Simple Connectedness of [0,1]^k

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Discussion Overview

The discussion centers on the preservation of simple connectedness in k-cubes, specifically whether the image of the k-cube [0,1]^k under a continuous function to R^n remains simply connected. The scope includes theoretical considerations and mathematical reasoning.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether k-cubes preserve simple connectedness, referencing that continuous maps do not necessarily maintain this property.
  • Another participant proposes a specific continuous function, c(x) = (cos(2πx), sin(2πx)), to explore its implications on the connectedness of the image.
  • A different participant suggests that if M is a k-manifold embedded in R^n and c:[0,1]^k-->M is homeomorphic onto its image, then c([0,1]^k) should be simply connected due to having a trivial fundamental group.

Areas of Agreement / Disagreement

The discussion remains unresolved, with multiple competing views on the preservation of simple connectedness in the context of k-cubes and continuous mappings.

Contextual Notes

Participants reference specific examples and properties of continuous functions and manifolds, but the discussion does not clarify the assumptions or definitions that might affect the conclusions drawn.

quasar987
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I am wondering if k-cubes preserve simple connectedness. I.e., is the image of [0,1]^k by a continuous function c:[0,1]^k-->R^n simply connected?

Wiki has shown me that continuous maps do not take simple connected sets to simple connected sets. For instance exp:C-->C takes C to C\{0}.

But is it true for [0,1]^k ?
 
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What about c(x) = (cos(2pix), sin(2pix)) for x in [0,1]?
 
Yeah. :P
 
On the other hand, if M is a k-manifold imbedded in R^n and c:[0,1]^k-->M is a k-cube homeomorphic onto its image, then c([0,1]^k) is simply connected because it has a trivial fundamental group. Correct?
 

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