# Spivak Differential Geometry 1 Chapter 1 Problem 8

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.

• 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... – user25959 Nov 20 '18 at 21:30
• 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 $\text{inside of } A \subset U$. – Paul Frost Dec 31 '18 at 15:43
• ^^ yes I agree, and when doing the problem I took that assumption for granted since It was needed to do the problem. – D. Brogan Dec 31 '18 at 17:38

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 $$D^3$$.