Lemma

Let X and Y be topological spaces. In order that a mapping $f: X\rightarrow Y$ be closed, it is necessary and sufficient that for every point $y$ of Y and every neighbourhood W of the fibre $\overset{-1}{f}(y)$, there exist a neighbourhood V

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).