A map \(f: X \to Y\) is said to be an open map if for every open set \(U\) of \(X\), the set \(f(U)\) is open in \(Y\). Show that \(\pi_1: X \times Y \to X\) and \(\pi_2: X \times Y \to Y\) are open maps.
Solution
Let \(U\subseteq X\times Y\) be open. For every \((x,y)\in U\) there is a basic product \(V_{x,y}\times W_{x,y}\) with \[ (x,y)\in V_{x,y}\times W_{x,y}\subseteq U. \] Then \[ \pi_1(U)=\bigcup_{(x,y)\in U}V_{x,y}, \] which is open in \(X\). Thus \(\pi_1\) is open. The same argument, interchanging the factors, gives \[ \pi_2(U)=\bigcup_{(x,y)\in U}W_{x,y}, \] so \(\pi_2\) is open as well.