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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).

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    $\begingroup$ 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. $\endgroup$ – avs Apr 18 at 6:39
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    $\begingroup$ Oh yeah that was a typo ill fix it now. $\endgroup$ – ipreferpi Apr 18 at 6:40
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    $\begingroup$ 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). $\endgroup$ – avs Apr 18 at 6:43
  • $\begingroup$ Thanks to point that out, I'll try to make it work when the domain and codomain have the same cardinality. $\endgroup$ – ipreferpi Apr 18 at 6:59

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