Bourbaki

Exercise 16, § 2

Algebra, X, p. 176

Let C be an A-complex, (FpC)p∈Z(F^p C)_{p \in \mathbf{Z}} a decreasing sequence of subcomplexes of C. a) For p,q,r∈Zp, q, r \in \mathbf{Z}, let MrpqM_r^{pq} be the set of elements xx of (FpC)p+q(F^p C)^{p+q} such that dx∈Fp+rCdx \in F^{p+r} C; for r⩾0r \geqslant 0, put Erpq=Mrpq/(d(Mr−1p−r+1,q+r−2)+Mrp+1,q−1)E_r^{pq} = M_r^{pq}/(d(M_{r-1}^{p-r+1,q+r-2}) + M_r^{p+1,q-1}) and denote by dr:Erpq→Erp+r,q−r+1d_r : E_r^{pq} \to E_r^{p+r,q-r+1} the homomorphism induced by dd by passing to quotients. Now prove that there exists for r⩾0r \geqslant 0 an isomorphism αr:H(Er)→Er+1\alpha_r : H(E_r) \to E_{r+1}, such that the sequence (Er,dr)r⩾2(E_r, d_r)_{r \geqslant 2}, endowed with the mappings αr\alpha_r, is a spectral sequence. One has

E0pq=(FpC)p+q/(Fp+1C)p+q. E_0^{pq} = (F^p C)^{p+q}/(F^{p+1} C)^{p+q} .

b) Suppose that ⋃pFpC=C\bigcup_p F^p C = C and that for every nn, there exists an integer pp such that (FpC)n=0(F^p C)^n = 0. Now prove that the spectral sequence defined in a) converges (exercise 14) to the graded A-module H(C)H(C), endowed with the sequence of graded submodules FpH(C)=Im⁡H(ip)F^p H(C) = \operatorname{Im} H(i_p) where ip:FpC→Ci_p : F^p C \to C is the canonical injection. c) If F0C=CF^0 C = C, the A-module ErpqE_r^{pq} is zero for p<0p < 0; if (FpC)n=0(F^p C)^n = 0 for p>np > n, one has Erpq=0E_r^{pq} = 0 for q<0q < 0. d) Let C’ be a second complex, (FpC′)p∈Z(F^p C')_{p \in \mathbf{Z}} a decreasing sequence of subcomplexes of C’, E’ the associated spectral sequence. Let u:C→C′u : C \to C' be a morphism of complexes such that u(FpC)⊂FpC′u(F^p C) \subset F^p C' for every pp. Now prove that uu induces a morphism of spectral sequences u~:E→E′\tilde{u} : E \to E'. e) With the notations of d), let v:C→C′v : C \to C' be a second morphism such that v(FpC)⊂FpC′v(F^p C) \subset F^p C' for every pp, and let hh be a homotopy relating uu to vv. Let kk be an integer ⩾0\geqslant 0 such that h(FpC)⊂Fp−kC′h(F^p C) \subset F^{p-k} C' for every pp. Now prove that u~r=v~r\tilde{u}_r = \tilde{v}_r for r>kr > k.

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.