Question Prove that a quotient of an n-dimensional ball by an equivalence relation, whose only non-trivial equivalence class is the n-1 dimensional sphere, is homeomorphic to an n-dimensional sphere.

Let $B_n = \{x \in \mathbb{R^n} : \|x\| \le 1\}$ and $S^n = \{x \in \mathbb{R^{n+1}} : \|x\| = 1\}$.


$B_n$ as n dimensional ball.

$S^n$ as n dimensional sphere.

$\|x\|=(x_1^2+ \dots+x_n^2)^{1/2}$

I have defined a equivalence relation of $B_n$ by $x \sim y \iff \|x\|=\|y\|$ so that $[x]=\{y \in \mathbb{R^n} : \|y\|=\|x\|\}$. Then we have equivalence class as n-1 dimensional sphere.

And I was thinking of using a theorem:

Let $X$ be a topological space with an equivalence relation ~. Let $f :X \rightarrow Y$ be a continuous map with the properties

  1. $f(a)=f(b)$ iff $a$~$b$
  2. $f$ is onto
  3. $U$ is open in $Y$ if $f^{-1}(U)$ is open in $X$.

Then the unique map $g:X/\sim \rightarrow Y$ is homeomorphism

  • 1
    $\begingroup$ I doubt this is meant that way. I read the question in such a way, that the boundary of the ball is identified to one point, while the rest of the ball is unaffected. That is, the equivalence class of a point $x$ such that $||x||<1$ consists exactly of $\{x\}$, while the equivalence class of any $x$ with $||x||=1$ is the sphere. $\endgroup$ – Thomas May 4 '14 at 17:40
  • $\begingroup$ I bet this question has been asked, try looking it up! $\endgroup$ – The very fluffy Panda May 4 '14 at 17:40
  • $\begingroup$ @Thomas Ah yea. Things I wrote down after the question is what I've done and I was confused. How could the question be solved? $\endgroup$ – MathsMy May 4 '14 at 17:44
  • $\begingroup$ @Thomas Ah I get what you mean. That's right! $\endgroup$ – MathsMy May 4 '14 at 17:45

We can find the homeomorphism explicitly. Note that for all points $\mathbf x=(x_1,\ldots,x_n)$ in $B\setminus\partial B$, we have $0\le \|\mathbf x\|<1$, and of course $\|\mathbf x\|=1$ for the points on $\partial B$. We may try to map all points of the boundary to the north pole - and the interior to the rest; symmetry suggests to map the origin to the south pole. Now how to find a point $\mathbf y=(y_0,y_1,\ldots ,y_n)$ with $\|\mathbf y\|=1$ from a given point $\mathbf x\in B$ such that $\|\mathbf x\|=1$ implies $y_0=1$ and $\|\mathbf x\|=0$ implies $y_0=-1$? This suggests $y_0=2\|\mathbf x\|-1$ and $y_k$ a multiple of $x_k$ such that $\|y\|$ becomes $1$.

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$(1)$ Let's look at the case $n=1$. Then $B$ is the interval $[-1,1]$. What we can do is identifiy the endpoints of the interval (the boundary points of $B$). This gives us a space homemorphic to the circle. We do the same in the case when $n >1 $.

$(2)$ When we have an onto map $ f: X \rightarrow Y$, where $X$ is compact and $Y$ is Hausdorff, then $f$ is an identification map. If we have an identification map, then the space $Y$ and that created under the partition by $f$ , say $Y^o$ are homemorphic.

$(3)$ Now find a one-one, onto map which identifies the boundary of $B_n$ to a single point. We know that $B_n$ without the boundary and $S^n$ without a single point is homeomorphic to $\mathbb{R^n}$. Use these two homeomorphisms to construct the desired map.

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