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Simple connectedness

  1. Aug 14, 2008 #1

    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 ???
     
  2. jcsd
  3. Aug 14, 2008 #2

    morphism

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    What about c(x) = (cos(2pix), sin(2pix)) for x in [0,1]?
     
  4. Aug 14, 2008 #3

    quasar987

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    Yeah. :P
     
  5. Aug 17, 2008 #4

    quasar987

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    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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