that pβ²=pβΞ». Denote the set of all such C by EAβ. Abusing the notation, we shall write C in place of (C,p). The set EAβ is directed. Indeed, if p1β:C1ββA, p2β:C2ββA are surjections then the pull-back C={(c1β,c2β):c1ββC1β,c2ββC2β,p1β(c1β)=p2β(c2β)} obviously surjects onto C1β, C2β and A, hence satisfies Cβ₯C1β, Cβ₯C2β.
This directed set has a minimal and a maximal elements. The minimal element is C=A, and the maximal element was constructed in [5], Section 2, under the name of universal extension.
Let CβEAβ, and let J=Kerp. Let EN(C,J;B) be the semigroup of homotopy equivalence classes of pairs (Ο+,Οβ) of asymptotic homomorphisms from C to B that agree on J, with the semigroup structure defined using stability of B. This semigroup need not be a group, as (Οβ,Ο+) may not be the inverse for (Ο+,Οβ). Set EN(A,B)=injlimCβEAββEN(C,J;B).
Lemma 6.1**.**
EN(A,B)* is a group.*
Proof.
Due to the standard rotation argument, it suffices to prove that (Ο,Ο) represents the zero element in EN(A,B), where CβEAβ, and Ο is an asymptotic homomorphism from C to B. Let Cone(C) be the cone over C, ev1β:Cone(C)βC the evaluation map at 1, then pβev1β:Cone(C)βA is a surjection, so Cone(C)βEAβ, Cone(C)β₯C. As Cone(C) is contractible, Οβev1β is homotopic to zero, hence the pair (Ο,Ο) is zero in EN(A,B).
β
Let CβEAβ, (Ο+,Οβ)βEN(C,J;B).
Let s:AβC be a C-homogeneous β-respecting continuous map such that p(s(a))=a for any aβA, which exists by [1]. Set Φ±(a)=ΟΒ±(s(a)), aβA.
Lemma 6.2**.**
The pair (Ξ¦+,Ξ¦β) is a compact asymptotic KK-cycle.
Proof.
The property (a1) is obvious, so we need to check (a3). Note that
[TABLE]
[TABLE]
[TABLE]
[TABLE]
and that s(ab)βs(a)s(b)βJ, so (s(ab)βs(a)s(b))s(c1β)β―s(cnβ)βJ, and as Ο+ and Οβ agree on J, so we are done.
β
If s,sβ²:AβC are two maps satisfying the above assumptions then the linear homotopy connecting s and sβ² gives a homotopy between the corresponding compact asymptotic KK-cycles, so the above construction gives a well-defined map
[TABLE]
Let ΞΊ:Cβ²βC be a surjection, pβ²=pβΞΊ:Cβ²βA, Jβ²=Kerpβ². Then we have Ξ³Cβ²β=ΞΊββΞ³Cβ, where the map ΞΊβ:EN(C,J;B)βEN(Cβ²,Jβ²;B) is induced by ΞΊ, and we may pass to the direct limit.
Let us check homotopy invariance of Ξ³. Let [(Οi+β,Οiββ)], i=0,1, where ΟiΒ±β are asymptotic homomorphisms from C to B, represent the same element in EN(A,B). This means that there exists some Cβ²β₯C in EAβ and a homotopy (Ο+,Οβ), where ΟΒ± are asymptotic homomorphisms from Cβ² to IB such that eviββΟΒ±=ΟiΒ±ββΞΊ, i=0,1, and Ο+ coinsides with Οβ on Ker(Cβ²βA). Let sβ²:CβCβ² be a continuous section for the surjection ΞΊ. Then Ξ³Cβ²β(Ο+,Οβ)=(Ο+βsβ²βs,Οββsβ²βs). As
[TABLE]
we see that the map Ξ³:EN(A,B)βEM(A,B) is well defined as the direct limit of the maps Ξ³Cβ.
For a compact asymptotic KK-cycle (Ξ¦+,Ξ¦β) there exists [Ο]βEN(A,B) such that Ξ³([Ο])=[(Ξ¦+,Ξ¦β)].
Proof.
Let q:EββEβ/Kβ, r:EββEβ/Dβ.
Note that the compositions
[TABLE]
are genuine β-homomorphisms.
Set
[TABLE]
This is a Cβ-algebra that surjects onto A, p(a,e+β,eββ)=a. It has also two surjections, p+β(a,e+β,eββ)=e+β, pββ(a,e+β,eββ)=eββ, onto Eβ+β and Eβββ respectively. Let Ο:EββE be a continuous BartleβGraves selection map. As pΒ±β are β-homomorphisms, ΟΒ±=ΟβpΒ±β:CΞ¦ββE are asymptotic homomorphisms.
Then s is a continuous right inverse for p, and ΟΒ±(s(a))βΌΞ¦Β±(a).
If CβEAβ, Cβ₯CΞ¦β, then there is a β-homomorphism ΞΊ:CβCΞ¦β, and we can define asymptotic homomorphisms ΟCΒ±β:CβB by ΟCΒ±β=ΟΒ±βΞΊ. As ΟC+ββ£Jβ=ΟCβββ£Jβ, where J=Kerp, p:CβA, so (ΟC+β,ΟCββ) represents an element of EN(C,J;B), and Ξ³([(Ο+,Οβ)])=[(Ξ¦+,Ξ¦β)]. β
β
Set Ξ΄(Ξ¦+,Ξ¦β)=(Ο+,Οβ), and let us check homotopy invariance of Ξ΄.
Let (Ξ¨+,Ξ¨β) be a compact asymptotic KK-cycle from A to IB, Ξ¦iΒ±β=eviββΨ±, i=0,1.
Apply the construction of Lemma 6.3 to this compact asymptotic KK-cycle. Let EIββCbβ(R+β;IB) be the Cβ-algebra generated by Ψ±(A),and let KIββEIβ be the ideal generated by Φ±(a+b)βΦ±(a)βΦ±(b) and Φ±(ab)βΦ±(a)Φ±(b), a,bβA. Similarly, Let KiββEiββCbβ(R+β;B) be the corresponding ideal and Cβ-subalgebra for eviββΦ±, and let underlining denotes the image of these Cβ-algebras in Cbβ(R+β;IB)/C0β(R+β;IB) and in Cbβ(R+β;B)/C0β(R+β;B) respectively.
Let qIβ:EβIββEβIβ/KβIβ, qiβ:EβiββEβiβ/Kβiβ, rIβ:EβIββEβIβ/JβIβ, riβ:EβiββEβiβ/Jβiβ be the quotient maps, and let, as in Lemma 6.3, Ξ¨βΒ±:AβEβIβ/KβIβ be the β-homomorphisms induced by Ψ±. Let CΞ¨β and CΞ¦iββ, i=0,1, be the Cβ-algebras in EAβ constructed as in Lemma 6.3, with surjections pIβ:CΞ¨ββA and piβ:CΞ¦iβββA. Note that eviβ:EβIββEβiβ is surjective, and eviβ(KβIβ)=Kβiβ, hence eviβ gives rise to a surjective β-homomorphism eviβ:EβIβ/KβIββEβiβ/Kβiβ.
This gives a commuting diagram
[TABLE]
where ΞΊiβ:CΞ¨ββCΞ¦iββ is given by ΞΊiβ(a,e+β,eββ)=(a,eviβ(e+β),eviβ(eββ)). Recall that (a,e+β,eββ)βCΞ¨β if aβA, eΒ±ββEβIβ, Ξ¨βΒ±(a)=qIβ(eΒ±β), rIβ(e+β)=rIβ(eββ).
Let ΟIβ:EβIββEIβ, Οiβ:EβiββEiβ. Then
Ξ΄(Ξ¨+,Ξ¨β)=(Ο+,Οβ), where ΟΒ±=pΒ±Ξ¨ββΟIβ, Ξ΄(Ξ¦i+β,Ξ¦iββ)=(Οi+β,Οiββ), where ΟiΒ±β=pΒ±Ξ¦iβββΟiβ. But the asymptotic homomorphisms eviββΟΒ± are asymptotically equivalent to ΟiΒ±ββΞΊ, i=0,1, so [(Ο0+β,Ο0ββ)]=[(Ο1+β,Ο1ββ)]. Thus, the map Ξ΄:EM(A,B)βEN(A,B) is well-defined.
Theorem 6.4**.**
The map Ξ³:EN(A,B)βEM(A,B) is an isomorphism.
Proof.
Lemma 6.3 proves surjectivity of Ξ³, so it suffices to show that Ξ΄βΞ³ is the identity map.
Let CβEAβ, ΟΒ± be asymptotic homomorphisms from C to B equal on J=Ker(CβA). Then Ξ³(Ο+,Οβ)=(Ξ¦+,Ξ¦β), where the latter is a compact asymptotic KK-cycle defined by Φ±=ΟΒ±βs, where s:AβC is a continuous section.
Using notation from Lemma 6.3, we have two commuting diagrams
[TABLE]
where [ΟΒ±] denotes the composition of ΟΒ± with the quotient map Q:EβEβ.
There exists a β-homomorphism ΞΊ:CβCΞ¦β given by ΞΊ(c)=(p(c),[Ο+](c),[Οβ](c)) (as Ο+ and Οβ are equal on J, r([Ο+](c))=r([Οβ](c))), so ΞΊ(c)βCΞ¦β. As p is surjective and [ΟΒ±] are surjective too, by the definition of EβΒ±β, ΞΊ is surjective. Thus, [ΟΒ±]=ΞΊβpΒ±β, hence, passing to their lifts, ΟΒ± is asymptotically equivalent to ΟΒ±=ΟβΞΊβpΒ±β, where Ο:EββE is a continuous lift for Q. But Ξ΄βΞ³(Ο+,Οβ)=(Ο+,Οβ).
β
7. Full surjections
Definition 7.1**.**
Let CβEAβ. We call it full if the canonical map EN(C,J;B)βEN(A,B) is an isomorphism for any B.
Note that EAβ has the minimal element A. For this minimal element, EN(A,0;B) is the semigroup of homotopy classes of pairs of asymptotic homomorphisms from A to B, hence EN(A,0;B)β [[A,B]]β[[A,B]], and thus A very rarely is full.
Lemma 7.2**.**
The cone extension CA and the universal extension are full.
Proof.
For the universal extension the statement follows from its universality, so let us prove that CA is full. Essentially, this is a special case of Theorem 5.3 of [12], but we give more details here.
We shall construct a map Ξ΄CAβ:EM(A,B)βEN(CA,SA;B) inverse to Ξ³CAβ. Given (Ξ¦+,Ξ¦β) in EM(A,B), let vβCbβ(R+β;B) be as before, i.e. let v satisfy
(v1β)
0β€vβ€1;
(v2β)
[f(v),Φ±(a)]βΌ0 for any fβC[0,1] and any aβA;
(v3β)
(f(v)β1)(Ξ¦+(a)βΞ¦β(a))βΌ0 for any fβC0β[0,1) and any aβA.
Define ΟΒ±:CAβCbβ(R+β;B) by ΟΒ±(fβa)=f(v)Φ±(a), where fβC0β[0,1), aβA.
It is easy to see that ΟΒ± are asymptotic homomorphisms from CA to B, and (v3β) provides that Ο+ and Οβ agree on SA, i.e. when fβC0β(0,1). Thus we get a well-defined map Ξ΄CAβ:EM(A,B)βEN(CA,SA;B), Ξ΄CAβ(Ξ¦+,Ξ¦β)=(Ο+,Οβ).
Then Ξ΄CAββΞ³CAβ(Ξ¦+,Ξ¦β)=(vΞ¦+,vΞ¦β), and, as in the proof of Lemma 2.2, [vΞ¦+,vΞ¦β]=[Ξ¦+,Ξ¦β], hence Ξ΄CAββΞ³CAβ=id.
Let us show that Ξ³CAββΞ΄CAβ=id.
Let (Ο+,Οβ)βEN(Cone(A),SA;B), i.e. ΟΒ± are two asymptotic homomorphisms from Cone(A) to B that agree on SA. Let s:AβCA be the map defined by s(a)(Ο)=Οa, aβA, Οβ[0,1]. Then Ξ³CAββΞ΄CAβ(Ο)=(Ο+,Οβ), where ΟΒ±(fβa)=f(v)ΟΒ±(s(a)), aβA, fβC0β[0,1).
Let us show that the pairs (Ο+,Οβ) and (Ο+,Οβ) of asymptotic homomorphisms are homotopic, i.e. that there exist two homotopies, connecting Ο+ with Ο+, and Οβ with Οβ, such that they agree on SA.
Define asymptotic homomorphisms ΞΎΒ± (resp. Ξ·Β±) from C0β([0,1)Γ[0,1])βA (resp. from C0β([0,1]Γ[0,1))βA) to B by
[TABLE]
[TABLE]
Note that on the common domain C0β([0,1)2βA), these asymptotic homomorphisms are homotopic via the linear homotopy
[TABLE]
and if gβC0β(0,1) then the two maps ΞΆΟ+β and ΞΆΟββ agree.
Let (x,y) be coordinates in [0,1]2, and let p1β,p2β,p:[0,1]2β[0,1] be defined by p1β(x,y)=x, p2β(x,y)=y, p(x,y)=\left\{\begin{array}[]{cc}x+y&\mbox{\ if\ }x+y\leq 1;\\
1&\mbox{\ if\ }x+y\geq 1.\end{array}\right.
There are obvious linear homotopies p1β(Ο) and p2β(Ο), Οβ[0,1, connecting p1β and p2β with p.
By qβ we denote the β-homomorphism C[0,1)βAβC[0,1]2βA induced by a map q:[0,1]2β[0,1].
Note that p1βββΞΎΒ±(fβa)=f(v)ΟΒ±(s(a))=ΟΒ±(fβa), and Imp1ββ=C0β([0,1)Γ[0,1]). The maps p1βββΞΎΒ± and p2βββΞΎΒ± are homotopic via the homotopy p1ββ(Ο)βΞΎΒ±, Οβ[0,1], and p1ββ(Ο)βΞΎ+ and p1ββ(Ο)βΞΎβ agree on fβa when fβC0β(0,1), for any Οβ[0,1]. Hence [(Ο+,Οβ)]=[(pββΞΎ+,pββΞΎβ)].
Similarly, p2βββΞ·Β±(fβa)=ΟΒ±(fβa), and [(Ο+,Οβ)]=[(pββΞ·+,pββΞ·β)].
As pβ(fβa)βC0β([0,1)2) for any fβC0β[0,1), we may use the homotopy ΞΆΒ± to connect pββΞΎΒ± with pββΞ·Β±.
β
Lemma 7.3**.**
Let CβEAβ. If there exists a splitting β-homomorphism Ξ»:AβC and if AβEAβ is not full then C is not full.
Proof.
Any continuous map s:AβC can be connected to Ξ» by a linear homotopy, therefore, any compact asymptotic KK-cycle is homotopic to a compact asymptotic KK-cycle, which is a pair of β-homomorphisms, hence EN(C,J;B)β EN(A,0;B).
β
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