Questions: Topological Invariants

Question 1

You compute that spaces X and Y are both compact, path-connected, and have trivial fundamental group. What can you conclude?

• A. X and Y are homeomorphic — they share all the key invariants
• B. X and Y are not homeomorphic — identical invariants on distinct spaces imply a contradiction
• C. Sharing these invariants is insufficient to establish homeomorphism; they may still differ on finer invariants
• D. X and Y are homotopy equivalent but not necessarily homeomorphic, and homotopy equivalence implies homeomorphism

Answer: (C)

Topological invariants work asymmetrically: a single differing invariant proves non-homeomorphism, but no finite collection of shared invariants proves homeomorphism. The disk D² and the sphere S² are both compact, path-connected, and simply connected (trivial fundamental group), yet they are not homeomorphic. Proving homeomorphism requires constructing an explicit homeomorphism, not just accumulating invariant evidence.

Question 2

To prove that the real line ℝ is not homeomorphic to the plane ℝ², the most direct argument is:

• A. Show that ℝ is simply connected while ℝ² is not
• B. Remove a single point from each: the punctured ℝ is disconnected, but the punctured ℝ² remains path-connected
• C. Compute the fundamental groups: π₁(ℝ) = ℤ while π₁(ℝ²) = 0
• D. Show that ℝ has Euler characteristic 1 while ℝ² has Euler characteristic 2

Answer: (B)

Removing a single point from ℝ leaves two disjoint open rays — the space is disconnected. Removing a single point from ℝ² leaves a punctured plane, which is still path-connected (you can route any path around the missing point). Since connectedness after point removal is a topological invariant (it depends only on the homeomorphism type), ℝ and ℝ² cannot be homeomorphic. This elegant argument requires no algebraic invariants. Note: both ℝ and ℝ² are simply connected with trivial fundamental group, so option C is actually wrong.

Question 3

If two topological spaces share the same fundamental group, they is expected to be homeomorphic.

• T. True
• F. False

Answer: False

Invariants establish necessary conditions for homeomorphism, not sufficient ones. Two spaces can share their fundamental group — and even all their homotopy groups — while still failing to be homeomorphic. A single differing invariant rules out homeomorphism; shared invariants only tell you the spaces might be homeomorphic. Proving homeomorphism requires constructing an explicit bijective continuous map with continuous inverse.

Question 4

The open interval (0,1) and the closed interval [0,1] are not homeomorphic because (0,1) is unbounded while [0,1] is bounded.

• T. True
• F. False

Answer: False

Boundedness is not a topological invariant — it depends on how the space is embedded in ℝ, which is not preserved by homeomorphisms. The correct reason is that [0,1] is compact (every open cover has a finite subcover) and (0,1) is not compact. Compactness IS a topological invariant because homeomorphisms preserve open sets in both directions and hence preserve the finite subcover property.

Question 5

Explain the asymmetry in using topological invariants: why can a single differing invariant immediately prove non-homeomorphism, yet no finite list of shared invariants proves homeomorphism?

Model answer: A homeomorphism must preserve every topological property without exception. If spaces differ on even one invariant, no homeomorphism can exist. But homeomorphism is stronger than sharing invariants: two spaces might agree on every invariant we have checked and still differ on one we haven't computed. Proving homeomorphism requires constructing the map itself.

This asymmetry is why topological classification is hard in the positive direction but often easy in the negative direction. Counterexample: the closed disk and the sphere S² agree on compactness, path-connectedness, and simple connectedness, yet are not homeomorphic (the disk has a boundary, the sphere does not). 'Boundary' is another topological invariant that distinguishes them. The lesson: every non-homeomorphism proof needs only one witness; every homeomorphism proof needs a complete construction.