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- Thread starter Goklayeh
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quasar987

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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?

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lavinia

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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.

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.

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