# Identification Space

1. Feb 23, 2010

### Mikemaths

Is this a cone after Identifaction?

Let A = (I × I)/J
where J = (I × {1}) ∪ ({0; 1} × I) ⊂ I × I

2. Feb 23, 2010

### quasar987

It's a closed disk. But if by "cone" you mean baseless cone (i.e. a child's birthday party hat : http://www.utterwonder.com/archives/images/happy_birthday_party_hat-thumb.jpg [Broken]), then one can also say that A is a cone, since the two are homeomorphic.

Last edited by a moderator: May 4, 2017
3. Feb 23, 2010

### Mikemaths

how would you construct that homeomorphism

4. Feb 23, 2010

### quasar987

Take C the cone with base a unit circle in the xy plane and vertex at say (0,0,1). Let also D be the unit disk centered at (0,0,0). Then the projection down on the xy-plane C-->D²:(x,y,z)-->(x,y,0) is the desired homeomorphism.

5. Feb 24, 2010

### Mikemaths

Could I do it using the circles
Ct = {(x; y) ∈ D2 | (x − t)2 + y2 = (1 − t)2} (t ∈ I)
to construct a homeomorphism f : A → D2

6. Feb 24, 2010

### quasar987

What do you mean?

7. Feb 24, 2010

### Mikemaths

Sorry about that I constructed it as follows is this ok?

A -> D^2

(s,t) in I x I -> (cos(2pi(1-t)s),sin(2pi(1-t)s)) in D^2

8. Feb 24, 2010

### quasar987

This is a map from I x I to D^2. I thought you were trying to construct a homeomorphism between C and D^2. I don't see the link.

9. Feb 24, 2010

### Mikemaths

I was trying to create a homeomorphism between the identified I x I and D^2 such that Ct is satisfied.