TITLE: The $n$-disk $D^n$ quotiented by its boundary $S^{n-1}$ gives $S^n$
QUESTION [15 upvotes]: Define $D^n = \{ x \in \mathbb{R}^n : |x| \leq 1 \}$. By identifying all the points of $S^{n-1}$ we get a topological space which is intuitively homeomorphic to $S^n$.
If $n = 2$, this can be visualised by pushing the centre of the disc $D^2$ down so you have a sack, then shrinking the boundry of the sack to a point which gives you a teardrop shaped object which is clearly homeomorphic to $S^2$.
I am new to algebraic topology. How do I prove that the quotient space is actually homeomorphic to $S^n$. I haven't been able to write down explicitly a continuous map between $D^n$ and $S^n$ which maps $S^{n-1}$ to a point on $S^n$, which at the moment is the only way I know how to begin showin that two spaces are homeomorphic.
Is more machinery needed? If so I am interested to hear what is needed. If not, please tell me how stupid I am and give me a hint!

REPLY [16 votes]: None of the answer give any map explicitely so I'll write one here. Note that $S^{n−1}\times I→D^n,(u,t)\mapsto tu$ (this is scalar product) is a quotient map and thus one can define (after verifying some things) the continuous map $D^n\to S^n,tu\mapsto (\sin⁡(tπ)u,\cos⁡(tπ))$ which can be verified to work.