Let E and F be modules of type C. The equality [E]C=[F]C holds if and only if there exist exact sequences of modules of type C
(E)0⟶L⟶P⟶M⟶0
(F)0⟶L⟶Q⟶M⟶0
such that E⊕Q is isomorphic to F⊕P.
The stated condition is sufficient because it implies the relations
[P]C=[L]C+[M]C,[Q]C=[L]C+[M]C,[E]C+[Q]C=[F]C+[P]C
which give [E]C=[F]C.
Now suppose that we have [E]C=[F]C. By the construction of the group K(C), there exist two finite families of exact sequences of modules of type C
(Gi)0⟶Gi′⟶Gi⟶Gi′′⟶0
for i∈I and
(Hj)0⟶Hj′⟶Hj⟶Hj′′⟶0
for j∈J such that in Z(C), we have
cl(E) − cl(F) =∑j∈JrHj−∑i∈IrGi.
More explicitly, this relation can be written as
(7) cl(E) +∑i∈I cl(Gi) +∑j∈J cl(Hj′) +∑j∈J cl(Hj′′)
= cl(F) +∑i∈I cl(Gi′) +∑i∈I cl(Gi′′) +∑j∈J cl(Hj).
Set G=⨁i∈IGi,G′=⨁i∈IGi′, etc. By passing to the direct sums, we obtain the exact sequences
(G) 0 // G′p // G q // G′′ // 0,
(H) 0 // H′r // H s // H′′ // 0
consisting of modules of type C.
Next, let M1,...,Mm,N1,...,Nn be A-modules of type C. If we have ∑i=1m cl(Mi) =∑j=1n cl(Nj) in the group Z(C), then we have m=n and there exists a permutation σ∈Sm such that cl(Mi) = cl(Nσ(i)) for every 1⩽i⩽m (I, §7, No. 9, p. 95, Proposition 11). Consequently, the modules ⨁i=1mMi and ⨁j=1nNj are isomorphic. In particular, from (7), we deduce the existence of an isomorphism from E⊕Q to F⊕P, where we have set
P=G′⊕G′′⊕H,Q=G⊕H′⊕H′′
We also set
L=G′⊕H′,M=G′′⊕H′′
The modules L, M, P, Q are of type C, and the sequence
(E) 0 // L λ // P μ // M // 0,
where we define λ and μ by
λ(g′,h′)=(g′,0,r(h′)),μ(g′,g′′,h)=(g′′,s(h))
is exact. We construct an exact sequence
(F)0⟶L⟶Q⟶M⟶0
in the same way; this concludes the proof.
The set C is a commutative monoid for the law of composition (E,E′)↦ cl(E⊕E′). We sometimes denote the group of differences of the commutative monoid C (I, §2, No. 4, p. 20) by K′(C) and call it the Grothendieck group of C for direct sums. For any module E of type C, we denote the image of cl(E) in K′(C) by [E]C′.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
Nothing.