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.

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    $\begingroup$ actually you can conclude that from the proof of the theorem. I'll write up an answer shortly $\endgroup$ – peek-a-boo Jun 17 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 \begin{equation} \pi_2(x,y) = y. \end{equation} 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.

  • $\begingroup$ Many thanks for your help, it is similar to the way he proves the uniqueness of the continuous map. I agree he should have added this to the conclusion. $\endgroup$ – Olof Rubin Jun 17 at 11:31

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