# Topology: homeomorphism between quotient spaces

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.