Lemma 10

Let $a,b$ be points of I such that $a\leqslant b$.

a) The set $F_{a,b}$ is closed in $\overset{-1}{p}(a)\times \overset{-1}{p}(b)$.

b) One has pr$_1(F_{a,b}) =\overset{-1}{p}(a)$ and pr$_2(F_{a,b}) =\overset{-1}{p}(b)$.

c) Let $c\in I$ such that $b\leqslant c$. If $(y, z)$ belongs to $F_{a,b}$ and $(z, t)$ belongs to $F_{b,c}$, then $(y, t)$ belongs to $F_{a,c}$.

The set $\overset{-1}{p}([a, b])$ is compact (TG, I, p. 77, Prop. 7). Hence, for a pair $(y, z)\in \overset{-1}{p}(a)\times \overset{-1}{p}(b)$ to belong to $F_{a,b}$, it is necessary and sufficient that every open and closed subset of $\overset{-1}{p}([a, b])$ which contains $y$ should contain $z$ (TG, II, p. 32, Prop. 6).

Put $Y =\overset{-1}{p}([a, b])$. For every open and closed subset U of Y, the set $((U\times U)\cup ((Y-U)\times (Y-U)))\cap (\overset{-1}{p}(a)\times \overset{-1}{p}(b))$ is closed in $\overset{-1}{p}(a)\times \overset{-1}{p}(b)$. The intersection of these sets is equal to $F_{a,b}$, whence a).

Let $y\in \overset{-1}{p}(a)$. Let $\mathscr{U}$ denote the set of open and closed neighbourhoods of $y$ in Y. The mapping of Y into $[a, b]$ deduced from $p$ by passing to subspaces is open and proper (I, p. 17, Prop. 8), hence also closed. It follows that, for every set U belonging to $\mathscr{U},p(U)$ is a nonempty open and closed subset of $[a, b]$; since the interval $[a, b]$ is connected, one has $p(U) = [a, b]$ and in particular $U\cap \overset{-1}{p}(b)=\not\emptyset$. Hence, $(U\cap \overset{-1}{p}(b))_{U\in\mathscr{U}}$ is a decreasing filtering family of nonempty closed subsets of the compact space $\overset{-1}{p}(b)$. The intersection of this family is not empty (TG, I, p. 59); let $z$ be one of its elements. One has $(y, z)\in F_{a,b}$. We have proved the relation pr$_1(F_{a,b}) =\overset{-1}{p}(a)$. The relation pr$_2(F_{a,b}) =\overset{-1}{p}(b)$ is deduced from it by replacing $p$, I$,a,b$ by $-p$, $-I,-b,-a$.

Under the hypotheses of c), the pair $(y, t)$ belongs to $\overset{-1}{p}(a)\times \overset{-1}{p}(c)$, the set $\{y, z\}$ is contained in a connected part C of $\overset{-1}{p}([a, b])$ and the set $\{z, t\}$ is contained in a connected part $C'$ of $\overset{-1}{p}([b, c])$. Then, $C\cup C'$ is a connected part of $\overset{-1}{p}([a, c])$ (TG, I, p. 81, prop. 2) and contains $\{y, t\}$, whence the relation $(y, t)\in F_{a,c}$.

Let us return to the proof of theorem 3. Each fibre of the mapping $p$ is compact (TG, I, p. 77, prop. 7). According to Tychonoff's theorem (TG, I, p. 63, th. 3), the product space $K =\prod_{a\in I}\overset{-1}{p}(a)$ is compact. Let $K'$ be the set of elements $(y_a)_{a\in I}$ of K such that $y_{p(x)}$ is equal to $x$ and such that one has $(y_a, y_b)\in F_{a,b}$ for every pair $(a, b)$ of elements of I such that $a < b$. Theorem 3 follows from the following lemma.