Corollary

Let X and Y be topological spaces and let $f: X\rightarrow Y$ be a strict and surjective mapping. Let T and $T'$ be spread Y-spaces. If there exists an X-isomorphism $X\times_YT\rightarrow X\times_YT'$ compatible with the canonical descent data, the Y-spaces T and $T'$ are isomorphic.

Let $\psi : X\times_YT\rightarrow X\times_YT'$ be an X-isomorphism of spread spaces. By proposition 4, there exists a unique morphism of Y-spaces $\varphi : T\rightarrow T'$ such that $\psi =f^*(\varphi )$. Since $f$ is surjective, the mapping $\varphi$ is bijective. It then follows from cor. 2 of I, p. 30 that $\varphi$ is an isomorphism, since the Y-spaces T and $T'$ are spread.