# Global Implicit Function Theorem - Necessary conditions

I was reading the this paper on Global Implicit function theorem. I am unable to reason one of the assumptions in the following statement:

Let $$f(\mathbf{x}, \mathbf{y}): U \times V \to W$$ be a continuously differentiable function. Here $$U$$ denotes an open subset of $$\mathbb{R}^n$$, $$V$$ an open convex set of $$\mathbb{R}^m$$ such that $$m \geq n$$ while $$W$$ is assumed to be the whole of $$\mathbb{R}^n$$. Furthermore, the rank of the matrix $$\partial f/ \partial \mathbf{y}$$ is $$n$$ for all points $$(\mathbf{x}, \mathbf{y})$$ such that $$f(\mathbf{x}, \mathbf{y}) = \mathbf{0}$$. If in addition to these, the following conditions are true

• For some $$\mathbf{y}_0 \in V$$, there is exactly one $$\mathbf{x}_0 \in U$$ such that $$f(\mathbf{x}_0, \mathbf{y}_0) = \mathbf{0}$$
• For each $$S \in A$$, there is a $$T \in B$$ such that $$\mathbf{y} \in S$$, $$\mathbf{x} \in U$$ and $$f(\mathbf{x}, \mathbf{y}) = \mathbf{0}$$ imply $$\mathbf{x} \in T$$
• Det$$(\partial f/ \partial \mathbf{x}) \neq 0$$ for all points $$(\mathbf{x}, \mathbf{y})$$ such that $$f(\mathbf{x}, \mathbf{y}) = \mathbf{0}$$

then there is a unique $$g: V \to U$$ such that $$f(g(\mathbf{y}), \mathbf{y}) = \mathbf{0}$$ for all $$\mathbf{y} \in V$$ and $$g$$ is continuously differentiable on $$V$$. In the above statement, $$A$$ refers to any family of compact subsets of the metric space $$V$$ such that for each compact set $$C \in V$$, there is $$S \in A$$ such that $$C \subseteq S$$ and similarly $$B$$ refers to any family of compact subsets of the metric space $$U$$ such that for each compact set $$D \in U$$, there is $$T \in B$$ such that $$D \subseteq T$$.

I don't quite understand why should $$\partial f/ \partial \mathbf{y}$$ have a rank $$n$$. I can see why Det$$(\partial f/ \partial \mathbf{x}) \neq 0$$ should hold but not the other thing. Even in the proof, they do not use the condition that $$\partial f/ \partial \mathbf{y}$$ has a rank $$n$$. Furthermore, even in the proof of implicit function theorem (local version), there is no such assumption. So I am not sure why this additional condition is necessary.

Any insight into this will be really helpful! Thank you.