Jordan curve and Schönflies theorems

Put an equivalence relation on the unit disk by making all points on the boundary circle equivalent to each other. Show that the resulting quotient space $D^2 / \sim$ is homeomorphic to $S^2$.

Construct an example of a simple closed curve in the plane where a horizontal line intersects the curve in an infinite number of points but the curve contains no horizontal line segments.

Show that the region in the figure from Exercise 1.9.95 (a polygonal annular region) is homeomorphic to the region $R$ enclosed between the squares $[-1, 1] \times [-1, 1]$ and $[-2, 2] \times [-2, 2]$. Do this both by a direct argument and by breaking each region into two regions to which we can apply the polygonal Schönflies theorem.

Use the previous exercise to show that a Möbius band cannot be embedded in the plane.

Theorem (strong form of the Schönflies theorem): let $C = f(S^{1})$ be a simple closed curve in the plane and $\mathbb {R}^{2} \setminus C = A \cup B$ as given by the Jordan curve theorem, with $A$ bounded. Then there is an ambient isotopy $G: \mathbb {R}^{2} \times I \to \mathbb {R}^{2} \times I$ with $G_{0} = \mathrm{id}$, so that $G_{1}(C) = S^{1}$ and $G_{1}(A) = D^{2}$. The ambient isotopy can be chosen to be the identity outside a large disk.

Lemma: any embedded disk $f: D^{2} \to \mathbb {R}^{2}$ is ambient isotopic to $g: D^{2} \to \mathbb {R}^{2}$ with $g(D^{2}) = D^{2}$ and $g(S^{1}) = S^{1}$, where the ambient isotopy can be chosen to be the identity outside a large disk.

Use the theorem above to prove the lemma.

The following facts are useful here and in the exercises below. If $C = f(\left\{ 0\right\} \times S^{1})$ is an embedded circle in an orientable surface $M$, then $f$ extends to an embedding of $D^{1} \times S^{1}$ into $M$, and $M \setminus C$ is connected iff $M \setminus f(D^{1} \times S^{1})$ is connected. If $C = f(\left\{ 0\right\} \times S^{1})$ is an embedded circle in a nonorientable surface, then either there is an extension of $f$ to an embedding of $D^{1} \times S^{1}$ into $M$ as above, or there is an embedding of the Möbius band $B$ into $M$ with $C$ corresponding to the center circle; in either case, the connectivity of $M \setminus C$ is equivalent to the connectivity of the complement of the neighborhood of $C$ (either $f(D^{1} \times S^{1})$ or $f(B)$). If $C_{1}, \ldots , C_{k}$ are disjoint embedded circles, then we can choose the extensions so that the neighborhoods $f_{i}(D^{1} \times S^{1})$ or $f_{i}(B)$ are disjoint.

Show that the genus (Exercise 2.9.99) of $S$ is $0$ and the genus of $T$ is $1$.

Use the reduced Mayer--Vietoris sequence of Exercise 6.13.6 to show that, if $A$ and $B$ are path connected, $A \cap B \neq \emptyset$, $A \cup B = X$, and the hypotheses giving the Mayer--Vietoris exact sequence apply, then $X$ is path connected.

We apply the reduced Mayer--Vietoris exact sequence to prove the Jordan curve theorem, working in $S^2$ rather than $\mathbb {R}^{2}$: there is a homeomorphism between $\mathbb {R}^{2}$ and $S^2 \setminus \left\{ N\right\}$, where $N$ is the north pole, and since there is a rotation sending any one point on the sphere to another (the sphere is homogeneous, as is any connected surface, see Exercise 2.9.34), $S^2 \setminus \left\{ p\right\}$ is homeomorphic to $\mathbb {R}^{2}$ for any point $p$.

Suppose that $C$ is a compact set in $\mathbb {R}^{2}$ and $h(C)$ its image under a homeomorphism between $\mathbb {R}^{2}$ and $S^2 \setminus \left\{ p\right\}$. Show that one of the path components of $\mathbb {R}^{2} \setminus C$ contains all points outside some large ball, and that there is a homeomorphism between this path component and $K \setminus \left\{ p\right\}$, where $K$ is the path component of $S^2 \setminus h(C)$ which contains $p$. Show that for all other path components of $\mathbb {R}^{2} \setminus C$, there is a homeomorphism to a corresponding path component of $S^2 \setminus h(C)$.

Exercise 6.14.1 allows the problem of a simple closed curve separating $\mathbb {R}^{2}$ into two path components to be transferred to a problem in the sphere. What is needed first is that an arc does not separate the sphere into multiple path components. By an arc we mean the homeomorphic image of an interval $[a,b]$. (Definition 6.14.1: a set $C$ disconnects $S^2$ if $S^2 \setminus C$ is not path connected, or, equivalently, $\bar{H}_0(S^2 \setminus C) \not\simeq 0$.)

Let $A$ be an arc $h([a,b])$ in $S^2$ and $c = \frac{1}{2}(a+b)$. Let $A_1 = h([a,c])$, $A_2 = h([c,b])$. Use the Mayer--Vietoris exact sequence with

to show that $\bar{H}_0(S^2 \setminus A_1) \oplus \bar{H}_0(S^2 \setminus A_2) \simeq \bar{H}_0(S^2 \setminus A)$.

Use Exercise 6.14.2 to show that if an arc $A$ disconnects $S^2$ (Definition 6.14.1), then there is a sequence of subarcs $\cdots \subset A_i \subset A_{i-1} \subset \cdots \subset A_1 \subset A$ so that the maps induced by inclusions $\bar{H}_0(S^2 \setminus A) \to \bar{H}_0(S^2 \setminus A_1) \to \cdots \to \bar{H}_0(S^2 \setminus A_i)$ each map a given nontrivial element $[p] - [q]$ to a nontrivial element. Moreover, show that these subarcs intersect in a point.

Use Exercise 6.14.3 to prove Theorem 6.14.1: if $A$ is an arc in $S^2$, then $S^2 \setminus A$ is path connected (i.e. an arc does not disconnect $S^2$, in the sense of Definition 6.14.1).

By using an analogous argument to the one used for $\bar{H}_0$ (Exercises 6.14.2--6.14.4), show that $H_1(S^2 \setminus A) \simeq 0$, where $A$ is an arc in $S^2$.

Consider a simple closed curve $C \subset S^2$. Since $C = h(S^1)$ and $S^1$ is the union of two subarcs, we can write $C = A_1 \cup A_2$, where $A_1 \cap A_2 = \left\{ p,q\right\}$ represent two points in the sphere.

Suppose that $p, q$ are points in different path components $K_1, K_2$ of $S^2 \setminus C$ (with $C$ a simple closed curve as in Exercise 6.14.6), and $r \in C$. Let $D$ be an open disk about $r$. Show that there is a point of each path component in $D$.

Prove Theorem 6.14.2 (the Jordan curve theorem): suppose $C$ is a simple closed curve in $\mathbb {R}^{2}$.

Prove Lemma 6.14.3: suppose there is a closed set $B$ which is expressed as the intersection of a nested family of closed sets $B_i$, $i = 0, \ldots , \infty$, with $B_{i+1} \subset B_i$, and a nonzero element $x \in \bar{H}_p(S^n \setminus B_0)$ which maps injectively to $\bar{H}_p(S^n \setminus B_i)$ for all $i$. Then $x$ maps injectively to a nonzero element of $\bar{H}_p(S^n \setminus B)$.

Use induction and the Mayer--Vietoris sequence to prove Theorem 6.14.4: suppose $h : D^k \to S^n$ is an embedding. Then $\bar{H}_i(S^n \setminus h(D^k)) \simeq 0$, $i \geq 0$.

Prove Theorem 6.14.5 (Alexander duality for spheres) when $k = 0$: suppose that $h : S^k \to S^n$, $0 \leq k < n$, is an embedding. Then

Under the hypotheses of Theorem 6.14.5 (Exercise 6.14.12) with $k = m+1 \geq 1$, use the Mayer--Vietoris exact sequence based on the decomposition

and the results on the homology of complements of embedded disks (Theorem 6.14.4, Exercise 6.14.11) to show that

Use Exercises 6.14.12 and 6.14.13 and induction to prove Theorem 6.14.5 (Alexander duality for spheres, stated in Exercise 6.14.12) in general.

Prove Theorem 6.14.6 (Jordan separation theorem): suppose that $h : S^{n-1} \to S^n$ is an embedding. Then $S^n \setminus h(S^{n-1})$ is the disjoint union of two open sets. The closure of each set is the union of the set and $h(S^{n-1})$.

Theorem 6.14.7 (Invariance of domain): suppose that $U$ is an open subset of $\mathbb {R}^{n}$ and $f : U \to \mathbb {R}^{n}$ is $1$--$1$ and continuous. Then $f$ is an open map; that is, it maps open sets to open sets. (By composing $f$ with a homeomorphism between $\mathbb {R}^{n}$ and $S^n \setminus \left\{ p\right\}$, we may assume the image is contained in $S^n$, and it suffices to show that the image of an open set in $U$ is an open set in $S^n$.)

Suppose $x \in U$, $y = f(x) \in S^n$, and let $V$ be an open set in $U$ containing $x$. Choose a small disk $D_x = D(x, \epsilon )$ about $x$ which is contained in $V$, with $S_x = \partial D_x$.