Algebra, VIII, p. 324
Let G be a group and Z[G] the algebra of G over Z; we identify G with the canonical basis of Z[G]. Let M be a Z[G]-module, that is, an abelian group endowed with an action of G respecting its group structure. For any integer n⩾0, denote by Cn(G,M) the group of mappings from Gn to M; set Cn(G,M)=0 for n<0. Define a homomorphism dn:Cn(G,M)→Cn+1(G,M) by the formula
(dnc)(g0,...,gn)=g0c(g1,...,gn)+i=0∑n−1(−1)i+1c(g0,...,gigi+1,...,gn)
+(−1)n+1c(g0,...,gn−1)
For any integer n, set Zn(G,M)= Ker dn and Bn(G,M)= Im dn.
a) Prove that Bn(G,M) is contained in Zn(G,M).
Set Hn(G,M)=Zn(G,M)/Bn(G,M). The group H0(G,M) is the set of elements of M fixed by the action of G. The group Z1(G,M) is the group of mappings f:G→M satisfying f(gg′)=f(g)+gf(g′), and B1(G,M) is the subgroup of mappings f for which there exists an m∈M such that f(g)=gm−m (cf. I, §6, p. 148, Exercise 7).
b) Let N be a Z[G]-module and u:M→N a Z[G]-linear mapping. For every integer n, the mapping c↦u∘c is a homomorphism from Cn(G,M) to Cn(G,N) that sends Zn(G,M) into Zn(G,N) and Bn(G,M) into Bn(G,N) and therefore induces a homomorphism from Hn(G,M) to Hn(G,N) that we denote by Hn(G,u) or simply Hn(u). If v:N→P is a homomorphism of Z[G]-modules, then we have Hn(v∘u)=Hn(v)∘Hn(u).
c) Let 0→M′→ιM→πM′′→0 be an exact sequence of Z[G]-modules. Let n be an integer and x′′ an element of Hn(G,M′′), the class of an element c′′∈ Zn(G,M′′). Let c∈Cn(G,M) be such that π∘c=c′′. Prove that there exists an element c′∈Zn+1(G,M′) such that ι∘c′=dn(c) and that the class of c′ in Hn+1(G,M′) depends only on x′′; we denote that class by ∂n(x′′). The mapping ∂n:Hn(G,M′′)→Hn+1(G,M′) is a homomorphism (“linking homomorphism”). Prove that the sequence
n′Hn(ι)nHn(π)n′′∂nn+1′
⋯⟶H(G,M)⟶H(G,M)⟶ H (G,M)⟶H(G,M)⟶⋯
is exact.