This one seems obvious to me, but I've been wrong enough lately to doubt myself. I refer to C.H. Edwards, Jr.'s Advanced Calculus of Several Variables, Theorem III-3.4, (the multivariable implicit mapping theorem.) Pages 190 and 191.
Let the mapping $\mathbf{G}:\mathbb{R}^{m+n}\rightarrow\mathbb{R}^{n}$ be $\mathscr{C}^{1}$ in a neighborhood of the point $(\mathbf{a},\mathbf{b})$ where $\mathbf{G}(\mathbf{a},\mathbf{b})=\mathbf{0}$. If the partial derivative matrix $\mathit{D}_{2}\mathbf{G}(\mathbf{a},\mathbf{b})$ is non-singular then there exists a neighborhood $\mathit{U}$ of $\mathbf{a}$ in $\mathbb{R}^{m}$, a neighborhood $\mathit{W}$ of $(\mathbf{a},\mathbf{b})$ in $\mathbb{R}^{m+n}$, and a $\mathscr{C}^{1}$ mapping $\mathbf{h}:\mathit{U}\rightarrow\mathbb{R}^{n}$, such that $\mathbf{y}=\mathbf{h}(\mathbf{x})$ solves the equation $\mathbf{G}(\mathbf{x},\mathbf{y})=\mathbf{0}$ in $\mathit{W}$.''
In the proof, a mapping $\mathbf{f}:\mathbb{R}^{m+n}\rightarrow\mathbb{R}^{m+n}$ defined by $\mathbf{f}(\mathbf{x},\mathbf{y})=(\mathbf{x},\mathbf{G}(\mathbf{x},\mathbf{y}))$ is introduced. The neighborhood $\mathit{V}$ of $(\mathbf{a},\mathbf{0})$ is also defined.
I'm including a link to my interpretation of Figure 3.8 on page 191. The notation in my diagram is somewhat different from what Edwards uses. The relevant differences are that I used $\{\mathbf{x}_{\alpha},\mathbf{y}_{\beta}\}=(\mathbf{a},\mathbf{b})$ and $\mathbf{y}(\mathbf{x})=\mathbf{h}(\mathbf{x})$. I labled the axes $\mathbf{x}$ and $\mathbf{y}$, which Edwards didn't do.
The proof then ensures the existence of a $\mathscr{C}^{1}$ mapping $\mathbf{g}:\mathbb{R}^{m+n}\rightarrow\mathbb{R}^{m+n}$, inverse to $\mathbf{f}:\mathbb{R}^{m+n}\rightarrow\mathbb{R}^{m+n}$. Edwards then states:
Since $\mathbf{f}(\mathbf{x},\mathbf{y})=\mathbf{0}$ if and only if $\mathbf{G}(\mathbf{x},\mathbf{y})=\mathbf{0}$, it is clear that $\mathbf{g}$ maps the set $\mathit{U}\times\mathbf{0}\subset\mathbb{R}^{m+n}$ one-to-one onto the intersection with $\mathit{W}$ of the zero set $G(\mathbf{x},\mathbf{y})=\mathbf{0}.$
I believe that should actually say "Since $\mathbf{f}(\mathbf{x},\mathbf{y})=(\mathbf{x},\mathbf{0})$ if and only if $\mathbf{G}(\mathbf{x},\mathbf{y})=\mathbf{0}$,...".
Is that a valid correction?
Notice that $\mathbf{x}=\mathbf{0}$ isn't necessarily in $\mathit{U}$. In that case
$\mathbf{f}(\mathbf{x},\mathbf{y})=(\mathbf{0},\mathbf{G}(\mathbf{0},\mathbf{y}))=(\mathbf{0},\mathbf{0})=\mathbf{0}$
may not even be defined.
Yes. There is a minor error in the statement of the proof, corrected as suggested.