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jjou
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I posted this earlier and thought I solved it using a certain definition, which now I think is wrong, so I'm posting this again:
Show that the quotient spaces R^2, R^2/D^2, R^2/I, and R^2/A are homeomorphic where D^2 is the closed ball of radius 1, centered at the origin. I is the closed interval [0,1]\in\mathbb{R}. A is a union of line segments with a common endpoint (without loss of generality, we can assume the common endpoint is the origin and the first line segment is the interval [0,1] on the x-axis).
I showed that R^2 ~ R^2/D^2.
Showing R^2 ~ R^2/I: We define R^2/I by the equivalence relation (x_1,0) ~ (x_2,0) iff x_1,x_2\in[0,1]. I think it is enough to show that R^2/[-1,1] ~ R^2/D^2 since changing the interval shouldn't make a drastic difference (I could just rescale & recenter the original segment I).
My proof of this is quite complicated (I think), so would somebody mind checking it / suggesting a more elegant proof? My proof is as follows:
It is enough to find a continuous, surjective map f:R^2/D^2\rightarrow R^2/[-1,1] to show the two spaces are homeomorphic. I define this function f to send each point in D^2 to its projection on [-1,1]: f(r,\theta)=(r\cos\theta,0) for r\leq1.
Then, for points outside of D^2, I consider the function r(\theta)=e^{\theta} for 0\leq\theta\leq\pi/2. This defines a portion of a spiral-like curve starting at the point (1,0). I reflect this curve over the y-axis for \pi/2<\theta\leq\pi and then reflect over the x-axis for \pi<\theta<2\pi.
For points outside of the "spiral," I define f(r,\theta)=(r\theta) (it is the identity map). For points (r,\theta) where 1<\theta\leq e^{\theta} (i.e. for points outside of D^2 but within the spiral), I in essence "stretch" the segment ((1,\theta),(e^{theta},\theta)] to cover ((0,\theta),(e^{\theta},\theta)]. (The actual formulation of this stretching is a bit convoluted, but I have it.)
Then this map f is clearly surjective. Furthermore, it is sequentially continuous (we only need to check the boundaries r=1 and r=e^{\theta}). The only "fishy" points are (1,0) and (\pi,0), but it should work out.
Can anyone find anything wrong with this proof or suggest a simpler method? (I'm almost certain the solution should not be this complicated!)
PS: I think, to show that R^2/A ~ R^2/I can be done by induction and through a method similar to the one described above but with even more reflections - thus, an extremely ugly process.
Show that the quotient spaces R^2, R^2/D^2, R^2/I, and R^2/A are homeomorphic where D^2 is the closed ball of radius 1, centered at the origin. I is the closed interval [0,1]\in\mathbb{R}. A is a union of line segments with a common endpoint (without loss of generality, we can assume the common endpoint is the origin and the first line segment is the interval [0,1] on the x-axis).
I showed that R^2 ~ R^2/D^2.
Showing R^2 ~ R^2/I: We define R^2/I by the equivalence relation (x_1,0) ~ (x_2,0) iff x_1,x_2\in[0,1]. I think it is enough to show that R^2/[-1,1] ~ R^2/D^2 since changing the interval shouldn't make a drastic difference (I could just rescale & recenter the original segment I).
My proof of this is quite complicated (I think), so would somebody mind checking it / suggesting a more elegant proof? My proof is as follows:
It is enough to find a continuous, surjective map f:R^2/D^2\rightarrow R^2/[-1,1] to show the two spaces are homeomorphic. I define this function f to send each point in D^2 to its projection on [-1,1]: f(r,\theta)=(r\cos\theta,0) for r\leq1.
Then, for points outside of D^2, I consider the function r(\theta)=e^{\theta} for 0\leq\theta\leq\pi/2. This defines a portion of a spiral-like curve starting at the point (1,0). I reflect this curve over the y-axis for \pi/2<\theta\leq\pi and then reflect over the x-axis for \pi<\theta<2\pi.
For points outside of the "spiral," I define f(r,\theta)=(r\theta) (it is the identity map). For points (r,\theta) where 1<\theta\leq e^{\theta} (i.e. for points outside of D^2 but within the spiral), I in essence "stretch" the segment ((1,\theta),(e^{theta},\theta)] to cover ((0,\theta),(e^{\theta},\theta)]. (The actual formulation of this stretching is a bit convoluted, but I have it.)
Then this map f is clearly surjective. Furthermore, it is sequentially continuous (we only need to check the boundaries r=1 and r=e^{\theta}). The only "fishy" points are (1,0) and (\pi,0), but it should work out.
Can anyone find anything wrong with this proof or suggest a simpler method? (I'm almost certain the solution should not be this complicated!)
PS: I think, to show that R^2/A ~ R^2/I can be done by induction and through a method similar to the one described above but with even more reflections - thus, an extremely ugly process.