Construct a parameterization of the inverse image of a regular value. 
Let $f: U \subset \Bbb{R}^n \to \Bbb{R}^m$ be a $C^k$ function. Suppose that $c \in \Bbb{R}^n$ is a regular value of $f$. Show that
(a) $M = f^{-1}(c)$ is locally the graph of a $C^k$ function.
(b) Construct a parameterization of $M$ using the previous item.

(a) Sketch of the proof. $\quad$ I can construct two sets $E_1, E_2$ with $E_1 \oplus E_2 = \Bbb{R}^n$ and use the Implicit Function theorem at any $p \in f^{-1}(c)$ to get a function $\varphi_p: U \to V$ where $U$ and $V$ are open satisfying $f(x,\varphi_p(x)) = c$, from where $f^{-1}(c)\cap(U \times V) = graph(\varphi_p)$.
(b) My idea is to show that $\psi_p: \Bbb{R}^{n-m} \to \Bbb{R}^n$ by $\psi_p(x) = (x,\varphi_p(x))$ is a parameterization and $f^{-1}(c) = \bigcup_p \psi_p$. I need to show that $\psi_p$ is a imersion and a homeomorphism. I can show that $\psi_p$ is a imersion, but I cannot show that $\psi_p^{-1}$ is continuous.
Also, a result follows that if $M$ is a $C^k$ manifold then $M$ is locally the graph of a $C^k$ map. Using this result, the item (a) is proved. The proof I know about this result, using indirectly the Implicit Function Theorem, it use some results about submersion. I would like how to prove this result using explicitly the Implicit Function Theorem, like in the item (a).

I dont know if this is characterized as "two question in one". If this is a problem, the first question is my priority.
Thanks for the advance.
 A: When trying to show claims as the above, I have always found it most clarifying to use the Inverse Function Theorem first.
Given a $C^k$ map $F\colon U \subset \mathbb{R}^n \to \mathbb{R}^n$ the Inverse Function Theorem states that if there is $x \in U$ such that $DF(x) \colon \mathbb{R}^n \to \mathbb{R}^n$ is an isomorphism, then there is a neighborhood $V$ of $x$ in $U$ such that the restriction $F|_V$ is an homeomorphism onto its image, that is, there is a continuous map $G\colon F(V) \to V$ such that $G \circ F|_V = id_{V},$ $F|_V \circ G = id_{F(V)},$ and moreover $G \in C^k(F(V)).$
In your problem, let us write $\mathbb{R}^n = \mathbb{R}^{n-m} \times \mathbb{R}^m \ni (x,y)$ and fix a point $(x_0, y_0) \in M = f^{-1}(c)$.
Assume without loss of generality that the map $D_yf(x_0, y_0) := D f(x_0, y_0)|_{\{0\}\times\mathbb{R^m}}\colon \mathbb{R}^m \to \mathbb{R}^m$ is invertible (we may always assume this using that $c$ is a regular value, together with a suitable relabelling of the variables).
Consider the map $F\colon U \subset \mathbb{R}^n \to \mathbb{R}^n$ defined by $F(x, y) = (x, f(x,y)).$ We have
\begin{equation}
DF(x_0, y_0) = 
\left(\begin{array}{}
id_{\mathbb{R}^{n-m}} & 0 \\
D_xf(x_0, y_0) & D_yf(x_0, y_0)
\end{array}\right)
\end{equation}
so $\det DF(x_0, y_0) = \det D_yf(x_0, y_0) \neq 0$ and $DF(x_0, y_0)$ is an isomorphism.
By the inverse function theorem, there is a neighborhood $V\subset \mathbb{R}^n$ of $(x_0, y_0)$ where $F$ is invertible. Notice that $F(V)$ is then a neighborhood of $(x_0, c).$
Let $G$ be the $C^k$ inverse function of $F$ as above and consider the map
\begin{equation}
\varphi \colon F(V)\subset \mathbb{R}^n = \mathbb{R}^{n-m} \times \mathbb{R}^m \to \mathbb{R}^m, \quad (x, t) \mapsto \pi_2 \circ G(x,t)
\end{equation} 
where $\pi_2(x,y) = y$ is just the projection onto the second coordinate.
By definition of $G$ it is easy to see that $(x, y) \in V \cap f^{-1}(t)$ (where $t \in \pi_2(F(V))$) if and only if $y = \varphi(x, t).$
In particular $(x, y) \in V\cap M$ if and only if $y = \varphi(x, c).$ 
If we write $\pi_1(x,y) = x,$ it is now clear that a local parametrization of $M$ around $(x_0, y_0)$ is given by 
\begin{equation}
\psi \colon \pi_1(V) \to \mathbb{R}^n, \quad x \mapsto (x, \varphi(x, c)).
\end{equation}
We notice that $\psi = G \circ \iota_c$ where $\iota_c \colon \pi_1(V) \to \mathbb{R}^n$ is defined by $\iota_c(x) = (x, c)$.
Since both $\iota_c$ and $G$ are homeomorphism onto their image (this is easy to show for $\iota_c$ and is clear for $G$ because of the Inverse Function Theorem), we conclude that the composition $\psi$ is also an homeomorphism onto its image, which is what you wanted to show.
P.S. Notice that I have simply rewritten the proof of the Implicit Function Theorem carefully, which shows that the functions involved are all nice (i.e. they are compositions of homeomorphisms onto their image).
