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Homeomorphism through cutting and pasting of manifolds

  1. Jul 1, 2013 #1
    I am independently working through the topology book called, "Introduction to Topology: Pure and Applied." I am currently in a chapter regarding manifolds. They attempt to show that a connected sum of a Torus and the Projective plane (T#P) is homeomorphic to the connected sum of a Klein Bottle and a Projective Plane (K#P). I can go through the detail if someone would like but the conclusion is, "Therefore, T#P is topologically equivalent to K#P.

    I am not having any trouble with the proof. I am having trouble with the conclusion. How can it be that through multiple quotient maps we can end up with a space that is topologically equivalent to what we started with?
     
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  3. Jul 1, 2013 #2
    I would like to see the details, do you know how to use latex, the site reads LaTeX all you have to do is [tex] write something[/tex/] *remove the last "/"*
     
  4. Jul 2, 2013 #3

    lavinia

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    Topologically equivalent just means homeomorphic
     
  5. Jul 2, 2013 #4
    Wait that is what he was asking?
     
  6. Jul 2, 2013 #5
    That is not what I was asking...

    I posted the exact page here...
    http://imgur.com/gallery/m6MhRjL/new

    I am not sure if how they do their quotient map is standard in all of topology but if it is not i will be willing to elaborate further. just ask.
     
  7. Jul 2, 2013 #6
    have you tried writing them like words I hear that helps... I mean it did to me when I looked at those two and compared their equivalence classes. like call side 2 a and side 1 b of the torus and the one side of the the projective plane c so you have the connected sum of the word##[aba^{-1}b^{-1}] \# [cc]=[aba^{-1}b^{-1}cc]## note that the first word is the tourus T and the second is the projective plane P once you follow a few circulation rules you will be able to see it, so all you gotta do is cut and paste to get ##ab[a^{-1}b^{-1}]c|c \implies abcbac##this move was 5 the next is 6 next we show that ##ab[c]b|ac \implies abbc^{-1}ac## okay now for 7 and the end product ##[a]bbc^{-1}a| \implies bbc^{-1}|a[c]a \implies bbc^{-1}c^{-1}aa## and that is it, do you get it? It is relativly difficult it took me three days and 4 nights to figure this out lol, I am pretty sure you are a faster learned then I... maybe equivalent classes isn't the right word for this... but you should note that you can write a klein bottle as the connected sum of two projective planes

    Edit: Also these circulation rules, you have them in your book correct?
    EDIT2: Also, writing them as words over some surface f will make the visualization easier :)
     
    Last edited: Jul 2, 2013
  8. Jul 2, 2013 #7
    Ahh yes. That seems to be a more efficient technique to handle these kinds of quotient maps.

    I am still a bit concerned.

    If we deal with the closed square [0,1] x [0,1] and want to use a quotient map to create a cylinder, it is fairly straight forward. It is obvious that the square is not homeomorphic to the cylinder.

    In conclusion of this little "thought experiment" and the recalcitrant data (understanding that T#P is homeomorphic to K#P) we realize that after the surjective map a homeomorphism between the prior and the latter is not guaranteed.

    After arriving at this conclusion, I am forced to ask, how can it be the case that the quotient mapping preserves the homeomorphism in our current underlying discussion?
     
  9. Jul 2, 2013 #8
    Okay, to deal with the closed square why not just identify two sides and leave the over two sides free? I mean it is kinda self evident that you can continuously deform a torus into a cylinder, even more so if you can change it into a coffee cup(but that involves homotopy or homology I believe. What do you mean that it isn't guaranteed? Of course you could take another transformations between T#P and produce a difference result how about a different object like M#M#M which is a 3 möbius strip without a torus after all P is homeomorphic to M ... but still I do not get your question.
     
  10. Jul 2, 2013 #9

    Office_Shredder

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    Every time they draw an arrow they claim that the two spaces are homeomorphic (so maybe a double arrow would be more appropriate). Obviously homeomorphic is a transitive property so the start and end result are homeomorphic.

    The fact that each consecutive space is homeomorphic is something that needs to be checked individually (but the steps are laid out so the homeomorphisms are quite simple).

    There's no claim that two different quotient topologies on, e.g. a hexagon, are always homeomorphic, it's just that the ones they propose happen to be and you should are supposed to be able to figure out why they give the same topology
     
  11. Jul 2, 2013 #10
    Thank you Tenshou and Shredder. The question has been resolved.
     
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