process

336 U. Bunke, T. Schick, M. Spitzweck, and A. ThomBecause of the presence of mixed terms this is only possible if everything vanishes.This implies that d k 2 F 1 H 3 .E.k/I Z/. We write P i;j x i ^x j ˝u i;j .k/ for its symbolin E 1;22 , where u i;j 2 H 1 .BI Z/. As above equation (63) now impliesXpr 1 .x i ^ x j / ˝ u i;j .k/ C X pr 2 .x l ^ x r / ˝ u l;r .m/i;jl;kD X .pr 1 x a C pr 2 x a/ ^ .pr 1 x b C pr 2 x b/ ˝ u a;b .k C m/:a;bAgain the presence of mixed terms implies that everything vanishes. This shows thatd k 2 F 2 H 3 .E.k/I Z/. The assertion of the lemma is the case k D 1.6.3.7 Let us now combine (60) and (59) into a single diagram. We get the followingweb of horizontal and vertical exact sequences:H 1 .BI Z n /˛H 1 .BI Z n / H 3 .BI Z/KsAhQ EfF 2 H 3 .EI Z/Oc H 2 .BI Z n / BNˇˇ H 4 .BI Z/ H 4 .BI Z/0 0 0.(65)The map f W Q E ! F 2 H 3 .EI Z/ associates to P 2 Q E the Dixmier–Douady classof the gerbe of the pair up.P / 2 P.B/ with underlying T n -bundle E. Here we useLemma 6.16. Surjectivity of f follows by a diagram chase once we have shown thefollowing lemma.Lemma 6.17. The diagram (65) commutes.Proof. We have to check that the left and the right squaresH 3 h.BI Z/ Q EfA F 2 H 3 .EI Z/andOcQ EH 2 .BI Z n /fF 2 H 3 .EI Z/(66)e Bcommute. Let us start with the left square. Let d 2 H 3 .BI Z/. Under the identificationH 3 .BI Z/ Š H 2 .BI T/ Š Ext 2 Sh Ab S=B.ZI T/