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Deformation retract

  1. Feb 24, 2010 #1
    Y = (X ∪ I)/{x0 ~ 0} (disjoint union) xo - base point in X

    I am trying to show that X is a deformation retract of Y.

    I understand I need to maps f:X -> Y and g:Y -> X and show homotopic equivalence

    Where f is the inclusion map, is the identity ok for f?
    i.e f:X->Y , x in X -> x in Y

    What is appropriate for g? Is the idea to map all of I in Y to x0 in X?
     
    Last edited: Feb 24, 2010
  2. jcsd
  3. Feb 24, 2010 #2

    quasar987

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    To show that X is a deformation retract of Y is not the same as to show that X and Y are homotopy equivalent.

    A deformation retract of Y onto X is an homotopy h:Y x [0,1] --> Y (that we note [itex]h(y,t)=h_t(y)[/itex]) such that
    (i) [itex]h_0=id_Y[/itex]
    (ii) [itex]h_t|_X=id_X[/itex]
    (iii) [itex]h_1(Y)=X[/itex]
    Ok, so that h does is that it take Y and over the course of time, shrinks it onto X. And condition (ii) says that h doesn't move the points of X "since they are already where we want them".

    Now if a deformation retract of Y to X exists, then it can be said that Y and X are homotopy equivalent. Indeed, for f:Y-->X given by [itex]f(y)=h_1(y)[/itex] and g:X-->Y given by [itex]g=\iota_X[/itex], the inclusion of X in Y, we have that [itex]f \circ g =id_X[/itex] and [itex]g\circ f = h_1\approx h_0=id_Y[/itex]. (where [itex]\approx[/itex] means homotopic)

    But to show that X and Y are homotopy equivalent is not the same as finding a deformation retract between the two. Do you see that?
     
  4. Feb 24, 2010 #3
    So if a deformation retract exists from Y to X this implies homotopic equivalence but not the converse yes?

    I cant see how to formulate the function?

    so that h(y,0) = y in Y
    and h(y,1) = y in X

    unless you map it all to x0?
     
  5. Feb 24, 2010 #4
    A deformation retract is a homotopy between a retraction and the identity map. Your retraction will be taking I and mapping it to xo, and the identity is, well, the identity. So without giving away the actual map....the points in X will remain fixed throughout, while the points in I will all be mapped to xo.
     
  6. Feb 24, 2010 #5
    I understand that and
    therefore that f:X->Y is defined by x->x
    But you cannot surely defined g:Y->X by whether is in X or I as member of X U I is still only (x) not (x,i)?
    Therefore the homotopy h: Y -> Y is fg?
    such that h(x,t) -> (1-t)x0 + x?????
     
  7. Feb 24, 2010 #6

    quasar987

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    The set Y is made up of a space X and of the interval I=[0,1], where the points 0 in I and x_0 in X are identified. For instance, if X is the upper hemisphere of the 2-sphere and x_0 is the south pole of that sphere, then Y looks just like an umbrella.

    The goal then, is to find a continuous deformation h:Yx[0,1]-->Y such that h_0=id_Y and such that as t goes from 0 to 1, the points of the "stick" in the umbrella are slid to the hemisphere, while the points already on the hemisphere don't move.

    The idea expressed by mrbohn1 (and perhaps by you in post 3?) is that the most natural way to do this is to slide all the points in the "stick" I onto 0=x_0.

    You just have to find the formula for a map that does this.
     
  8. Feb 25, 2010 #7

    WWGD

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    Quasar 987:

    Shouldn't the point to be collapsed be in the space being collapsed?.

    The south pole is not contained in the northern hemisphere.


    (Sorry, I have not posted in a while; I forgot how to quote from others'

    messages.)
     
  9. Feb 25, 2010 #8

    quasar987

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    Thanks for pointing this out. I tend to confuse north and south the way some people mix up left and right!
     
  10. Mar 1, 2010 #9

    WWGD

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    " Thanks for pointing this out. I tend to confuse north and south the way some people mix up left and right! "

    No problem. Just let me know if you decide to drive here in NYC; I will then take

    a sick day and stay at home. :)
     
  11. Mar 1, 2010 #10

    quasar987

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    As a matter of fact, I did visit NYC back in november. Amazing city...
     
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