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.
Best Answer
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.