Question

Let \(X\) and \(Y\) be topological spaces and \(f: X \to Y\) be a continuous and bijective mapping. Which one of the following statements is TRUE?

• A. \(f\) is a homeomorphism if \(X\) and \(Y\) are compact.
• B. \(f\) is a homeomorphism if \(X\) is Hausdorff and \(Y\) is compact.
• C. \(f\) is a homeomorphism if \(X\) is compact and \(Y\) is Hausdorff.
• D. \(f\) is a homeomorphism if \(X\) and \(Y\) are Hausdorff.

Hint:

Remember the standard theorem: a continuous bijection from a compact space onto a Hausdorff space is always a homeomorphism, because compactness of the domain plus Hausdorff-ness of the codomain force the map to be closed.

Answer 1

The Correct Option is C

Step 1: Understanding the Question.
A continuous bijection \(f: X \to Y\) is a homeomorphism only when its inverse \(f^{-1}: Y \to X\) is also continuous. A continuous bijection need not automatically have a continuous inverse, so we need the extra condition on \(X\) and \(Y\) that forces \(f^{-1}\) to be continuous.

Step 2: Recall the key theorem.
\(f^{-1}\) is continuous exactly when \(f\) is a closed map, that is, \(f\) sends closed sets of \(X\) to closed sets of \(Y\). The standard theorem is: if \(X\) is compact and \(Y\) is Hausdorff, then every continuous bijection \(f: X \to Y\) is automatically a homeomorphism.

Step 3: Prove why compact X plus Hausdorff Y works.
Take any closed set \(C \subseteq X\). Since \(X\) is compact, \(C\) is a closed subset of a compact space, so \(C\) itself is compact. A continuous image of a compact set is compact, so \(f(C)\) is compact in \(Y\). Since \(Y\) is Hausdorff, every compact subset of \(Y\) is closed. So \(f(C)\) is closed in \(Y\). This shows \(f\) sends closed sets to closed sets, so \(f\) is a closed map, so \(f^{-1}\) is continuous, and \(f\) is a homeomorphism. This matches option (C).

Step 4: Why the other combinations fail.
Option (D), both \(X\) and \(Y\) Hausdorff with no compactness anywhere, is not enough: take \(X = [0, 2\pi)\) with the usual topology (Hausdorff but not compact) and \(Y\) the unit circle \(S^1\) (Hausdorff), with \(f(t) = (\cos t, \sin t)\). This \(f\) is continuous and bijective, but \(f^{-1}\) is not continuous at \((1,0)\), since points on \(S^1\) just below angle \(0\) come from \(t\) near \(2\pi\), far from \(t=0\) in \([0,2\pi)\). So \(f\) is not a homeomorphism, even though both spaces are Hausdorff, ruling out (D). The same example has \(Y\) compact and \(X\) Hausdorff but not compact, so it also rules out option (B), since (B) needs \(X\) compact, not just Hausdorff.
Option (A) drops the Hausdorff requirement on \(Y\) and keeps only compactness on both sides, which is also not enough, since the proof in Step 3 genuinely needs \(Y\) Hausdorff. For instance, the identity map from \([0,1]\) with the usual topology to \([0,1]\) with the cofinite topology (in which every infinite space is compact but not Hausdorff) is a continuous bijection between two compact spaces, yet it is not a homeomorphism, since the cofinite topology is not Hausdorff while the usual topology is. This rules out (A).

Final Answer:
The correct requirement is \(X\) compact and \(Y\) Hausdorff, which is option (C).