(i) The diagonal $\Delta_X$ of the fibred product $X\times_YX$ is a closed subspace;

(ii) For every topological space W and every pair of continuous mappings $(g_1, g_2)$ from W into X such that $f\circ g_1=f\circ g_2$, the set of points $w\in W$ such that $g_1(w) =g_2(w)$ is closed in W;

(iii) For every pair $(x_1, x_2)$ of points of X such that $x_1=\not x_2$ and $f(x_1) =f(x_2)$, there exists a neighbourhood $V_1$ of $x_1$ in X and a neighbourhood $V_2$ of $x_2$ in X such that $V_1\cap V_2=\emptyset$.

(i)$\Rightarrow$(ii) : Let $g_1,g_2$ be continuous mappings from W into X such that $f\circ g_1=f\circ g_2$, and let $g: W\rightarrow X\times_YX$ be the mapping deduced from $g_1$ and $g_2$. The set of points $w\in W$ such that $g_1(w) =g_2(w)$ is $\overset{-1}{g}(\Delta_X)$. Since $g$ is continuous, it is therefore closed if the diagonal $\Delta_X$ is closed.

(ii)$\Rightarrow$(i) : The diagonal $\Delta_X$ is the set of points $z\in X\times_YX$ such that pr$_1(z) =$ pr$_2(z)$. It follows from (ii) applied to $W = X\times_YX$ and to the pair of mappings (pr$_1$, pr$_2$) that the diagonal $\Delta_X$ is closed in $X\times_YX$.

(i)$\Leftrightarrow$(iii) : Let $(x_1, x_2)$ be a point of $X\times_YX$ and let $V_1$ and $V_2$ be neighbourhoods of $x_1$ and $x_2$ respectively. The condition $V_1\cap V_2=\emptyset$ is equivalent to the condition $(V_1\times V_2)\cap \Delta_X=\emptyset$, that is, to $(V_1\times V_2)\cap$ $(X\times_YX)\cap \Delta_X=\emptyset$. Since the sets $(V_1\times V_2)\cap (X\times_YX)$ form a basis of neighbourhoods of $(x_1, x_2)$ in $X\times_YX$, this proves the equivalence of (i) and (iii).

a) If $f$ and $g$ are separated, then $g\circ f$ is separated.

b) If $g\circ f$ is separated, then $f$ is separated.

c) Suppose moreover that the mapping $f$ is proper and surjective. Then, if $g\circ f$ is separated, $g$ is separated.

Consider, in $X\times X$, the subspaces $\Delta_X$ (diagonal), $X\times_YX$ and $X\times_ZX$. These are respectively the sets of points $u$ of $X\times X$ such that pr$_1(u) =$ pr$_2(u),f\circ$pr$_1(u) =f\circ$pr$_2(u),g\circ f\circ$pr$_1(u) =g\circ f\circ$pr$_2(u)$. If $g\circ f$ is separated, $\Delta_X$ is closed in $X\times_ZX$, hence also in $X\times_YX$, whence b).

By prop. 1, (ii), applied to $W = X\times_ZX,g_1=f\circ$ pr$_1$ and $g_2=f\circ$ pr$_2,X\times_YX$ is closed in $X\times_ZX$ if $g$ is separated. If moreover $f$ is separated, $\Delta_X$ is closed in $X\times_YX$ (prop. 1, (i)), hence in $X\times_ZX$, whence a).

Finally, let us prove c). The mapping $(f, f): X\times X\rightarrow Y\times Y$ is proper (TG, I, p. 73, prop. 4). The subspace $X\times_ZX$ is the inverse image of $Y\times_ZY$ by $(f, f)$ ; by TG, I, p. 72, prop. 3, the mapping $u: X\times_Z$ $X\rightarrow Y\times_ZY$ deduced from $(f, f)$ is proper. Since $g\circ f$ is separated, the diagonal $\Delta_X$ is closed in $X\times_ZX$. Consequently, $u(\Delta_X)$ is closed in $Y\times_ZY$. Since $f$ is surjective, $u(\Delta_X)$ is the diagonal of $Y\times_ZY$. This shows that $g$ is separated.

Let $s: Y\rightarrow X$ be a continuous section of $f$. The mapping $s$ induces a homeomorphism of Y onto the subspace $s(Y)$ of X (TG, I, p. 22, prop. 9). The identity mapping of X and the mapping $s\circ f$ are continuous and one has $f\circ$ Id$_X=f\circ (s\circ f)$. Since the mapping $f$ is separated, the set $s(Y)$, which is the set of points $x$ of X such that $x=s\circ f(x)$, is closed in X (prop. 1, (ii)).

${X'}^{f'}$ X

$p'p$

${B'}^f$ B

be a cartesian square. If the mapping $p$ is separated, the same is true of $p'$. Conversely, if $p'$ is separated and if $f$ is universally strict and surjective, then $p$ is separated.

Consider the square

$q'q$

${B'}^f$ B

where the mapping $\varphi$ is induced by the mapping $f'\times f': X'\times X'\rightarrow X\times X$. Recall (I, p. 13, example 1) that this square is cartesian and that one has $\overset{-1}{\varphi}(\Delta_X) = \Delta_{X'}$.

If the mapping $p$ is separated, the set $\Delta_X$ is closed in $X\times_BX$; consequently, $\Delta_{X'}$ is a closed subset of $X'\times_{B'}X'$, which proves that the mapping $p'$ is separated.

If the mapping $f$ is surjective, the mapping $\varphi$ is surjective (I, p. 10, cor.) ; if $f$ is universally strict, $\varphi$ is strict (I, p. 20, def. 6). Suppose the mapping $p'$ separated. Then $\Delta_{X'}$ is closed in $X'\times_{B'}X'$. Since $\Delta_{X'}=\overset{-1}{\varphi}(\Delta_X)$ and since $\varphi$ is surjective and strict, $\Delta_X$ is closed in $X\times_BX$, which proves that the mapping $p$ is separated.

Consider the following cartesian square:

$X\times_ZY^{pr_2}$ Y

pr$_1g$

X $^{g\circ f}Z$.

Let $s: X\rightarrow X\times_ZY$ be the mapping $x\mapsto (x, f(x))$. It is a continuous section of pr$_1: X\times_ZY\rightarrow X$. By prop. 4, the mapping pr$_1$ is separated; it follows that the mapping $s$ is proper (I, p. 27, prop. 3 and TG, I, p. 72, prop. 2). On the other hand, the mapping pr$_2$ is proper (I, p. 17, prop. 8). Hence the mapping $f$, which is equal to pr$_2\circ s$, is proper (TG, I, p. 73, prop. 5).

a) Suppose that $f$ and $g$ are etale; then $g\circ f$ is etale.

b) Suppose that $g$ is étale, $f$ is continuous and $g\circ f$ is open. Then $f$ is open.

c) Suppose that $g\circ f$ and $g$ are étale and that $f$ is continuous. The mapping $f$ is then étale.

d) Suppose that $g\circ f$ is étale and that the mapping $f$ is continuous and open; then $f$ is étale and $g$ is étale at every point of $f(X)$.

Let us prove a). Suppose the mappings $f$ and $g$ are étale. Then they are continuous and open, hence the mapping $g\circ f$ is continuous and open. Let $x$ be a point of X. There exists a neighbourhood W of $f(x)$ in Y such that the mapping $g|W$ is injective, and a neighbourhood V of $x$

in X, contained in $\overset{-1}{f}(W)$, such that the mapping $f|V$ is injective; the mapping $(g\circ f)|V$ is then injective. This proves that the mapping $g\circ f$ is étale (remark 2).

Let us prove b). Let $x$ be a point of X and W an open neighbourhood of $f(x)$ such that $g$ induces a homeomorphism of W onto the open set $g(W)$. Let V be a neighbourhood of $x$ such that $f(V)$ is contained in W; then $g\circ f(V)$ is a neighbourhood of $g\circ f(x)$, hence $f(V)$ is a neighbourhood of $f(x)$ in the open set W, and also in Y. This proves that the mapping $f$ is open (TG, I, p. 33, prop. 5).

Let us prove c). By b$),f$ is open. Let $x$ be a point of X. Since $g\circ f$ is étale, there exists a neighbourhood V of $x$ in X such that $g\circ f|V$ is injective. Thus, $f|V$ is injective and $f$ is étale (I, p. 29, remark 2).

Let us prove d). Let $x$ be a point of X and put $y=f(x)$. There exists an open neighbourhood V of $x$ such that $g\circ f$ induces a homeomorphism of V onto the open set $g\circ f(V)$. Since the mapping $f$ is open, $f(V)$ is an open neighbourhood of $y$. The mapping $f|V: V\rightarrow f(V)$ deduced from $f$ by passing to subspaces is continuous, open and bijective; hence it is a homeomorphism, so that $f$ is étale at $x$. Moreover, the mapping $g$ induces a homeomorphism of $f(V)$ onto $g\circ f(V)$, hence is étale at $y$.

This follows from assertion c) of prop. 6.

By corollary 1, such a morphism is étale; it is therefore open (I, p. 29, remark 2). If it is bijective, it is a B-isomorphism.

In fact, such a section is étale (corollary 1), hence open.

Let $s$ be a section of $p$; since $p$ is étale, the image of $s$ is open in E (Corollary 3); it is also closed, since $p$ is separated (I, p. 27, Prop. 3). Since E is connected, we have $s(B) = E$ and $p$ is a homeomorphism.

Suppose first that the mapping $p$ is étale. Then it is open and every point of E possesses a neighbourhood V such that $p|V$ is injective, which amounts to saying that one has $(V\times V)\cap (E\times_BE)\subset \Delta_E$. Hence $\Delta_E$ is an open subset of $E\times_BE$.

Conversely, suppose that $p$ is an open mapping and that $\Delta_E$ is an open subset of $E\times_BE$. Let $x$ be a point of E. Let V be an open neighbourhood of $x$ in E such that $(V\times V)\cap (E\times_BE)$ is contained in $\Delta_E$. Then $p|V$ is injective. By Remark 2, I, p. 29, the mapping $p$ is étale.

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

$p'p$

${B'}^f$ B

be a cartesian square. If the mapping $p$ is étale, the mapping $p'$ is étale. Conversely, if the mapping $p'$ is étale and the mapping $f$ is universally strict and surjective, the mapping $p$ is étale.

Suppose that the mapping $p$ is étale. It is in particular open and $p'$ is open (I, p. 17, Prop. 8). Consider then the cartesian square (I, p. 13, Example 1)

$q'q$

${B'}^fB$. One has $\Delta_{E'}=\overset{-1}{\varphi}(\Delta_E)$ (loc. cit.). Moreover, the diagonal $\Delta_E$ is open in $E\times_BE$ (Prop. 7), hence the diagonal $\Delta_{E'}$ is open in $E'\times_{B'}E'$. This proves that the mapping $p'$ is étale (loc. cit.).

Suppose now that the mapping $f$ is surjective and universally strict and that the mapping $p'$ is étale. Then $p'$ is open, hence $p$ is open (I, p. 21, Prop. 11, a)). On the other hand $\Delta_{E'}$ is an open subset of $E'\times_{B'}E'$. Since $\Delta_{E'}=\overset{-1}{\varphi}(\Delta_E)$ and the mapping $\varphi$ is surjective and strict (I, p. 20, Def. 6), $\Delta_E$ is open in $E\times_BE$. By Prop. 7, the mapping $p$ is étale.

Let $(E, p)$ and $(E', p')$ be étale B-spaces. The mapping pr$_1: E\times_B$ $E'\rightarrow E$ is étale (prop. 8), hence the mapping $p\circ$ pr$_1: E\times_BE'\rightarrow B$ is étale (prop. 6, a)).

Let $z\in Z$ and $x\in E$ be points such that $f(z) =p(x)$. There exists a neighbourhood W of $z$ in Z and a continuous lifting $g$ of $f$ to E, defined on W, such that $g(z) =x$.

There exists an open neighbourhood V of $p(x)$ in B and a continuous section $s$ of $p$ over V such that $s(p(x)) =x$ (prop. 9). The set

$W =\overset{-1}{f}(V)$ is an open neighbourhood of $z$ and the mapping $g=s\circ (f|W)$ is a continuous lifting of $f$ to E, defined on W and such that $g(z) =x$.

a) If the mapping $p$ is separated, W is closed.

b) If the mapping $p$ is étale, W is open.

Let us denote by $h$ the continuous mapping $(g, g'): Z\rightarrow E\times E$. The image of $h$ is contained in $E\times_BE$ since $p\circ g=f=p\circ g'$ and the set W of points of Z where $g$ and $g'$ coincide is the inverse image under $h$ of the diagonal $\Delta_E$.

If the mapping $p$ is étale, the diagonal $\Delta_E$ is open in $E\times_BE$ (I, p. 31, prop. 7), therefore the set W is open in Z.

If the mapping $p$ is separated, the diagonal $\Delta_E$ is closed in $E\times_BE$ (I, p. 25, def. 1) and the set W is closed in Z.

In fact, if $f: E\rightarrow E$ is a B-morphism, the set of points where $f$ and Id$_E$ coincide is equal to E or to the empty set by Cor. 1.

By Corollary 1, the restrictions of $g_1$ and $g_2$ to $U_1\cap U_2$ are equal. The mapping $g: U_1\cup U_2\rightarrow E$ defined by $g(z) =g_i(z)$ for $z\in U_i(i= 1$, 2) is a continuous lifting of $f$ defined on $U_1\cup U_2$ (TG, I, p. 19, Prop. 4).

In the preceding results, the particular case where Z = B and $f=$ Id$_B$ is important: the continuous liftings of $f$ are then the continuous sections of $p$.

The condition is necessary (Proposition 11, I, p. 34). Let us prove that it is sufficient. Let $b$ be a point of B, let $x$ and $x'$ be two distinct points of E such that $p(x) =p(x') =b$. Let V be an open neighbourhood of $b$ and $s,s'$ continuous sections of $p$ over V such that $s(V)$ and $s'(V)$ are open neighbourhoods of $x$ and $x'$ respectively (Prop. 9). By hypothesis, the set W of points $x\in V$ such that $s(x)=\not s'(x)$ is open in V, hence in B. The sets $s(W)$ and $s'(W)$ are open neighbourhoods of $x$ and $x'$ respectively; they are disjoint by construction. This proves that the mapping $p$ is separated (I, p. 25, Definition 1).

Since the restriction of $s$ to $U\times \{y_1\}$ is continuous and the mapping $f$ is étale, there exists a neighbourhood $U'$ of $x_1$ such that $s(x, y_1) =\sigma (x, y_1)$ for every $x\in U'($I, p. 34, prop. 11, b)). Let $x\in U'$; since the restriction of $s$ to $\{x\} \times V$ is continuous and the mapping $f$ is étale and separated, it follows from Corollary 1, I, p. 34, that $\sigma (x, y) =s(x, y)$ for every $y\in V$. Thus $s$ and $\sigma$ coincide on $U'\times V$.

We now prove the theorem. First let us prove that the mapping $s$ is continuous in a neighbourhood of every point of $X\times \{y_0\}$. Let $x_0\in$ X; one can choose an open neighbourhood U of $x_0$ in X, a connected open neighbourhood V of $y_0$ in Y, and a continuous section $\sigma : U\times V\rightarrow Z$ of $f$ such that $\sigma (x_0, y_0) =s(x_0, y_0)$. By the above lemma, $s$ is continuous in a neighbourhood of $(x_0, y_0)$.

Let us now prove that the mapping $s$ is continuous at every point of $X\times Y$. For $x_0\in X$, let us denote by $C(x_0)$ the set of points $y\in$ Y such that $s$ is continuous in a neighbourhood of $(x_0, y)$. The set $C(x_0)$ is open in Y by definition, and contains $y_0$ by what precedes.

Let us prove that it is closed in Y. Consider a point $y_1$ of Y adherent to $C(x_0)$, an open neighbourhood U of $x_0$ in X, an open neighbourhood V of $y_1$ in Y, and a continuous section $\sigma : U\times V\rightarrow Z$ of $f$ such that $\sigma (x_0, y_1) =s(x_0, y_1)$. Since the restriction of $s$ to $\{x_0\} \times V$ is continuous and $f$ is étale, there exists an open neighbourhood $V'$ of $y_1$ in Y such that $\sigma (x_0, y) =s(x_0, y)$ for every $y\in V'$ (prop. 11 of I, p. 34). Since Y is locally connected, we may suppose that $V'$ is connected. Since $y_1$ is adherent to $C(x_0)$, the set $C(x_0)\cap V'$ is not empty; let $y_2$ be a point of $C(x_0)\cap V'$. By the lemma applied to $(x_0, y_2)$, there exists a neighbourhood $U'$ of $x_0$ contained in U such that $s=\sigma$ over $U'\times V'$. This proves that $C(x_0)$ is closed, for $U'\times V'$ is a neighbourhood of $(x_0, y_1)$. Since Y is connected, we have $C(x_0) = Y$, and this for every $x_0\in X$, whence the theorem.

Let us denote by Z the fibre product $(X\times Y)\times_BE$ and by $f: Z\rightarrow X\times Y,h': Z\rightarrow E$ the canonical projections. The mapping $f$ is étale (I, p. 31, prop. 8) and separated (I, p. 27, prop. 4). The mapping $s: X\times Y\rightarrow$ Z defined by $s(x, y) = ((x, y), g(x, y))$ is a section of $f$ satisfying the hypotheses of theorem 1. It is therefore continuous, as is $g=h'\circ s$.

(i) A admits a fundamental system of paracompact neighbourhoods;

(ii) B is paracompact and A closed;

(iii) B is metrizable;

(iv) A is compact and two distinct points of A possess disjoint neighbourhoods in B.

Then every continuous section over A of an étale mapping $p: E\rightarrow B$ extends to a continuous section of $p$ over a neighbourhood of A.

Consider the following property (PCV) of the pair $(B,A)$:

(PCV) For every covering $(U_i)_{i\in I}$ of A by open subsets

of B, there exists a neighbourhood V of A and a locally

finite family $(F_j)_{j\in J}$ of closed subsets of V covering V, such that

each of the $F_j$ is contained in one of the $U_i$.

Theorem 2 follows from Lemmas 1 and 3 below.

Let $(U_i)_{i\in I}$ be an open covering of A by open sets of B. Under hypothesis (i), there exists a paracompact neighbourhood V of A contained in the union of the $U_i$, an open covering $(U'_j)_{j\in J}$ of the space V, locally finite and finer than the covering $(V\cap U_i)_{i\in I}$, and an open covering $(U''_j)_{j\in J}$ of the space V such that for every $j\in J$, the closure $F_j$ of $U''_j$ in V is contained in $U'_j$ (TG, IX, p. 49, prop. 4 and p. 48, cor. 1 to th. 3). Hence property (PCV) in this case.

Condition (ii) implies condition (i) by Corollary 2 of TG, IX, p. 50. Analogously, condition (iii) implies condition (i): in fact, if B is metrisable, every neighbourhood of A is metrisable, hence paracompact (TG, IX, p. 51, th. 4).

Lastly, suppose that condition (iv) is satisfied, and let us show that property (PCV) holds. Since A is compact, it is enough to consider the case of a finite covering $(U_i)_{0\leqslant i\leqslant n}$. We then construct by induction open sets $V_0, . . . ,V_n$ of B satisfying the following conditions for $0\leqslant i\leqslant n:$

Suppose that $V_0, . . . ,V_{r-1}$ have been constructed satisfying the above conditions for $0\leqslant i\leqslant r-1$, and let us construct $V_r$. The sets K = $A\cap \complement U_r$ and $L = A\cap \complement (V_0\cup \cdots \cup V_{r-1}\cup U_{r+1}\cup \cdots \cup U_n)$ are closed in A, hence compact. By virtue of $(\alpha )$, they are disjoint. By hypothesis, for every point $a$ of L and every point $b$ of K, there exist disjoint neighbourhoods of $a$ and $b$ in B. By Lemma 2 below, there exist disjoint open sets $V_r$ and W in B such that $L\subset V_r$ and $K\subset W$. From the inclusions $L\subset V_r$ and $A\subset V_0\cup \cdots \cup V_{r-1}\cup U_r\cup \cdots \cup U_n$ and from the definition of L, it follows that $(\alpha )$ holds for $i=r$. On the other hand, we have $\overline{V_r}\cap K =\emptyset$, hence $\overline{V_r}\cap A\subset U_r$.

The set $M =\bigcup_{0\leqslant i\leqslant n}(V_i\cap \complement U_i)$ is closed and does not meet A by $(\beta )$. By $(\alpha )$, the set $V =((\bigcup_{0\leqslant i\leqslant n}\overline{V_i})-$ M is a neighbourhood of A in B. For $i= 0, . . . , n$, put $F_i= V\cap \overline{V_i}:$ this is a closed subset of V, contained in $U_i$. The family $(F_i)_{0\leqslant i\leqslant n}$ is a covering of V, whence property (PCV).