Discussion
Let \(f : X \to Y\) be a continuous map from a Hausdorff topological space \(X\) to a metric space \(Y\). Consider the following two statements:
P: \(f\) is a closed map and the inverse image \(f^{-1}(y) = \{x \in X : f(x) = y\}\) is compact for each \(y \in Y\).
Q: For every compact subset \(K \subset Y\), the inverse image \(f^{-1}(K)\) is a compact subset of \(X\).
Which one of the following is true?
Answer: C
P and Q are equivalent.
P implies Q: a closed continuous map with compact fibres (a perfect map) pulls compact sets back to compact sets. This is a standard theorem and needs no condition on \(Y\) beyond what is given.
Q implies P: the fibres \(f^{-1}(y)\) are compact because \(\{y\}\) is compact. For closedness, let \(C\) be closed in \(X\) and \(f(c_{n}) \to y\) with \(c_{n} \in C\); sequences suffice because \(Y\) is a metric space. The set \(K = \{f(c_{n})\} \cup \{y\}\) is compact, so the \(c_{n}\) lie in the compact set \(f^{-1}(K) \cap C\) and have a cluster point \(c \in C\). By continuity \(f(c)\) is a cluster point of \(f(c_{n})\), so \(f(c) = y\) and \(y \in f(C)\). Hence \(f\) is closed.
Since each statement implies the other, the options asserting a one-way implication or none are wrong.
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