Searching a particular map $D^n \to \mathbb{S}^n$ Let $f:D^n\to \mathbb{S}^n,x\mapsto (2\lVert x\rVert^2-1,2x\sqrt{1-\lVert x\rVert^2})$ where I'm considering $\mathbb{S}^n\subseteq \mathbb{R}\times\mathbb{R}^n$, and $D^n=\{x\in \mathbb{R}^n:\lVert x \rVert ^2\leq1\}$.
Then $f(\partial D^n)=(1,0)$ and $f:$Int$D^n\to \mathbb{S}^n-\{(1,0)\}$ is an homeomorphism.

Take $y\in \mathbb{S}^n$ arbitrary. How can I construct a map $f:D^n\to \mathbb{S}^n$ such that $f(\partial D^n)=y$ and $f:$Int$D^n\to \mathbb{S}^n-\{y\}$ is an homeomorphism ?

I was trying to find an homeomorphism $g:\mathbb{S}^n\to \mathbb{S}^n$ such that $g((1,0))=y$, so that $g\circ f$ would be the map I'm searching for, but I'm not able to do it.
Any help is welcomed.

And, what is the intuition/ geometrical construction behind the expression of $f$ ?

 A: The geometric idea behind the function $f$ is to take the disc $D^n$ embed it as $\{-1\}\times D^n$  into $\mathbb R \times \mathbb R^n$ and then wrap it arround the sphere $S^n$ and glue the boundary to the northpole $N=(1,0)\in\mathbb R \times \mathbb R^n$. 
We do it in such a way that the distance $\|x\|$ of a point of $D^n$ to the midpoint of $D^n$ corresponds  to the height of which the point gets wrapped up: for this we take the value $2\|x\|^2-1$ (i think we take this value instead of simply $2\|x\|-1$ to even get a smooth map). Here by height we mean the value of the first coordinate.
So first we move a point  $x$ to $(2\|x\|^2-1,x)$ and then scale it by $\lambda\geq 0$ in the second variable to land on $S^n$, this is we have to solve $\|(2\|x\|^2-1,\lambda x)\|=1$ which leads to $\lambda=2\sqrt{1-\lVert x\rVert^2}$ and hence to the formula for $f$.
$\bullet$ Now to your first question:
The idea of finding such a $g$ is good! For this you can reflect the sphere about the hyperplane which intersects the line connecting $N$ and $y$ horizontally. For an explixit formula set $v=\frac{y-N}{\|y-N\|}\in\mathbb R^{n+1}$ (or $v=0$ if $y=N$) then $g$ is given  by the Householder transformation $$g(x)=x-2\langle x,v\rangle v$$ or written as a matrix $g=Id-2vv^t$. 
