Bourbaki

Exercise 10, § 6

Algebra, X, p. 190

Let GG and HH be two groups, u:H→Gu:H\to G a homomorphism, MM a Z(G)\mathbf Z^{(G)}-module. For every q≥0q\geq 0, one deduces from the homomorphisms α\alpha and β\beta of Exercise 1 homomorphisms

uqM:Hq(H,M)⟶Hq(G,M)uMq:Hq(G,M)⟶Hq(H,M) u_q^M:H_q(H,M)\longrightarrow H_q(G,M) \qquad u_M^q:H^q(G,M)\longrightarrow H^q(H,M)

which are simply denoted by uqu_q and uqu^q if there is no ambiguity concerning MM.

a) For f∈Cq(G,M)f\in C^q(G,M), denote by fuf_u the element of Cq(H,M)C^q(H,M) such that

fu(h1,…,hq)=f(u(h1),…,u(hq)) f_u(h_1,\ldots,h_q)=f\bigl(u(h_1),\ldots,u(h_q)\bigr)

for h1,…,hq∈Hh_1,\ldots,h_q\in H. Prove that the mapping f↦fuf\mapsto f_u is a morphism of complexes from Cq(G,M)C^q(G,M) into Cq(H,M)C^q(H,M), which induces on homology the homomorphism ⨁quq\bigoplus_q u^q. Define analogously a morphism from Cq(H,M)C^q(H,M) into Cq(G,M)C^q(G,M) which induces (⨁uq)(\bigoplus u_q).

b) Suppose that H=GH=G and take for uu an inner automorphism of GG. Prove that the homomorphisms uqu_q and uqu^q are equal to the identity (reason by induction on qq).

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.