Algebra, X, p. 175
Bourbaki marks this one as harder than the rest.
We call spectral sequence of A-modules the datum of a sequence of A-complexes (Er,dr)r≥2 and of mappings αr:H(Er)→Er+1, such that:
(i) The grading of the complex Er is the total grading associated with a bigrading
Er=p,q∈Z⨁Erp,q(r≥2); one has
dr(Erpq)⊂Erp+r,q−r+1.
(ii) The mappings αr are isomorphisms of bigraded A-modules. We shall also use the descending bigrading of Er; we shall denote Erpq=Er−p,−q.
a) We shall denote here by Zr+1(Er) (resp. Br+1(Er)) the bigraded A-module Ker dr (resp. Im dr). Let
pr:Er→Er/Br+1(Er) be the mapping of passing to the quotient and
ir:Er+1→Er/Br+1(Er) the mapping composed of
αr−1 and the natural injection. For all
r≥2, define by induction on
k the bigraded submodules
Zr+k(Er) and
Br+k(Er) of
Er by putting:
Zr+k(Er)=pr−1(ir(Zr+k(Er+1)));Br+k(Er)=pr−1(ir(Br+k(Er+1))). Prove that one has the inclusions:
0⊂Br+1(Er)⊂Br+2(Er)⊂…⊂Zr+2(Er)⊂Zr+1(Er)⊂Er and that
Ek is isomorphic to
Zk(Er)/Bk(Er) for
k≥r+1.
We put Z∞(E2)=⋂r≥3Zr(E2), B∞(E2)=⋃r≥3Br(E2) and E∞=Z∞(E2)/B∞(E2). The spectral sequence is said to be regular if the decreasing sequence (Zr(E2pq))r≥2 is stationary for every pair (p,q).
b) Let E=(Er,dr,αr)r≥2 and E′=(Er′,dr′,αr′)r≥2 be two spectral sequences of A-modules; we call a morphism from E into E′ a sequence of morphisms of A-complexes ur:Er→Er′, compatible with the bigradations, such that αr′∘H(ur)=ur+1∘αr for all r≥2. Prove that for all r≥2 and all k≥r+1, ur induces a homomorphism from Zk(Er) (resp. Bk(Er)) into Zk(Er′) (resp. Bk(Er′)). In particular, u2 induces homomorphisms Z∞(u2):Z∞(E2)→Z∞(E2′), B∞(u2):B∞(E2)→B∞(E2′) and u∞:E∞→E∞′; all these homomorphisms are compatible with the bigradations.
c) Suppose that u2 is an isomorphism. Now prove that ur is an isomorphism for all r≥2, as well as B∞(u2). If, moreover, the sequences E and E′ are regular, Z∞(u2) and u∞ are isomorphisms.
d) Let G=⨁n∈ZGn be a graded A-module, endowed with a decreasing sequence (FpG)p∈Z of graded sub-A-modules, such that ⋂pFpG=0 and ⋃pFpG=G. We say that the spectral sequence E approaches (G,(FpG)p∈Z) (or simply G) if there exist isomorphisms βpq:E∞pq→(FpG)p+q/(Fp+1G)p+q for all
(p,q)∈Z×Z.
We say that the spectral sequence E converges to G if it approaches G and if, moreover: (i) The spectral sequence E is regular; (ii) For every integer n, there exists an integer p such that (FpG)n=0.
Suppose that E (resp. E') converges to G (resp. G'). Let u:E→E′ be a morphism of spectral sequences, v:G→G′ a homomorphism of degree 0, such that v(FpG)⊂FpG′ for all p and that the induced homomorphisms vˉp,n:(FpG)n/(Fp+1G)n→(FpG′)n/(Fp+1G′)n satisfy vˉp,p+q∘βp,q=β′p,q∘u∞p,q for every pair (p,q). Now prove that if u2pq is an isomorphism for every pair (p,q), then v is an isomorphism.