# Question regarding the use of the Implicit function theorem

Let $$f: U \subseteq \mathbb{R}^m \rightarrow \mathbb{R}^{m+1}$$ be a $$C^k$$ function. Suppose $$f(x)=(f_1(x), f_2(x), \cdots, f_{m+1}(x)) \quad \text{where} \quad det \; \left ( \frac{\partial f_i}{\partial x_j} (x)\right ) \ne 0$$ for $$1\leq i,j \leq m$$ for all $$x \in U$$.

Prove that for every $$x \in U$$, there exist $$W\subseteq U$$ such that $$f(W)$$ is the graph of a function $$y_{m+1}=\varphi(y_1,y_2, \cdots, y_m)$$ of class $$C^k$$.

I'm almost certainly sure we need the implicit theorem to answer this, but since $$f: U \subseteq \mathbb{R}^m \rightarrow \mathbb{R}^{m+1}$$ i have two questions:

1) Is it possible to resolve this exercise just the way it is?

2) What if I change $$f: U \subseteq \mathbb{R}^{m} \rightarrow \mathbb{R}^{m+1}$$ to $$f: U \subseteq \mathbb{R}^m \rightarrow \mathbb{R}^{m}$$ (just trying to see if the exercise was written incorrectly).

• Looks like the dimensions of the domain and co-domain of $f$ should match (what you have under 2)), so that we are taking the determinant of a square matrix. – avs Apr 18 at 6:39
• Oh yeah that was a typo ill fix it now. – ipreferpi Apr 18 at 6:40
• If the following helps any further: If we tried the special case $m = 2$, we would have $U = ]a, b[$ (for simplicity, a connected open interval), the function $$f(x) = (f_{1}(x), f_{2}(x))$$ would parameterize a smooth curve in the $y_1y_2$-plane, and the condition on the determinant would boil down to... what? It should probably be something like $$\left.{df_{1} \over dx}\right|_{x = x_{0}} \neq 0,$$ which is not analogous to the determinant of the total derivative of $f$ (which wouldn't even be a square matrix). – avs Apr 18 at 6:43
• Thanks to point that out, I'll try to make it work when the domain and codomain have the same cardinality. – ipreferpi Apr 18 at 6:59