Proposition 4

Suppose that the mapping $f$ is strict and surjective and that T is a spread Y-space. Then the mapping $f^*:\mathscr{C}_Y(T; T')\rightarrow \mathscr{C}_f(X\times_YT; X\times_YT')$ is bijective.

Let $\tau$ (resp. $\tau '$) denote the equivalence relation on $X\times_YT$ (resp. on $X\times_YT'$) associated with the canonical descent datum. Since the mapping $f$ is surjective, the projection pr$_2: X\times_YT\rightarrow T$ is surjective and the canonical mapping $(X\times_YT)/R_{\tau}\rightarrow T$ is bijective. In particular, the mapping $f^*$ is injective. Let us prove that it is surjective. Let $\varphi : X\times_YT\rightarrow X\times_YT'$ be an X-morphism compatible with the canonical descent data. Then for $(x, t)\in X\times_YT$, one has $\varphi (x, t) = (x, \varphi (t))$, where $\overline{\varphi}$ is a mapping of T into $T'$.

By definition of $\overline{\varphi}$, the mapping $\overline{\varphi}\circ$ pr$_2: X\times_YT\rightarrow T'$ is equal to pr$_2\circ \varphi$, and is therefore continuous. Since the mapping $f$ is surjective and strict and T is a spread Y-space, the projection pr$_2: X\times_YT\rightarrow T$ is strict (I, p. 32, remark 3). By proposition 9 of I, p. 18, the mapping $\overline{\varphi}$ is therefore continuous. It is a Y-morphism such that $f^*(\overline{\varphi}) =\varphi$, which proves the proposition.