Let B be a topological space, F a presheaf on B,F the associated sheaf and σF:F→F the canonical morphism. For F to be a sheaf, it is necessary and sufficient that σF be an isomorphism.
If σF is an isomorphism, F is a sheaf. Conversely, if F is a sheaf, there exists by proposition 3 a morphism φ:F→F such that φ∘σF= IdF. Since IdF is the unique morphism ψ:F→F such that ψ∘σF=σF, one has therefore σF∘φ= IdF.
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.