Bourbaki

Exercise 2, § 16

Tag 00JX, permanent. Cite it as [Bourbaki, Tag 00JX].

Algebra, VIII, p. 324

Let φ:H→G\varphi : H\rightarrow G be a group homomorphism, M a Z\mathbf{Z}[G]-module, N a Z\mathbf{Z}[H]-module, and u:M→Nu: M\rightarrow N a homomorphism of abelian groups. We say that φ\varphi and uu are compatible if we have u(φ(h)x)=hu(x)u(\varphi (h)x) =hu(x) for every h∈Hh\in H and x∈Mx\in M.

a) For n⩾0n\geqslant 0, let φn:Hn→Gn\varphi^n: H^n\rightarrow G^n be the mapping deduced from φ\varphi; if φ\varphi and uu are compatible, then the homomorphism c↦u∘c∘φnc\mapsto u\circ c\circ \varphi^n from Cn(G,M)C^n(G,M) to Cn(H,N)C^n(H,N) sends Zn(G,M)Z^n(G,M) into Zn(H,N)Z^n(H,N) and Bn(G,M)B^n(G,M) into Bn(H,N)B^n(H,N) and therefore induces a homomorphism Hn(φ,u)H^n(\varphi , u) from Hn(G,M)H^n(G,M) to Hn(H,N)H^n(H,N). b) Let ψ:K→H\psi : K\rightarrow H be another group homomorphism, P aZ\mathbf{Z}[K]-module, and v:N→Pv: N\rightarrow P a homomorphism of abelian groups. If φ\varphi and uu are compatible, and ψ\psi and vv also are, then we have Hn(φ∘ψ,v∘u)=Hn(ψ,v)∘Hn(φ,u)H^n(\varphi \circ \psi , v\circ u) = H^n(\psi , v)\circ H^n(\varphi , u).

c) Consider a commutative diagram

0 // M′M' // M // M′′M'' // 0

u′uu′′u'uu''

0 // N′N' // N // N′′N'' // 0 ,

where the first (resp. second) line is an exact sequence of Z\mathbf{Z}[G]-modules (resp. of Z\mathbf{Z}[H]-modules) and where u′,u,u′′u', u, u'' are compatible with φ\varphi. Prove that the diagram

⋯\cdots // Hn(G,M′)H_n(G,M') // Hn(G,M)H_n(G,M) // Hn(G,M′′)H_n(G,M'') // Hn+1(G,M′)H_{n+1}(G,M') //⋯/\cdots

Hn(φ,u′)Hn(φ,u)Hn(φ,u′′)Hn+1(φ,u)H^n(\varphi ,u')H^n(\varphi ,u)H^n(\varphi ,u'')H^{n+1}(\varphi ,u)

⋯\cdots // Hn(H,N′)H_n(H,N') // Hn(H,N)H_n(H,N) // Hn(H,N′′)H_n(H,N'') // Hn+1(H,N′)H_{n+1}(H,N') //⋯/\cdots

(whose horizontal lines are the long exact sequences defined in Exercise 1, c)) is commutative.

d) Examples: If H = G, then the mapping uu is compatible with IdG_G if and and only if it is Z\mathbf{Z}[G]-linear, and we have HnH^n(IdG,u_G, u) =Hn(G,u)= H^n(G, u). If N is equal to M endowed with the Z\mathbf{Z}[G]-module structure deduced from φ\varphi, then the homomorphisms φ\varphi and 1M1_M are compatible; we denote by Resq:Hq(G,M)→Hq(H,M)^q: H^q(G,M)\rightarrow H^q(H,M) the homomorphism Hq(φ,1M)H^q(\varphi ,1_M) (“restriction homomorphism”).

e) Let H be a normal subgroup of G, and let MHM^H be the subgroup of M consisting of the elements invariant under H, endowed with the action of G/HG/H deduced from that of G by passing to the quotient. The canonical surjection p:G→p: G\rightarrow G/HG/H and the canonical injection j:MH→Mj: M^H\rightarrow M are compatible; we denote by Infq:Hq(G/H,MH)→Hq(G,M)^q: H^q(G/H,M^H)\rightarrow H^q(G,M) the homomorphism Hq(p,j)H^q(p, j) (“inflation homomorphism”). Prove that the sequence

0→H1(G/H,MH)⟶Inf1H1(G,M)⟶Res1H1(H,M) 0\rightarrow H^1(G/H,M^H)\longrightarrow^{Inf^1}H^1(G,M)\longrightarrow^{Res^1}H^1(H,M)

is exact. If, moreover, H1(H,M)H^1(H,M) is zero, then the sequence

0→H2(G/H,MH)⟶Inf2H2(G,M)⟶Res2H2(H,M) 0\rightarrow H^2(G/H,M^H)\longrightarrow^{Inf^2}H^2(G,M)\longrightarrow^{Res^2}H^2(H,M)

is exact.

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