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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!

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  • $\begingroup$ One way to do it would be to make a homeomorphism of $D^n$ with $\mathbb{R}^n$ and then use stereographic projection. $\endgroup$ Mar 3, 2011 at 10:08
  • $\begingroup$ $D^n$ and $\mathbb{R}^n$ are not homeomorphic. Do you mean the interior of $D^n$? $\endgroup$
    – DBr
    Mar 3, 2011 at 10:24

5 Answers 5

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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.

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    $\begingroup$ If we denote the quotient map from $ \mathbb{D}^{n} $ to $ \mathbb{D}^{n}/\mathbb{S}^{n - 1} $ by $ q $, and the map above from $ \mathbb{D}^{n} $ to $ \mathbb{S}^{n} $ by $ f $, then, by the universal property of quotient topologies, there exists a continuous bijection $ h: \mathbb{D}^{n}/\mathbb{S}^{n - 1} \to \mathbb{S}^{n} $ that satisfies $ h \circ q = f $. As $ \mathbb{D}^{n}/\mathbb{S}^{n - 1} $ is compact (because $ \mathbb{D}^{n} $ is itself compact) and $ \mathbb{S}^{n} $ is Hausdorff, it follows that $ h $ is a homeomorphism. $\endgroup$ Apr 7, 2022 at 21:58
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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.

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Since $(D^n\setminus S^{n-1}) \cong \mathbb{R}^n \cong (S^n-\{p\})$, we have a homeomorphism $k:D^n\setminus S^{n-1} \rightarrow S^n-\{p\}$. Then we can define $f(x)=k(x)$ for $x \in D^n\setminus S^{n-1}$ and $f(x)=p$ for $x \in S^{n-1}$.

The continuity of $f$ can be shown as follows. Let $U$ be an open proper subset of $S^n$ containing $p$. Then $U \cong \mathbb{R}^n \cong (D^n\setminus S^{n-1})$. Simultaneously removing a closed subset of each of the spaces yields $U\setminus\{p\} \cong \mathbb{R}^n\setminus D^n \cong (D^n\setminus S^{n-1})\setminus C$, where $C\subset D^n\setminus S^{n-1}$ is closed. So we can say $f^{-1}(U\setminus \{ p \} ) = (D^n\setminus S^{n-1})\setminus C$, since on the domain $U\setminus \{ p \}$, $f$ is a homeomorphism. Noting that any open neighborhood of $S^{n-1}$ must be of the form $D^n\setminus C$, it is easy to see that $f^{-1}(U)$ is of such form and open. Then $f$ is the desired map.

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    $\begingroup$ How do you say that if $U$ is a proper open set containing $p$, it is homeomoprhic to $\mathbb{R}^n$? For all we know, $U$ may not be connected, but $\mathbb{R}^n$ is. Also, even if we consider connectedness, the stereographic projection would not give this conclusion. So, what are we using here? $\endgroup$ Sep 13, 2021 at 6:08
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Well, in the case $D^1 = [0, 1]$ and $S^1 = $ unit circle around $0$, can you write down a continuous map?

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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 $S^n$.

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