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**ILikeDaisies** Let Y be the subspace {(x,0)|x $\displaystyle \in $ $\displaystyle \mathbb{Q} $, 0 $\displaystyle \leq $ x $\displaystyle \leq $ 1} of $\displaystyle \mathbb{R}^2 $ and let X be the cone on Y with vertex (0,1), i.e. the set of all points on straight line segments joining points of Y to (0,1). Show X is contractible, but that in any homotopy H between the identity map on X and the constant map with value (0,0), the point (0,0) must 'move' (i.e. there exists t with H((0,0),t) $\displaystyle \neq$ (0,0)).

Can you find a contractible space Z such that every point in Z has to move in the course of a contracting homotopy?

For the first part of the question (i.e. to show X is contractible), I think it's enough to take the retraction to (0,1), i.e. H(x,t) = t(0,1) + (1-t)x, where x is any point in X. Is this correct? Yes.

For the second part (show for any homotopy, (0,0) has to move), I think that (0,0) has to move to (0,1) with all the other points in X before it can move back to (0,0) otherwise our 'homotopy' won't be continuous (and thus not valid). If we try to devise some homotopy where (0,0) stays fixed, all the points in X must pass through (0,1), and I would expect some sort of singularity to happen then..?