Library / Algebra / Chapter X / § 3
Proposition 4
Tag 0DU0, permanent. Cite it as [Bourbaki, Tag 0DU0].
Algebra, chapter X, § 3. Résolutions
L ( f ) : L ( M ) → L ( N ) L(f) : L(M) \to L(N) L ( f ) : L ( M ) → L ( N ) p M ∘ L ( f ) = f ∘ p N p_M \circ L(f) = f \circ p_N p M ∘ L ( f ) = f ∘ p N
It is a matter of proving, for every integer n > 0 n > 0 n > 0
d n N ∘ L n ( f ) = L n − 1 ( f ) ∘ d n M .
d_n^N \circ L_n(f) = L_{n-1}(f) \circ d_n^M .
d n N ∘ L n ( f ) = L n − 1 ( f ) ∘ d n M . One has first
d 1 N ∘ L 1 ( f ) = i N ∘ p Z 0 ( N ) ∘ L 0 ( Z 0 ( f ) ) (by (3) and (6)) = i N ∘ Z 0 ( f ) ∘ p Z 0 ( M ) (by (4)) = L 0 ( f ) ∘ i M ∘ p Z 0 ( M ) (by (5)) = L 0 ( f ) ∘ d 1 M (by (3)) .
\begin{align*}
d_1^N \circ L_1(f) &= i_N \circ p_{Z_0(N)} \circ L_0(Z_0(f)) & \text{(by (3) and (6))} \\
&= i_N \circ Z_0(f) \circ p_{Z_0(M)} & \text{(by (4))} \\
&= L_0(f) \circ i_M \circ p_{Z_0(M)} & \text{(by (5))} \\
&= L_0(f) \circ d_1^M & \text{(by (3))} .
\end{align*}
d 1 N ∘ L 1 ( f ) = i N ∘ p Z 0 ( N ) ∘ L 0 ( Z 0 ( f )) = i N ∘ Z 0 ( f ) ∘ p Z 0 ( M ) = L 0 ( f ) ∘ i M ∘ p Z 0 ( M ) = L 0 ( f ) ∘ d 1 M (by (3) and (6)) (by (4)) (by (5)) (by (3)) . When n > 1 n > 1 n > 1
d n N ∘ L n ( f ) = i Z n − 2 ( N ) ∘ p Z n − 1 ( N ) ∘ L 0 ( Z n − 1 ( f ) ) (by (3) and (6)) = i Z n − 2 ( N ) ∘ Z n − 1 ( f ) ∘ p Z n − 1 ( M ) (by (4)) = i Z n − 2 ( N ) ∘ Z 0 ( Z n − 2 ( f ) ) ∘ p Z n − 1 ( M ) (by (6)) = L 0 ( Z n − 2 ( f ) ) ∘ i Z n − 2 ( M ) ∘ p Z n − 1 ( M ) (by (5)) = L n − 1 ( f ) ∘ d n M (by (3) and (6)) .
\begin{align*}
d_n^N \circ L_n(f) &= i_{Z_{n-2}(N)} \circ p_{Z_{n-1}(N)} \circ L_0(Z_{n-1}(f)) & \text{(by (3) and (6))} \\
&= i_{Z_{n-2}(N)} \circ Z_{n-1}(f) \circ p_{Z_{n-1}(M)} & \text{(by (4))} \\
&= i_{Z_{n-2}(N)} \circ Z_0(Z_{n-2}(f)) \circ p_{Z_{n-1}(M)} & \text{(by (6))} \\
&= L_0(Z_{n-2}(f)) \circ i_{Z_{n-2}(M)} \circ p_{Z_{n-1}(M)} & \text{(by (5))} \\
&= L_{n-1}(f) \circ d_n^M & \text{(by (3) and (6))} .
\end{align*}
d n N ∘ L n ( f ) = i Z n − 2 ( N ) ∘ p Z n − 1 ( N ) ∘ L 0 ( Z n − 1 ( f )) = i Z n − 2 ( N ) ∘ Z n − 1 ( f ) ∘ p Z n − 1 ( M ) = i Z n − 2 ( N ) ∘ Z 0 ( Z n − 2 ( f )) ∘ p Z n − 1 ( M ) = L 0 ( Z n − 2 ( f )) ∘ i Z n − 2 ( M ) ∘ p Z n − 1 ( M ) = L n − 1 ( f ) ∘ d n M (by (3) and (6)) (by (4)) (by (6)) (by (5)) (by (3) and (6)) . One has at once
(7)
L ( 1 M ) = 1 L ( M ) .
L(1_M) = 1_{L(M)} .
L ( 1 M ) = 1 L ( M ) . On the other hand, if g : N → P g : N \to P g : N → P
(8)
L ( g ∘ f ) = L ( g ) ∘ L ( f ) .
L(g \circ f) = L(g) \circ L(f) .
L ( g ∘ f ) = L ( g ) ∘ L ( f ) . In fact, for m ∈ M m \in M m ∈ M
L 0 ( g ∘ f ) ( e m ) = e g ∘ f ( m ) = L 0 ( g ) ( e f ( m ) ) = L 0 ( g ) ∘ L 0 ( f ) ( e m ) ,
L_0(g \circ f)(e_m) = e_{g \circ f(m)} = L_0(g)(e_{f(m)}) = L_0(g) \circ L_0(f)(e_m) ,
L 0 ( g ∘ f ) ( e m ) = e g ∘ f ( m ) = L 0 ( g ) ( e f ( m ) ) = L 0 ( g ) ∘ L 0 ( f ) ( e m ) , hence
L 0 ( g ∘ f ) = L ( g ) ∘ L ( f ) L_0(g \circ f) = L(g) \circ L(f) L 0 ( g ∘ f ) = L ( g ) ∘ L ( f ) ; consequently
Z 0 ( g ∘ f ) = Z 0 ( g ) ∘ Z 0 ( f ) Z_0(g \circ f) = Z_0(g) \circ Z_0(f) Z 0 ( g ∘ f ) = Z 0 ( g ) ∘ Z 0 ( f ) ; whence at once
L n ( g ∘ f ) = L n ( g ) ∘ L n ( f ) L_n(g \circ f) = L_n(g) \circ L_n(f) L n ( g ∘ f ) = L n ( g ) ∘ L n ( f ) , for
n ⩾ 0 n \geqslant 0 n ⩾ 0 , by induction on
n n n , whence (8).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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Translated with machine help, not yet reviewed by a person.