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This may be a duplicate of Inverse image of a subset of the codomain with elements without corresponding elements in domain , but the answer given seems to contradict what is intended by CH Edwards in his Advanced Calculus of Several Variables.

I shall frame the question using the following example.

Let there be a mapping $\vec{r}:\mathbb{R}^{2}\to\mathbb{R}^{3}$ defined by $\vec{r}\left(\mathbf{R}\right)=\left\{ x,y,1\right\} $ with $\mathbf{R}=\left\{ x,y\right\} ,$ and let $\mathcal{B}_{\epsilon}\left(\vec{r}\left(\mathbf{R}\right)\right)\subset\mathbb{R}^{3}$ be an open ball of radius $\epsilon>0$ centered on some specific $\vec{r}\left(\mathbf{R}\right).$

Is it proper to call $\vec{r}^{-1}\left(\mathcal{B}_{\epsilon}\left(\vec{r}\left(\mathbf{R}\right)\right)\right)$ the inverse image of $\mathcal{B}_{\epsilon}\left(\vec{r}\left(\mathbf{R}\right)\right)?$

I am specifically interested in those points $\left\{ x,y,z\right\} =\mathcal{B}_{\epsilon}\left(\vec{r}\left(\mathbf{R}\right)\right)$ for which $z\ne1.$ That is, those points not in the image of $\vec{r}.$

The following abbreviated except from Edwards's book is the reason I ask:

Theorem 8.2 The mapping $f:\mathbb{R}^{n}\to\mathbb{R}^{m}$ is continuous if and only if, given any open set $\mathcal{U}\subset\mathbb{R}^{m},$ the inverse image $f^{-1}\left(\mathcal{U}\right)$ is open in $\mathbb{R}^{n}.$ $\dots$

PROOF The inverse image $f^{-1}\left(\mathcal{U}\right)$ is the set of points in $\mathbb{R}^{n}$ that map under $f$ into $\mathcal{U},$ that is, $$ f^{-1}\left(\mathcal{U}\right)=\left\{ \mathbf{x}\in\mathbb{R}^{n}:f\left(\mathbf{x}\right)\in\mathcal{U}\right\} . $$

We prove the "only if" part of the Theorem, and leave the converse as Exercise 8.4.

Suppose $f$ is continuous. If $\mathcal{U}\subset\mathbb{R}^{m}$ is open, and $\mathbf{a}\in f^{-1}\left(\mathcal{U}\right),$ then there exists an open ball $\mathcal{B}_{r}\left(f\left(\mathbf{a}\right)\right)\subset\mathcal{U}.$ Since $f$ is continuous, there exists an open ball $\mathcal{B}_{\delta}\left(\mathbf{a}\right)$ such that $f\left(\mathcal{B}_{\delta}\left(\mathbf{a}\right)\right)\subset\mathcal{B}_{r}\left(f\left(\mathbf{a}\right)\right)\subset\mathcal{U}.$ Hence $\mathcal{B}_{\delta}\left(\mathbf{a}\right)\subset f^{-1}\left(\mathcal{U}\right);$ this shows that $f^{-1}\left(\mathcal{U}\right)$ is open.

Steven Thomas Hatton
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    Yes, for any function $f$ and any subset $S$ of the codomain of $f$, we do call $f^{-1}(S)$ the inverse image of $S$ regardless of whether some (or all) points of $S$ are not in the range of $f$. – Greg Martin May 02 '22 at 19:26
  • $f^{-1}S={x:f(x)\in S}$ regardless of what $S$ is. E.g. in the topological definition of continuity: $f:A\to B$ is continuous iff $f^{-1}S$ is open in $A$ whenever $S$ is open in $B$. – DanielWainfleet May 02 '22 at 19:40
  • @DanielWainfleet The problem with the latter part is that Edwards states your definition as a theorem, and defines continuity as the limit and the value of $f$ existing and equaling one another everywhere on the domain. – Steven Thomas Hatton May 02 '22 at 19:48
  • From what I read in your abbreviated excerpt, the definition of $f^{-1} S$ given in the comment of @DanielWainfleet agrees exactly with the definition of $f^{-1}(\mathcal U)$ given in your second highlighted portion of that excerpt. – Lee Mosher May 03 '22 at 02:53
  • @LeeMosher I was referring to the definition of continuity. The theorem 8.2 of Edwards is the definition of continuity used by others. – Steven Thomas Hatton May 03 '22 at 03:53
  • Okay, then it looks like your question is answered? – Lee Mosher May 03 '22 at 12:37
  • @LeeMosher Not in a way that I can check off. It would be nice to have a second authoritative source. Edwards defines inverse image in one sentence of a proof without giving any examples. – Steven Thomas Hatton May 03 '22 at 14:37
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    Okay, so I'll just say that this is the universal definition of the inverse image of a subset of $Y$ with respect to any function $X \mapsto Y$, regardless of whether that subset is contained in the image of the function. Check any topology book, that's what you'll find. – Lee Mosher May 03 '22 at 16:51
  • @LeeMosher I don't doubt that in the least bit. If someone posts that as an answer with one, or preferably two authoritative references, I will happily accept it. In lieu of that, I will have to set this aside until I have time to examine other sources for myself. – Steven Thomas Hatton May 03 '22 at 17:28
  • https://en.wikipedia.org/wiki/Image_(mathematics)#Inverse_image – Moishe Kohan May 03 '22 at 18:35

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