21. The Jordan Curve Theorem

If $p_{*}$ is an isomorphism then $H_{*}(X)$ is isomorphic to $H_{*}(1)$ i.e. $H_0(X)\simeq \mathbb{Z}$ and $H_i(X)\simeq 0$ for $i>0$, which means that $X$ is acyclic. Suppose instead we start withthe assumption that $X$ is acyclic. For $i>0$ the groups $H_i(X)$ and $H_i(1)$ are both zero, so the map $p_{*}:H_i(X)\to H_i(1)$ is automatically an isomorphism. Next, we know that $H_0(X)\simeq \mathbb{Z}\{{\pi }_0(X)\}$. As $X$ is acyclic we have $H_0(X)\simeq \mathbb{Z}$ so $|{\pi }_0(X)|=1$ so $X$ is nonempty and path connected. We can therefore choose $a\in X$ and we have $H_0(X)=\mathbb{Z}.[a]$. We also have $p(a)=0$ and $H_0(1)=\mathbb{Z}.[0]$ so $p_{*}:H_0(X)\to H_0(1)$ is an isomorphism. ∎

Put $V=S^n\setminus Y$ and $W=S^n\setminus Z$. These are open subsets of $S^n$ with $V\cup W=S^n\setminus (Y\cap Z)$ and $V\cap W=S^n\setminus (Y\cup Z)$. Thus, our assumptions are that $V$, $W$ and $V\cup W$ are acyclic, and we need to prove that $V\cap W$ is acyclic. We have a Mayer-Vietoris sequence

If $m>0$ then the first and third terms are zero so $H_m(V\cap W)=0$ as expected. If $m=0$ we instead have an exact sequence

and it is not hard to deduce that $H_0(V\cap W)\simeq \mathbb{Z}$. ∎

We use the same notation and the same Mayer-Vietoris sequence as in Lemma 21.8:

We again have $H_{m+1}(V\cup W)=0$, so $\alpha$ is injective, so $\alpha (u)\ne 0$. However, $\alpha (u)$ is just $(i_{*}(u),-j_{*}(u))$, so either $i_{*}(u)$ or $j_{*}(u)$ must be nonzero. ∎

We can express $w$ as $m_1{w}_1+\mathrm{\cdots }+m_r{w}_r$ for some integers $m_i$ and continuous maps $w_i:{\mathrm{\Delta }}_p\to U$. Note that the sets $w_i^{-1}(U_j)$ are open in ${\mathrm{\Delta }}_p$, and the union of all these sets is $w_i^{-1}(U)={\mathrm{\Delta }}_p$. As ${\mathrm{\Delta }}_p$ is compact, it must be covered by a finite subcollection of the sets $w_i^{-1}(U_j)$. As these sets are nested inside each other, this just means that there exists $j_i$ with $w_i^{-1}(U_{j_i})={\mathrm{\Delta }}_p$. Equivalently, we have $w_i({\mathrm{\Delta }}_p)\subseteq U_{j_i}$ or $w_i\in C_p(U_{j_i})$. Thus, if we put $j=\max (j_1,\mathrm{\dots },j_r)$ then we have $w_i\in C_p(U_j)$ for all $i$, and so $w\in C_p(U_j)$. ∎

Choose a cycle $z\in Z_m(U_0)$ such that $u=[z]$. As $u$ maps to zero in $H_m(U)$, there must be a chain $w\in C_{m+1}(U)$ with $z=\partial (w)$ in $C_m(U)$. By the lemma, the element $w$ lies in $C_m(U_j)$ for some $j$. We can therefore interpret the equation $\partial (w)=z$ as an equation in $C_m(U_j)$, showing that $[z]=0$ in $H_m(U_j)$ as required. ∎

If $k=0$ then $X$ is just a single point, so $U$ is homeomorphic to ${ℝ}^n$ by stereographic projection. This means that $U$ is contractible and therefore acyclic. For $k>0$ we will argue by induction. Choose a homeomorphism $f:{[0,1]}^k\to X$.

Suppose (for a contradiction) that we have a nonzero element $u\in H_m(S^n\setminus X)$ for some $m\ge 0$. If $m=0$ we also assume that $p_{*}(u)=0$ in $H_0(1)=\mathbb{Z}$. We will define a sequence of slices $X(j)$ of width $2^{-j}$ such that $u$ has nonzero image in $H_m(S^n\setminus X(j))$ for all $j$.

We start with $X(0)=X$. Suppose we have already defined $X(j)$. We can write $X(j)$ as $X(j)=X{(j)}_{+}\cup X{(j)}_{-}$, where $X{(j)}_{+}$ and $X{(j)}_{-}$ are slices of width $2^{-j-1}$, and the set $X{(j)}_0=X{(j)}_{+}\cap X{(j)}_{-}$ has width $0$ and so is homeomorphic to ${[0,1]}^{k-1}$. Our induction hypothesis says that $X{(j)}_0$ has acyclic complement, so Lemma 21.11 tells us that $u$ must have nonzero image in $H_m(S^n\setminus X{(j)}_{+})$ or in $H_m(S^n\setminus X{(j)}_{-})$. We choose $X(j+1)=X{(j)}_{+}$ or $X(j+1)=X{(j)}_{-}$ as appropriate to ensure that $u$ has nonzero image in $H_m(S^n\setminus X(j+1))$.

By construction we have $X(j)=f({[0,1]}^{k-1}\times [a_j,a_j+2^{-j}])$ for some sequence $(a_j)$ with $a_{j+1}\in \{a_j,a_j+2^{-j-1}\}$. It follows that the numbers $a_j$ converge to a limit $a_{\mathrm{\infty }}$, and that the set $X(\mathrm{\infty })={\bigcap }_{j}X(j)$ is just $f({[0,1]}^{k-1}\times \{a_{\mathrm{\infty }}\})$. This is homeomorphic to ${[0,1]}^{k-1}$ and so has acyclic complement by our induction hypothesis. In particular, the element $u$ must map to zero in $H_m(S^n\setminus X(\mathrm{\infty }))$. (In the case $m=0$, we are using the assumption $p_{*}(u)=0$ here.) However, we can regard $S^n\setminus X(\mathrm{\infty })$ as the union of the nested open sets $S^n\setminus X(j)$, so $u$ must map to zero in $H_m(S^n\setminus X(j))$ for some $j$, which contradicts our construction of $X(j)$.

This contradiction shows that no element $u$ as described above can exist. In other words, for $m>0$ we have $H_m(S^n\setminus X)=0$, and also the kernel of the map $p_{*}:H_0(S^n\setminus X)\to \mathbb{Z}$ is zero, so $p_{*}$ is injective. On the other hand, $X$ is contractible but $S^n$ is not, so $X$ cannot be equal to $S^n$, so we can choose a point $a\in S^n\setminus X$, and then $p_{*}[a]=1$. This shows that $p_{*}$ is also surjective, so it is an isomorphism as required. ∎

Suppose that $X\subseteq S^n$ is homeomorphic to $S^k$ for some $k\le n$. If $k=0$ then this just means that $X$ consists of two points. If $n=0$ this clearly means that $X=S^0=\{1,-1\}$ as claimed. If $n>0$ then we recall that $S^n\setminus \{\text{point}\}$ is homeomorphic to ${ℝ}^n$, so removing two points gives ${ℝ}^n\setminus \{\text{point}\}$ which is homotopy equivalent to $S^{n-1}$ and so has the same homology as $S^{n-1}$, as claimed.

We now suppose that $k>0$, and argue by induction on $k$. We can write $S^k$ as the union of two hemispheres, with intersection $S^{k-1}$. Correspondingly, we can write $X$ as $Y\cup Z$, where $Y$ and $Z$ are homeomorphic to $B^k$ (or ${[0,1]}^k$) and $Y\cap Z$ is homeomorphic to $S^{k-1}$. Theorem 21.7 tells us that the sets $V=S^n\setminus Y$ and $W=S^n\setminus Z$ are acyclic. The induction hypothesis tells us that the space $V\cup W=S^n\setminus (Y\cap Z)$ has the same homology as $S^{n-(k-1)-1}=S^{n-k}$. We need to show that the space $V\cap W=S^n\setminus X$ has the same homology as $S^{n-1-k}$ (or that $V\cap W=\mathrm{\varnothing }$ if $k=n$). For $m>0$ we note that $H_m(V)=H_m(W)=H_{m+1}(V)=H_{m+1}(W)=0$ so the Mayer-Vietoris sequence

shows that $\delta :H_{m+1}(V\cup W)\to H_m(V\cap W)$ is an isomorphism. This means that $H_m(V\cap W)=0$ for all $m>0$ with $m\ne n-1-k$, but that if $n-1-k>0$ (or equivalently ) then $H_{n-1-k}(V\cap W)\simeq \mathbb{Z}$.

This just leaves a few exceptional cases to consider. First suppose that , so $n-k>1$, so $H_1(V\cup W)=H_1(S^{n-k})=0$. We then have a Mayer-Vietoris sequence

and it follows easily that $H_0(V\cap W)\simeq \mathbb{Z}$ as required.

Now suppose instead that $k=n-1$. In this case we must show that $V\cap W$ has the same homology as $S^0$, or in other words that $H_0(V\cap W)\simeq {\mathbb{Z}}^2$ and $H_m(V\cap W)=0$ for $m>0$. The case $m>0$ is covered by our main discussion above. The induction hypothesis tells us that $V\cup W$ has the same homology as $S^1$, so the Mayer-Vietoris sequence

becomes

We can describe $H_0$ as the free abelian group generated by ${\pi }_0$. Using this we see that $\beta$ is essentially the addition map $(n,m)\mapsto n+m$, with kernel generated by $(1,-1)$. We can choose an element $u\in H_0(V\cap W)$ with $\alpha (u)=(1,-1)$, then it is not hard to deduce that $\{\delta (1),u\}$ is a basis for $H_0(V\cap W)$. This means that $H_0(V\cap W)\simeq {\mathbb{Z}}^2$, or in other words that $V\cap W$ has precisely two path components.

Finally suppose that $k=n$. Here the induction hypothesis shows that $V\cup W$ has the same homology as $S^0$, so in particular it has two path components. We must show that $V\cap W=\mathrm{\varnothing }$. The Mayer-Vietoris sequence

becomes

Choose $a\in V$ and $b\in W$, so $H_0(V)=\mathbb{Z}.[a]$ and $H_0(W)=\mathbb{Z}.[b]$. The sequence shows that $\beta$ is surjective, but that is only possible if $a$ lies in one path component of $V\cup W$ and $b$ lies in the other. That implies that $\beta$ is actually an isomorphism, and then exactness shows that $H_0(V\cap W)=0$. However, we know that $H_0(V\cap W)$ is the free abelian group on ${\pi }_0(V\cap W)$, so ${\pi }_0(V\cap W)=\mathrm{\varnothing }$, so $V\cap W=\mathrm{\varnothing }$ as claimed. ∎

First let $U$ be open in $S^n$, and suppose we have a point $a\in U$, with path component $A$ say. Suppose that $b\in A$, so there is a path $u$ from $a$ to $b$ in $U$. As $U$ is open, we can find a radius $ϵ>0$ such that $OB(b,ϵ)\cap S^n\subseteq U$. By reducing $ϵ$ if necessary, we can assume that . For any $c\in OB(b,ϵ)\cap S^n$ we have a linear path $v$ from $b$ to $c$ in ${ℝ}^n$. As we see that this does not pass through $0$, so we can define $w(t)=v(t)/\parallel v(t)\parallel$; this gives a path from $b$ to $c$ in $S^n$. This stays within $OB(b,ϵ)$ so it stays within $U$. This means we have a path $u*w$ from $a$ to $c$ in $U$, so $c\in A$. This proves that $OB(b,ϵ)\subseteq A$. As $b$ was arbitrary, this proves that $A$ is open as claimed.

The argument for ${ℝ}^n$ is similar but easier. ∎

Let $f:S^{n-1}\to {ℝ}^n$ be an injective continuous map, where $n\ge 2$. First note that $f(S^{n-1})$ is compact, so it is bounded and closed in ${ℝ}^n$. It will be harmless to multiply $f$ by a small positive constant, so we can assume that for all $x\in S^{n-1}$.

We now recall a few more details about stereographic projection. We identify ${ℝ}^{n+1}$ with ${ℝ}^n\times ℝ$, so $S^n=\{(y,z)|{\parallel y\parallel }^2+z^2=1\}$. We put $a=(0,1)$ and $A_{+}=\{(y,z)\in S^n|z>0\}$. We have a homeomorphism $g:{ℝ}^n\to S^n\setminus \{a\}$ given by $g(u)=(u,{\parallel u\parallel }^2-1)/({\parallel u\parallel }^2+1)$. This also gives a homeomorphism ${ℝ}^n\setminus f(S^{n-1})\to U\setminus \{a\}$, where $U=S^n\setminus g(f(S^{n-1}))$. Theorem 21.1 tells us that $U$ has the same homology as $S^0$, so in particular $H_0(U)\simeq {\mathbb{Z}}^2$, so $U$ has precisely two path components. Let $A$ be the path component containing $a$, and let $B$ be the other path component, so $U$ is the disjoint union of $A$ and $B$. For $m>0$ we therefore have $0=H_m(U)=H_m(A)\oplus H_m(B)$, so $H_m(A)=H_m(B)=0$. This shows that both $A$ and $B$ are acyclic.

Because for all $x\in S^{n-1}$ we see that the last coordinate of $g(f(x))$ is always negative. Thus, the whole upper hemisphere $A_{+}$ of $S^n$ is contained in $S^n\setminus g(f(S^n))$. Moreover, $A_{+}$ is clearly path connected and contains $a$ so $A_{+}\subseteq A$. It follows that $B$ is contained in the lower hemisphere, and so the corresponding subset of ${ℝ}^n$ is bounded.

We next claim that the set $A^{\prime }=A\setminus \{a\}$ is path connected. We will prove this using the Mayer-Vietoris sequence

Here $A_{+}\cup A^{\prime }$ is just $A$, and $A$ is acyclic. Thus, the first and fourth groups above are $0$ and $\mathbb{Z}$. The space $A_{+}\cap A^{\prime }$ is the same as $A_{+}\setminus \{a\}$, which is homeomorphic to $(0,1)\times S^{n-1}$ and so homotopy equivalent to $S^{n-1}$. In particular, as we are assuming that $n\ge 2$, we know that $A_{+}\cap A^{\prime }$ is connected. Thus, the second group in our sequence is $\mathbb{Z}$. It is also clear that $A_{+}$ is contractible and so $H_0(A_{+})=\mathbb{Z}$. We therefore have an exact sequence

This is only consistent if $H_0(A^{\prime })\simeq \mathbb{Z}$, which means that $A^{\prime }$ is path connected.

We now see that $U\setminus \{a\}$ is the disjoint union of path connected sets $A^{\prime }$ and $B$, so these are the path components. It is clear that $g^{-1}(B)$ is bounded and $g^{-1}(A^{\prime })$ is unbounded. ∎

Put $U=S^n\setminus f(S^{n-1})$. As $f(S^{n-1})$ is compact, it is closed in $S^n$, so $U$ is open. Theorem 21.1 tells us that $H_{*}(U)\simeq H_{*}(S^0)$, so $U$ has two path components. Let $V$ be the path component containing $f(0)$, and let $W$ be the other one. The sets $V$ and $W$ are open in $S^n$ by Lemma 21.14. Now put $V^{\prime }=f(O{B}^n)$ and $W^{\prime }=S^n\setminus f(B^n)$. Using the injectivity of $f$ we see that $V^{\prime }$ and $W^{\prime }$ are disjoint and $U=V^{\prime }\cup W^{\prime }$. Given $x,y\in O{B}^n$ we have a path $t\mapsto f((1-t)x+ty)$ from $f(x)$ to $f(y)$ in the set $f(O{B}^n)=V^{\prime }$; this shows that $V^{\prime }$ is path connected. As $f(0)\in V^{\prime }$ we see that $V^{\prime }\subseteq V$. Next, Theorem 21.7 tells us that $W^{\prime }$ is acyclic and therefore also path connected. If there was a path in $U$ joining some point in $V^{\prime }$ to some point in $W^{\prime }$, then we could conclude that the whole space $U=V^{\prime }\cup W^{\prime }$ was path connected. However, we know that $U$ has two path components, so no such path can exist.

Now consider a point $x\in V$. As $V$ is the path component of $f(0)$, we can find a path $u:[0,1]\to U$ with $u(0)=f(0)$ and $u(1)=x$. Here $u(0)\in V^{\prime }$ and no path in $U$ can cross from $V^{\prime }$ to $W^{\prime }$ so the point $u(1)=x$ must also lie in $V^{\prime }$. This proves that $V=V^{\prime }=f(O{B}^n)$. We have already remarked that $V$ is open, and it follows that $f(O{B}^n)$ is open, as claimed. ∎

As usual we can identify ${ℝ}^n$ with the complement of a point in $S^n$, which is an open subset of $S^n$. It will therefore be enough to show that $f(U)$ is open in $S^n$. Consider a point $a\in U$. As $U$ is open, we can choose $ϵ>0$ such that $OB(a,ϵ)$ is contained in $U$. We then have a continuous injective map $g_a:B^n\to S^n$ given by $g_a(x)=f(a+ϵx/2)$. The proposition tells us that the set $V_a=g_a(O{B}^n)$ is open, and it is clear that $f(a)\in V_a=f(OB(a,ϵ/2))\subseteq f(U)$. It follows that $f(U)$ is the union of all these open sets $V_a$, and thus that $f(U)$ is open. ∎