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

This question tests when a continuous bijection between topological spaces is guaranteed to be a homeomorphism. A homeomorphism needs both $f$ and $f^{-1}$ continuous, and continuity of $f^{-1}$ is the same as $f$ being a closed map (closed sets go to closed sets). Let's go through each option.

1. X and Y compact: This is not enough on its own. Compactness of $X$ lets us say a closed (hence compact) subset of $X$ maps to a compact subset of $Y$, but without $Y$ Hausdorff, a compact subset of $Y$ need not be closed. So $f$ need not be a closed map, and this option can fail.
2. X Hausdorff and Y compact: Hausdorff-ness of $X$ says nothing about images of closed sets in $Y$, and compactness of $Y$ alone does not help either, since we need compactness on the domain side. The classic map from $[0,2\pi)$ onto the circle $S^1$ (continuous, bijective, $X$ Hausdorff, $Y$ compact) fails to be a homeomorphism, so this option is false.
3. X compact and Y Hausdorff: Take any closed set $C$ in $X$. Since $X$ is compact, $C$ is compact. Continuous maps send compact sets to compact sets, so $f(C)$ is compact in $Y$. Since $Y$ is Hausdorff, compact subsets of Hausdorff spaces are always closed, so $f(C)$ is closed. This makes $f$ a closed map, so $f^{-1}$ is continuous, and $f$ is a homeomorphism. This option is always true.
4. X and Y both Hausdorff: Hausdorff-ness alone never forces compactness anywhere, so the closed-map argument has no compact set to work with. The same $[0,2\pi) \to S^1$ map is a counterexample here too, since both spaces are Hausdorff but $f$ is not a homeomorphism. This option is false.

Only the combination "domain compact, codomain Hausdorff" is strong enough to guarantee a continuous bijection is a homeomorphism, so the correct choice is option (C).

Let's summarize:

• A continuous bijection is a homeomorphism exactly when it is also a closed map.
• Compactness of the domain plus the Hausdorff property on the codomain is exactly the condition that forces every continuous bijection to be closed.
• Swapping which space is compact and which is Hausdorff, or dropping compactness altogether, breaks the argument, as the $[0,2\pi) \to S^1$ example shows.

So the correct answer is option (C): $f$ is a homeomorphism if $X$ is compact and $Y$ is Hausdorff.