Lemma 11

a) The set $K'$ is not empty.

b) Let $(s_a)_{a\in I}$ be an element of $K'$. The mapping $s:a\mapsto s_a$ from I into X is a continuous section of $p$ such that $s(p(x)) =x$.

For every finite subset S of I containing the point $p(x)$, denote by $K_S$ the set of elements $(y_a)_{a\in I}$ of K satisfying the relation $y_{p(x)}=x$ and the relations $(y_a, y_b)\in F_{a,b}$ for every pair $(a, b)$ of elements of S such that $a < b$. The sets $K_S$ are closed in K (lemma 10, a)) and form a decreasing filtering family of subsets of K, whose intersection is $K'$. To prove that $K'$ is not empty, it is enough to prove that, for every finite subset S of I containing the point $p(x)$, the set $K_S$ is not empty (TG, I, p. 59).

Let S be such a subset; order its elements in a strictly increasing sequence $(a_1, . . . , a_n)$ and denote by $i$ the integer such that $p(x) =a_i$. Put $y_{a_i}=x$. Lemma 10, b), permits us to construct by induction elements $y_{a_j}\in \overset{-1}{p}(a_j)$, for $i < j\leqslant n$, and by descending induction elements $y_{a_j}\in \overset{-1}{p}(a_j)$ for $1\leqslant j < i$, in such a way that one has $(y_{a_j}, y_{a_{j+1}})\in F_{a_j,a_{j+1}}$ for every integer $j$ such that $1\leqslant j < n$. According to lemma 10, c), one has $(y_a, y_b)\in F_{a,b}$ for every pair $(a, b)$ of elements of S such that $a < b$. Since the mapping $p$ is surjective, we can choose for every $a\in I-S$ an element $y_a\in \overset{-1}{p}(a)$. The family $(y_a)_{a\in I}$ thus constructed belongs to $K_S$, hence $K_S$ is not empty.

Let us prove b). By definition of $K',s$ is a section of $p$ such that $s(p(x)) =x$. Let $a\in I$; let us prove the continuity of $s$ at the point $a$. Let U be an open neighbourhood of $s_a$ in X. Since $\overset{-1}{p}(a)$ is a compact space (TG, I, p. 77, prop. 7) and totally disconnected, $s_a$ possesses in $\overset{-1}{p}(a)$ an open and closed neighbourhood C, contained in U (TG, II, p. 32, corollary). The sets C and $\overset{-1}{p}(a)-C$ are closed in $\overset{-1}{p}(a)$, hence compact, and they are disjoint. Since X is separated, they possess in X open and disjoint neighbourhoods V and $V'$ (TG, I, p. 61, prop. 3). The set $(V\cap U)\cup V'$ is an open neighbourhood of the fibre $\overset{-1}{p}(a)$ in X; since the mapping $p$ is closed (TG, I, p. 72, prop. 1), $(V\cap U)\cup V'$ contains a set of the form $\overset{-1}{p}(J)$, where $J\subset I$ is an open interval containing $a($I, p. 75, lemma). Put $W = V\cap U\cap \overset{-1}{p}(J)$. The set W is open in $\overset{-1}{p}(J)$; it is also closed in $\overset{-1}{p}(J)$ since one has $\overset{-1}{p}(J)-W = V'\cap \overset{-1}{p}(J)$. Let $b\in J$. The closed interval of I with endpoints $a$ and $b$ is contained in J. By assumption, $(s_a, s_b)$ belongs to $F_{a,b}$ if $a\leqslant b$ and $(s_b, s_a)$ belongs to $F_{b,a}$ if $b\leqslant a$. There therefore exists a connected subset of $\overset{-1}{p}(J)$ containing $\{s_a, s_b\}$. Consequently, the point $s_b$ belongs to every open and closed subset of $\overset{-1}{p}(J)$ which contains $s_a$, hence in particular to W; fortiori, $s_b$ belongs to U. We therefore have $s(J)\subset U$, which proves the continuity of $s$ at the point $a$.