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Apparent Contradiction: Every Map from a Contractible Space to any X is trivi

  1. Apr 20, 2010 #1
    Hi, everyone:

    I am confused about the result that every map from a contractible space X into
    any topological space Y is contractible.

    I think the caveat here is that the homotopy between any f:X-->Y and c:X-->Y

    with c(X)={pt.} is that the homotopy is free, i.e., the endpoints are not fixed.

    Still, I find the example of f:I-->S^1 , with I the unit interval,
    particularly confusing:

    Consider the path f(t)=e^2iPit. According to this result, since
    I is contractible ( I am contractible?:smile) :, f(t) is trivial, i.e., f(t) is homotopic
    to a point.

    *But* since e^2iPit does a full loop around S^1 , i.e,
    f(I)= (S^1), we must tear f(I) open , to be able to homotope it
    to a point. Isn't this tearing necessarily discontinuous, even if we do not fix the
    endpoints in S^1.?.
    If not, what would be an actual homotopy between e^2iPit and {.pt.}?

  2. jcsd
  3. Apr 20, 2010 #2
    Hi Bacle. The easy answer is that if a space retracts onto a point, i.e. we have a map

    [tex] F:X \times I \rightarrow X [/tex]

    such that F restricted to [tex]X \times {0} [/tex] is the identity, and F restricted to [tex] X \times {1} [/tex] is constant, then a homotopy is constructed using [tex] f(F) [/tex].
  4. Apr 20, 2010 #3


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    Saying that a map f:X --> Y is homotopic to a constant map is not the same thing as saying that f(X) is contractible.
  5. Apr 20, 2010 #4
    If that were true, every space would be contractible.
  6. Apr 20, 2010 #5
    Thanks to Both.

    Hurkyl wrote:

    " Saying that a map f:X --> Y is homotopic to a constant map is not the same thing as saying that f(X) is contractible. "

    I understand, and that is the whole point. I think I know
    how to show this formally (see below) , but I cannot see it geometrically/intuitively.
    I fail to see intuitively/geometrically how we could deform the circle into a point in a
    continuous way.

    I had been working along the lines of what Zhentil suggested. I had
    constructed a homotopy F(x,t) , from the unit interval I to {1}:

    F(x,t)=(1-t)x+t , then H(x,0)=x ; H(x,1)=1 . Using f(t)=ei2Pit

    Then f(F(x,t))=f( (1-t)x+t)= ei2Pi((1-t)x+t)

    and f(F(x,0))=ei2Pix , and f(F(x,1))=ei2Pi=(1,0)

    Which I think is formally correct, tho showing the continuity of f(F) is kind of
    a hassle.

    What seems confusing is that in this case f(X)=Y . So f(X) coincides with the
    whole space Y . How isn't then the contractibility of f(X) not equivalent to
    f:X-->Y being inessential.?. If f(x) were not onto, the question would be trivial
    to me. What is confusing is that f(X)=Y in this case.

  7. Apr 21, 2010 #6


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    You can't deform a circle to a point -- but you can deform a map from the interval to a circle to a map from the interval to a point.

    Maybe your problem is that you're confusing the map with its image?

    0 and 1 are different points of the interval, so a homotopy g-->h is allowed to "move" the images of those points independently.

    But by focusing on the image of f, you've "forgotten" that 0 and 1 are different, which makes it difficult for you to picture how your homotopy works.
  8. Apr 21, 2010 #7
    Thanks, Hurkyl. Just one more please. (if I don't get it after this one, I will

    just meditate on it some more.)

    Please tell me if this is correct: this homotopy actually _does_ tear the circle

    open. But because of what you said ( 1 and 0 are not close by.) , tearing the

    circle (which is, I think, what we do when go from:

    f(F(x,0))=ei2Pix, to:

    ei2Pi((1-eps.)t+eps. , with eps.->0+

    is not discontinuous, because it is not the case that points nearby are not sent

    (via the tearing) to points "far-away" , i.e., tearing does not violate a del.-eps

    argument (which we can use for X,Y metric).

    Sorry if this is contrived.

    Thanks Again.
  9. Apr 21, 2010 #8
    Re: Apparent Contradiction: Rewrite. Please Ignore Previous.

  10. Apr 21, 2010 #9
    Think of it this way: take a shoe string, and line up the ends so it forms a circle. You can pull it apart without tearing, i.e. on the interval, 0 and 1 are distinct points, but on the circle, they aren't (if we view the circle as the quotient of the interval).
  11. Apr 21, 2010 #10


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    Maybe the words you are using invoke a different mental picture to me than they do to you. It really sounds like the words you use are talking about functions and homotopies from the circle, rather than from the interval.

    Now, I could imagine thinking about the following:

    We can interpret the inverse image of f:I --> S1 as a sheaf on S1 -- over each point of S1 we have a single point, except for 1, over which we have two points.

    There is probably a sense in which we could compose with the homotopy F, giving us a picture of the two points lying over 1 separating; one staying in place and the other moving around the circle.
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