Proposition 4

The product of two simply connected spaces, one of which is locally connected, is a simply connected space.

Let X and Y be simply connected spaces, and suppose that Y is locally connected. The space $X\times Y$ is connected, since X and Y are so (I, p. 124). Let $(Z, f)$ be a nonempty covering of $X\times Y$; let us prove that it is trivializable. By corollary 2 of I, p. 70, it is enough to prove that for every point $z_0$ of Z, there exists a continuous section $s$ of $f$ such that $s(f(z_0)) =z_0$. Let therefore $z_0$ be a point of Z. Put $(x_0, y_0) =f(z_0)$. The subspace $X\times \{y_0\}$ of $X\times Y$ is homeomorphic to X. Hence the covering deduced from Z over $X\times \{y_0\}$ is trivializable and has a continuous section $\sigma$ such that $\sigma (x_0, y_0) =z_0$. Analogously, for every point $x$ of X, there exists a continuous section $\tau_x$ of $f$ over $\{x\}\times Y$ such that $\tau_x(x, y_0) =\sigma (x, y_0)$. For $(x, y)\in X\times Y$, put $s(x, y) =\tau_x(x, y)$. The mapping $s$ is a section of $f$, and one has $s(x_0, y_0) =z_0$. Since the space Y is connected and locally connected, it follows from theorem 1 of I, p. 35 that the mapping $s$ is continuous.