Continuity and Homeomorphism Questions and Answers

Q: Let \(p : ([0,1], T_{1}) \to (\{0,1\}, T_{2})\) be the quotient map, arising from the characteristic function on \([\frac{1}{2}, 1]\), where \(T_{1}\) is the subspace topology induced by the Euclidean topology on \(\mathbb{R}\). Which of the following statements is TRUE?

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.

Q: 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?

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.

Q: Let\(X = \{(x, y) \in \mathbf{R}^{2} : 0 \le |x| = |y| \le 1\}\)\(Y = \{(x, y) \in \mathbf{R}^{2} : |x| + |y| = 1\}\) and\(Z = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} = 1\}\)be subspaces of the Euclidean space \(\mathbf{R}^{2}\). Then :

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.

Q: Let \(X = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} \lt 1\}\) \(\cup \{(x, y) \in \mathbf{R}^{2} : (x - 2)^{2} + y^{2} \lt 1\}\) and \(Y = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} \lt 1\}\) U \(\{(x, y) \in \mathbf{R}^{2} : (x - 3)^{2} + y^{2} \lt 1\}\) and \(Z = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} \lt 1, x \ne 0\}\) be subspaces of the Euclidean space \(\mathbf{R}^{2}\). Then :

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.

Consider \(\mathbb{R}^{n}\) with the usual topology for \(n = 1, 2, 3\). Each of the following options gives topological spaces \(X\) and \(Y\) with respective induced topologies. In which option is \(X\) homeomorphic to \(Y\)?

A.\(X = \{(x, y, z) \in \mathbb{R}^{3} : x^{2} + y^{2} = 1\}\), \(Y = \{(x, y, z) \in \mathbb{R}^{3} : z = 0, x^{2} + y^{2} \neq 0\}\)
B.\(X = \{(x, y) \in \mathbb{R}^{2} : y = \sin(1/x), 0 \lt x \le 1\}\) \(\cup \{(x, y) \in \mathbb{R}^{2} : x = 0, -1 \le y \le 1\}\), \(Y = [0, 1] \subseteq \mathbb{R}\)
C.\(X = \{(x, y) \in \mathbb{R}^{2} : y = x\sin(1/x), 0 \lt x \le 1\}\), \(Y = [0, 1] \subseteq \mathbb{R}\)
D.\(X = \{(x, y, z) \in \mathbb{R}^{3} : x^{2} + y^{2} = 1\}\), \(Y = \{(x, y, z) \in \mathbb{R}^{3} : x^{2} + y^{2} = z^{2} \neq 0\}\)

Exam Year(s)

GATE Mathematics - 2018,

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.

Discuss in Forum

Let \(p : ([0,1], T_{1}) \to (\{0,1\}, T_{2})\) be the quotient map, arising from the characteristic function on \([\frac{1}{2}, 1]\), where \(T_{1}\) is the subspace topology induced by the Euclidean topology on \(\mathbb{R}\). Which of the following statements is TRUE?

A.p is an open map but not a closed map
B.p is a closed map but not an open map
C.p is a closed map as well as an open map
D.p is neither an open map nor a closed map

Exam Year(s)

GATE Mathematics - 2022,

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.

Discuss in Forum

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.

Exam Year(s)

GATE Mathematics - 2026,

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.

Discuss in Forum

Let

\(X = \{(x, y) \in \mathbf{R}^{2} : 0 \le |x| = |y| \le 1\}\)
\(Y = \{(x, y) \in \mathbf{R}^{2} : |x| + |y| = 1\}\) and
\(Z = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} = 1\}\)

be subspaces of the Euclidean space \(\mathbf{R}^{2}\). Then :

A.X is homeomorphic to Y and Z
B.X is homeomorphic to Y, but not Z
C.Y is homeomorphic to Z, but not X
D.Z is homeomorphic to X, but not Y

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

Let \(X = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} \lt 1\}\) \(\cup \{(x, y) \in \mathbf{R}^{2} : (x - 2)^{2} + y^{2} \lt 1\}\) and \(Y = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} \lt 1\}\) U \(\{(x, y) \in \mathbf{R}^{2} : (x - 3)^{2} + y^{2} \lt 1\}\) and \(Z = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} \lt 1, x \ne 0\}\) be subspaces of the Euclidean space \(\mathbf{R}^{2}\). Then :

A.X and Y are homeomorphic to Z
B.X is homeomorphic to Y, but not to Z
C.X is homeomorphic to Z, but not to Y
D.none of X or Y or Z is homeomorphic to the other two

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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