Lemma 5

a) Let $c= (U_0, . . . ,U_n)$ and $c'= (U'_0, . . . ,U'_m)$ be caterpillars such that $U_0= U'_0$ and $p(c) =p(c')$. Then $c=c'$.

b) Let U and $U'$ be elements of $\mathscr{U}$. Then there exists a caterpillar $c= (U_0, . . . ,U_n)$ such that $U_0= U$ and $U_n= U'$.

a) Necessarily, $m=n$. For every integer $i$ such that $0\leqslant i\leqslant n$, put $V_i=p(U_i)$ and let $s_i$ and $s'_i$ denote the continuous sections of $p$ over $V_i$ with respective images $U_i$ and $U'_i$. To prove the equality $c=c'$, we shall prove, by induction on $i$, the equality $s_i=s'_i$ for every integer $i$ such that $0\leqslant i\leqslant n$. By hypothesis, one has $s_0=s'_0$. Let $i$ be such that $1\leqslant i\leqslant n$ and $s_{i-1}=s'_{i-1}$. Since $U_{i-1}\cap U_i$ contains a point $x$, the continuous sections $s_{i-1}$ and $s_i$ coincide at $p(x)$, hence at every point of $V_{i-1}\cap V_i$, since it is connected (Corollary 1 of I, p. 34). The same is true of $s'_{i-1}$ and $s'_i$. For the same reason, $s_i$ and $s'_i$, which coincide on $V_{i-1}\cap V_i$, also coincide on $V_i$, whence the result.

b) If $x$ and $y$ are points of E, we say that a caterpillar $c= (U_0, . . . ,U_n)$ joins $x$ to $y$ if $x\in U_0$ and $y\in U_n$. In the set E, let R be the relation “there exists a caterpillar which joins $x$ to $y$”. The relation R is reflexive since $\mathscr{U}$ is a covering of E. It is obviously symmetric. Let us prove that it is transitive: let $x,y,z$ be three points of E, $(U_0, . . . ,U_n)$ and $(U'_0, . . . ,U'_m)$ caterpillars joining $x$ to $y$ and $y$ to $z$ respectively. Let $U\in \mathscr{U}$ be a neighbourhood of $y$ contained in $U_n\cap U'_0$; we have $p(U_n)\cap p(U) =p(U'_0)\cap p(U) =p(U)$, and, since U is connected , the sequence $(U_0, . . . ,U_n,U,U'_0, . . . ,U'_m)$ is a caterpillar which joins $x$ to $z$. The equivalence classes under R are open, hence also closed. Since E is connected, it follows that any two points of E are always joined by a caterpillar.

Let now U and $U'$ be elements of $\mathscr{U},x$ a point of U and $x'$ a point of $U'$. There exists a caterpillar $(U_2, . . . ,U_{n-2})$ joining $x$ to $x'$. Let $U_1$ and $U_{n-1}$ be open sets of $\mathscr{U}$ such that $x\in U_1,x'\in U_{n-1}$, $U_1\subset U\cap U_2,U_{n-1}\subset U'\cap U_{n-2}$. Put $U_0= U$, $U_n= U'$. Then the sequence $(U_0, . . . ,U_n)$ is a caterpillar.

Let us now prove the theorem. Let U be an element of $\mathscr{U}$ and let C be the set of caterpillars $(U_0, . . . ,U_n)$ such that $U_0= U$. The mapping of C into S which, to a caterpillar $c= (U_0, . . . ,U_n)$ of C, associates the sequence $p(c) = (p(U_0), . . . , p(U_n))$ is injective (lemma 5, a)), hence

(1) Card(C) $\leqslant$ Card(S).

The mapping of C into $\mathscr{U}$ which, to $c= (U_0, . . . ,U_n)\in C$, associates the open set $U_n$ is surjective by the second part of lemma 5, hence (2) Card($\mathscr{U}$)$\leqslant$ Card(C).

For every point $x$ of E, let us choose an open set $U_x$ of $\mathscr{U}$ such that $x\in U_x$. If $x$ and $y$ are distinct points of one and the same fibre $E_a$ of $p$, we have $U_x= U\not_y$ since $p|U_x$ is injective. Hence, for $a\in B$, (3) Card(E$_a$)$\leqslant$ Card($\mathscr{U}$).

Finally, if $\mathscr{V}$ is a finite set, the set S of finite sequences of elements of $\mathscr{V}$ is countable (E, III, p. 49, prop. 1), hence

(4) Card(S) $\leqslant$ Card($\mathbf{N}$).

If $\mathscr{V}$ is infinite, we have

(5) Card(S) = Card($\mathscr{V}$)$\leqslant$ Card($\mathscr{W}$)$^2=$ Card($\mathscr{W}$)

by the corollary of E, III, p. 50 and Corollary 1 of E, III, p. 49. The theorem follows from relations (1) to (5) above.