I want to prove that $S=\{(x,y,z)|x^2+y^2+z^2=1\}$(Unit Sphere) is a surface of $\mathbb{R}^3$. I'm following below definition,
Surface: A subset S of $\mathbb{R}^3$ is called surface if for every point $p\in S$ there is open subset W of $\mathbb{R}^3$ containing and open subset U of $\mathbb{R}^2$ such that $S\cap W$ and $U$ are homeomorphic.
Here, I have two open subsets of $\mathbb{R}^3$ and one subset of $\mathbb{R}^2$
$W_1=\{(x,y,z)|x<0\}\cup\{(x,y,z)|y\neq 0\}=A\cup B $
$W_2=\{(x,y,z)|x>0\}\cup\{(x,y,z)|z\neq 0\}=C \cup D$
$U=(-\frac{\pi}{2},\frac{\pi}{2})\times(0,2\pi)$
Clearly, $W_1,W_2$ and U are open subsets of in $\mathbb{R}^3$ and $\mathbb{R}^2$ respectively.(why?)
Ans. For $W_1$, if we define a map $f_i: \mathbb{R}^3 \to \mathbb{R}^2$,i=1,2 by
$f_1(x,y,z)=x$, $f_2(x,y,z)=y$ then $f_i$ are continuous map and $f_1^{-1}(0,\infty)=A, (f_2^{-1}(0))^{c}=B $, hence, A and B are open it implies that $W_1$ too.
Similarly, $W_2$ is open in $\mathbb{R}^3$ and U is nothing but cartesian product of open interval therefore it is open.
I have particular parametrisation for sphere For i=1,2 define $\sigma_i:U \to S\cup W_i$ by
$\sigma_1(u,v)=(\cos u \cos v,\cos u \sin v,\sin v)$
$\sigma_2(u,v)=(-\cos u \cos v, -\sin v, -\cos v \sin v )$
How can I prove that $\sigma_1$ and $\sigma_2$ are homeomorphism and $\sigma_1(U) \cup \sigma_2(U)=S$? or Can I have another way to explain these thing using some mathematical software ? If yes which software is useful for me? I need to approach Geometric as well as mathematical to explain this thing to students. Give me some hint regarding this. Thanks in advance. (If there is any mistake in above question kindly point out me.)