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:(adsbygoogle = window.adsbygoogle || []).push({});

Show that the quotient spaces [tex]R^2, R^2/D^2, R^2/I,[/tex] and [tex]R^2/A[/tex] are homeomorphic where [tex]D^2[/tex] is the closed ball of radius 1, centered at the origin. [tex]I[/tex] is the closed interval [tex][0,1]\in\mathbb{R}[/tex]. [tex]A[/tex] 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 [tex]R^2[/tex] ~ [tex]R^2/D^2[/tex].

Showing [tex]R^2[/tex] ~ [tex]R^2/I[/tex]: We define [tex]R^2/I[/tex] by the equivalence relation [tex](x_1,0)[/tex] ~ [tex](x_2,0)[/tex] iff [tex]x_1,x_2\in[0,1][/tex]. I think it is enough to show that [tex]R^2/[-1,1][/tex] ~ [tex]R^2/D^2[/tex] since changing the interval shouldn't make a drastic difference (I could just rescale & recenter the original segment [tex]I[/tex]).

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 [tex]f:R^2/D^2\rightarrow R^2/[-1,1][/tex] to show the two spaces are homeomorphic. I define this function [tex]f[/tex] to send each point in [tex]D^2[/tex] to its projection on [-1,1]: [tex]f(r,\theta)=(r\cos\theta,0)[/tex] for [tex]r\leq1[/tex].

Then, for points outside of [tex]D^2[/tex], I consider the function [tex]r(\theta)=e^{\theta}[/tex] for [tex]0\leq\theta\leq\pi/2[/tex]. This defines a portion of a spiral-like curve starting at the point (1,0). I reflect this curve over the y-axis for [tex]\pi/2<\theta\leq\pi[/tex] and then reflect over the x-axis for [tex]\pi<\theta<2\pi[/tex].

For points outside of the "spiral," I define [tex]f(r,\theta)=(r\theta)[/tex] (it is the identity map). For points [tex](r,\theta)[/tex] where [tex]1<\theta\leq e^{\theta}[/tex] (i.e. for points outside of D^2 but within the spiral), I in essence "stretch" the segment [tex]((1,\theta),(e^{theta},\theta)][/tex] to cover [tex]((0,\theta),(e^{\theta},\theta)][/tex]. (The actual formulation of this stretching is a bit convoluted, but I have it.)

Then this map [tex]f[/tex] is clearly surjective. Furthermore, it is sequentially continuous (we only need to check the boundaries [tex]r=1[/tex] and [tex]r=e^{\theta}[/tex]). The only "fishy" points are [tex](1,0)[/tex] and [tex](\pi,0)[/tex], 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 [tex]R^2/A[/tex] ~ [tex]R^2/I[/tex] 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.

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# Homework Help: Topology: homeomorphism between quotient spaces

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