Let M 1 , M 2 M_1, M_2 M 1 , M 2 be two g g g -modules. The canonical K K K -linear mappings (Algebra, Chapter II, § 4, no. 2, Proposition 2 and no. 1, Proposition 1):
M 1 ∗ ⊗ K M 2 → ϕ L K ( M 1 , M 2 ) , L K ( M 1 , M 2 ∗ ) → ψ ( M 1 ⊗ K M 2 ) ∗
M_1^* \otimes_K M_2 \xrightarrow{\phi} \mathscr{L}_K(M_1, M_2), \quad \mathscr{L}_K(M_1, M_2^*) \xrightarrow{\psi} (M_1 \otimes_K M_2)^*
M 1 ∗ ⊗ K M 2 ϕ L K ( M 1 , M 2 ) , L K ( M 1 , M 2 ∗ ) ψ ( M 1 ⊗ K M 2 ) ∗
(where the second is bijective) are g g g -module homomorphisms.
We write
N = M 1 ∗ ⊗ M 2 , P = L ( M 1 , M 2 ) , Q = L ( M 1 , M 2 ∗ ) , R = ( M 1 ⊗ M 2 ) ∗ .
N = M_1^* \otimes M_2, \quad P = \mathscr{L}(M_1, M_2), \quad Q = \mathscr{L}(M_1, M_2^*), \quad R = (M_1 \otimes M_2)^*.
N = M 1 ∗ ⊗ M 2 , P = L ( M 1 , M 2 ) , Q = L ( M 1 , M 2 ∗ ) , R = ( M 1 ⊗ M 2 ) ∗ .
Then, for x ∈ g , f ∈ M 1 ∗ , m 1 ∈ M 1 , m 2 ∈ M 2 x \in g, f \in M_1^*, m_1 \in M_1, m_2 \in M_2 x ∈ g , f ∈ M 1 ∗ , m 1 ∈ M 1 , m 2 ∈ M 2 ,
( ( ϕ x N ) ( f ⊗ m 2 ) ) ⋅ m 1 = ( ϕ ( x M 1 ∗ f ⊗ m 2 + f ⊗ x M 2 m 2 ) ) ⋅ m 1 = ⟨ x M 1 ∗ f , m 1 ⟩ m 2 + ⟨ f , m 1 ⟩ x M 2 m 2
((\phi x_N)(f \otimes m_2)) \cdot m_1 = (\phi(x_{M_1^*} f \otimes m_2 + f \otimes x_{M_2} m_2)) \cdot m_1 \\
= \langle x_{M_1^*} f, m_1 \rangle m_2 + \langle f, m_1 \rangle x_{M_2} m_2
(( ϕ x N ) ( f ⊗ m 2 )) ⋅ m 1 = ( ϕ ( x M 1 ∗ f ⊗ m 2 + f ⊗ x M 2 m 2 )) ⋅ m 1 = ⟨ x M 1 ∗ f , m 1 ⟩ m 2 + ⟨ f , m 1 ⟩ x M 2 m 2
( ( x P ϕ ) ( f ⊗ m 2 ) ) ⋅ m 1 = x M 2 ( ϕ ( f ⊗ m 2 ) ⋅ m 1 ) − ϕ ( f ⊗ m 2 ) ( x M 1 m 1 ) = ⟨ f , m 1 ⟩ x M 2 m 2 − ⟨ f , x M 1 m 1 ⟩ m 2
((x_P \phi)(f \otimes m_2)) \cdot m_1 = x_{M_2}(\phi(f \otimes m_2) \cdot m_1) - \phi(f \otimes m_2)(x_{M_1} m_1) \\
= \langle f, m_1 \rangle x_{M_2} m_2 - \langle f, x_{M_1} m_1 \rangle m_2
(( x P ϕ ) ( f ⊗ m 2 )) ⋅ m 1 = x M 2 ( ϕ ( f ⊗ m 2 ) ⋅ m 1 ) − ϕ ( f ⊗ m 2 ) ( x M 1 m 1 ) = ⟨ f , m 1 ⟩ x M 2 m 2 − ⟨ f , x M 1 m 1 ⟩ m 2
and hence ϕ x N = x P ϕ \phi x_N = x_P \phi ϕ x N = x P ϕ . On the other hand, for x ∈ g , u ∈ L ( M 1 , M 2 ∗ ) , m 1 ∈ M 1 , m 2 ∈ M 2 x \in g, u \in \mathscr{L}(M_1, M_2^*), m_1 \in M_1, m_2 \in M_2 x ∈ g , u ∈ L ( M 1 , M 2 ∗ ) , m 1 ∈ M 1 , m 2 ∈ M 2 :
( ψ x Q u ) ( m 1 ⊗ m 2 ) = ⟨ ( x Q u ) ⋅ m 1 , m 2 ⟩ = ⟨ x M 2 ∗ u m 1 − u x M 1 m 1 , m 2 ⟩ ( x R ψ u ) ( m 1 ⊗ m 2 ) = − ⟨ ψ u , x M 1 m 1 ⊗ m 2 + m 1 ⊗ x M 2 m 2 ⟩ = − ⟨ u x M 1 m 1 , m 2 ⟩ − ⟨ u m 1 , x M 2 m 2 ⟩
(\psi x_Q u)(m_1 \otimes m_2) = \langle (x_Q u) \cdot m_1, m_2 \rangle = \langle x_{M_2^*} u m_1 - u x_{M_1} m_1, m_2 \rangle \\
(x_R \psi u)(m_1 \otimes m_2) = -\langle \psi u, x_{M_1} m_1 \otimes m_2 + m_1 \otimes x_{M_2} m_2 \rangle \\
= -\langle u x_{M_1} m_1, m_2 \rangle - \langle u m_1, x_{M_2} m_2 \rangle
( ψ x Q u ) ( m 1 ⊗ m 2 ) = ⟨( x Q u ) ⋅ m 1 , m 2 ⟩ = ⟨ x M 2 ∗ u m 1 − u x M 1 m 1 , m 2 ⟩ ( x R ψ u ) ( m 1 ⊗ m 2 ) = − ⟨ ψ u , x M 1 m 1 ⊗ m 2 + m 1 ⊗ x M 2 m 2 ⟩ = − ⟨ u x M 1 m 1 , m 2 ⟩ − ⟨ u m 1 , x M 2 m 2 ⟩
and hence ψ x Q = x R ψ \psi x_Q = x_R \psi ψ x Q = x R ψ , which completes the proof.
The g g g -modules L ( M 1 , M 2 ∗ ) \mathscr{L}(M_1, M_2^*) L ( M 1 , M 2 ∗ ) and ( M 1 ⊗ M 2 ) ∗ (M_1 \otimes M_2)^* ( M 1 ⊗ M 2 ) ∗ are identified under the isomorphism ψ \psi ψ . If M 1 M_1 M 1 and M 2 M_2 M 2 have finite bases, ϕ \phi ϕ is an isomorphism (Algebra, Chapter II, § 4, no. 2, Proposition 2), which allows us to identify the g g g -modules M 1 ∗ ⊗ M 2 M_1^* \otimes M_2 M 1 ∗ ⊗ M 2 and L ( M 1 , M 2 ) \mathscr{L}(M_1, M_2) L ( M 1 , M 2 ) ; in that case, we can therefore identify the g g g -modules M 1 ∗ ⊗ M 2 ∗ , L ( M 1 , M 2 ∗ ) M_1^* \otimes M_2^*, \mathscr{L}(M_1, M_2^*) M 1 ∗ ⊗ M 2 ∗ , L ( M 1 , M 2 ∗ ) and ( M 1 ⊗ M 2 ) ∗ (M_1 \otimes M_2)^* ( M 1 ⊗ M 2 ) ∗ .
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.