Does Edwards need to explicitly state $a\in\mathcal{W}$ in his Inverse Mapping Theorem?

Question

In view of the answer to my question: A function is continuously differentiable in an open neighborhood of $x_o$ and differentiable at $x_o$, is it continuously differentiable at $x_o$? it seems to me that in the theorem quoted below, Edwards should explicitly state $a\in\mathcal{W}$. So I ask: is the situation any different in multivariable calculus? That is, without explicitly stating that $a\in\mathcal{W},$ is Edwards's theorem correct? Is it not possible to have a nonsingular $f^{\prime}\left(a\right),$ and still have $f$ not $\mathscr{C}^1$ at $a$.

C.H. Edwards's Advanced Calculus of Several Variables

Theorem III 3.3 Suppose that the mapping $f:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is $\mathscr{C}^{1}$ in a neighborhood $\mathcal{W}$ of the point $a$, with the matrix $f^{\prime}\left(a\right)$ being nonsingular. Then $f$ is locally invertible--- there exist neighborhoods of $a$ $\mathcal{U}\subset\mathcal{W}$ and $\mathcal{V}$ of $b=f\left(a\right)$, and a one-to-one mapping $g:\mathcal{V}\to\mathcal{W}$ such that

$$ g\left(f\left(x\right)\right)=x\text{ for }x\in\mathcal{U}, $$

and

$$ f\left(g\left(y\right)\right)=y\text{ for }y\in\mathcal{V}. $$

In particular, the local inverse of $g$ is the limit of the sequence $\left\{ g_{k}\right\} _{0}^{\infty}$ of successive approximations defined inductively by

$$ g_{0}\left(y\right)=a,g_{k+1}\left(y\right)=g_{k}\left(y\right)-f^{\prime}\left(a\right)^{-1}\left[f\left(g_{k}\left(y\right)\right)-y\right] $$

for $y\in\mathcal{V}.$

Answer 1

In the definitive section I-7, Edwards states:

Let $\mathcal{D}$ be a subset of $\mathbb{R}^{n},$ and $f$ a mapping of $\mathcal{D}$ into $\mathbb{R}^{n}$ (that is, a rule that associates with each point $x\in\mathcal{D}$ a point $f\left(x\right)\in\mathbb{R}^{n}$). We write $f:\mathcal{D}\to\mathbb{R}^{n},$ and call $\mathcal{D}$ the domain (of definition) of $f.$

In order to define $\lim_{\mathbf{x}\to\mathbf{a}}f\left(\mathbf{x}\right),$ the limit of $f$ at $\mathbf{a},$ it will be necessary that $f$ be defined at points arbitrary close to $\mathbf{a},$ that is, that $\mathcal{D}$ contains points arbitrarily close to $\mathbf{a}.$ However we do not want to insist that $\mathbf{a}\in\mathcal{D},$ that is, that $f$ be defined at $\mathbf{a}.$ For example, when we define the derivative $f^{\prime}\left(a\right)$ of a real-valued single-variable function as the limit of its difference quotient at $a.$

$f^{\prime}\left(a\right)=\lim_{x\to a}\frac{f\left(x\right)-f\left(a\right)}{x-a},$

this difference quotient is not defined at $a.$

To my knowledge, Edwards gives no formal definition of neighborhood. There are a number of similar examples where we assume something to be true for points arbitrarily close to a point $\mathbf{a}$, but not at the point $\mathbf{a}.$ So I left open the possibility that open neighborhood $\mathcal{U}$ of $\mathbf{a}$ allows $\mathbf{a}\notin\mathcal{U}.$

In the Mathematical Appendix to Gravitation and Inertia, by Ignazio Ciufolini and John Archibald Wheeler is found:

A neighborhood of a point $P$ is a set of $\mathcal{S}$ containing an open set containing $P.$

At first, I thought this addresses my question. But, while composing this answer, it occurred to me that the adjective open does not appear in the term defined as neighborhood.

So my answer is that Edwards should define what he means by open neighborhood $\mathcal{U}$ of $\mathbf{a},$ and if, by that definition, open neighborhood, does not require $\mathbf{a}\in\mathcal{U},$ the Theorem, as stated, is incorrect. Otherwise, it is acceptable.