Proposition 5

A connected and locally connected topological space such that the intersection of any two connected open subsets is connected is a simply connected space.

Let B be such a topological space. Let $(E, p)$ be a covering of B, let $x$ be a point of E and write $b=p(x)$. We have to prove that there exists a continuous section $s$ of $p$ such that $s(b) =x$ (cor. 2, I, p. 70). Let $\mathscr{S}$ be the set of pairs $(U, s_U)$ where U is a connected open subset of B containing $b$ and $s_U$ a continuous section of $p_U:\overset{-1}{p}(U)\rightarrow U$ such that $s_U(b) =x$. The set $\mathscr{S}$ is not empty ( I, p. 34, prop. 10). Let $(U, s_U)$ and $(V, s_V)$ be elements of $\mathscr{S}$. Then $s_U|U\cap V$ and $s_V|U\cap V$ are continuous sections of $p_{U\cap V}$ taking the value $x$ at $b$. By hypothesis, $U\cap V$ is connected; hence $s_U|U\cap V =s_V|U\cap V$ (I, p. 34, cor. 1 of prop. 11). Let A be the union of the open sets U when $(U, s_U)$ ranges over $\mathscr{S}$ and let $s: A\rightarrow E$ be the unique mapping such that $s|U =s_U$ for every pair $(U, s_U)\in \mathscr{S}$. The set A is open and connected (TG, I, p. 81, prop. 2), it contains $b$, and $s$ is a continuous section of $p_A$ such that $s(b) =x$. It now suffices to prove that the set A is closed, which will imply that it is equal to B.

Let $a$ be a point of B adherent to A and let V be a connected open neighbourhood of $a$ such that the covering E is trivializable over V. There exists a point $c$ in $A\cap V$ and a continuous section $s_V$ of $p_V$ such that $s_V(c) =s(c)$. Let $A'$ be the open set $A\cup V$. Since $A\cap V$ is connected, there exists a continuous section $s'$ of $p_{A'}$ extending $s$ and $s_V($I, p. 35, cor. 3 of prop. 11); the pair $(A', s')$ belongs to $\mathscr{S}$ and therefore $A'$ is contained in A. Hence $a$ belongs to A and A is closed.