Questions: Homeomorphisms and Topological Equivalence

Question 1

The map f: [0,1) → S¹ defined by f(t) = (cos 2πt, sin 2πt) is a continuous bijection. Is it a homeomorphism?

• A. Yes — any continuous bijection between topological spaces is a homeomorphism
• B. No — the map is not continuous at t = 0
• C. No — the inverse f⁻¹ is not continuous: a small open arc near the seam of S¹ pulls back to a disconnected set near both 0 and 1 in [0,1), so open sets in S¹ do not correspond to open sets in [0,1)
• D. No — bijectivity requires both spaces to have the same cardinality, which fails here

Answer: (C)

f is continuous and bijective, but f⁻¹ fails to be continuous at the identification point. A small open arc near the 'seam' of S¹ pulls back to a set near t = 0 and t approaching 1 in [0,1) — a disconnected preimage. A homeomorphism requires both f and f⁻¹ to be continuous, so f is not a homeomorphism. The key lesson: continuous bijection ≠ homeomorphism.

Question 2

To show that [0,1] and S¹ are not homeomorphic, a topologist removes a point from each space and compares the results. What does this argument show?

• A. Removing an interior point from [0,1] disconnects it, while removing any point from S¹ leaves it connected — so they cannot be homeomorphic
• B. Removing a point from S¹ always creates a disconnected arc
• C. Removing a point from [0,1] always leaves a connected space since intervals are path-connected
• D. The argument only works if you remove corresponding points from both spaces

Answer: (A)

Removing an interior point from [0,1] splits it into two disjoint open intervals — the space becomes disconnected. But removing any single point from S¹ leaves a connected arc. Since homeomorphisms preserve connectedness, and the 'remove a point' operation yields different connectivity results, the two spaces cannot be homeomorphic. This is the template: find a topological invariant they disagree on.

Question 3

A continuous bijection f: X → Y is generally a homeomorphism.

• T. True
• F. False

Answer: False

A homeomorphism requires both f and f⁻¹ to be continuous. A continuous bijection can fail to be a homeomorphism when the inverse is discontinuous. The standard example: f: [0,1) → S¹ wrapping the interval onto the circle is continuous and bijective, but the inverse is not continuous at the identification point. Continuity of f and bijectivity together do not force f⁻¹ to be continuous.

Question 4

If f: X → Y is a homeomorphism, then X is compact if and only if Y is compact.

• T. True
• F. False

Answer: True

Compactness is a topological property — preserved by homeomorphisms in both directions. The continuous image of a compact space under f is compact (so Y is compact if X is), and the continuous image under f⁻¹ is compact (so X is compact if Y is). Any property defined purely in terms of the topology is preserved by homeomorphisms.

Question 5

Why must the inverse of a homeomorphism also be continuous — what goes wrong if we only require f to be continuous and bijective?

Model answer: A homeomorphism is supposed to establish that X and Y have identical topological structure — the same open sets, just relabeled. Continuity of f means open sets in Y pull back to open sets in X. Without continuity of f⁻¹, open sets in X need not push forward to open sets in Y, so the topology of X and Y could be entirely different despite f being continuous. A bijection with only one-way continuity can collapse topological structure rather than preserve it.

This is why homeomorphism is defined as a continuous bijection with continuous inverse: equivalence requires the relationship to be symmetric, which is what continuity of f⁻¹ provides.