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.