Suppose that \(X\) is compact, \(Y\) is Hausdorff, \(f : X \to Y\) is a continuous surjective map, and \(g : Y \to Z\) is a function. Show that if \(g \circ f : X \to Z\) is continuous, then \(g\) is continuous.
Solution
1. The map\[ f:X\longrightarrow Y \]is closed.
Let \(A\subseteq X\) be closed. Since \(X\) is compact, \(A\) is compact. Continuity of \(f\) implies that \[ f(A) \] is compact in \(Y\). Because \(Y\) is Hausdorff, compact subsets of \(Y\) are closed. Hence \(f(A)\) is closed, and therefore \(f\) is a closed map.
2. The map \(f\) is a quotient map.
The map \(f\) is continuous and surjective by hypothesis. It remains to verify the quotient condition.
Let \(V\subseteq Y\) and suppose \[ f^{-1}(V) \] is closed in \(X\). By 1, \[ f\bigl(f^{-1}(V)\bigr) \] is closed in \(Y\). Surjectivity gives \[ f\bigl(f^{-1}(V)\bigr)=V. \] Thus \(V\) is closed. Equivalently, a subset of \(Y\) is open exactly when its inverse image under \(f\) is open, so \(f\) is a quotient map.
3. For every closed set \(F\subseteq Z\), the inverse image\[ g^{-1}(F) \]is closed in \(Y\).
Let \(F\subseteq Z\) be closed. Since \(g\circ f\) is continuous, \[ (g\circ f)^{-1}(F) = f^{-1}\bigl(g^{-1}(F)\bigr) \] is closed in \(X\). By 1, its image under \(f\) is closed in \(Y\). Using surjectivity of \(f\), \[ f\left(f^{-1}\bigl(g^{-1}(F)\bigr)\right) = g^{-1}(F). \] Hence \(g^{-1}(F)\) is closed.
4. Therefore\[ g:Y\longrightarrow Z \]is continuous.
By 3, the inverse image under \(g\) of every closed subset of \(Z\) is closed in \(Y\). This is the closed-set characterization of continuity.