Proposition 6

Let X be a topological space. Let $U_1$ and $U_2$ be open ( resp. closed) subspaces of X such that $X = U_1\cup U_2$. Suppose that $U_1$ and $U_2$ are simply connected and that their intersection $U_1\cap U_2$ is connected and nonempty. Then the space X is simply connected.

Let $(E, p)$ be a covering of X and let $y$ be a point of E. It is enough to prove that there exists a continuous section $s$ of $p$ such that $s(p(y)) =y$ (I, p. 70, cor. 2 of prop. 1). Suppose for example that $p(y)$ belongs to $U_1$. There then exists a continuous section $s_1: U_1\rightarrow E$ of $p_{U_1}$ such that $s_1(p(y)) =y$. Let $x$ be a point of $U_1\cap U_2$; there exists a continuous section $s_2$ of $p_{U_2}$ such that $s_2(x) =s_1(x)$. By corollary 3 of proposition 11 of I, p. 34, there exists a continuous section $s$ of $p$ extending both $s_1$ and $s_2$.