Question

Suppose $h:S^1 \to S^1$ is nullhomotopic, and $f:S^1 \to S^1$ defined by $f(x)=-x$ Then $f\circ h$ is nullhomotopic

score 116 · Answer 1 · 2026-04-02 09:52:46

116

Views

Suppose $h:S^1 \to S^1$ is nullhomotopic, and $f:S^1 \to S^1$ defined by $f(x)=-x$ Then $f\circ h$ is nullhomotopic

Published on 02 Apr 2026 - 9:52

score 70 · Answer 2 · 2026-03-25 20:35:30

70

Views

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28

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1k

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Topology: Homeomorphism between finite complement topology in $\mathbb{R}$ and one of its subspaces

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score 277 · Answer 5 · 2026-03-27 13:25:52

277

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score 146 · Answer 6 · 2026-04-08 19:42:15

146

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score 800 · Answer 7 · 2018-07-01 19:46:14

800

Views

Subset is Open and Closed implies Subset is Entire Space

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score 132 · Answer 8 · 2026-03-28 18:13:24

132

Views

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score 243 · Answer 9 · 2018-07-01 22:29:54

243

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Set that is closed under limits but not closed topologically?

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score 167 · Answer 10 · 2018-07-02 00:46:33

167

Views

The set $A=\{0\} \cup \{\frac 1n \mid n \in \mathbb N\}$

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score 93 · Answer 11 · 2018-07-02 08:02:09

93

Views

Prove $f$ is continuous

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score 579 · Answer 12 · 2018-07-02 09:03:10

579

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score 311 · Answer 13 · 2018-07-02 09:27:47

311

Views

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score 193 · Answer 14 · 2018-07-02 13:51:58

193

Views

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score 247 · Answer 15 · 2026-04-08 02:28:01

247

Views

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List Question

Suppose $h:S^1 \to S^1$ is nullhomotopic, and $f:S^1 \to S^1$ defined by $f(x)=-x$ Then $f\circ h$ is nullhomotopic

Cartan-Hadamard and universal covers

Topology Problem with metric space

Topology: Homeomorphism between finite complement topology in $\mathbb{R}$ and one of its subspaces

Definition of integration of forms over manifolds, Spivak.

Topology: continuity on product of (metric) spaces

Subset is Open and Closed implies Subset is Entire Space

Compactness Definition Concerning Accumulation Points

Set that is closed under limits but not closed topologically?

The set $A=\{0\} \cup \{\frac 1n \mid n \in \mathbb N\}$

Prove $f$ is continuous

Euclidean and arrow topology

Prove: $\overline{U}=\overline{A\cap U}$

If $A$ dense $A$ is not connected

Explicit Map from $\mathbb{T}^{3}$ to $SU(2)$

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