(Jan ’08 # B8) Suppose that \(X\) is an arbitrary topological space and \(Y\) is a compact space. Consider the projection map \(\pi:X\times Y\to X\) defined by \(\pi(x,y)=x\). Prove that if \(X\times Y\) has the product topology, then \(\pi\) is a closed map.
Solution
Let \(C\subseteq X\times Y\) be closed. We show \(\pi(C)\) is closed. Take \(x_0\notin\pi(C)\). Then \[ \{x_0\}\times Y\subseteq (X\times Y)\setminus C. \] For each \(y\in Y\), choose basic open sets \(U_y\subseteq X\), \(V_y\subseteq Y\) such that \[ (x_0,y)\in U_y\times V_y\subseteq (X\times Y)\setminus C. \] The sets \(V_y\) cover the compact space \(Y\), so choose \(y_1,\dots,y_n\) with \[ Y=V_{y_1}\cup\cdots\cup V_{y_n}. \] Set \(U=U_{y_1}\cap\cdots\cap U_{y_n}\). Then \(U\) is an open neighborhood of \(x_0\), and for every \(x\in U\) and \(y\in Y\), some \(i\) has \(y\in V_{y_i}\), hence \[ (x,y)\in U_{y_i}\times V_{y_i}\subseteq (X\times Y)\setminus C. \] Thus \(U\cap\pi(C)=\varnothing\). Therefore \(X\setminus\pi(C)\) is open, so \(\pi(C)\) is closed.