Lemma

If U is a connected component of $E'$, the restriction of $f$ to U induces a homeomorphism of U onto a connected component of E.

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.