Continuity and Homeomorphism Questions and Answers

Answer: A

\(X\) is homeomorphic to \(Y\) in the first option: the cylinder \(x^{2} + y^{2} = 1\) in \(\mathbb{R}^{3}\) and the punctured plane \(z = 0\), \(x^{2} + y^{2} \neq 0\).

The cylinder is \(S^{1} \times \mathbb{R}\). In polar form the punctured plane is \(S^{1} \times (0, \infty)\), and \((0, \infty)\) is homeomorphic to \(\mathbb{R}\) by \(r \mapsto \ln r\). Explicitly, \((\cos\theta, \sin\theta, z) \mapsto (e^{z}\cos\theta, e^{z}\sin\theta, 0)\) is a homeomorphism.

The other pairs differ in a topological property. In the second, \(X\) is the topologist's sine curve, which is connected but not path connected, while \([0, 1]\) is path connected. In the third, \(X\) is the graph of a continuous function on \((0, 1]\), so it is homeomorphic to \((0, 1]\), which is not compact, while \([0, 1]\) is. In the fourth, the cone \(x^{2} + y^{2} = z^{2}\) without its vertex splits into the parts \(z \gt 0\) and \(z \lt 0\), so it is disconnected, while the cylinder is connected.

Answer: D

The map p is neither an open map nor a closed map.

p sends [0, 1/2) to 0 and [1/2, 1] to 1. In the quotient topology a set is open when its preimage is open in [0, 1]. The preimage of {0} is [0, 1/2), open in [0, 1]; the preimage of {1} is [1/2, 1], not open. So \(T_{2}\) = {∅, {0}, {0, 1}}, whose closed sets are {0, 1}, {1} and ∅.

Not open: (1/2, 1] is open in [0, 1] but its image {1} is not open. Not closed: [0, 1/4] is closed in [0, 1] but its image {0} is not closed.

So the options claiming p is open, closed, or both are all wrong.

Answer: C

The true statement is f is a homeomorphism if X is compact and Y is Hausdorff.

It suffices to show f is a closed map. A closed subset of the compact space X is compact; its continuous image is compact; and a compact subset of the Hausdorff space Y is closed. So f maps closed sets to closed sets, which makes \(f^{-1}\) continuous.

The other conditions are not enough. The identity from [0,1] with the usual topology onto [0,1] with the indiscrete topology is a continuous bijection between compact spaces that is not a homeomorphism; the same map shows that a Hausdorff domain with a compact codomain fails. The map \(t \mapsto (\cos t, \sin t)\) from \([0, 2\pi)\) onto the circle is a continuous bijection between Hausdorff spaces whose inverse is not continuous.

Answer: C

Y is homeomorphic to Z, but not to X.

Y, the square \(|x| + |y| = 1\), is a simple closed curve; radial projection \(p \mapsto p/|p|\) is a homeomorphism from it onto the circle Z.

X is the set where \(|x| = |y| \le 1\): the two diagonals of the square \([-1, 1]^{2}\), a cross meeting at the origin. Removing the origin splits X into four pieces, while removing any one point from a circle leaves it connected. A homeomorphism preserves such cut-point behaviour, so X is homeomorphic to neither Y nor Z.

Answer: A

The answer is X and Y are homeomorphic to Z.

Each set is a union of two disjoint open pieces, each homeomorphic to an open disc. X is two open unit discs centred at (0, 0) and (2, 0); they touch only at (1, 0), which belongs to neither, so they are disjoint. Y is two disjoint open discs centred at (0, 0) and (3, 0). Z is the open unit disc with the segment \(x = 0\) removed, leaving two open half-discs, and an open half-disc is homeomorphic to an open disc.

A homeomorphism can be built piece by piece, so all three spaces are homeomorphic. The distance between the pieces, or whether their closures touch, does not affect the topology of the subspaces themselves.

Answer: C

\(f([0, 1] \cap Q)\) is dense in S¹.

The map \(f(x) = e^{2\pi i x}\) is continuous from [0, 1] onto the circle. The rationals in [0, 1] are dense in [0, 1], and a continuous surjection carries a dense set onto a dense subset of the image, so the points \(e^{2\pi i q}\), the roots of unity, are dense in \(S^{1}\).

The image is countable, so it cannot be all of \(S^{1}\) or any closed set containing an arc; being dense but not everything, it is not closed. It is not open, because every arc of the circle contains points \(e^{2\pi i x}\) with \(x\) irrational, which are not in the image. And a countable subset of the circle with more than one point is not connected: removing two points with irrational angles, one on each arc between two of its points, separates it into two relatively open pieces.

Answer: D

The quotient is homeomorphic to the circle \(s^{1} = \{(x, y) : x^{2} + y^{2} = 1\}\).

The relation glues the two endpoints 0 and 1 together and leaves every other point alone. The map \(g(t) = (\cos 2\pi t, \sin 2\pi t)\) from [0, 1] onto the circle is continuous and identifies exactly 0 and 1. It therefore induces a continuous bijection from \(X/\sim\) onto \(S^{1}\). The quotient of a compact space is compact and the circle is Hausdorff, so this continuous bijection is a homeomorphism.

It is not [0, 1], [0, 1) or (0, 1): removing any single point from the circle leaves it connected, while removing an interior point from any interval disconnects it. [0, 1) and (0, 1) are also not compact, while the quotient is.

Answer: B

The true statement is S¹ is not homeomorphic to D.

D is the filled square, which is two-dimensional, while \(S^{1}\) is a curve. Removing two points from \(S^{1}\) disconnects it, but removing any two points from the filled square leaves it connected, so the two are not homeomorphic.

\(S^{1}\) is compact while R is not, so they are not homeomorphic. Removing one point leaves \(S^{1}\) connected, but removing an interior point disconnects any interval, so \(S^{1}\) is not homeomorphic to an interval. The square D is not homeomorphic to an interval either, by the same cut-point argument.

Answer: B

The homeomorphic pair is square and ellipse.

The boundary of a square and an ellipse are both simple closed curves. Radial projection from the centre gives a homeomorphism between them, so each is homeomorphic to the circle.

A parabola is homeomorphic to R, which is not compact, so it cannot match the compact square or ellipse. A hyperbola has two separate branches, so it is disconnected, while a parabola is connected; they are not homeomorphic either.

Answer: C

X is homeomorphic to Z but not to Y.

Z is the image of X under the reflection \((x, y) \mapsto (x, 1 - y)\), which sends [0, 1) to (0, 1] in the second coordinate. A reflection is a homeomorphism of the plane, so X and Z are homeomorphic.

X and Y differ in their edges. The edge points of X (where it is not locally like an open disc) form the two segments \(\{0\} \times [0, 1)\) and \([0, 1) \times \{0\}\), which meet at the corner, a connected set. The edge points of Y form \(\{0\} \times (0, 1)\) and \(\{1\} \times (0, 1)\), two disjoint segments. A homeomorphism carries edge points to edge points, so it would have to carry a connected set onto a disconnected one, which is impossible. Hence X is not homeomorphic to Y.