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Non-contractibility of a set

  1. Aug 1, 2012 #1
    Consider the unit ball [itex]B:=B_1(0)\subset \mathbb{R}^2[/itex]. How can one prove that the set [itex]B\times B \setminus D[/itex], where [itex]D:=\left\{(x, x)\biggr| x \in B\right\}[/itex] is the diagonal, is non-contractible? Is it even disconnected? Thank you in advance.
  2. jcsd
  3. Aug 1, 2012 #2


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    It is not hard to see that this set is path connected: take two points in it and find a path between them by varying only the first coordinate until you reach the desired value, and then varying only the second coordinate. The fact that this procedure can be carried out amounts to the fact that in B, you can find a path btw any two points that avoids a third one.

    Now about your actual question, the usual way to prove non-contractibility of a space is to compute a homotopy invariant of it (like the fundamental group) that does not coincides with the value of that invariant for the one-point space. Have you tried doing this?
  4. Aug 1, 2012 #3


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    Here is an idea.

    If x and y are different points in the unit disc then one can draw a line segment from x through y and extend it until it touches the boundary circle. This defines a map,f(x,y), from BxB\D onto the circle. Is this map continuous?

    If so then, the continuous map (x,y) ->(0,f(x,y)) maps is the identity on the circle, (0,e[itex]^{i\theta}[/itex]).

    If BxB\D were contactible then the compositions

    ((0,e[itex]^{i\theta}[/itex]),t) -> BxB\D X I -> BxB\D -> (0,f(x,y)) where the first arrow is inclusion, and the second a contraction homotopy, would show that the circle is contractible.

    But the circle is not contractible because a contraction homotopy would make the circle into a retract of the unit disc.
    Last edited: Aug 1, 2012
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