(i) Let(Ei)i∈I,(Fi)i∈Ibe two families ofA*-modules with the same indexing set* I; for every family of linear mappingsfi:Ei→Fi (i∈I), the restriction to⨁i∈IEiof the linear mapping(xi)↦(fi(xi))is a linear mappingf:⨁i∈IEi→⨁i∈IFidenoted by⨁i∈Ifior⨁ifi(wheref=fp⊕fp+1⊕⋯⊕fqifI=[p,q]is an interval inZ*).*
(ii) Let(Gi)i∈Ibe a third family ofA*-modules with* Ias indexing set and, for alli∈I, letgi:Fi→Gibe a linear mapping; we writeg=⨁igi. For each of the sequencesEifiFigiGito be exact, it is necessary and sufficient that the sequence
i⨁Eifi⨁Figi⨁Gi
be exact.
Obviously, for all (xi)∈⨁iEi, the family (fi(xi)) has finite support, whence (i). On the other hand, to say that an element y=(yi) of ⨁iFi belongs to Ker(g) means that yi∈Ker(gi) for all i∈I (no. 5, Proposition 5); similarly, if yi∈Im(fi) for all i∈I, there exists for each i∈I an xi∈Ei such that yi=fi(xi) and when yi=0, it may be supposed that xi=0; hence y∈Im(f) and the converse is obvious.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.