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I would like to prove that the map $f:B^n\to S^n$ given by $x\mapsto \cos(\pi \Vert x\Vert)u+\sin(\pi \Vert x\Vert)\frac{x}{\Vert x\Vert}$ is surjective? Where $u=(0,\cdots,1)$ and $B^n=\{x\in \Bbb{R}^n: \Vert x\Vert\le 1\}$ and $S^{n-1}=\{x\in \Bbb{R}^n: \Vert x\Vert= 1\}$.

Of course I look at $B^n=\{(x_1,\cdots,x_n,0):\sum x_i^2\le 1\}\subset \Bbb{R}^{n+1}$

The context is to prove that $B^n/S^{n-1}$ is homeomorphic to $B^n$.

After thinking a lot, I tried the following function. Everything is "fine", it's continuous and $f(S^{n-1})=\{-u\}$, so I get the result using some theorems.

But I forgot to prove that $f$ is actually surjective; not sure how can I do that.

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    What is $B^n$ and $S^n$ ? – 2017-02-21
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    @VikrantDesai I edited. – 2017-02-21

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Guidance without explicit solution: Pick a point $z = (z_1, \ldots, z_{n+1}) \in \Bbb S^n \subset R^{n+1}$. Note that $-1 \le z_{n+1} \le 1$. Observe that you need $cos(\pi \| x \| ) = z_{n+1}$. That tells you what $\|x\|$ must be. (Why?)

Now look at $(z_1, \ldots, z_n)$; that points in the direction of $x$. Why? What's the angle between $u$ and $x$?

So scale $(z_1, \ldots, z_n)$ until it has the right norm, and you've found an appropriate $x$. What if $(z_1, \ldots, z_n)$ has norm 0 and cannot be scaled? Hmm. That's a special case you'll need to consider and understand.

(Note: you have to check that the norm is no more than $1$, and that the resulting $x$ really does map to $z$.)