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Problem $8$ in Chapter $1$ of Spivak's A Comprehensive Introduction to Differential Geometry, Vol. 1 reads:

8. For this problem, assume

  1. (The Generalized Jordan Curve Theorem) If $A\subset \mathbb{R}^n$ is homeomorphic to $S^{n-1},$ then $\mathbb{R}^n-A$ has $2$ components, and $A$ is the boundary of each.
  2. If $B\subset\mathbb{R}^n$ is homeomorphic to $D^n=\{x\in\mathbb{R}^n:d(x,0)\le 1\},$ then $\mathbb{R}^n-B$ is connected.

(a) One component of $\mathbb{R}^n-A$ (the "outside of $A$") is unbounded, and the other (the "inside of $A$") is bounded.

(b) If $U\subset\mathbb{R}^n$ is open, $A\subset U$ is homeomorphic to $S^{n-1}$ and $f:U\to \mathbb{R}^n$ is one-to-one and continuous (so that $f$ is a homeomorphism on $A$), then $f(\text{inside of }A)=\text{inside of }f(A)$.

(c) Prove Invariance of Domain, i.e. if $U\subset \mathbb{R^n}$ is open and $f:U\to\mathbb{R}^n$ is one-to-one and continuous, then $f(U)\subset \mathbb{R}^n$ is open.

I have solved (a) and (c), which I will present here.

Proof of (a): $A$ is homeomorphic to $S^{n-1},$ and so is compact. It follows that $A$ is bounded. Let $B$ be a closed ball containing $A.$ Then $\mathbb{R}^n-B\subset \mathbb{R}^n-A$ is connected and clearly unbounded. It must lie in a connected component of $\mathbb{R}^n-A,$ and so one of the components of the latter set is unbounded. It must then be that the other connected component of $\mathbb{R}^n-A$ is contained in $B,$ so that this connected component is bounded.

Proof of (c): Let $y\in f(U)$ and consider the preimage of $y$ under $f$, call it $x$, in $U$. Then there is some closed ball $B$ about $x$ in $U$. The boundary $A$ of $B$ is homeomorphic to $S^{n-1}$ and $x$ lies in the inside of $A$, so by part (b) $f(\text{inside of }A)=\text{inside of }f(A)$ is an open ball which contains $y$. We have thus found a neighborhood of $y$ contained in $f(U)$, so $f(U)$ is open in $\mathbb{R}^n$.

I do not know how to solve (b). I am completely and utterly stuck. I'm clearly missing something obvious. Any hint or even just the solution would be appreciated.

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  • $\begingroup$ I think I can show this if the following "fact" is true: If $U\subset $ inside of $A$ is open with $\partial U = A$, then $U = $inside of $A$. The problem is I don't know if this is true... $\endgroup$ – user25959 Nov 20 '18 at 21:30
  • $\begingroup$ As it is stated in your question, (b) does not make sense. As an example consider $U = \mathbb{R}^2 \setminus \{ 0 \}$, $A = S^1$. The inside of $A$ is not contained in $U$, hence $f(\text{inside of } A)$ is not defined. So you will need the additional assumption that $\text{inside of } A \subset U$. $\endgroup$ – Paul Frost Dec 31 '18 at 15:43
  • $\begingroup$ ^^ yes I agree, and when doing the problem I took that assumption for granted since It was needed to do the problem. $\endgroup$ – D. Brogan Dec 31 '18 at 17:38
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For $n = 1$ both 1. and 2. are wrong, so we must assume $n > 1$.

Let us first analyze the proof of (a). Here we only need 1. That the complement of a standard closed ball $B = D(0,r)$ is connected and unbounded is a trivial fact which does not use the full strength of 2.

Based on (a), for $A\subset \mathbb{R}^n$ homeomorphic to $S^{n-1}$, let $\text{ins}(A)$ denote the inside of $A$ and $\text{outs}(A)$ the outside of $A$.

As it is stated in your question, (b) does not make sense. As an example consider $U = \mathbb{R}^n \setminus \{ 0\}$ and $A = S^{n-1}$. The inside of $A$ is not contained in $U$, hence $f(\text{ins}(A))$ is not defined. So you will need the additional assumption that $\text{ins}(A) \subset U$.

We shall prove only a special case of (b) which is sufficient to show (c):

Let $B = D(x,r)$ be a closed ball contained in $U$ and $A$ its boundary (so that $\text{ins}(A)$ is the open ball with radius $r$ around $x$). Then $f(\text{ins}(A)) = \text{ins}(f(A))$.

Obviously $f(B)$ is compact, hence $C = \mathbb{R}^n \setminus f(B) \subset \mathbb{R}^n \setminus f(A)$ is unbounded. Using 2. we see that $C \subset \text{outs}(f(A))$.

We can write $\mathbb{R}^n \setminus f(A) = C \cup f(\text{ins}(A)) = \text{outs}(f(A)) \cup \text{ins}(f(A))$. These are two (potentially different) decompositions of $\mathbb{R}^n \setminus f(A)$ into disjoint sets.

$f(\text{ins}(A))$ is a connected subset of $\mathbb{R}^n \setminus f(A)$, hence contained either in $\text{ins}(f(A))$ or in $\text{outs}(f(A))$. Assume it is contained in $\text{outs}(f(A))$. Then $\mathbb{R}^n \setminus f(A) = C \cup f(\text{ins}(A)) \subset\text{outs}(f(A))$ which is impossible. Therefore $f(\text{ins}(A)) \subset \text{ins}(f(A))$. But now $$C \cup f(\text{ins}(A)) = \text{outs}(f(A)) \cup \text{ins}(f(A)) ,$$ $$C \subset \text{outs}(f(A)), f(\text{ins}(A)) \subset \text{ins}(f(A)) .$$ This implies $$C = \text{outs}(f(A)), f(\text{ins}(A)) = \text{ins}(f(A)) .$$

Edited:

I would suggest to replace (b) by its essence:

Let $f : D^n \to \mathbb{R}^n$ be a continuous injection. Then $f(\text{ins}(S^{n-1})) = \text{ins}(f(S^{n-1}))$.

Note that for an arbitrary $A\subset \mathbb{R}^n$ homeomorphic to $S^{n-1}$, the set $B(A) = \overline{\text{ins}(A)} = \text{ins}(A) \cup A$ is closed and bounded, i.e. compact, but in general not homeomorphic to $D^n$. Thus the above proof breaks down. To get an example for this phenomenon take the Alexander horned sphere $S \subset \mathbb{R}^3$. Define $A = \phi(S)$, where $\phi : \mathbb{R}^3 \setminus \{ 0 \} \to \mathbb{R}^3 \setminus \{ 0 \}, \phi(x) = \dfrac{x}{\lVert x \rVert ^2}$. Then $B(A)$ is not homeomorphic to $D^3$.

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