Skip to content
Tags

, , , , , , , , , , , ,

Continuous deformation of a disc (with center point removed) into a circle

by Octavian on Tuesday, June 30, 2009

If we have a closed disc D^2 = \{ (x,y) | x^2 + y^2 \leq 1 \}, then we can show that this disc with its center point removed, D^2 - \{ 0 \}, can be deformed continuously into a circle S^1. Indeed, we can find the homotopy between an identity function id : (D^2 - \{ 0 \}) \to (D^2 - \{ 0 \}) and g : (D^2 - \{ 0 \}) \to S^1, where g (r e^{i \theta}) = e^{i \theta}, and 0 < r \leq 1.

Define a map H : (D^2 - \{ 0 \}) \times I \to (D^2 - \{ 0 \}), where I = [0,1], such that

H (r e^{i \theta}, t) = (1-(1-r)t) e^{i \theta}, for t \in I

Since H (r e^{i \theta}, 0) = e^{i \theta} = g (r e^{i \theta}), and H (r e^{i \theta}, 1) = r e^{i \theta} = id (r e^{i \theta}), we conclude that a map H is a homotopy between g and id. Hence, circle S^1 is a deformation retract of disc D^2 - \{ 0 \}, so we can get S^1 from D^2 - \{ 0 \} by using continuous deformation.

For arbitrary dimension n \geq 2, we also conclude that the unit sphere S^n is a deformation retract of D^{n+1} - \{ 0 \}. Indeed, we must have a homotopy between an identity id : (D^{n+1} - \{ 0 \}) \to (D^{n+1} - \{ 0 \}) and a map g (r \phi) = \phi, where \phi is the angular coordinate of a point considered, and r = \sqrt{\sum_{i=1}^{n+1} x_i^2}, where \{ x_i \} is the coordinate of the points of D^{n+1} - \{ 0 \} which is generated by a homeomorphism \psi : D^{n+1} \to \mathbb{R}^{n+1} such that \psi (p) = x_i (p) for every p \in D^{n+1}.

Analog with the previous result, we can define a map H : (D^{n+1} - \{ 0 \}) \times I \to (D^{n+1} - \{ 0 \}), where I = [0,1], such that

H (r \phi, t) = (1-(1-r)t) \phi, for t \in I

Clearly, H (r \phi, 0) = \phi = g (r \phi), and H (r \phi, 1) = r\phi = id (r \phi). Hence H is a homotopy between g and id, and the conclusion follows.

From → Topology

No comments yet

Leave a Reply

Note: XHTML is allowed. Your email address will never be published.

Subscribe to this comment feed via RSS