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Topology: simple connectedness and fundamental groups

  1. Mar 6, 2007 #1

    quasar987

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    I read on wiki* that a (pointed) topological space is simply connected iff its fundamental group is trivial. But I don't see how this in accordance with the R² caracterisation that U is simply connected iff it is path-connected and has no holes in it.

    Take the closed unit-disk with a point of its boundary as base point for example. Then what are the elements of the fundamental group? There is the trivial loop, which is the identity element, and then there are the homotopy equivalence classes of the loops that circles the boundary S^1 once, twice, three times, etc. such that the fundamental group is isomorphic to Z, not {e}.



    *http://en.wikipedia.org/wiki/Simply_connected#Formal_definition_and_equivalent_formulations
     
  2. jcsd
  3. Mar 6, 2007 #2

    StatusX

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    All loops in the disk are homotopic to the constant loop. Also, it is usually specified that simply connected spaces are path connected (this does not follow from having trivial fundamental group).
     
  4. Mar 6, 2007 #3

    matt grime

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    they are all trivially trivial.

    rubbish. those maps are all trivially homotopic to the constant map. The space is the whole disk, not S^1.
     
  5. Mar 6, 2007 #4

    matt grime

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    But it does from having trivial fundamental groupoid, right?
     
  6. Mar 6, 2007 #5

    StatusX

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    I don't know, it depends what you mean by "trivial". The fundamental groupoid of a simply connected space is certainly not the trivial groupoid (which I'm taking to be the single element groupoid), but it is about as trivial as a fundamental groupoid can be (ie, every pair of points has a unique homotopy class of paths between them).
     
  7. Mar 6, 2007 #6

    Hurkyl

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    Ah, a debate on whether we should mean "trivial up to isomorphism" or "trivial up to equivalence". :tongue:
     
  8. Mar 6, 2007 #7

    StatusX

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    What's the difference?
     
  9. Mar 7, 2007 #8

    Hurkyl

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    The groupoids G and H are isomorphic iff there are functors

    P : G --> H
    Q : H --> G

    such that PQ and QP are equal to the appropriate identity functors.


    The groupoids G and H are equivalent iff there are functors

    P : G --> H
    Q : H --> G

    such that PQ and QP are naturally isomorphic to the appropriate identity functors.



    An equivalent way of stating it is that two groupoids are equivalent iff they become isomorphic after we identify isomorphic points.


    In particular, a groupoid is equivalent to the groupoid you stated if and only if, for any ordered pair of points (A, B), there is exactly one arrow A --> B.
     
    Last edited: Mar 7, 2007
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