So $S^n$ in $\Bbb R^{n+1}$ can be described by the equation $x_1^2+\ldots+x_{n+1}^2=1$. Now consider two subsets $U_N:=S^n-\{(0,0,\ldots,1)\}$ and $U_S:=S^n-\{(0,0,\ldots,-1)\}$, the sphere less it's north and south pole respectively. Now if we use stereographic projection onto the plane $x_{n+1}=0$ we get two sets of local co-ordinates say $\phi_i=\large\frac{x_i}{1-x_{n+1}}$ and $\pi_i=\large\frac{x_i}{1+x_{n+1}}$ say for the respective subsets with $i=1,\ldots,n$. These can be inverted as follows:
$x_{n+1}=\large\frac{(\sum_i \phi_i^2)-1}{(\sum_i \phi_i^2)+1}$ and $x_{n+1}=\large\frac{1-(\sum_i \pi_i^2)}{1+(\sum_i \pi_i^2)}$
$x_i=\large\frac{2\phi_i}{1+(\sum_j \phi_j^2)}$ and $x_i=\large\frac{2\pi_i}{1+(\sum_j \pi_j^2)}$ with $i,j=1,\ldots n$
We can express one in terms of the other via $\pi_i=\large\frac{\phi_i}{\sum_j \phi_j^2}$ or with $\phi_i=\large\frac{\pi_i}{\sum_j \pi_j^2}$.
Now if I wanted to show this was a submanifold, I could try show the jacobian of the previous line in either case is non-zero, but I'm not sure I can simplify the determinant.
Is there a clever way?
I was also thinking say to name maps $\phi=(\phi_1,\ldots,\phi_n)$ and $\pi=(\pi_1,\ldots,\pi_n)$ mapping $U_N$ and $U_S$ respectively to $\Bbb R^n$. Then defining a map $f:S^n\rightarrow S^n$ with $(x_1,\ldots, x_n, x_{n+1})\mapsto (x_1,\ldots,x_n,-x_{n+1})$, so that $\phi=\pi \circ f$ in which case $\phi\circ \pi^{-1}=\pi \circ f \circ \pi^{-1}$ and I could work out the differentials separately, $d\phi\cdot d(\pi^{-1})$ or $d\pi \cdot df\cdot d(\pi^{-1})$ and then take a determinant, But again it's not much better.
Since I was asked to use these stereographic projections I would assume something like the above is the methodology.
I've read that you could also say that $f(\vec x)=x_1^2+\ldots+x_{n+1}^2$ is a submersion and that would prove it either, but it doesn't correspond to stereographic projection$\ldots$
There's also this as well here about a homeomorphism using stereographic projection but does that work to show its a submanifold?