a) The following diagram, where h θ ( α ) = θ ∘ α h_\theta(\alpha) = \theta \circ \alpha h θ ( α ) = θ ∘ α
T o r k + n A ( P , M ′ ) → h θ T o r k A ( P , M ) ψ k + n ( P , R ′ ) ↓ ↓ ψ k ( P , R ) H k + n ( P ⊗ A R ′ ) → H k + n ( 1 ⊗ φ ) H k ( P ⊗ A R )
\begin{array}{ccc}
\mathrm{Tor}_{k+n}^A(P, M') & \xrightarrow{h_\theta} & \mathrm{Tor}_k^A(P, M) \\
\psi_{k+n}(P, R') \downarrow & & \downarrow \psi_k(P, R) \\
\mathrm{H}_{k+n}(P \otimes_A R') & \xrightarrow{H_{k+n}(1 \otimes \varphi)} & \mathrm{H}_k(P \otimes_A R)
\end{array}
Tor k + n A ( P , M ′ ) ψ k + n ( P , R ′ ) ↓ H k + n ( P ⊗ A R ′ ) h θ H k + n ( 1 ⊗ φ ) Tor k A ( P , M ) ↓ ψ k ( P , R ) H k ( P ⊗ A R ) b) The following diagram, where δ θ ( β ) = β ∘ θ \delta_\theta(\beta) = \beta \circ \theta δ θ ( β ) = β ∘ θ
H k ( H o m g r A ( R , N ) ) → H k + n ( H o m g r A ( φ , 1 ) ) H k + n ( H o m g r A ( R ′ , N ) ) φ k ( R , N ) ↓ φ k + n ( R ′ , N ) ↓ E x t A k ( M , N ) → δ θ E x t A k + n ( M ′ , N ) .
\begin{array}{ccc}
H^k(\mathrm{Homgr}_A(R, N)) & \xrightarrow{H^{k+n}(\mathrm{Homgr}_A(\varphi, 1))} & H^{k+n}(\mathrm{Homgr}_A(R', N)) \\
\varphi^k(R, N) \downarrow & & \varphi^{k+n}(R', N) \downarrow \\
\mathrm{Ext}_A^k(M, N) & \xrightarrow{\delta_\theta} & \mathrm{Ext}_A^{k+n}(M', N).
\end{array}
H k ( Homgr A ( R , N )) φ k ( R , N ) ↓ Ext A k ( M , N ) H k + n ( Homgr A ( φ , 1 )) δ θ H k + n ( Homgr A ( R ′ , N )) φ k + n ( R ′ , N ) ↓ Ext A k + n ( M ′ , N ) . Let α : L ( M ) → R \alpha : L(M) \to R α : L ( M ) → R a ∘ α = p M a \circ \alpha = p_M a ∘ α = p M
L n ( M ′ ) ⟶ L n − 1 ( M ′ ) ⟶ … ⟶ L 0 ( M ′ ) ⟶ M ′ ⟶ 0 u n ↓ u n − 1 ↓ u 0 ↓ 1 ↓ 0 ⟶ M → f K n ⟶ … ⟶ K 1 ⟶ M ′ ⟶ 0
\begin{array}{ccccccccc}
L_n(M') & \longrightarrow & L_{n-1}(M') & \longrightarrow & \ldots & \longrightarrow & L_0(M') & \longrightarrow & M' \longrightarrow 0 \\
u_n \downarrow & & u_{n-1} \downarrow & & & & u_0 \downarrow & & 1 \downarrow \\
0 & \longrightarrow & M & \xrightarrow{f} & K_n & \longrightarrow & \ldots & \longrightarrow & K_1 \longrightarrow M' \longrightarrow 0
\end{array}
L n ( M ′ ) u n ↓ 0 ⟶ ⟶ L n − 1 ( M ′ ) u n − 1 ↓ M ⟶ f … K n ⟶ ⟶ L 0 ( M ′ ) u 0 ↓ … ⟶ ⟶ M ′ ⟶ 0 1 ↓ K 1 ⟶ M ′ ⟶ 0 a commutative diagram, let us choose a homomorphism v n : L n ( M ′ ) → L 0 ( M ) v_n : L_n(M') \to L_0(M) v n : L n ( M ′ ) → L 0 ( M ) p M ∘ v n = ( − 1 ) n u n p_M \circ v_n = (-1)^n u_n p M ∘ v n = ( − 1 ) n u n X , p. 47, Proposition 1, a), v n v_n v n v : L ( M ′ ) → L ( M ) v : L(M') \to L(M) v : L ( M ′ ) → L ( M ) ( − n ) (-n) ( − n ) θ \theta θ H n ( H o m g r A ( L ( M ′ ) , L ( M ) ) ) → E x t A n ( M ′ , M ) H^n(\mathrm{Homgr}_A(L(M'), L(M))) \to \mathrm{Ext}_A^n(M', M) H n ( Homgr A ( L ( M ′ ) , L ( M ))) → Ext A n ( M ′ , M ) v v v X , p. 117, Remark 1). We define a morphism of complexes β : L ( M ′ ) → R ′ \beta : L(M') \to R' β : L ( M ′ ) → R ′ β p = u p \beta_p = u_p β p = u p p ≤ n − 1 p \leq n - 1 p ≤ n − 1 β p = α p − n ∘ v p \beta_p = \alpha_{p-n} \circ v_p β p = α p − n ∘ v p p ≥ n p \geq n p ≥ n
φ ∘ β = α ( − n ) ∘ v .
\varphi \circ \beta = \alpha(-n) \circ v.
φ ∘ β = α ( − n ) ∘ v . On the other hand, by definition of the mappings φ \varphi φ ψ \psi ψ
ψ k ( P , R ) = H k ( p P ⊗ α ) , φ k ( R , N ) = H k ( H o m g r A ( α , e N ) ) , ψ k + n ( P , R ′ ) = H k + n ( p P ⊗ β ) , φ k + n ( R ′ , N ~ ) = H k + n ( H o m g r A ( β , e N ) ) .
\begin{aligned}
& \psi_k(P, R) = H_k(p_P \otimes \alpha), \quad \varphi^k(R, N) = H^k(\mathrm{Homgr}_A(\alpha, e_N)), \\
& \psi_{k+n}(P, R') = H_{k+n}(p_P \otimes \beta), \quad \varphi^{k+n}(R', \tilde{N}) = H^{k+n}(\mathrm{Homgr}_A(\beta, e_N)).
\end{aligned}
ψ k ( P , R ) = H k ( p P ⊗ α ) , φ k ( R , N ) = H k ( Homgr A ( α , e N )) , ψ k + n ( P , R ′ ) = H k + n ( p P ⊗ β ) , φ k + n ( R ′ , N ~ ) = H k + n ( Homgr A ( β , e N )) . Finally, by definition of the composition product, we have
h θ = H ( 1 L ( P ) ⊗ v ) , δ θ = H ( H o m g r A ( v , 1 L ( N ) ) ) .
h_\theta = H(1_{L(P)} \otimes v), \quad \delta_\theta = H(\mathrm{Homgr}_A(v, 1_{L(N)})) .
h θ = H ( 1 L ( P ) ⊗ v ) , δ θ = H ( Homgr A ( v , 1 L ( N ) )) . Consequently, we have the equalities
ψ k ( P , R ) ∘ h θ = H k ( p P ⊗ α ) ∘ H k ( 1 L ( P ) ⊗ v ) = H k ( p P ⊗ ( α ∘ v ) ) = H k + n ( p P ⊗ ( φ ∘ β ) ) = H k + n ( 1 ⊗ φ ) ∘ H k + n ( p P ∘ β ) = H k + n ( 1 ⊗ φ ) ∘ ψ k + n ( P , R ′ ) ,
\begin{aligned}
\psi_k(P, R) \circ h_\theta &= H_k(p_P \otimes \alpha) \circ H_k(1_{L(P)} \otimes v) = H_k(p_P \otimes (\alpha \circ v)) = H_{k+n}(p_P \otimes (\varphi \circ \beta)) \\
&= H_{k+n}(1 \otimes \varphi) \circ H_{k+n}(p_P \circ \beta) = H_{k+n}(1 \otimes \varphi) \circ \psi_{k+n}(P, R'),
\end{aligned}
ψ k ( P , R ) ∘ h θ = H k ( p P ⊗ α ) ∘ H k ( 1 L ( P ) ⊗ v ) = H k ( p P ⊗ ( α ∘ v )) = H k + n ( p P ⊗ ( φ ∘ β )) = H k + n ( 1 ⊗ φ ) ∘ H k + n ( p P ∘ β ) = H k + n ( 1 ⊗ φ ) ∘ ψ k + n ( P , R ′ ) , from which a) follows; the proof of b) is analogous.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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