The $n$-disk $D^n$ quotiented by its boundary $S^{n-1}$ gives $S^n$

Question

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!

Answer 1

Well, in the case $D^1 = [0, 1]$ and $S^1 = $ unit circle around $0$, can you write down a continuous map?

Answer 2

Yes, you only need to find a suitable map between $D^n$ and $S^n$, and to do that, your visualisation should help you a lot. For this kind of problem, getting a visualisation is most of the work, and you did that already. The rest is just describing what you see.

Let's pick $n=2$. You said that if you push "down" the center of $D^2$ and "close" the sack, you would get $S^2$. This should tell you that you are looking at a map $f : D^2 \rightarrow S^2$ such that the $f(0,0) = (0,0,-1)$ and is $f(x,y) = (0,0,+1)$ for $(x,y) \in S^1$. Then you need to pick a way to choose for example the $z$ coordinate for intermediate points in a continuous way. Any continuous function should work.

Do you visualise your sack with a round shape all the time ? What should be the inverse image of the equator ? A circle of center $(0,0)$ in $D^1$ ? Then your map should be rotationnaly invariant around $(0,0)$ and you can settle to search for $f$ only on a diameter of $D^2$. What is the image of a diameter in your visualisation ? A circle of $S^2$ ? So $f(x,y)$ should be $(kx,ky,z)$ for some functions $z$ and $k$ ? etc.

Answer 3

You know that $S^n$ is a one point compactification of $R^n$. That is $S^n = R^n \cup \{\infty\}$ with appropriate topology. Now define a map $f\colon D^n \to S^n$ as follows if $x$ is in interior of $D^n$ then $f(x) = \frac{x}{1 -\mod{x}}$ but if $x$ is in boundary then $f(x) = \infty$. Now it is easy to see that $f$ is continuas in the interior and as well as in boundary. Sequential criterion may help you to conclude the statement. Now $D^n$ is compact so the map is a closed map and therefore a quotient map. Hence it induces a homeomorphism from $D^n$ to $S^n$.