a) For every f∈C′(SpA′(x)), we have SpA′(f(x))=f(SpA′(x)) ;
b) For every f∈C′(SpA′(x)) and for every g∈C′(SpA′(f(x))), we have (g∘f)(x)=g(f(x)) ;
c) Let A′ be a stellar algebra and let π be a morphism from A into A′; then π(x) is normal in A′, we have SpA′′(π(x))⊂ SpA′(x) and π(f(x))=f(π(x)) for every f∈C′(SpA′(x));
d) If A is commutative, and if f∈C′(SpA′(x)), then GA′(f(x))=f∘GA′(x).
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.