Let ( C , d ) (C,d) ( C , d ) ( C ′ , d ′ ) (C',d') ( C ′ , d ′ ) A A A n , p , q n,p,q n , p , q p ≥ q p \geq q p ≥ q p ≥ i ≥ q − 1 p \geq i \geq q-1 p ≥ i ≥ q − 1 f i : C i ⟶ C i + n + 1 ′ f_i:C_i\longrightarrow C'_{i+n+1} f i : C i ⟶ C i + n + 1 ′ A A A f p ∘ d = 0 f_p\circ d=0 f p ∘ d = 0 f i ∘ d = d ′ ∘ f i + 1 f_i\circ d=d'\circ f_{i+1} f i ∘ d = d ′ ∘ f i + 1 p > i ≥ q − 1 p>i\geq q-1 p > i ≥ q − 1 d ′ ∘ f q − 1 = 0 d'\circ f_{q-1}=0 d ′ ∘ f q − 1 = 0
C p + 1 → d C p → d C p − 1 → d ⋯ → d C q → d C q − 1 → d C q − 2 0 ↓ ↓ f p ↓ f p − 1 ↓ f q ↓ f q − 1 ↓ 0 C p + n + 2 ′ → d ′ C p + n + 1 ′ → d ′ C p + n ′ → d ′ ⋯ → d ′ C q + n + 1 ′ → d ′ C q + n ′ → d ′ C q + n − 1 ′
\begin{array}{ccccccccccccc}
C_{p+1}&\xrightarrow{\ d\ }&C_p&\xrightarrow{\ d\ }&C_{p-1}&\xrightarrow{\ d\ }&\cdots&\xrightarrow{\ d\ }&C_q&\xrightarrow{\ d\ }&C_{q-1}&\xrightarrow{\ d\ }&C_{q-2}\\
0\downarrow&&\downarrow f_p&&\downarrow f_{p-1}&&&&\downarrow f_q&&\downarrow f_{q-1}&&\downarrow 0\\
C'_{p+n+2}&\xrightarrow{\ d'\ }&C'_{p+n+1}&\xrightarrow{\ d'\ }&C'_{p+n}&\xrightarrow{\ d'\ }&\cdots&\xrightarrow{\ d'\ }&C'_{q+n+1}&\xrightarrow{\ d'\ }&C'_{q+n}&\xrightarrow{\ d'\ }&C'_{q+n-1}
\end{array}
C p + 1 0 ↓ C p + n + 2 ′ d d ′ C p ↓ f p C p + n + 1 ′ d d ′ C p − 1 ↓ f p − 1 C p + n ′ d d ′ ⋯ ⋯ d d ′ C q ↓ f q C q + n + 1 ′ d d ′ C q − 1 ↓ f q − 1 C q + n ′ d d ′ C q − 2 ↓ 0 C q + n − 1 ′ Fig. 2.
Set α = f p − 1 ∘ d = d ′ ∘ f p \alpha=f_{p-1}\circ d=d'\circ f_p α = f p − 1 ∘ d = d ′ ∘ f p β = f q − 1 ∘ d = d ′ ∘ f q \beta=f_{q-1}\circ d=d'\circ f_q β = f q − 1 ∘ d = d ′ ∘ f q a a a b b b A A A n n n C C C C ′ C' C ′ α \alpha α β \beta β a ∈ Z n ( Homgr A ( C , C ′ ) ) a\in Z_n(\operatorname{Homgr}_A(C,C')) a ∈ Z n ( Homgr A ( C , C ′ )) b ∈ Z n ( Homgr A ( C , C ′ ) ) b\in Z_n(\operatorname{Homgr}_A(C,C')) b ∈ Z n ( Homgr A ( C , C ′ ))
a − ( − 1 ) ( n + 1 ) ( p − q ) b ∈ B n ( Homgr A ( C , C ′ ) ) .
a-(-1)^{(n+1)(p-q)}b\in B_n(\operatorname{Homgr}_A(C,C')).
a − ( − 1 ) ( n + 1 ) ( p − q ) b ∈ B n ( Homgr A ( C , C ′ )) . One has d ′ ∘ α = d ′ ∘ f p − 1 ∘ d = 0 d'\circ\alpha=d'\circ f_{p-1}\circ d=0 d ′ ∘ α = d ′ ∘ f p − 1 ∘ d = 0 α ∘ d = f p − 1 ∘ d ∘ d = 0 \alpha\circ d=f_{p-1}\circ d\circ d=0 α ∘ d = f p − 1 ∘ d ∘ d = 0
a ∈ Z n ( Homgr A ( C , C ′ ) ) ;
a\in Z_n(\operatorname{Homgr}_A(C,C')) ;
a ∈ Z n ( Homgr A ( C , C ′ )) ; Analogously b ∈ Z n ( Homgr A ( C , C ′ ) ) b\in Z_n(\operatorname{Homgr}_A(C,C')) b ∈ Z n ( Homgr A ( C , C ′ )) ε = ( − 1 ) n + 1 \varepsilon=(-1)^{n+1} ε = ( − 1 ) n + 1 Homgr A ( C , C ′ ) \operatorname{Homgr}_A(C,C') Homgr A ( C , C ′ )
D f p − 1 = d ′ ∘ f p − 1 − ε f p − 1 ∘ d = f p − 2 ∘ d − ε α
Df_{p-1}=d'\circ f_{p-1}-\varepsilon f_{p-1}\circ d=f_{p-2}\circ d-\varepsilon\alpha
D f p − 1 = d ′ ∘ f p − 1 − ε f p − 1 ∘ d = f p − 2 ∘ d − ε α D f i = d ′ ∘ f i − ε f i ∘ d = f i − 1 ∘ d − ε f i + 1 ∘ d ( p − 1 > i > q )
Df_i=d'\circ f_i-\varepsilon f_i\circ d=f_{i-1}\circ d-\varepsilon f_{i+1}\circ d\qquad (p-1>i>q)
D f i = d ′ ∘ f i − ε f i ∘ d = f i − 1 ∘ d − ε f i + 1 ∘ d ( p − 1 > i > q ) D f q = d ′ ∘ f q − ε f q ∘ d = β − ε f q ∘ d .
Df_q=d'\circ f_q-\varepsilon f_q\circ d=\beta-\varepsilon f_q\circ d .
D f q = d ′ ∘ f q − ε f q ∘ d = β − ε f q ∘ d . thence
∑ i = 1 p − q ε i D f p − i = ε p − q β − α ,
\sum_{i=1}^{p-q} \varepsilon^i Df_{p-i} = \varepsilon^{p-q} \beta - \alpha,
i = 1 ∑ p − q ε i D f p − i = ε p − q β − α , which proves the lemma.
Consider two left A-modules M and N and an exact sequence of A-modules
(4)
0 → N → R n → R n − 1 → … → R 1 → M → 0.
0 \to N \to R_n \to R_{n-1} \to \ldots \to R_1 \to M \to 0.
0 → N → R n → R n − 1 → … → R 1 → M → 0. By prop. 3 and 3 bis of X, p. 49, there exists a commutative diagram:
(5)
0 ⟶ N → e N I 0 ( N ) ⟶ … ⟶ I n − 1 ( N ) → δ n − 1 I n ( N ) → δ n I n + 1 ( N ) ↑ ↑ ↑ ↑ ↑ 0 ⟶ N ⟶ R n ⟶ … ⟶ R 1 ⟶ M ⟶ 0 ↑ ↑ ↑ ↑ ↑ L n + 1 ( M ) → d n + 1 L n ( M ) → d n L n − 1 ( M ) ⟶ … → d 1 L 0 ( M ) → p M M ⟶ 0.
\begin{array}{ccccccccccccc}
0 & \longrightarrow & N & \xrightarrow{e_N} & I^0(N) & \longrightarrow & \ldots & \longrightarrow & I^{n-1}(N) & \xrightarrow{\delta^{n-1}} & I^n(N) & \xrightarrow{\delta^n} & I^{n+1}(N) \\
& & \uparrow & & \uparrow & & & & \uparrow & & \uparrow & & \uparrow \\
0 & \longrightarrow & N & \longrightarrow & R_n & \longrightarrow & \ldots & \longrightarrow & R_1 & \longrightarrow & M & \longrightarrow & 0 \\
& & \uparrow & & \uparrow & & & & \uparrow & & \uparrow & & \uparrow \\
L_{n+1}(M) & \xrightarrow{d_{n+1}} & L_n(M) & \xrightarrow{d_n} & L_{n-1}(M) & \longrightarrow & \ldots & \xrightarrow{d_1} & L_0(M) & \xrightarrow{p_M} & M & \longrightarrow & 0.
\end{array}
0 0 L n + 1 ( M ) ⟶ ⟶ d n + 1 N ↑ N ↑ L n ( M ) e N ⟶ d n I 0 ( N ) ↑ R n ↑ L n − 1 ( M ) ⟶ ⟶ ⟶ … … … ⟶ ⟶ d 1 I n − 1 ( N ) ↑ R 1 ↑ L 0 ( M ) δ n − 1 ⟶ p M I n ( N ) ↑ M ↑ M δ n ⟶ ⟶ I n + 1 ( N ) ↑ 0 ↑ 0. Consider the two elements b b b a a a H o m g r A ( L ( M ) , I ( N ) ) \mathrm{Homgr}_A(L(M), I(N)) Homgr A ( L ( M ) , I ( N ))
b n = e N ∘ u n : L n ( M ) → I 0 ( N ) and a n = v n ∘ p M : L 0 ( M ) → I n ( N )
b^n = e_N \circ u_n : L_n(M) \to I^0(N) \quad \text{and} \quad a^n = v^n \circ p_M : L_0(M) \to I^n(N)
b n = e N ∘ u n : L n ( M ) → I 0 ( N ) and a n = v n ∘ p M : L 0 ( M ) → I n ( N ) respectively.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.
Nothing.