*Let M be a free A-module of dimension n; let e∈∧n(M) be an element forming a basis of ∧n(M) and let e∗ be the element of ∧n(M∗) such that {(−1)n(n−1)/2θ∧(e∗)} is the dual basis of {e} in (∧n(M))∗. Let ϕ:∧(M∗)→∧(M) be the mapping z↦z∧e∗ and ϕ′:∧(M∗)→∧(M) the mapping z∗↦z∗∧e. Let ϕp (resp. ϕp′) be the restriction of ϕ (resp. ϕ′) to ∧p(M) (resp. ∧p(M∗)). Then:
(i) The mapping ϕ is a left ∧(M)-module isomorphism and the mapping ϕ′ is a left ∧(M∗)-module isomorphism; moreover the mappings ϕ and ϕ′ are inverses of one another.
(ii) *The mapping ϕp is an isomorphism of the A-module ∧p(M) onto the A-module ∧n−p(M∗) and the mapping ϕp′ is an isomorphism of the A-module ∧p(M∗) onto the A-module ∧n−p(M).
(iii) *If we write B(u,v∗)=⟨u,θ∧(v∗)⟩ for u∈∧(M) and v∗∈∧(M∗) then, for u∗∈∧p(M∗) and v∗∈∧n−p(M∗),
B(ϕp′(u∗),v∗)=(−1)p(n−p)B(u∗,ϕn−p′(v∗)).
The fact that ϕ is ∧(M)-linear and ϕ′ is ∧(M∗)-linear follows from the formulae (u∧v)∧e∗=u∧(v∧e∗) and (u∗∧v∗)∧e=u∗∧(v∗∧e) (no. 6, formula (37)), using the fact that θ∧ is an isomorphism of ∧(M∗) onto the opposite algebra to ∧(M)∗. On the other hand there exists a basis (ei)1≤i≤n of M such that
e=e1∧e2∧⋯∧enande∗=(−1)n(n−1)/2e1∗∧e2∗∧⋯∧en∗,
where (ei∗) is the basis dual to (ei). We write I={1,n}; it follows from (76) that, for every subset J of I with p elements,
{ϕ(eJ)=(−1)n(n−1)/2+p(p−1)/2ρJ,I−JeI∗Jϕ′(eJ∗)=(−1)p(p−1)/2ρJ,I−JeI−J.
This proves that ϕ and ϕ′ are bijective; moreover ρJ,I−JρI−J,J=(−1)p(n−p) (§ 7, no. 8, formula (21)); as the number
2n(n−1)+2p(p−1)+2(n−p)(n−p−1)+p(n−p)=n(n−1)
is even, it follows that ϕ and ϕ′ are inverses of one another. Finally, to prove (77), it suffices to take u∗=eJ∗ and v∗=eI−J∗; the verification also follows from the definition of θ∧, formulae (78) and the relation ρJ,I−JρI−J,J=(−1)p(n−p) (§ 7, no. 8, formula (21)). Note that, for u∗∈∧p(M∗) and v∗∈∧n−p(M∗),
B(\phi_p'(u^*), v^*) is, to within a sign, the coefficient of u∗∧v∗ with respect to the basis {e∗} of Λn(M∗).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.