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

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

THEOREM:

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

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

I am having difficulty showing that [tex]\mathbb{R}^2/I[/tex] is homeomorphic to either [tex]\mathbb{R}^2[/tex] or [tex]\mathbb{R}^2/D^2[/tex]. First, I do not understand the notation [tex]\mathbb{R}^2/I[/tex] since I is traditionally a subset of R, not R^2. Does [tex]\mathbb{R}^2/I[/tex] 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 [tex]x,w\in[0,1][/tex], OR

(3) (x,y)~(w,z) iff [tex]x,w\in[0,1][/tex] 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 [tex]x\leq0[/tex] and f(x,y)=(x+1,y) for x>0.

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

Thanks. :)