Consider the maps (page 169, M. Nakahara) $$x=a\sin\theta\cos\phi, \hspace{1cm} y=a\sin\theta\sin\phi, \hspace{1cm} z=a\cos\theta\tag{1}$$ from the surface of a sphere $S^2$ of radius $a$ to $\mathbb{R}^2$.
From the maps (1), how can one understand that surface of a sphere $S^2$ is homeomorphic to $\mathbb{R}^2$ locally? I want to unerstand, how does the maps (1) help us understand that surface of a sphere (or any manifold) locally Euclidean?
Why is (1) a map from $S^2\to \mathbb{R}^2$ but not $S^2\to \mathbb{R}^3$? Is it because of the constraint $x^2+y^2+z^2=a^2$?
Firstly (re post v1), the sphere is NOT homeomorphic to $\mathbb{R}^2$.
Suppose a homeomorphism existed. Then we could find a continuous $f:S^2\to \mathbb{R}^2$. $S^2$ is compact, and the continuous image of a compact set is compact, so $\mathbb{R}^2$ would be compact. Contradiction. (If you include the single point at infinity, you get a more interesting situation, the Riemann sphere, which is $\mathbb{C}\bigcup \{\infty\}$ with a suitable topology, and is in fact homeomorphic to $S^2$)
To answer question 1, the statement in question would be to prove that "every point in $S^2$ has a neighbourhood which is homeomorphic to the Euclidean space $\mathbb{R}^2$". To do this, consider the point $a=(x,y,z)$ on the sphere. Consider the stereographic projection which sends $a$ to $(0,0)$ and sends $-a$ to infinity (ie, leaves it out of the domain). This is a continuous map with continuous inverse, and so it is a homeomorphism between a neighborhood of $a$ and $\mathbb{R}^2$.
To answer question 2, (1) is certainly not a map to $\mathbb{R}^2$, it is a map to $\mathbb{R}^3$, plain and simple.