A quotient of a compact space is compact

Let \(X\) be a compact topological space, and let \(q: X \to Y\) be a quotient map (so \(Y\) has the quotient topology induced by the surjective map \(q\)).

1. Continuity of the quotient map: By the definition of the quotient topology, a subset \(V \subseteq Y\) is open in \(Y\) if and only if \(q^{-1}(V)\) is open in \(X\). In particular, for every open set \(V \subseteq Y\), the preimage \(q^{-1}(V)\) is open in \(X\), which means the quotient map \(q: X \to Y\) is continuous and surjective.

2. Continuous image of a compact space is compact: Let \(\mathcal{U} = \{V_\alpha\}_{\alpha \in A}\) be an open cover of \(Y\): \[ Y = \bigcup_{\alpha \in A} V_\alpha. \] Since \(q\) is continuous, for each \(\alpha \in A\), the preimage \(U_\alpha = q^{-1}(V_\alpha)\) is open in \(X\). Since \(q\) is surjective, the preimages cover \(X\): \[ X = q^{-1}(Y) = q^{-1}\left( \bigcup_{\alpha \in A} V_\alpha \right) = \bigcup_{\alpha \in A} q^{-1}(V_\alpha) = \bigcup_{\alpha \in A} U_\alpha. \] Thus, \(\{U_\alpha\}_{\alpha \in A}\) is an open cover of the compact space \(X\).

3. Extracting a finite subcover: Since \(X\) is compact, there exists a finite subcover \(\{U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_k}\}\) of \(X\), so: \[ X = \bigcup_{i=1}^k U_{\alpha_i} = \bigcup_{i=1}^k q^{-1}(V_{\alpha_i}). \] Applying the surjective map \(q\) to both sides: \[ Y = q(X) = q\left( \bigcup_{i=1}^k q^{-1}(V_{\alpha_i}) \right) = \bigcup_{i=1}^k q(q^{-1}(V_{\alpha_i})) = \bigcup_{i=1}^k V_{\alpha_i}. \] Thus, \(\{V_{\alpha_1}, V_{\alpha_2}, \ldots, V_{\alpha_k}\}\) is a finite subcover of \(Y\).

Therefore, the quotient space \(Y\) is compact.