(i) The mapping $p$ is continuous and the B-space $(E, p)$ is a trivializable covering;

(ii) There exists a partition $(V_i)_{i\in I}$ of E consisting of open sets such that, for every $i\in I$, the mapping $p|V_i: V_i\rightarrow B$ is a homeomorphism.

Suppose that condition (i) is satisfied, and let $g: E\rightarrow B\times F$ be a trivialization of the B-space $(E, p)$, of fibre type F. Then F is a discrete topological space and the sets $V_i=\overset{-1}{g}(B\times \{i\})$, for $i\in F$, form a partition of E consisting of open sets. For every $i$, the mapping $p|V_i: V_i\rightarrow B$ is a homeomorphism, with reciprocal homeomorphism the mapping $x\mapsto g^{-1}(x, i)$, whence condition (ii).

Conversely, suppose that condition (ii) is satisfied. The mapping $p$ is then continuous. Consider the mapping from E into $B\times I$ which associates to $x\in E$ the pair $(p(x), i)$, where $i$ is the unique element of I such that $x\in V_i$. If I is endowed with the discrete topology, condition (ii) means that this mapping is a B-isomorphism and the B-space $(E, p)$ is a trivializable covering.

This is obviously necessary. Conversely, for every continuous section $s$ of $p$, the set $s(B)$ is open and the mapping $p$ induces a homeomorphism of $s(B)$ onto B (I, p. 30, Cor. 3 of Prop. 6). Moreover, when $s$ ranges over the set $\mathscr{S}(B; E)$ of sections of $p$, the sets $s(B)$ are pairwise disjoint, since B is connected (I, p. 34, Cor. 1 of Prop. 11). It then follows from proposition 1 that if the sets $s(B)$ cover E, the space E is a trivializable covering of B.

Suppose first that the B-space E and the $B'$-space $E'$ admit trivializations $g: E\rightarrow B\times F,g': E'\rightarrow B'\times F'$; the composite of the homeomorphism $g\times g'$ of $E\times E'$ onto $(B\times F)\times (B'\times F')$ and of the canonical homeomorphism of $(B\times F)\times (B'\times F')$ onto $(B\times B')\times (F\times F')$ is a trivialization of the $B\times B'$-space $E\times E'$, of fibre-type $F\times F'$. If F and $F'$ are discrete spaces, the space $F\times F'$ is also discrete.

In the general case, for every point $(a, a')$ of $B\times B'$, there exists a neighbourhood U of $a$ in B and a neighbourhood $U'$ of $a'$ in $B'$ such that the U-space $E_U$ and the $U'$-space $E'_{U'}$ are trivializable. It follows from the foregoing that the $B\times B'$-space $E\times E'$ is trivializable over $U\times U'$. Hence it is a locally trivial fibre space with base $B\times B'$. Its fibres are discrete if E and $E'$ are coverings, whence the proposition.

It should be observed that, if $(p_i: E_i\rightarrow B_i)_{i\in I}$ is a family of locally trivial fibre spaces, or even of coverings, the product space $\prod_{i\in I}E_i$ endowed with the mapping $(p_i)_{i\in I}$ is not necessarily a locally trivial fibre space with base $\prod_{i\in I}B_i($I, p. 145, Exercise 3).

Since the set $E\times_AE'$ is the inverse image of $B\times_AB'$ by the mapping $p\times p': E\times E'\rightarrow B\times B'$, the corollary follows from proposition 2 and remark 1 (I, p. 68).

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

$p'p$

${B'}^f$ B

be a cartesian square. If $(E, p)$ is a locally trivial fibre B-space ( resp. a covering), then $(E', p')$ is a locally trivial fibre space ( resp. a covering) with base $B'$, trivializable if $(E, p)$ is.

This follows from corollary 1, applied with B = A and $E'= B'$.

This follows from corollary 1, applied to the case where A, B and $B'$ are equal.

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

${B'}^f$ B

be a cartesian square. Suppose that in a neighbourhood of every point of B, the mapping $f$ has a continuous section. Then, if $(E', p')$ is a locally trivial fibre $B'$-space ( resp. a covering), the B-space $(E, p)$ is a locally trivial fibre space ( resp. a covering).

It is a question of proving that every point $a$ of B possesses a neighbourhood U such that the U-space induced by $(E, p)$ over U is a locally trivial fibre space (resp. a covering). Take for U a neighbourhood of $a$ over which there exists a continuous section $s$ of $f$. Let $i: U\rightarrow B$ and $j:\overset{-1}{p}(U)\rightarrow E$ denote the canonical injections. Since square (1) is cartesian, there exists a unique continuous mapping $s':\overset{-1}{p}(U)\rightarrow E'$ such that $f'\circ s'=j$ and $p'\circ s'=s\circ p_U$. The square

U $^sB'$

is thus commutative and its composite with square (1) is the cartesian square

$\overset{-1}{p}(U)^j$ E

$p_Up$

U $^iB$.

By proposition 7 of I, p. 15, square (2) is cartesian. Corollary 2 permits one to conclude.

Let G be a topological space, let $(F, g)$ be a covering of G and let $(E, f)$ be a covering of F; suppose that these coverings have a degree. If the G-space $(E, g\circ f)$ is a covering, then it has a degree and one has deg(E$, g\circ f$) $=$ deg(F$, g$) deg(E$, f$), which can also be written

In fact, if $z$ is a point of G, all the fibres of the mapping

have for cardinal deg(E$, f$), and $\overset{-1}{g}(z)$ has for cardinal deg(F$, g$). The assertion therefore follows from the shepherds' principle (E, III, p. 41, prop. 9).

Let B and $B'$ be topological spaces, let $(E, p)$ be a covering of B and let $(E', p')$ be a covering of $B'$. Suppose that these coverings have a degree; then the covering $(E\times E', p\times p')$ of $B\times B'$ (I, p. 71, prop. 2) has a degree and one has:

deg(E $\times E', p\times p'$) $=$ deg(E$, p$) deg(E$', p'$).

In fact, for every pair $(b, b')\in B\times B'$, the fibre $(E\times E')_{(b,b')}$ is the product $E_b\times E'_{b'}$.

Let B and $B'$ be topological spaces, $(E, p)$ a covering of B, $(E', p')$ a covering of $B'$ and let

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

$p'p$

${B'}^f$ B

a Cartesian square. If $B'$ is non-empty and if the covering $(E, p)$ has a degree, then the covering $(E', p')$ has a degree, equal to that of E, after I, p. 10, cor. of prop. 4.

(i) The space E, endowed with the mapping $p$, is a locally finite covering of B;

(ii) The mapping $p$ is étale, proper and separated;

(iii) The mapping $p$ is continuous, open and separated, its fibres are finite, and the numerical function $b\mapsto$ Card($\overset{-1}{p}(b)$) is upper semicontinuous on B (TG, IV, p. 28).

We shall use the following lemma in the proof:

of $y$ such that W contain $\overset{-1}{f}(V)$.

In this statement, one may consider only the neighbourhoods W of $\overset{-1}{f}(y)$ which are open. Denoting by F the complement of W in Y, one may then reformulate the statement in the following way: in order that a mapping $f: X\rightarrow Y$ be closed, it is necessary and sufficient that, for every closed subset F of X, every point $y$ of Y which does not belong to $f(F)$ possess a neighbourhood V disjoint from $f(F)$. Now this assertion results immediately from the definition of a closed mapping (TG, I, p. 30, def. 1).

Let us now prove theorem 1. Each of the three conditions implies that the mapping $p$ is continuous, open and separated, and also that the fibres of $p$ are finite. This is clear under assumptions (i) and (iii); under assumption (ii), the fibres of $p$ are discrete (I, p. 29, remark 2) and quasi-compact (TG, I, p. 75, th. 1), hence finite (TG, I, p. 60, example 1).

(i)$\Rightarrow$(ii) : it is enough to prove that $p$ is proper and, for this, that for every open set U over which the covering $(E, p)$ is trivializable, the mapping $p_U:\overset{-1}{p}(U)\rightarrow U$ is proper (TG, I, p. 72, prop. 3). Since the fibres of $p_U$ are finite, the latter assertion results from Corollary 5 of TG, I, p. 77.

(ii)$\Rightarrow$(iii) : let $b$ be a point of B and, for every $x\in \overset{-1}{p}(b)$, let $W_x$ be an open neighbourhood of $x$ in E such that $p|W_x$ be injective. The set $W =\bigcup_{x\in\overset{-1}{p}(b)}W_x$ is an open neighbourhood of $\overset{-1}{p}(b)$. Since the mapping $p$ is closed, there exists by the above lemma an open neighbourhood U of $b$ such that $\overset{-1}{p}(U)\subset W$. For every $a\in U$, one has $\overset{-1}{p}(a)\subset W$; since the restriction of $p$ to each $W_x$ is injective, it follows that Card($\overset{-1}{p}(a)$)$\leqslant$ Card($\overset{-1}{p}(b)$), which proves the upper semicontinuity of the mapping $a\mapsto$ Card($\overset{-1}{p}(a)$).

(iii)$\Rightarrow$(i): let $b$ be a point of B. Since the fibre $E_b=\overset{-1}{p}(b)$ is finite and the mapping $p$ is separated, one can choose, for each $x\in E_b$, an open neighbourhood $V'_x$ of $x$ such that the $V'_x$ are pairwise disjoint (I, p. 26, remark 4). Since the mapping $p$ is open and the set $E_b$ finite, the set $U'=\bigcap_{x\in E_b}p(V'_x)$ is an open neighbourhood of $b$ in B. Let U be an open neighbourhood of $b$ in B, contained in $U'$ and such that for every $a\in U$, Card(E$_a$)$\leqslant$ Card(E$_b$). For each $x\in E_b$, put $V_x= V'_x\cap \overset{-1}{p}(U)$. Let $a$ be a point of U; the sets $E_a\cap V_x$, for $x\in E_b$, are non-empty and pairwise disjoint. Hence each of these sets contains a unique element and they form a partition of $E_a$. This shows that, for each $x\in E_b$, the mapping $p|V_x$ is injective and that one has $\overset{-1}{p}(U) =\bigcup_{x\in E_b}V_x$. Since the mapping $p$ is open, it induces a homeomorphism of $V_x$ onto U and consequently, $(E, p)$ is a covering of B (I, p. 70, cor. 1 of prop. 1).

The set F of points of E at which the mapping $p$ is not étale is closed in E (I, p. 29, remark 1). Its image $p(F)$ is closed since the mapping $p$ is proper; the complement U of $p(F)$ is therefore open. The mapping $p_U$ is separated (I, p. 27, prop. 4) and proper (TG, I, p. 72, prop. 3). By construction, the mapping $p_U$ is étale; it therefore satisfies condition (ii) of theorem 1.

The mapping $p$ is separated (I, p. 26, remark 2) and proper (TG, I, p. 76, cor. 2). By corollary 1, E is therefore a locally finite covering of B. Since E is compact, $p(E)$ is a quasi-compact subset of B; the mapping of $p(E)$ into $\mathbf{N}$ given by $b\mapsto$ Card $\overset{-1}{p}(b)$ being locally constant, it is bounded above (TG, IV, p. 30, corollary).

a) If the mapping $p$ is étale and separated, the E-space $(E', f)$ is a covering.

b) If the mapping $f$ is surjective and makes the space $E'$ a covering of E, then E is a covering of B.

Under assumption a), the mappings $p$ and $p'$ are etale and separated, and the mapping $p'$ is proper (theorem 1). The mapping $f$ is therefore etale (I, p. 30, cor. 1 of prop. 6), separated (I, p. 25, prop. 2 b)) and proper (I, p. 28, prop. 5). By theorem 1, $(E', f)$ is a covering of E.

Now suppose that $f$ is surjective and that $(E', f)$ is a covering of E. Then it is locally finite, hence the mapping $f$ is proper and surjective (th. 1). By I, p. 25, prop. 2, c), the mapping $p$ is separated, since $p'=p\circ f$ and $p'$ is separated. The mapping $p$ is etale (I, p. 29, prop. 6, d)), proper (TG, I, p. 73, prop. 5, b)) and its fibres are finite. By th. 1, the B-space $(E, p)$ is then a covering of B.

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

$p'p$

${B'}^f$ B be a cartesian square. Suppose that $(E', p')$ is a locally finite covering of $B'$ and that the mapping $f$ is universally strict and surjective. Then $(E, p)$ is a locally finite covering of B.

In fact, the mapping $p$ is etale (I, p. 31, prop. 8), proper (I, p. 21, prop. 11) and separated (I, p. 27, prop. 4).

The result is true if B is empty; henceforth we assume it nonempty.

If X is an open and closed subset of E, the restriction of $p$ to X is a separated mapping (I, p. 25, prop. 2, a) and I, p. 26, remark 1), etale and proper (TG, I, p. 74, corollary 1); by theorem 1, $(X, p)$ is a locally finite covering of B. Since B is connected, this covering has finite degree. If X is distinct from E, there exists a point $b\in B$ such that $X_b(E_b$, whence $[X : B]<[E : B]$ because B is connected. It follows that every decreasing sequence of open and closed sets in E is stationary.

Let $x\in E$; then there exists a smallest open and closed subset X of E containing $x$ (E, III, p. 51, prop. 6). Such a set X is connected; it is therefore the connected component of $x$ in E. This shows that the connected components of E are open and closed and that each of them is a covering of B. Since B is connected, each connected component of E meets each fibre of $p$; the connected components of E are therefore finite in number.

Let us denote by $p$ and $p'$ the respective projections of the B-spaces E and $E'$. Let $b$ be a point of B. There exists an open neighbourhood U of $b$ in B and a finite partition $(V_i)_{i\in I}$ of $\overset{-1}{p}(U)$ formed of open sets of E such that, for every $i\in I$, the mapping $p$ induces a homeomorphism of

$V_i$ onto U. Put $V'_i=\overset{-1}{f}(V_i)$. The sets $V'_i$ are open in $E'$,

form a partition of $(^-{p'}^1)(U)$ and the mapping from $V'_i$ to U deduced from $p'$ by passing to the subspaces makes $V'_i$ into a locally trivial fibred U-space. It follows that the space $(^-{p'}^1)(U)$, endowed with the mapping

$p'_U:(^-{p'}^1)(U)\rightarrow U$, is a locally trivial fibred U-space (I, p. 69, remark 2). If $(E', f)$ is a covering of E, the fibres of $p'_U$ are discrete because their intersections with each of the open sets $V'_i$ are so. This completes the proof.

Let $b$ be a point of B. If E is a covering of B, every open neighbourhood U of $b$ which is connected and over which the covering E is trivializable satisfies the conditions stated in the proposition. Conversely, let U be an open neighbourhood of $b$ satisfying these conditions. The set $\overset{-1}{p}(U)$ is open; its connected components are open subsets of E (TG, I, p. 85, prop. 11) and constitute a partition of $\overset{-1}{p}(U)$. The proposition therefore follows from corollary 1 (I, p. 70) to proposition 1.

The spaces E and $E'$ are locally connected. For every open subset U of B, the set $E'\cap \overset{-1}{p}(U)$ is open and closed in $\overset{-1}{p}(U)$, hence is a union of connected components of $\overset{-1}{p}(U)$. The open subsets U of B such that $p$ induce a homeomorphism of each connected component of $\overset{-1}{p}(U)$ onto U cover B. By the proposition, $p|E'$ makes $E'$ into a covering of B. The second assertion follows from this (cf. I, p. 68).

Let X be the set of points $x$ of $E'$ such that $f(x) =g(x)$. It is the set of points where the liftings $f$ and $g$ of $p'$ to E coincide; therefore X is open and closed in $E'$ (prop. 11 of I, p. 34). The complement Y of X is also open and closed; consequently $p(Y)$ is open and closed in B (corollary 1). Its complement, which is the set of points $b$ of B such that $f_b=g_b$, is therefore so as well.

Let $b$ be a point of B. Let $f$ and $g$ be B-morphisms from $E'$ into E such that $f_b=g_b$. The set of points $a$ of B such that $f_a=g_a$ is open and closed in B (Corollary 2) and contains $b$. It is therefore equal to B, and $f$ is equal to $g$.

The condition is necessary. Let $E'$ be the union of the images of the continuous sections of $p$, and let $E''$ be its complement in E. The set $E'$ is open and closed in E (cf. I, p. 79). Hence $(E', p|E')$ is a covering of B (Corollary 1), and this covering is trivializable by virtue of Corollary 2 (I, p. 70). Suppose that $E'$ contains $E_b$ and let us prove that $E''$ is empty. Since $E''$ is open and closed in E, $p(E'')$ is open and closed in B (Corollary 1). Since B is connected and $b$ does not belong to $p(E''),p(E'') =\emptyset$, hence $E''=\emptyset$.

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

$p'p$

${B'}^f$ B

be a cartesian square. Suppose that the B-space $(E, p)$ is a covering and that the space $B'$ is connected and locally connected. Let $b'$ be a point of $B'$. In order that the $B'$-space $(E', p')$ be a trivializable covering, it is necessary and sufficient that for every point $x$ of E such that $p(x) =f(b')$, there exist a continuous lifting $g: B'\rightarrow E$ of $f$ such that $g(b') =x$.

Let us recall that $(E', p')$ is a covering of $B'($I, p. 71, Cor. 2) and that the mapping $f'$ induces a bijection $E'_{b'}\rightarrow E_{f(b')}($I, p. 10, Cor.). By Prop. 3 of I, p. 9, the mapping $s\mapsto f'\circ s$ defines a bijection between the set of continuous sections of $p'$ and the set of continuous liftings of $f$ to E. The corollary therefore follows from Corollary 4.

a) If the projection of the B-space E is an étale and separated mapping, $(E', f)$ is a covering of E.

b) If $f$ is surjective, then E is a covering of B.

Let $p$ and $p'$ denote the projections of the B-spaces E and $E'$ respectively. Under assumption a$),f$ is étale (I, p. 29, Prop. 6, c)). Under assumption b$),p$ is étale (loc. cit., d)). Hence, under either one of these two assumptions, E is locally connected; in particular, its connected components are open and closed in E (TG, I, p. 85, Prop. 11).

We shall first prove Proposition 7 under the additional assumption that B is connected, locally connected, and that $E'$ is the trivial covering $(B\times F'$, pr$_1)$.

Let $x\in F'$ be such that $U = B\times \{x\}($cf. I, p. 79). The mapping from B into E which to $b\in B$ assigns $f(b, x)$ is a continuous section of $p$. Its image X is therefore connected and open in E, since $p$ is etale (I, p. 30, Corollary 3). It is moreover closed under assumption a), since $p$ is separated (I, p. 27, Proposition 3). It is also closed under assumption b) by Corollary 1 of I, p. 80, since U is open and closed in $E'$. Consequently, X is a connected component of E.

Since $f|U: U\rightarrow X$ is bijective and open, it is a homeomorphism onto its image, which proves the lemma.

Let us retain the assumptions preceding the lemma and now prove assertion a). Let V be a connected component of E. It is a

set open and closed in E and $\overset{-1}{f}(V)$ is a union of connected components of $E'$ which $f$ maps homeomorphically onto V by the lemma. It follows from Proposition 6 of I, p. 79 that $(E', f)$ is a covering.

Let us prove b). Since $f$ is surjective, it follows from the lemma that every connected component V of E is the homeomorphic image of a connected component $U = B\times \{x\}$ of $E'$. The mapping $p$ then induces a homeomorphism of V onto B. By Proposition 6, E is a covering of B.

This therefore proves the proposition, under the additional assumption that B is connected and $E'$ a trivializable covering of B. Let us prove it in the general case.

There exists a covering $(U_i)_{i\in I}$ of B, consisting of connected open sets over which the covering $E'$ is trivializable. Let $i\in I$.

Let us prove a). If the mapping $p$ is etale and separated, the same is true of the mapping $p_{U_i}: U_i\times_BE\rightarrow U_i$, for $i\in I$, by Propositions 8 of I, p. 31 and 4 of I, p. 27. It therefore follows from the particular case treated that,

for every $i\in I$, the $U_i$-space ($(^-{p'}^1)(U_i), f_{U_i}$) is a covering. Let A denote the sum space of the $U_i$ and $q: A\rightarrow B$ the canonical mapping. Then the space $A\times_BE'$, endowed with the mapping $f_A: A\times_BE'\rightarrow A\times_BE$ is a covering of $A\times_BE$. The mapping $q$ admits a continuous section in a neighbourhood of every point of B and Proposition 3 of I, p. 72, applied to the cartesian square

$A\times_BE$ E

implies that $E'$ is a covering of E.

Let us prove b). Suppose that $f$ is surjective. Then, for every element $i$ of I, the mapping $f_{U_i}: U_i\times_BE'\rightarrow U_i\times_BE$ is surjective and the space $U_i\times_BE'$, endowed with the mapping $f_{U_i}$, is a covering of $U_i\times_BE$ (I, p. 71, Corollary 2 of Proposition 2). It follows from the particular case treated above that the $U_i$-space $(\overset{-1}{p}(U_i), p_{U_i})$ is a covering. Consequently, E is a covering of B.

Let B be a topological space, let E and $E'$ be coverings of B, and let $f: E\rightarrow E'$ be a B-morphism We have seen that $(E, f)$ is a covering under each of the two following assumptions: 1) the covering E is locally of finite degree (I, p. 76, Corollary 1) ; 2) the space B is locally connected (I, p. 81, Proposition 7). This phenomenon may be explained as follows.

Let F and $F'$ be discrete topological spaces and let $f: B\times F\rightarrow$ $B\times F'$ be a B-morphism. The mapping $\widetilde{f}:a\mapsto$ pr$_2\circ f(b,\cdot )$ of B into the space $\mathscr{C}_c(F; F')$ is continuous (TG, X, p. 28, th. 3). If U is an open set of B such that the mapping $\widetilde{f}$ is constant on U, the mapping $f_U: U\times$ $F\rightarrow U\times F'$ deduced from $f$ is a trivializable covering (I, p. 69, prop. 1).

The space $\mathscr{C}_c(F; F')$ is none other than the set $\mathscr{F}(F; F')$ of mappings of F into $F'$ endowed with the topology deduced from the topology of the product space $(F')^F$ by the canonical identification. For this topology, the space $\mathscr{F}(F; F')$ is totally disconnected (I, p. 84, prop. 10). Consequently, if the space B is connected, the mapping $\widetilde{f}$ is constant (I, p. 82, prop. 4); if the space B is locally connected, the mapping $\widetilde{f}$ is locally constant. When the set F is finite, the space $\mathscr{F}(F; F')$ is discrete and the mapping $\widetilde{f}$ is locally constant.

Let us denote by $p$ the projection of the B-space E. By the proposition, $(E', f)$ is a covering of E; its image $f(E')$ is open and closed in E. Let U denote the complement of $f(E')$ in E, so that the B-space $(U, p|U)$ is a covering of B (I, p. 80, corollary 1). If the mapping $f_b$ is surjective, the fibre at $b$ of this covering is empty. Since B is a connected space, U is then empty, hence $f$ is surjective.

The set V of points of E at which the fibre of $f$ has exactly one element is open and closed. The B-spaces $(V, p|V)$ and $(f(E'), p|f(E'))$ are coverings of B (loc. cit.) and the canonical mapping $i: V\rightarrow$ $f(E')$ is a B-morphism. If the mapping $f_b$ is injective, the mapping $i_b$ is surjective, hence $i$ is surjective by what precedes, and one has $V =f(E')$. Consequently, the mapping $f$ is injective.

Let us first prove a lemma.

Let $(W_j)_{j\in J}$ be an open covering of E; we shall prove that there exists an open locally finite covering $(A_k)_{k\in K}$ of E which is finer than the covering $(W_j)_{j\in J}$. For each $i\in I$, let $(A'_{\ell})_{\ell\in K_i}$ be a locally finite covering of $\overline{V_i}$ by open sets of $\overline{V_i}$, finer than the covering $(W_j\cap \overline{V_i})_{j\in J}$. Let K be the sum of the family $(K_i)_{i\in I}$ (E, II, p. 30, Def. 8). For each element $k= (\ell , i)$ of K, put $A_k= A'_{\ell}\cap V_i$. Then $A_k$ is open in E and one has $\bigcup_{k\in K}A_k=\bigcup_{i\in I}V_i= E$; moreover, for each $k\in K$, there exists an index $j\in J$ such that $A_k\subset W_j$. Thus, the family $(A_k)_{k\in K}$ is an open covering of E finer than $(W_j)_{j\in J}$. It remains to prove that the family $(A_k)_{k\in K}$ is locally finite. Let $x\in E$; there exists an open neighbourhood U of $x$ which meets $V_i$ only for $i$ belonging to a finite subset $I'$ of I. For each $i\in I',x$ has an open neighbourhood $U_i\subset U$ which meets only a finite number of the open sets $A_k$, for $k\in K_i\times \{i\}:$ this is obvious if $x$ does not belong to $\overline{V_i}$, and if $x$ belongs to $\overline{V_i}$, this follows from the property of local finiteness of the covering $(A'_{\ell})_{\ell\in K_i}$ of $\overline{V_i}$. It follows that $V'=\bigcap_{i\in I'}U_i$ is an open neighbourhood of $x$ which meets only a finite number of the $A_k,k\in K$, and the covering $(A_k)_{k\in K}$ is locally finite.

Let us prove the proposition. Let E be a covering of B, let $p$ denote its projection, and suppose that the space B is paracompact. Let $(A_i)_{i\in I}$ be an open locally finite covering of B such that, for each $i\in I$, the covering E is trivializable over $A_i$. For each $i\in I$, let $F_i$ be a discrete topological space and let $g_i:\overset{-1}{p}(A_i)\rightarrow A_i\times F_i$ be a trivialization of E over $A_i$. Let $(B_i)_{i\in I}$ be an open covering of B such that, for each $i\in I$, one has $\overline{B_i}\subset A_i$ (TG, IX, p. 49, Prop. 4 and p. 48, Cor. 1). For each $i\in I$, put $V_i=\overset{-1}{p}(B_i)$; one has $\overline{V_i}\subset \overset{-1}{p}(B_i)\subset \overset{-1}{p}(A_i)$ and $V_i=\overset{-1}{g_{i}}(B_i\times F_i)$, whence $\overline{V_i}\subset \overset{-1}{g_{i}}(B_i\times F_i)$. Since B is paracompact, $\overline{B_i}$ is paracompact (TG, I, p. 69, Prop. 16) and $\overline{B_i}\times F_i$ is paracompact (TG, I, p. 70, Prop. 18). It follows that $\overset{-1}{g_{i}}(B_i\times F_i)$ is paracompact, hence so is $\overline{V_i}$ (TG, I, p. 69, Prop. 16). The family $(V_i)_{i\in I}$ is, by construction, an open locally finite covering of E. Finally, the space E is separated (I, p. 26, Remark 3). It therefore satisfies the hypotheses of the lemma, whence the proposition.