TOPOLOGY: homeomorphism between quotient spaces

[jjou]

Show the following spaces are homeomorphic: \mathbb{R}^2, \mathbb{R}^2/I, \mathbb{R}^2/D^2.

Note: D^2 is the closed ball of radius 1 centered at the origin. I is the closed interval [0,1] in \mathbb{R}.

THEOREM:
It is enough to find a surjective, continuous map f:X\rightarrow Y to show that X/S(f)\approx Y.

I can easily show R^2/D^2 \approx R^2 by sending every point in D^2 to the origin and, for every point outside D^2, letting f(r,\theta)=(r-1,\theta).

I am having difficulty showing that \mathbb{R}^2/I is homeomorphic to either \mathbb{R}^2 or \mathbb{R}^2/D^2. First, I do not understand the notation \mathbb{R}^2/I since I is traditionally a subset of R, not R^2. Does \mathbb{R}^2/I mean:

(1) points on the x-axis are equivalent iff the x coordinate falls in [0,1] and all other points (any point x,y where y=/=0) are not equivalent, OR
(2) (x,y)~(w,z) iff y=z and x,w\in[0,1], OR
(3) (x,y)~(w,z) iff x,w\in[0,1] regardless of y and z?

If it means (2), then I simply take f so that f(x,y)=(0,y) for x in [0,1], f(x,y)=(x-1,y) for x>1 and f(x,y)=(x,y) for x<0.

If it is (1), I can find a surjective (but not continous!) map f that takes [0,1]x{0} to (0,0), (x,0) to (x-1,0) for x>1, and leaves all other points where they are.

For (3), I can find a continuous surjective map from R^2 to R^2/I where f(x,y)=(x,y) for x\leq0 and f(x,y)=(x+1,y) for x>0.

Can someone clarify what \mathbb{R}^2/I means? If it means (1), any hints?

Thanks. :)

 

[jjou]

Asked a friend - I think, we're using definition (2) for the problem, in which case I've got a solution.

Thanks. :)

 

[Bacle]

Sorry if this is too late for your needs.

A general result that will help you here is that if A is contractible in X, then

X/A is homeo to X , which kind of makes sense, since you may consider A as

a single point in X, and then X/{pt.} is a trivial quotient. In this case, D^2 is

contractible in R^2 , and I is ( I am.?) also contractible, both in R, and in R^2.