Bourbaki

Proposition 3

Tag 022E, permanent. Cite it as [Bourbaki, Tag 022E].

Topologie algébrique, chapter IV, § 4. Théorie de la descente

Let X and Y be topological spaces and let f:X→f: X\rightarrow Y be a continuous mapping. Let (Z,p)(Z, p) and (Z′,p′)(Z', p') be X-spaces endowed with descent data relative to ff, denoted respectively by τ\tau and τ′\tau '. If the descent datum τ′\tau ' is effective, the mapping φ↦φ‾\varphi \mapsto \overline{\varphi} is a bijection of Cτ,τ′(Z;Z′)\mathscr{C}_{\tau ,\tau'}(Z; Z') onto CY(Z/Rτ;Z′/Rτ′)\mathscr{C}_Y(Z/R_{\tau}; Z'/R_{\tau'}).

For every X-morphism φ\varphi of Z into Z′Z' compatible with the descent data, the mapping φ‾\overline{\varphi} is a Y-morphism. Conversely, let g:Z→Z/Rτg: Z\rightarrow Z/R_{\tau} and g′:Z′→Z′/Rτ′g': Z'\rightarrow Z'/R_{\tau'} denote the canonical mappings, and let ψ:Z/Rτ→Z′/Rτ′\psi : Z/R_{\tau}\rightarrow Z'/R_{\tau'} be a Y-morphism. The mappings p:Z→Xp: Z\rightarrow X and ψ∘g:Z→Z′/Rτ′\psi \circ g: Z\rightarrow Z'/R_{\tau'} are Y-morphisms. The hypothesis that the descent datum τ′\tau ' is effective means that the diagram

Z′g′Z′/Rτ′ {Z'}^{g'}Z'/R_{\tau'}
p′q′p'q'

X f^f Y

is a cartesian square. Hence there exists a unique continuous mapping φ:Z→Z′\varphi : Z\rightarrow Z' such that p′∘φ=pp'\circ \varphi =p and g′∘φ=ψ∘gg'\circ \varphi =\psi \circ g. The first equality means that φ\varphi is an X-morphism, the second equality means that φ\varphi is compatible with the descent data and that φ‾=ψ\overline{\varphi}=\psi, whence the proposition.

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.

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Cites

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Machine translation of the English, by gpt-5-6-mini, gpt-5.4, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.