# Find a surjective function $f:B_n \rightarrow S^n$ such that $f(x)=f(y) \iff \|x\|=\|y\|$

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

Find a surjective function $f:B_n \rightarrow S^n$ such that $f(x)=f(y) \iff \|x\|=\|y\|$.

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

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.

Define

$B_n$ as n dimensional ball.

$S^n$ as n dimensional sphere.

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

• what are the two norms? May 4, 2014 at 16:52
• @ellya Euclidean norms May 4, 2014 at 16:52
• So it's just a bijective function $[0,1] \to S^n$? May 4, 2014 at 16:54
• @NajibIdrissi Kind of yes. That's where I'm heading to and this is a subquestion of what I'm working on May 4, 2014 at 16:57
• Is the function required to be continuous? Because then it's not possible (the spaces are compact Hausdorff, so it would a homeomorphism, and $[0,1]$ and $S^n$ aren't homeomorphic). May 4, 2014 at 16:58

The condition "$f(x) = f(y)$ if and only if $\|x\| = \|y\|$" looks suspicious; perhaps you mean "$\|x\| = \|y\| = 1$", so that $f$ identifies points on the boundary of the ball?
Assuming this is the case, write $B_{n}$ in "generalized polar coordinates", $$x = \|x\|\, \frac{x}{\|x\|} = ru,$$ and define $$f(r, u) = \sin (\pi r)\, u - \cos (\pi r)\, e_{n+1}.$$ It remains to check that $f$ is continuous at the origin, that $f$ is surjective, and that $f$ is injective in the open unit ball, and maps the boundary sphere to the point $e_{n+1} = (0, \dots, 0, 1)$.
It will probably help to notice that the mapping $\tilde{f}: [0, 1] \times S^{n-1} \to S^{n}$ defined by the same formula as above is the composition of $f$ with the map $\Pi:[0, 1] \times S^{n-1} \to B_{n}$ defined by $\Pi(r, u) = ru$.
• How could we check that $f$ is continuous at the origin? May 4, 2014 at 17:55
• @Dosomemaths: The origin maps to the "south pole" $-e_{n+1}$ under $f$, and a "radially symmetric" neighborhood of the south pole pulls back to a ball centered at the origin. May 4, 2014 at 18:36