Proving Homeomorphism: D^n / S^{n-1} to S^n using n=1,2,3 cases

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The discussion focuses on proving that the quotient space D^n / S^{n-1} is homeomorphic to the n-sphere S^n for n=1, 2, and 3. The participants suggest starting with specific cases, particularly emphasizing the n=1 scenario where D^1 minus a point is homeomorphic to the unit circle S^1. A proposed homeomorphism function is f(y) = (cos((y+1)π), sin((y+1)π), which requires verification for bijectiveness and continuity. The conversation highlights the importance of visualizing the relationship between the circle minus a point and the real line.

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If D^n is the unit n ball in Euclidean n-space. i.e.
D^n = \{ x \in \mathbb{R}^n : ||x|| \leq 1 \}
and S^n is an n-sphere.

how do i show that D^n / S^{n-1} is homeomorphic to S^n?
there's a hint suggesting i first of all try the n=1,2,3 cases. where X/Y= X \backslash Y \cup \{ t \} where t \in X is a single distinguished point.

i'm not really sure how to start.
i can visualise for example the n=1 case where D^1 \backslash S^0=[x-1,x) \cup (x,x+1] and S^1 is the unit circle so if we take t=x then D^1/S^0=[x-1,x+1]and i was thinking a homeomorphism here would be some sort of function along the lines of f(y)= \left( \cos{(y+1) \pi},\sin{(y+1) \pi} \right). i think this works as a homeomorphism but do i have to run through the steps of showing it's bijective and continuous and that the inverse is continuous?
 
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Follow the hint. Think about for example, the The circle minis one point being homeomorphic to the line . You should be able to be explicit about this homeomorphism. It makes sense right? take away one point, and then "fold"" the surface out. Show that R is homeomorphic to any open interval (this is also easy and explicit).

Think about this for a while.
 


is the homeomorphism i gave above wrong and if so why? also, why am i considering the circle minus one point? surely i want the line minus one point unioned with one point to be homeomorphic to the enitre circle?
 

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