Let B be a topological space, let (E,p) and (E′,p′) be B-spaces and let f:E→E′ be a B-morphism. For every open subset U of B, define the mapping fU:S(U;E)→S(U;E′) by fU(s)=f∘s. The family S(f)=(fU) is a morphism of presheaves from S(B;E) into S(B;E′). If (E′′,p′′) is a B-space and g:E′→E′′ is a B-morphism, one has S(g∘f)=S(g)∘S(f).
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.