# Thread: Contraction mapping of a cone, show origin must move.

1. ## Contraction mapping of a cone, show origin must move.

Let Y be the subspace {(x,0)|x $\in$ $\mathbb{Q}$, 0 $\leq$ x $\leq$ 1} of $\mathbb{R}^2$ 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) $\neq$ (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?

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..?

I'm very new to this, so I'm sorry in advance if I'm frustratingly slow at understanding any answers.

2. ## Re: Contraction mapping of a cone, show origin must move.

Originally Posted by ILikeDaisies
Let Y be the subspace {(x,0)|x $\in$ $\mathbb{Q}$, 0 $\leq$ x $\leq$ 1} of $\mathbb{R}^2$ 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) $\neq$ (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..?
You have the right idea for the second part. To make it rigorous, I think you need to use the fact that a continuous function on a compact space is uniformly continuous. Thus the function H is uniformly continuous. So for any $\varepsilon>0$ there exists $\delta>0$ such that $d(H(X), H(Y))<\varepsilon$ whenever $d(X,Y)<\delta$ (d denotes the euclidean distance). Apply this definition with $\varepsilon = 1/2$, choose $x_0\in\mathbb{Q}$ with $0 and take $X=(x_0,0,t)$ and $Y=(0,0,t).$ On the assumption that $H(0,0,t) = (0,0)$ for all t, that tells you that $d(H(x_0,0,t),(0,0))<1/2$ for all t. But that contradicts the fact that $(x_0,0)$ has to go round the point (0,1) in order to reach the origin.

3. ## Re: Contraction mapping of a cone, show origin must move.

Yes, thank you!