(May 2013) Let \(X_\alpha\) be non-empty topological spaces and suppose that \(X=\prod_\alpha X_\alpha\) is endowed with the product topology.
Prove that each projection map \(\pi_\alpha\) is continuous and open.
Prove that \(X\) is Hausdorff if and only if each space \(X_\alpha\) is Hausdorff.
Solution
- A subbasis for the product topology consists of sets \[ \pi_\alpha^{-1}(U_\alpha), \] with \(U_\alpha\subseteq X_\alpha\) open. Hence every projection \(\pi_\alpha\) is continuous.
To prove openness, it suffices to consider a basic open set \[ B=\prod_\beta U_\beta, \] where each \(U_\beta\) is open and \(U_\beta=X_\beta\) except for finitely many indices. Since every factor is nonempty, \[ \pi_\alpha(B)=U_\alpha, \] which is open. Arbitrary open sets are unions of basic ones, and projections preserve unions, so \(\pi_\alpha\) is open.
- If every \(X_\alpha\) is Hausdorff and \(x\ne y\) in \(X\), then \(x_\alpha\ne y_\alpha\) for some \(\alpha\). Choose disjoint open neighborhoods \(U,V\subseteq X_\alpha\) of these coordinates. Then \[ \pi_\alpha^{-1}(U),\qquad\pi_\alpha^{-1}(V) \] are disjoint open neighborhoods of \(x,y\). Thus \(X\) is Hausdorff.
Conversely, suppose \(X\) is Hausdorff. Fix \(\alpha\) and choose one basepoint \(a_\beta\in X_\beta\) for every \(\beta\ne\alpha\). The map \[ s_\alpha:X_\alpha\to X, \qquad s_\alpha(x)_\alpha=x,\quad s_\alpha(x)_\beta=a_\beta\ (\beta\ne\alpha), \] is an embedding, with inverse on its image given by \(\pi_\alpha\). Hence \(X_\alpha\) is homeomorphic to a subspace of the Hausdorff space \(X\), and is therefore Hausdorff.