# Regarding the Implicit function theorem

The formulation in James R. Munkres' book Analysis on Manifolds of the Implicit Function Theorem is the following

(Implicit Function Theorem). Let $$A$$ be open in $$\mathbb{R}^{k+n}$$; let $$f:A\rightarrow \mathbb{R}^n$$ be of class $$C^r$$. Write $$f$$ in the form $$f(\boldsymbol{x},\boldsymbol{y})$$, for $$\boldsymbol{x}\in \mathbb{R}^k$$ and $$\boldsymbol{y}\in \mathbb{R}^n$$. Suppose that $$(\boldsymbol{a},\boldsymbol{b})$$ is a point of $$A$$ such that $$f(\boldsymbol{a},\boldsymbol{b}) = 0$$ and $$\det \frac{\partial f}{\partial \boldsymbol{y}}(\boldsymbol{a},\boldsymbol{b})\neq 0.$$ Then there is a neighborhood $$B$$ of $$\boldsymbol{a}$$ in $$\mathbb{R}^k$$ and a unique continuous function $$g:B\rightarrow \mathbb{R}^n$$ such that $$g(\boldsymbol{a}) = \boldsymbol{b}$$ and $$f(\boldsymbol{x},g(\boldsymbol{x})) = 0$$ for all $$\boldsymbol{x}\in B$$. The function $$g$$ is in fact of class $$C^r$$.

Now what I wonder is if one can conclude from this theorem that there is a neighborhood $$U$$ of $$(\boldsymbol{a},\boldsymbol{b})$$ in $$\mathbb{R}^{k+n}$$ such that for $$(\boldsymbol{x},\boldsymbol{y})\in U$$ we have that $$f(\boldsymbol{x},\boldsymbol{y}) = 0\Leftrightarrow \boldsymbol{y} = g(\boldsymbol{x}).$$

I feel instinctively that the latter does not necessarily follow directly but am curious how one can show it. I tried arguing by contradiction: that there is a sequence of points $$\{(\boldsymbol{x}_n,\boldsymbol{y}_n)\}_n$$ converging to $$(\boldsymbol{a},\boldsymbol{b})$$ such that $$f(\boldsymbol{x}_n,\boldsymbol{y}_n) = 0$$ and $$g(\boldsymbol{x}_n) \neq \boldsymbol{y_n}$$ however I could not complete the argument and would therefore be grateful for any help.

• actually you can conclude that from the proof of the theorem. I'll write up an answer shortly – peek-a-boo Jun 17 '19 at 10:58

I'll try to adhere to the notation of the book as much as possible. Recall that $$B$$ is an open neighbourhood of $$a$$, and $$V$$ is an open neighbourhood of $$b$$ (which contains $$g(B)$$). In the book Munkres already showed explicitly that for every $$x \in B$$, we have $$f(x,g(x)) = 0$$. So all that remains to be shown is that for any $$(x,y) \in B \times V$$, if $$f(x,y) = 0$$, then $$y=g(x)$$.
Before we do so, define the projection map $$\pi_2: \Bbb{R}^k \times \Bbb{R}^n \to \Bbb{R}^n$$, by $$$$\pi_2(x,y) = y.$$$$ Now, pick any arbitrary $$(x,y) \in B \times V$$, and suppose that $$f(x,y) = 0$$. Then, \begin{align} g(x) &:= h(x,0) \\ &:= (\pi_2\circ G)(x,0) \\ &= (\pi_2\circ G)(x,f(x,y)) \tag{since f(x,y) = 0} \\ &:= (\pi_2 \circ G)[F(x,y)] \\ &= \pi_2(x,y) \tag{since G := F^{-1}} \\ &:= y \end{align} This completes the proof. As you can see, this really follows mostly by definition of the maps constructed. But I think this result should have been included as part of the conclusion of the theorem.