Notes on Spectral Sequences

making the triangle exact at each vertex. Given this, the homomorphism d = jk : E → E satisfies dd = jkjk = 0 since kj = 0 The derived couple of (A,E, i, j, k) is (A′, E ′, i′, j′, k′), where E ′ = Ker(d)/Im(d), A′ = iA, i′(a) = ia, j′(ia) = [ja], k′[e] = ke. To see that j′ is well defined, suppose ia = ib: then exactness implies that b = a+ ke for some e ∈ E, and then [jb] = [ja] + [jke] = [ja] + [de] = [ja]. To see that k′ is well defined we first note that for [e] ∈ E ′, de = jke = 0, so ke ∈ Ker(j) = Im(i) = A′. If [e] = [e + df ], then k(e+df) = ke+kjk(f) = ke since kj = 0. We have defined the groups and maps in the triangle