The group with operatorsG=∏i∈IGidefined above is the product group with operators of theGi. The mappingspri:G→Giare called the projection homomorphisms.
A particular case of the product of groups with operators is the group GE consisting of the mappings of a set E into a group with operators G, the laws being defined by:
(fg)(x)=f(x)g(x)(f,g∈GE,x∈E)
fα(x)=f(x)α(f∈GE,α∈Ω,x∈E).
Let (ϕi:H→Gi)i∈I be a family of homomorphisms of groups with operators. The mapping h↦(ϕi(h))i∈I of H into ∏i∈IGi is a homomorphism of groups with operators. It is the only homomorphism Φ:H→∏i∈IGi satisfying pri∘Φ=ϕi for all i. This justifies the name "product group with operators" (Set Theory, IV, § 2, no. 4).
Let (ϕi:Hi→Gi)i∈I be a family of homomorphisms of groups with operators. The mapping ∏i∈Iϕi:(hi)i∈I↦(ϕi(hi))i∈I of ∏i∈IHi into ∏i∈IGi is a homomorphism of groups with operators.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.