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Here is one of my homework in the topology class: Show that $S^n=\{x_0\}\cup_{C_{x_0}} e^n$, where $C_{x_0}:S^{n-1}=\partial e^n\rightarrow \{x_0\}$ is the constant map.

My attempt: We construct a map

\begin{align}\varphi: \{x_0\}\sqcup e^n&\rightarrow S^{n}\\ x_0&\mapsto S=(0,\dots,0,1)\\ \rho(y_1,\cdots,y_n)&\mapsto (\sin (\rho\pi) y_1,\dots, \sin (\rho \pi) y_n,\cos \rho\pi ) \quad ((y_1,\cdots, y_n) \ is \ a \ unit \ vector ) \end{align}

i.e. $\varphi$ maps every radius of $e^n$ to a meridian of $S^n$. Then the decomposition space $\{\varphi^{-1}(p):p\in S^n\}$ is exactly the adjunction space $S^n=\{x_0\}\cup_{C_{x_0}} e^n$. So if one can show that $\varphi$ is a quotient map, by some argument in the quotient space, one can conclude that $\{x_0\}\cup_{C_{x_0}} e^n$ is homeomorphic to $S^n$.

But I cannot precisely write down the argument to show $\varphi$ is a quotient map. Are there any hints? Or perhaps there is a more direct way to show this?

Any helps or hints are welcomed, thanks!

Zoudelong
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    Hint: use compactness of the source. – Vincent Boelens Dec 02 '23 at 10:58
  • "$\rho(y_1,\cdots,y_n) \mapsto (\sin (\rho\pi) y_1,\dots, \sin (\rho \pi) y_n,\cos \rho\pi ) )$" . It can't be the same $\rho$ on the LHS and the RHS . – Paul Frost Dec 02 '23 at 14:21
  • @Paul Frost Sorry, I don't understand what the problem is? – Zoudelong Dec 03 '23 at 09:42
  • @VincentBoelens Thank you! Do you mean that we could use the fact that a continuous surjection btw. compact Hausdorff spaces is a quotient map? – Zoudelong Dec 03 '23 at 09:44
  • @Zoudelong On the LHS of the equation $\rho$ denotes a function of $n$ real variables, on the RHS it must be something else. But what is it? – Paul Frost Dec 03 '23 at 12:22
  • @PaulFrost Sorry for the ambiguity, I mean I define $\varphi$ by sending a point $\rho\cdot (y_1,\dots,y_n)$ in the disk $e^n$ to a point on $S^n$. Here $\rho$ is a scalar, and $(y_1,\dots,y_n)$ is a unit vector. i.e. $\varphi$ maps every radius of $e^n$ to a meridian of $S^n$. – Zoudelong Dec 03 '23 at 12:57
  • @VincentBoelens Thank you, I have written out the formal proof. – Zoudelong Dec 03 '23 at 13:41

1 Answers1

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I prefer to write $D^n$ instead of $e^n$. We attach $D^n$ to $x_0$ by the constant map $c : S^{n-1} = \partial D^n \to \{x_0\}$. The adjunction space $\{x_0\} \cup_c D^n$ is clearly homeomorphic to the quotient $D^n/S^{n-1}$. That $D^n/S^{n-1} \approx S^n$ has been adressed several times in this forum, for example in $B^n/S^{n-1}$ is homeomorphic to $S^n$.

Your map $\varphi : D^n \to S^n $ is fine, but to prove that is a quotient map identiyfing $S^{n-1}$ to the point $(0,\ldots,0,-1)$ you must verify the following:

  1. $\varphi$ is continuous in $0$. The "problem" here is that your definition amounts to $$\varphi(y) = \begin{cases} \left(\sin (\lVert y \rVert \pi) \frac{y}{\lVert y \rVert}, \cos(\lVert y \rVert \pi)\right) & y \ne 0 \\ (0,\ldots,0,1) & y = 0 \end{cases}$$ Continuity on $D^n \setminus \{0\}$ is obvious. To show continuity in $0$, consider any sequence $(y_n)$ in $D^n \setminus \{0\}$ such that $y_n \to 0$. Then $\sin (\lVert y_n \rVert \pi) \to 0$ and $\cos (\lVert y_n \rVert \pi) \to 1$. Since all $\frac{y_n}{\lVert y_n \rVert}$ have norm $1$, we see that $\sin (\lVert y_n \rVert \pi)\frac{y_n}{\lVert y_n \rVert} \to 0 \in \mathbb R^n$. Thus $\varphi(y_n) \to \varphi(0)$.

  2. $\varphi$ is onto. I leave this to you.

  3. Since $D^n$ is compact and $S^n$ is Hausdorff, $\varphi$ is a closed map and hence a quotient map.

  4. $\varphi(S^{n-1}) = \{(0,\ldots, 0,-1) \}$. This is obvious from the definition.

  5. $\varphi$ maps $D^n \setminus S^{n-1}$ bijectively onto $S^n \setminus \{(0,\ldots, 0,-1)$. I leave this to you.

Paul Frost
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