Proving that a map between abstract manifolds is smooth 
Let $M=\{(x,y)\in\mathbf{R}^n\times\mathbf{R}^p\mid ||x||^2-||y||^2=1\}$ be a submanifold of $\mathbf{R}^{n+p}$. Prove that $M$ is diffeomorphic to $\mathbf{S}^{n-1}\times \mathbf{R}^p$.

The map $f:M\to \mathbf{S}^{n-1}\times \mathbf{R}^p:(x,y)\mapsto \left(\dfrac{x}{\sqrt{1+||y||^2}},y\right)$ is easily seen to be a well-defined bijection. I am having trouble proving it is a smooth map.
The definition of a smooth map $f:M\to N$ between manifolds given in my syllabus is: there exists a map $(U,\phi)$ in $a$ and a map $(V,\psi)$ in $f(a)$ such that $\psi\circ f\circ\phi^{-1}$ is smooth. Obivously, we don't want to use explicit maps in this case.
My professor gave us two theorems without proof:

(1) If $f:X\to Z$ is a map and $i:Y\to Z$ is an immersion. If $\operatorname{Im}(f)\subset Y$ and $\overline{f}$ is $f$ with the image restricted to $Y$, then $i\circ \overline{f}=f$ is smooth iff $\overline{f}$ is smooth.
(2) If $f:Z\to X$ is a map and $p:Y\to Z$ is a submersion. Then $f\circ p$ is smooth iff $f$ is smooth.

Then he argued in the following way:
Let $i$ be the inclusion of $\mathbf{S}^{n-1}\times \mathbf{R}^p$ in $\mathbf{R}^n\times\mathbf{R}^p$. Then $f$ is smooth iff $i\circ f$ is smooth. Since $i\circ f$ lands in $\mathbf{R}^{n}\times\mathbf{R}^p$, we "got rid using this criterium of one manifold".
Let $I$ be the inclusion of $M$ in $\mathbf{R}^n\times \mathbf{R}^p$ and $\tilde{f}$ be $ f$ extended to $\mathbf{R}^n\times \mathbf{R}^p$. Then $i\circ f=I\circ \tilde{f}$ is smooth iff $\tilde{f}$ is smooth. Since $\tilde{f}$ is now a map from $\mathbf{R}^n\times \mathbf{R}^p$ to itself which is obviously smooth, we are done and we conclude that $f$ was smooth.

From the way I see it, the statement of the theorems have nothing to do with the application to this problem.
I also don't see why they are true. Is someone familiar by the way my professor attacks this problem using the inclusions and extensions? Is there any literature in which this is clearly explained?
Thanks for any answer which might be helpful.
 A: Note that $M$ is a submanifold of $\Bbb R^n \times \Bbb R^p$, whose dimension is $n+p - 1$. Coordinate maps on $M$ can be created by treating $\Bbb R^n = \Bbb R \times \Bbb R^{n-1}, x = (x_1, \bar x)$. For instance, defining $$U =\{(x_1, \bar x, y) \in M \mid x_1 > 0\}$$ and $$\phi(x_1, \bar x, y) = (\bar x, y)$$ forms a chart on $M$ with inverse
$$\phi^{-1}(\bar x, y) = \left(\sqrt{1 + \|y\|^2 - \|\bar x\|^2}, \bar x, y\right)$$
$\Bbb S^{n-1}$ is a submanifold of $\Bbb R^n$, with similar charts.
You express $$f : M \to \Bbb S^{n-1}\times \Bbb R^p : (x,y)\mapsto \left(\dfrac{x}{\sqrt{1+||y||^2}},y\right)$$
But note that this is carrying $\Bbb R^{n+p}$ coordinates to $\Bbb R^{n+p}$ coordinates. It is not expressed in terms of coordinate systems on $M$ or $\Bbb S^{n-1} \times \Bbb R^p$. To prove smoothness per the definition, you need to use coordinates on the two submanifolds. 
However, rather than resorting to charts, your professor is making use of the fact that the identity map restricted to $M$ is an immersion of $M$ into $\Bbb R^n \times \Bbb R^p$. This is generally true of any sub-manifold which is also a subset of the original manifold (and follows almost directly from the definition of "submanifold").
And noting that $x \ne 0$ for all $(x,y) \in M$, there is an obvious submersion $$\Bbb R^n\setminus \{0\} \to \Bbb S^{n-1} : x \mapsto \dfrac x{\|x\|}$$
which induces a submersion $p$ of $(\Bbb R^n-\{0\})\times\Bbb R^p$ onto $\Bbb S^{n-1} \times \Bbb R^p$.
As a map from  $\Bbb R^{n+p}$ to itself, $f$ is easily seen to be smooth: just differentiate. By applying the two theorems with the indicated immersion and submersion, your professor is leveraging the fact that this "full" $f$ is smooth to show that its restriction to $M$ and $\Bbb S^{n-1}\times \Bbb R^p$ is also smooth.

As for why the two theorems are true, if follows from the definition of "immersion" and "submersion". That the smoothness of $\bar f$ implies that of $f = i \circ \bar f$ is a just an application of the chain rule. That the smoothness of $f$ implies that of $\bar f$ is the gist of the Inverse Function Theorem, except translated to manifolds.
The submersion theorem is true because submersions have right inverses: If $p : X \to Y$ is a submersion, then there is an immersion $q : Y \to X$ with $p \circ q$ geing the identity map on $Y$. This is the gist of the Implicit Function Theorem, except translated to manifolds.
