Show that if \(Y\) is compact, then the projection \(\pi_1: X \times Y \to X\) is a closed map.
Solution
Goal: Prove that if \(Y\) is a compact topological space, then the canonical projection \(\pi_1: X \times Y \to X\) is a closed map (Kuratowski’s projection theorem / Tube Lemma).
1. Setting and reduction to complement:Let \(C \subseteq X \times Y\) be a closed set.We must show that \(\pi_1(C)\) is closed in \(X\), or equivalently, that \(X \setminus \pi_1(C)\) is open in \(X\).
2. Application of the Tube Lemma:Let \(x_0 \in X \setminus \pi_1(C)\).Then the entire slice \(\{x_0\} \times Y\) is contained in the open set \(W = (X \times Y) \setminus C\).There exists an open neighborhood \(U \subseteq X\) of \(x_0\) such that \(U \times Y \subseteq W\).Proof:
2.1. For each \(y \in Y\), \((x_0, y) \in W\).
2.2. By definition of the product topology, there exist open sets \(U_y \subseteq X\) and \(V_y \subseteq Y\) such that \((x_0, y) \in U_y \times V_y \subseteq W\).
2.3. The family \(\{V_y\}_{y \in Y}\) is an open cover of the compact space \(Y\).
2.4. Hence there exists a finite subcover \(\{V_{y_1}, \dots, V_{y_k}\}\) of \(Y\).
2.5. Define \(U = \bigcap_{i=1}^k U_{y_i}\). As a finite intersection of open neighborhoods of \(x_0\), \(U\) is an open neighborhood of \(x_0\) in \(X\).
2.6. For each \(i \in \{1, \dots, k\}\), \(U \times V_{y_i} \subseteq U_{y_i} \times V_{y_i} \subseteq W\).
2.7. Taking the union over \(i = 1, \dots, k\):\[U \times Y = U \times \left( \bigcup_{i=1}^k V_{y_i} \right) = \bigcup_{i=1}^k (U \times V_{y_i}) \subseteq W.\]
3. Openness of the complement:Proof:
3.1. By 2, \((U \times Y) \cap C = \varnothing\).
3.2. Projecting to \(X\), no point of \(U\) can be the first coordinate of any point in \(C\), so \(U \cap \pi_1(C) = \varnothing\).
3.3. Thus \(x_0 \in U \subseteq X \setminus \pi_1(C)\).
3.4. Since \(x_0\) was an arbitrary point of \(X \setminus \pi_1(C)\), the set \(X \setminus \pi_1(C)\) is open in \(X\).
4. Conclusion:\(\pi_1(C)\) is closed in \(X\), so \(\pi_1\) is a closed map. Q.E.D.