The following is a question about the nature of the domain of definition when we wish to apply the Implicit Function Theorem.
My question has to do with the nature of the domain of definition, say $ S \subset \mathbb R^{n+m}.$ In most sources I have looked, it is made the assumption that we take $S$ to be an open subset of the Euclidean space $ \mathbb R^{n+m}.$
What if instead of asking $S$ to be an open set, we assume that $S$ is the cartesian product of an open set with an arbitrary set.
Setting: Suppose that $S$ is of the shape $$ S = S_1 \times S_2, \quad \text{where} \quad S_1 \subset \mathbb R^n \quad \text{is open, and} \quad S_2 \subset \mathbb R^m \quad \text{is arbitrary}. $$
Suppose that $ f = (f_1, \ldots, f_m) : S \to \mathbb R^m$ is a vector-valued function defined for $ (x,y) \in S,$ where $ x$ is an $n$-tuple and $y$ is an $m$-tuple. We suppose that $f$ is continuously differentiable in $S.$ Let $(a,b) = (a_1, \ldots, a_n, b_1, \ldots, b_m) \in \text{int} (S)$ such that $ f(a,b) =0.$ We suppose that the Jacobian matrix
$$ J_{f,y} (a,b) = \left[ \frac{ \partial f_i}{\partial y_j} (a,b) \right]_{ \substack{ 1 \leq i \leq m \\ 1 \leq j \leq m}} $$ is invertible.
Question 1: Does there exist an open set $ U \subset \mathbb R^n$ containing the point $ a= (a_1, \ldots, a_n)$ and a continuous function $ g : U \to \mathbb R^m$ such that one has $ g(a)=b $ and $ f(x, g(x)) =0$ for all $ x \in U ?$
Question 2: What can we say about the uniqueness and the differentiability of the function $g ?$
Any reference would be really appreciated.
Edit : Following the comment of nullUser below, let me give the motivation behind my question. As I wrote in the beginning of the post, my question is related to the set of definition of the functions. For this reason, instead of having functions defined everywhere in $ \mathbb R $ let us suppose that our functions are defined only for non-negative reals, namely in the set $[0, \infty).$
Let $f_1, \ldots, f_m: [0, \infty)^{n+m} \to \mathbb R$ be real-valued functions defined for $ (x,y) = (x_1, \ldots, x_n, y_1, \ldots, y_{m}) \in [0, \infty)^{n+m}.$ We set
$$ f = (f_1, \ldots, f_m) : [0, \infty)^{n+m} \to \mathbb R^m.$$
Now we make the following assumptions.
The functions $f_1, \ldots, f_m$ are continuously differentiable on $[0, \infty)^{n+m}.$
There exists a point $(a,b) = (a_1, \ldots, a_n, b_1, \ldots, b_m) \in [0, \infty)^{n+m}$ with $b_1, \cdots, b_m > 0,$ such that $ f_1(a,b)= \cdots = f_m (a,b)=0.$
The Jacobian matrix
$$ J_{f,y} (a,b) = \left[ \frac{ \partial f_i}{ \partial y_j} (a,b) \right]_{ 1\leq i,j \leq m} $$ is invertible.
Question 3: Can we apply the Implicit Function Theorem in order to deduce that there exist an open set $ U \subset \mathbb R^n$ containing the point $a = (a_1, \ldots, a_n)$ and a continuous (uniqueness or differentiability ??) function $ g : U \to \mathbb R^m,$ so that the following hold:
(i). One has $ g(a) = b.$
(ii). For any $ x =(x_1, \ldots, x_n) \in U$ one has $$ f_1 (x, g(x)) = \cdots = f_m (x, g(x)) =0.$$
The point of confusion is that the set $[0, \infty)$ is not an open subset of $\mathbb R.$