Isn't there a better way to put the canonical smooth structure on the $n$-sphere? My differential geometry notes put a smooth structure on the $n$-sphere (denoted $S_n$) as follows. Firstly, $S_n$ is taken as a subset of $\mathbb{R}^{n+1}.$ Secondly, we define $U_0 = S_n \setminus \{(1,0,\ldots,0)\}$ and $U_1 = S_n \setminus \{(-1,0,\ldots,0).$ Charts with domain $U_0$ and $U_1$ ere then defined by stereographic projection.
I have a couple of issues with this:


*

*Its non-systematic; stereographic projection is "pulled out of hat" arbitrarily.

*Its non-explicit; it remains to figure out which charts on $S_n$ are actually compatible with this smooth structure.
Isn't there a better way of doing this?
 A: You can apply to $f(x)=\sum x_i^2-1$ the theorem that the zero set $S$ of a smooth function $f$ defined on $\mathbb R^N$ is a smooth manifold as soon as $grad (f)$ is never zero on $S$.    
In general the corresponding charts are deduced from the implicit function theorem and are thus not very explicit :-) But:
In the case of the sphere $S_n$ you have an atlas consisting of $2n+2$ very explicit charts.
Namely define (for $i=1,\cdots,n+1$):   $$U_i^+=\{x \in S_n|x_i> 0\} \quad (\operatorname {resp.} U_i^{-\vphantom{|}}=\{x \in S_n|x_i\lt 0\})$$  and put $$\phi_i^+(x)=(x_1,\cdots,\hat {x_i},\cdots, x_n) \quad (\operatorname {resp.} \phi_i^{-\vphantom{|}}(x)=(x_1,\cdots,\hat {x_i},\cdots, x_n))$$ .  
A subtle point
Given a subset $A\subset \mathbb R^N$ the question "Is $A$ a smooth submanifold of $\mathbb R^N ?$" is perfectly well posed and has an unambigous answer, which is (of course) "yes!" or "no!"
Let me emphasize that the answer has nothing to do with charts and atlases and is perhaps paradoxically easier to explain to someone who has never heard of abstract manifolds, defined as topological spaces endowed with a family of charts blah blah blah.
This point of view is explained on page 1 (!) of Topology from the Differentiable Viewpoint, a 64 page pamphlet by John Milnor, one of the most remarkable differential/algebraic topologists of all times.
That said, if the answer to the above question is "yes" (and the first two lines of my answer describe an example where this is the case) there is of course  a canonical way to endow the smooth submanifold $A\subset \mathbb R^N$ with the structure of an abstract manifold with its required charts.
Still I find it healthy to be aware that there is a very low-tech notion of differentiable submanifold of $\mathbb R^n$, which logically and historically predates the abstract concept of manifold.  
Edit: just watch!
Here is a wonderfully nostalgic video where Milnor defines the concept of submanifold at 13:53.
A: Given a set of charts $\mathcal{U} = \{(U_{\alpha}, \phi_{\alpha})\}_{\alpha \in A}$ that covers a topological manifold $M^m$ (that is, any $x \in M$ is in some $U_{\alpha}$) and such that each transition map $\phi_{\alpha} \circ \phi_{\beta}^{-1} : \phi_{\beta}\left( U_{\alpha} \cap U_{\beta} \right) \to \phi_{\alpha}\left( U_{\alpha} \cap U_{\beta} \right)$ is smooth (as a function from an open set of $\mathbb{R}^m$ to another), we know from general principles that there exists a unique maximal atlas $\mathcal{A}$ for $M$ such that $\mathcal{U} \subseteq \mathcal{A}$. It is in fact the set of all pairs $(U, \phi)$ which are compatible with the charts in $\mathcal{U}$ (in the sense that all of the transition functions $\phi_{\alpha} \circ \phi^{-1} : \phi\left( U_{\alpha} \cap U \right) \to \phi_{\alpha}\left( U_{\alpha} \cap U \right)$ are diffeomorphisms). Indeed, compatibility between all of these added pairs is automatic. Depending on our knowledge and on the simplicity of $\mathcal{U}$, this might give an effective criterion to determine whether a given pair $(U, \phi)$ is in $\mathcal{A}$ or not.

In the context of your question, things are somewhat even better.
Let's think of $S^n$ as the unit sphere in $\mathbb{R}^{n+1}$. For $P \in S^n$, define $U_P = S^n \backslash \{ P \}$ and write $\sigma_P : U_P \to \mathbb{R}^n$ for the stereographic projection from the "P-pole" onto the tangent plane to $S^n$ at $-P$ (some smooth and standarized identification between this tangent plane and $\mathbb{R}^n$ is chosen).
You took $\mathcal{U} = \{(U_{N}, \sigma_{N}), (U_{S}, \sigma_{S}) \}$ where $N$ and $S$ denote the north pole and the south pole respectively. We want to identify what is the maximal atlas $\mathcal{A}$ associated to $\mathcal{U}$. Some computation shows that the set $\mathcal{V} = \{(U_{P}, \sigma_{P})\}_{P \in S^n}$ is in $\mathcal{A}$.
Now, any $(U_{\alpha}, \phi_{\alpha}) \in \mathcal{A}$ is such that $U_{\alpha} \subset S^n$ is an open but not closed subset of $S^n$ ; In particular, it is not all of $S^n$. Hence, there exists $P_{\alpha} \in S^n \backslash U_{\alpha}$ ; This means that $U_{\alpha} \subseteq U_{P_{\alpha}}$. We conclude that the transition map
$$ \phi_{\alpha} \circ \sigma_{P_{\alpha}}^{-1} : \sigma_{P_{\alpha}} \left( U_{\alpha} \right) \to \phi_{\alpha}\left( U_{\alpha} \right)   $$
is a diffeomorphism between open sets of $\mathbb{R}^n$ (equipped with its standard smooth structure). It is not hard to see that this transition map determines completely $(U_{\alpha}, \phi_{\alpha})$ as a function ; What's more is that any such diffeomorphism between open sets of $\mathbb{R}^n$ determines an element $(U_{\alpha}, \phi_{\alpha}) \in \mathcal{A}$ (here, we use the fact that $\mathcal{V} \subset \mathcal{A}$). This yields the following criterion :
"A function $\phi : U (\subset S^n) \to \mathbb{R}^n$ is an element of $\mathcal{A}$ if and only if there exists $P \in S^n \backslash U$ such that $\phi \circ \sigma_P^{-1} : \sigma_P(U) \to \phi(U)$ is a diffeomorphism with respect to the standard smooth structure on $\mathbb{R}^n$."
This shows that the knowledge of a "good" atlas such as $\mathcal{V}$ gives an even simpler criterion for compatibility.
