Identification Space: Is A Cone?

[Mikemaths]

Is this a cone after Identifaction?

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

 

[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 ), then one can also say that A is a cone, since the two are homeomorphic.

 

[Mikemaths]

how would you construct that homeomorphism

 

[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.

 

[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

 

[quasar987]

What do you mean?

Also, for "x quared", put x^2 instead of x2 please.

 

[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

 

[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.

 

[Mikemaths]

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