Let $(Z, p)$ be an X-space and let $\tau$ be a descent datum relative to $f$ on $(Z, p)$. The assertion that $\tau$ is an effective descent datum is local in Y, which allows us to suppose that the mapping $f$ possesses a continuous section $s$. Let $g: Z\rightarrow Z/R_{\tau}$ and $q: Z/R_{\tau}\rightarrow Y$ denote the canonical mappings, and let $h: Z\rightarrow X\times_Y(Z/R_{\tau})$ be the mapping given by $z\mapsto (p(z), g(z))$. The mapping $h$ is bijective and continuous; it is enough to prove that it is a homeomorphism.

The mapping from Z into Z which, to $z$, associates $\tau (z, s(f(p(z))))$ is continuous and maps every element of Z to the unique element $z'$ of Z which is equivalent to it for the relation $R_{\tau}$ and such that $p(z')$ belongs to the image of $s$. It therefore defines, by passing to the quotient, a continuous mapping $t: Z/R_{\tau}\rightarrow Z$ which is a section of the mapping $g$. In particular, one has $f\circ p\circ t=q\circ g\circ t=q$.

For every $(x, u)\in X\times_Y(Z/R_{\tau})$, one has $f(p(t(u))) =f(x)$; let us then put $h'(x, u) =\tau (t(u), x)$. The resulting mapping $h'$ from $X\times_Y(Z/R_{\tau})$ into Z is continuous. For every $(x, u)\in X\times_Y(Z/R_{\tau})$, one has

because $\tau (t(u), x)$ is equivalent to $t(u)$ for the relation $R_{\tau}$. This proves that the mapping $h\circ h'$ is the identity mapping of $X\times_Y(Z/R_{\tau})$. For $z\in Z$, one then has $t(g(z)) =\tau (z, s(f(p(z)))$ and $f(p(t(g(z)))) =f(p(z))$, whence $z=\tau (t(g(z)), p(z))$, by definition of the equivalence relation $R_{\tau}$. Thus one has $h'(h(z)) =z$ and $h'\circ h=$ Id$_Z$. The mapping $h$ is therefore a homeomorphism, which was to be proved.

The mapping $\widetilde{\tau}: Z\times_YX\rightarrow$ X $\times_Y$ Z defined by $(z, x)\mapsto$ $(p(z), \tau (z, x))$ is a homeomorphism, with inverse mapping $(x, z)\mapsto (\tau (z, x), p(z))$. We have $\tau$ = pr$_2\circ \widetilde{\tau}$, where pr$_2: X\times_YZ\rightarrow$ Z is the second projection. If $f$ is proper, pr$_2$ is proper; if $f$ is open, pr$_2$ is open (I, p. 17, prop. 8). It follows that $\tau$ is proper (resp. open) if $f$ is. The saturation of a subset A of Z with respect to the relation $R_{\tau}$ is the image of $A\times_YX$ under $\tau$. Consequently, if $f$ is proper, the saturation of a closed subset is closed; if $f$ is open, the saturation of an open subset is open.

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

$p'q'$

X $^f$ Y

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

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.

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.

Let $(Z, p)$ be an X-étalé space and let $\tau$ be a descent datum relative to $f$ on $(Z, p)$. Let us denote by $R_{\tau}$ the equivalence relation in Z associated with $\tau$, let $g: Z\rightarrow Z/R_{\tau}$ be the canonical surjection and let $h$ denote the mapping of Z into $X\times_Y(Z/R_{\tau})$ defined by $z\mapsto (p(z), g(z))$. It is continuous and bijective; let us prove that it is a homeomorphism.

a) Suppose first that the mapping $f$ is open.

Let $z_0$ be a point of Z; put $x_0=p(z_0)$. Since $p$ is étale, there exists a neighbourhood U of $x_0$ in X and a continuous section $s: U\rightarrow Z$ of $p$ over U such that $s(x_0) =z_0$. The mapping $p\times p: Z\times_YZ\rightarrow$ $X\times_YX$ is étale, and the mappings of $U\times_YU$ into $Z\times_YZ$ defined by $(x, x')\mapsto (s(x), s(x'))$ and $(x, x')\mapsto (s(x), \tau (s(x), x'))$ are continuous sections of it over $U\times_YY$. Since they coincide at every point of $U\times_YU$ of the form $(x, x)$, they therefore coincide on an open neighbourhood V of $\Delta_U$ contained in $U\times_YU$. Let $U_0$ be a neighbourhood of $x_0$ in X such that $U_0\times_YU_0$ is contained in V. Let $(x, u)$ be a point of $U_0\times_Yg(s(U_0))$; let $x'\in U_0$ be such that $u=g(s(x'))$. We have therefore $s(x) =\tau (s(x'), x)$, whence $R_{\tau}\{s(x), s(x')\}$ and $(x, u) = (x, g(s(x'))) =$ $(x, g(s(x))) =h(s(x))$.

Since the mapping $g$ is open (IV, p. 386, prop. 2), the set $U_0\times_Yg(s(U_0))$ is an open subset of $X\times_Y(Z/R_{\tau})$ on which the mapping $h^{-1}$ is equal to the continuous mapping $s\circ$ pr$_1$. The mapping $h$ is therefore a homeomorphism.

b) Suppose now that the mapping $f$ is proper and separated.

Let $z_0$ be a point of Z; put $x_0=p(z_0)$ and $y_0=f(x_0)$. The mapping $s$ given by $x\mapsto \tau (z_0, x)$ is a section of $p$ over $\overset{-1}{f}(y_0)$.

The set $\overset{-1}{f}(y_0)$ is compact (TG, I, p. 75, Theorem 1 and I, p. 26, Remark 2), and two distinct points of $\overset{-1}{f}(y_0)$ have disjoint neighbourhoods in X, since $f$ is separated (I, p. 25, Proposition 1). By Theorem 2 of I, p. 37, there therefore exists a neighbourhood $U_0$ of $\overset{-1}{f}(y_0)$ in X and a section $s_0$ of $p$ over $U_0$ extending $s$. The set $s_0(U_0)$ is an open set of Z, since $p$ is étale (I, p. 30, Corollary 3), and contains the saturated set of $\{z_0\}$ for the relation $R_{\tau}$. The mapping $g$ is closed (IV, p. 386, Proposition 2). There then exists an open set V of $Z/R_{\tau}$ such that $W =\overset{-1}{g}(V)\cap \overset{-1}{p}(U_0)$ is a neighbourhood of $z_0$ contained in $s_0(U_0)$ (I, p. 75, Lemma).

Let now $(x, u)\in U_0\times_YV$, and let $z$ be the unique point of Z such that $h(z) = (x, u)$; by definition, one has $z\in W$. Since $W\subset s_0(U_0)$, one has $z=s_0(p(x))$. This shows that the restriction of $h^{-1}$ to the open set $U_0\times_YV$ of $X\times_Y(Z/R_{\tau})$ is equal to $s_0\circ p\circ$ pr$_1$. Consequently, $h$ is a homeomorphism.

Thus we have shown that the descent datum $\tau$ is effective. The mapping $f$ is universally strict (I, p. 20, corollary). Under the assumption that $f$ is surjective, it then follows from Proposition 8 of I, p. 31 that $q: Z/R_{\tau}\rightarrow Y$ is étale.

Let $a$ be a point of $B'$ and let V be a neighbourhood of $a$. Suppose that the fibre square $B'\times_BB'$ is locally connected and let W be a connected neighbourhood of $(a, a)$ in $B'\times_BB'$ which is contained in $V\times V$. Put U = pr$_1(W)$. The set U is contained in V, and is connected since the image of a connected set under a continuous mapping is connected. If $\Delta_{B'}$ denotes the diagonal of $B'\times_BB'$, the mapping pr$_1|\Delta_{B'}: \Delta_{B'}\rightarrow B'$ is a homeomorphism. Since U contains pr$_1(W\cap \Delta_{B'})$ and $W\cap \Delta_{B'}$ is a neighbourhood of $(a, a)$ in $\Delta_{B'}$, U is a neighbourhood of $a$ in $B'$. This proves the lemma.

${E'}^{f'}$ E

$p'p$

${B'}^f$ B

be a cartesian square. Suppose that the mapping $f$ is proper, separated and surjective, and make one of the following assumptions:

(i) The fibres of $f$ are locally connected and the fibre square $B'\times_BB'$ is locally connected.

(ii) The fibres of $f$ are finite, the diagonal $\Delta_{B'}$ of $B'\times_BB'$ is open in $B'\times_BB'$ and $B'\times_BB'-\Delta_{B'}$ is a locally connected space.

Then, if $(E', p')$ is a covering, $(E, p)$ is a covering.

Since the mapping $f$ is universally strict (I, p. 20, corollary), the mapping $p$ is étale (I, p. 31, prop. 8) and separated (I, p. 27, prop. 4). We shall suppose, which is permissible, that $E'= B'\times_BE$.

Let $a$ be a point of B; it is a question of proving that the point $a$ has a neighbourhood W such that the W-space $(E_W, p_W)$ is a covering

which is trivializable. Put $B'_a$ = $\overset{-1}{f}(a)$ and denote by $E'_a$ = $(^-{p'}^1)(E_a)$. The mapping $t_a: E'_a\rightarrow B'_a\times E_a$ defined by $t_a(y) = (p'(y), f'(y))$ is a $B'_a$-isomorphism (I, p. 9, prop. 4), hence $E'_a$ is a trivializable covering of $B'_a$ and $t_a$ is a trivialization of this covering.

Let us prove that there exists a neighbourhood $V'$ of $B'_a$ in $B'$ and a continuous trivialization $t$ of the covering $(E'_{V'}, p_{V'})$ extending $t_a$. Under assumption (ii), $B'_a$ is finite and its points have pairwise disjoint open neighbourhoods over which the covering $E'$ is trivializable, whence the assertion in this case. Under assumption (i), $B'_a$ is locally connected, and so is $B'($IV, p. 390, lemma 1); since the mapping $f$ is proper and separated, $B'_a$ is compact and two distinct points have disjoint neighbourhoods in $B'$, so that the pair $(B',B'_a)$ satisfies property (PCV) ( I, p. 37, lemma 1). The assertion therefore follows from Cor. 2 of I, p. 90.

As $f$ is proper, there exists a neighbourhood V of $a$ in B such that $V'$ contains $\overset{-1}{f}(V)$ (lemma, I, p. 75). We may thus suppose that

Let us endow the $V'$-spaces $V'\times E_a$ and $E'_{V'}= V'\times_BE$ with their canonical descent data relative to $f_V: V'\rightarrow V$. We shall now prove that, after shrinking V and $V'$, the isomorphism of $V'$-spaces $t: E'_{V'}\rightarrow$ $V'\times E_a$ which we have just defined is compatible with the descent data, that is to say that one has $t(b'_1, x) =t(b'_2, x)$ if $(b'_1, b'_2)\in V'\times_VV'$ and $x\in E_{f(b'_1)}$. Let $\widetilde{t}$ denote the mapping pr$_2\circ t: E'_{V'}\rightarrow E_a$.

Put $V''= V'\times_VV'$; let us consider it as a $V'$-space by means of the first projection. The mapping $((b'_1, b'_2), x)\mapsto ((b'_1, b'_2),(b'_1, x))$ of $V''\times_VE$ into $V''\times_{V'}E'$ is an isomorphism of $V''$-spaces; this shows that $V''\times_VE$ is a covering of $V''$.

For $i= 1$, 2, define a mapping $u_i: V''\times_VE\rightarrow V''\times E_a$ by setting $u_i(b'_1, b'_2, x) = (b'_1, b'_2,\widetilde{t}(b'_i, x))$; these are trivializations of the covering $V''\times_VE$. Let $W''$ be the set of points $w\in V''$ over which these trivializations $u_1$ and $u_2$ coincide; it contains $B'_a\times B'_a$, as well as the diagonal $\Delta_{V'}$. Let us prove that $W''$ is a neighbourhood of $B'_a\times B'_a$. Under assumption (i), this follows from Cor. 2 of I, p. 80, since $B'\times B'$ is locally connected. Under assumption (ii), $W''$ contains a neighbourhood of $(B'_a\times B'_a)-\Delta_{B'_a}$ in $(B'\times_BB')-\Delta_{B'}$, for this set is locally connected (loc. cit.). As $\Delta_{V'}$ is open in $B'\times_BB',W''$ is a neighbourhood of $B'_a\times B'_a$ in $V''$.

As $f$ is proper, the canonical mapping $f''$ of $B'\times_BB'$ into B is proper, for it is the composite of the projection pr$_1: B'\times_BB'\rightarrow B'$ and the mapping $f$.

By the lemma of I, p. 75, there exists a neighbourhood W of $a$ in V

such that $(f\overset{-1}{''})(W)$ is contained in $W''$; let us put $W'=(\overset{-1}{f}')(W)$; it is a subset of $V'$ and the isomorphism of étale spaces $t: E'_{W'}\rightarrow W'\times E_a$ is compatible with the canonical descent data relative to the mapping $f_W: W'\rightarrow W$. It follows from corollary (IV, p. 388) that the W-étale spaces $E_W$ and $W\times E_a$ are isomorphic. In particular, $E_W$ is a trivializable covering, whence the proposition.