Let \(X=\prod_{\mu\in M}X_\mu\) and \(Y=\prod_{\mu\in M}Y_\mu\) be the Cartesian products of the topological spaces \((X_\mu)_{\mu\in M}\) and \((Y_\mu)_{\mu\in M}\) and let \(X\) and \(Y\) have the product topologies, respectively. Prove that if for each \(\mu\in M\) the maps \(f_\mu:X_\mu\to Y_\mu\) are continuous then \(f:X\to Y\) defined by \(f(x)_\mu=f_\mu(x_\mu)\) is continuous.
Solution
For each \(\mu\in M\), write \[ p_\mu:X\to X_\mu, \qquad q_\mu:Y\to Y_\mu \] for the coordinate projections.
1. For every \(\mu\in M\) and every open set \(V\subseteq Y_\mu\),\[ f^{-1}\bigl(q_\mu^{-1}(V)\bigr) =p_\mu^{-1}\bigl(f_\mu^{-1}(V)\bigr). \]
For \(x\in X\), \[ x\in f^{-1}\bigl(q_\mu^{-1}(V)\bigr) \] if and only if \[ q_\mu(f(x))\in V. \] By the definition of \(f\), \[ q_\mu(f(x))=f(x)_\mu=f_\mu(x_\mu)=f_\mu(p_\mu(x)). \] Thus the preceding condition is equivalent to \[ p_\mu(x)\in f_\mu^{-1}(V), \] which is exactly \[ x\in p_\mu^{-1}\bigl(f_\mu^{-1}(V)\bigr). \]
2. The inverse image under \(f\) of every subbasic open set of \(Y\) is open in \(X\).
The product topology on \(Y\) has subbasis \[ \mathcal S =\{q_\mu^{-1}(V):\mu\in M,\ V\subseteq Y_\mu\text{ open}\}. \] Take \(q_\mu^{-1}(V)\in\mathcal S\). Because \(f_\mu\) is continuous, \[ f_\mu^{-1}(V) \] is open in \(X_\mu\). Because \(X\) has the product topology, the coordinate projection \(p_\mu\) is continuous, so \[ p_\mu^{-1}\bigl(f_\mu^{-1}(V)\bigr) \] is open in \(X\). By 1 this set is precisely \[ f^{-1}\bigl(q_\mu^{-1}(V)\bigr). \]
3. Therefore \(f:X\to Y\) is continuous.
Every open set in the topology generated by the subbasis \(\mathcal S\) is a union of finite intersections of members of \(\mathcal S\). Inverse images commute with arbitrary unions and finite intersections. By 2, the inverse image of each member of \(\mathcal S\) is open in \(X\), so the inverse image of every open subset of \(Y\) is open in \(X\). Hence \(f\) is continuous.