Corollary 1

Let Y be a regular space whose topology has a countable base (*). Let X be a connected and locally connected space, and let p : X \to Y be a continuous mapping with the following property: for each x \in X there is a closed neighbourhood V of x in X such that the restriction of p to V is a homeomorphism of V onto a closed subspace of Y. Then X is regular and the topology of X has a countable base.

First, the hypotheses imply that X is regular (§ 8, no. 4, Proposition 13). Let us show that the conditions of Theorem 1 are satisfied if we take \mathfrak{B} to be the set of all closed subsets V of X such that the restriction of p to V is a homeomorphism of V onto a closed subspace of Y. By hypothesis, the interiors of the sets of \mathfrak{B} cover Y and, by virtue of the assumption on Y, each V \in \mathfrak{B} has a countable base and therefore contains a countable dense subset (§ 1, no. 6, Proposition 6). Furthermore, if x \in p^{-1}(y), there is a neighbourhood V \in \mathfrak{B} of x in X such that V con-

() \ It can be shown that these conditions imply that the topology of Y is metrizable (Chapter IX, § 4, Exercise 22). \* tains no point of $\overline{p}^{-1}(y)$ other than $x$, and hence $\overline{p}^{-1}(y)$ is a discrete space. We may therefore apply Theorem 1, which shows that $X$ is the union of a countable family $(T_n)_{n \geq 0}$ of open sets, such that each subspace $T_n$ has a countable base $(U_{mn})_{m \geq 0}$. Then the $U_{mn}$ ($m \geq 0, n \geq 0$) form a base of the topology of $X$ ($§ 3$, no. 1, Remark).