a) Let P P P b : S → P b : S \to P b : S → P P P P
(3)
0 → S ⊗ A R ′ → 1 S ⊗ u ~ S ⊗ A R → 1 S ⊗ v ~ S ⊗ A R ′ ′ → 0
0 \to S \otimes_A R' \xrightarrow{1_S \otimes \tilde{u}} S \otimes_A R \xrightarrow{1_S \otimes \tilde{v}} S \otimes_A R'' \to 0
0 → S ⊗ A R ′ 1 S ⊗ u ~ S ⊗ A R 1 S ⊗ v ~ S ⊗ A R ′′ → 0 is exact. Then the following diagram is commutative:
Tor A ( P , M ′ ′ ) → ∂ ( P , ( 2 ) ) Tor A ( P , M ′ ) ↓ ψ ( S , R ′ ′ ) ↓ ψ ( S , R ′ ) H ( S ⊗ A R ′ ′ ) → ∂ ( 3 ) H ( S ⊗ A R ′ ) .
\begin{array}{ccc}
\operatorname{Tor}^{A}(P,M'')&\xrightarrow{\partial(P,(2))}&\operatorname{Tor}^{A}(P,M')\\
\Big\downarrow\vcenter{\rlap{$\psi(S,R'')$}}&&\Big\downarrow\vcenter{\rlap{$\psi(S,R')$}}\\
H(S\otimes_A R'')&\xrightarrow{\partial(3)}&H(S\otimes_A R').
\end{array}
Tor A ( P , M ′′ ) ↓ ⏐ ψ ( S , R ′′ ) H ( S ⊗ A R ′′ ) ∂ ( P , ( 2 )) ∂ ( 3 ) Tor A ( P , M ′ ) ↓ ⏐ ψ ( S , R ′ ) H ( S ⊗ A R ′ ) . b) Let N N N A A A c : N → E c:N\to E c : N → E N N N k k k
(4)
0 ⟶ Homgr A ( R ′ ′ , E ) → Homgr A ( ı ~ , 1 ) Homgr A ( R , E ) → Homgr A ( u ~ , 1 ) Homgr A ′ ( R ′ , E ) ⟶ 0
0\longrightarrow \operatorname{Homgr}_A(R'',E)
\xrightarrow{\operatorname{Homgr}_A(\tilde{\imath},1)}
\operatorname{Homgr}_A(R,E)
\xrightarrow{\operatorname{Homgr}_A(\tilde{u},1)}
\operatorname{Homgr}'_A(R',E)\longrightarrow 0
0 ⟶ Homgr A ( R ′′ , E ) Homgr A ( ~ , 1 ) Homgr A ( R , E ) Homgr A ( u ~ , 1 ) Homgr A ′ ( R ′ , E ) ⟶ 0 is exact. Then the following diagram is commutative:
H ( Homgr A ( R ′ , E ) ) → δ ( ( 4 ) ) H ( Homgr A ( R ′ ′ , E ) ) ↓ φ ( R ′ , E ) ↓ φ ( R ′ ′ , E ) Ext A ( M ′ , N ) → ∂ ( ( 2 ) , N ) Ext A ( M ′ ′ , N ) .
\begin{array}{ccc}
H(\operatorname{Homgr}_A(R',E))
&\xrightarrow{\delta((4))}
&H(\operatorname{Homgr}_A(R'',E))
\\
\Big\downarrow\vcenter{\rlap{$\varphi(R',E)$}}
&&
\Big\downarrow\vcenter{\rlap{$\varphi(R'',E)$}}
\\
\operatorname{Ext}_A(M',N)
&\xrightarrow{\partial((2),N)}
&\operatorname{Ext}_A(M'',N).
\end{array}
H ( Homgr A ( R ′ , E )) ↓ ⏐ φ ( R ′ , E ) Ext A ( M ′ , N ) δ (( 4 )) ∂ (( 2 ) , N ) H ( Homgr A ( R ′′ , E )) ↓ ⏐ φ ( R ′′ , E ) Ext A ( M ′′ , N ) . Let us prove a), for example. Let β : L ( P ) → S \beta:L(P)\to S β : L ( P ) → S t ∘ β = p P t\circ\beta=p_P t ∘ β = p P k k k
0 ⟶ S ⊗ A R ′ → 1 ⊗ u ~ S ⊗ A R → 1 ⊗ v ~ S ⊗ A R ′ ′ ⟶ 0 ↑ β ⊗ 1 R ↑ β ⊗ 1 R ↑ β ⊗ 1 R ′ 0 ⟶ L ( P ) ⊗ A R ′ → 1 ⊗ u ~ L ( P ) ⊗ A R → 1 ⊗ v ~ L ( P ) ⊗ A R ′ ′ ⟶ 0 ↓ 1 ⊗ a ′ ↓ 1 ⊗ a ↓ 1 ⊗ a ′ ′ 0 ⟶ L ( P ) ⊗ A M ′ → 1 ⊗ u L ( P ) ⊗ A M → 1 ⊗ v L ( P ) ⊗ A M ′ ′ ⟶ 0.
\begin{array}{ccccccccc}
0&\longrightarrow&S\otimes_A R'&\xrightarrow{1\otimes\tilde{u}}&S\otimes_A R&\xrightarrow{1\otimes\tilde{v}}&S\otimes_A R''&\longrightarrow&0\\
&&\Big\uparrow\vcenter{\rlap{$\beta\otimes1_R$}}&&\Big\uparrow\vcenter{\rlap{$\beta\otimes1_R$}}&&\Big\uparrow\vcenter{\rlap{$\beta\otimes1_{R'}$}}&&\\
0&\longrightarrow&L(P)\otimes_A R'&\xrightarrow{1\otimes\tilde{u}}&L(P)\otimes_A R&\xrightarrow{1\otimes\tilde{v}}&L(P)\otimes_A R''&\longrightarrow&0\\
&&\Big\downarrow\vcenter{\rlap{$1\otimes a'$}}&&\Big\downarrow\vcenter{\rlap{$1\otimes a$}}&&\Big\downarrow\vcenter{\rlap{$1\otimes a''$}}&&\\
0&\longrightarrow&L(P)\otimes_A M'&\xrightarrow{1\otimes u}&L(P)\otimes_A M&\xrightarrow{1\otimes v}&L(P)\otimes_A M''&\longrightarrow&0.
\end{array}
0 0 0 ⟶ ⟶ ⟶ S ⊗ A R ′ ⏐ ↑ β ⊗ 1 R L ( P ) ⊗ A R ′ ↓ ⏐ 1 ⊗ a ′ L ( P ) ⊗ A M ′ 1 ⊗ u ~ 1 ⊗ u ~ 1 ⊗ u S ⊗ A R ⏐ ↑ β ⊗ 1 R L ( P ) ⊗ A R ↓ ⏐ 1 ⊗ a L ( P ) ⊗ A M 1 ⊗ v ~ 1 ⊗ v ~ 1 ⊗ v S ⊗ A R ′′ ⏐ ↑ β ⊗ 1 R ′ L ( P ) ⊗ A R ′′ ↓ ⏐ 1 ⊗ a ′′ L ( P ) ⊗ A M ′′ ⟶ ⟶ ⟶ 0 0 0. This is a commutative diagram with exact rows (by the hypothesis for the first row, and by the fact that L ( P ) L(P) L ( P )
H ( S ⊗ A R ′ ′ ) → ∂ ( 3 ) H ( S ⊗ A R ′ ) ↑ H ( β ⊗ 1 ) ↑ H ( β ⊗ 1 ) H ( L ( P ) ⊗ A R ′ ′ ) ⟶ H ( L ( P ) ⊗ A R ′ ) ↓ H ( 1 ⊗ a ′ ′ ) ↓ H ( 1 ⊗ a ′ ) H ( L ( P ) ⊗ A M ′ ′ ) ⟶ H ( L ( P ) ⊗ A M ′ ) ↑ ψ P ( M ′ ′ ) ↑ ψ P ( M ′ ) Tor ( P , M ′ ′ ) → ∂ ( P , ( 2 ) ) Tor ( P , M ′ ) .
\begin{array}{ccc}
H(S\otimes_A R'')&\xrightarrow{\partial(3)}&H(S\otimes_A R')\\
\Big\uparrow\vcenter{\rlap{$H(\beta\otimes1)$}}&&\Big\uparrow\vcenter{\rlap{$H(\beta\otimes1)$}}\\
H(L(P)\otimes_A R'')&\longrightarrow&H(L(P)\otimes_A R')\\
\Big\downarrow\vcenter{\rlap{$H(1\otimes a'')$}}&&\Big\downarrow\vcenter{\rlap{$H(1\otimes a')$}}\\
H(L(P)\otimes_A M'')&\longrightarrow&H(L(P)\otimes_A M')\\
\Big\uparrow\vcenter{\rlap{$\psi_P(M'')$}}&&\Big\uparrow\vcenter{\rlap{$\psi_P(M')$}}\\
\operatorname{Tor}(P,M'')&\xrightarrow{\partial(P,(2))}&\operatorname{Tor}(P,M').
\end{array}
H ( S ⊗ A R ′′ ) ⏐ ↑ H ( β ⊗ 1 ) H ( L ( P ) ⊗ A R ′′ ) ↓ ⏐ H ( 1 ⊗ a ′′ ) H ( L ( P ) ⊗ A M ′′ ) ⏐ ↑ ψ P ( M ′′ ) Tor ( P , M ′′ ) ∂ ( 3 ) ⟶ ⟶ ∂ ( P , ( 2 )) H ( S ⊗ A R ′ ) ⏐ ↑ H ( β ⊗ 1 ) H ( L ( P ) ⊗ A R ′ ) ↓ ⏐ H ( 1 ⊗ a ′ ) H ( L ( P ) ⊗ A M ′ ) ⏐ ↑ ψ P ( M ′ ) Tor ( P , M ′ ) . By X, p. 67, prop. 4, H(1 ⊗ a'') and H(1 ⊗ a') are bijective; on the other hand, by definition of the homomorphisms ψ, we have H(β ⊗ 1) ◦ ψ(L(P), R'') = ψ(S, R'') and H(1 ⊗ a'') ◦ ψ(L(P), R'') = ψ(L(P), M'') = ψ_p(M''), therefore
ψ ( S , R ′ ′ ) = H ( β ⊗ 1 ) ◦ H ( 1 ⊗ a ′ ′ ) − 1 ◦ ψ p ( M ′ ′ ) ;
\psi(S, R'') = H(\beta \otimes 1) ◦ H(1 \otimes a'')^{-1} ◦ \psi_p(M'');
ψ ( S , R ′′ ) = H ( β ⊗ 1 ) ◦ H ( 1 ⊗ a ′′ ) − 1 ◦ ψ p ( M ′′ ) ; analogously, ψ ( S , R ′ ) = H ( β ⊗ 1 ) ◦ H ( 1 ⊗ a ′ ) − 1 ◦ ψ p ( M ′ ) \psi(S, R') = H(\beta \otimes 1) ◦ H(1 \otimes a')^{-1} ◦ \psi_p(M') ψ ( S , R ′ ) = H ( β ⊗ 1 ) ◦ H ( 1 ⊗ a ′ ) − 1 ◦ ψ p ( M ′ ) ∂ ( ( 3 ) ) ◦ ψ ( S , R ′ ′ ) = ψ ( S , R ′ ) ◦ ∂ ( P , ( 2 ) ) \partial((3)) ◦ \psi(S, R'') = \psi(S, R') ◦ \partial(P, (2)) ∂ (( 3 )) ◦ ψ ( S , R ′′ ) = ψ ( S , R ′ ) ◦ ∂ ( P , ( 2 ))
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.
Nothing.