# 21. The Jordan Curve Theorem

Video (Statement of Theorem 21.1 and Remark 21.2)

The main aim of this section will be to prove the following theorem.

###### Theorem 21.1.

Suppose that $X\subseteq S^{n}$ and that $X$ is homeomorphic to $S^{k}$ for some $k\leq n$.

• (a)

If $k=n$ then $X$ is just equal to $S^{n}$ and so $S^{n}\setminus X$ is empty and $H_{*}(S^{n}\setminus X)=0$.

• (b)

If $k=n-1$ then $H_{0}(S^{n}\setminus X)\simeq{\mathbb{Z}}^{2}$ and $H_{m}(S^{n}\setminus X)=0$ for all $m>0$.

• (c)

If $k then $H_{0}(S^{n}\setminus X)\simeq H_{n-k-1}(S^{n}\setminus X)\simeq{\mathbb{Z}}$ and all other homology groups are trivial.

###### Remark 21.2.

We can combine cases (b) and (c) in Theorem 21.1 by saying that $H_{*}(S^{n}\setminus X)\simeq H_{*}(S^{n-1-k})$. For the most obvious case of the theorem, we can express ${\mathbb{R}}^{n+1}$ as ${\mathbb{R}}^{k+1}\oplus{\mathbb{R}}^{n-k}$, so

 $S^{n}=\{(y,z)\;|\;\|y\|^{2}+\|z\|^{2}=1\}.$

The space $X=\{(y,0)\;|\;\|y\|=1\}\subseteq S^{n}$ is then homeomorphic to $S^{k}$, with

 $S^{n}\setminus X=\{(y,z)\;|\;\|y\|^{2}+\|z^{2}\|=1,\;z\neq 0\}.$

We can define maps $S^{n-k-1}\xrightarrow{i}S^{n}\setminus X\xrightarrow{r}S^{n-k-1}$ by $i(z)=(0,z)$ and $r(y,z)=z/\|z\|$. We then have $r\circ i=\operatorname{id}$, and we have a homotopy between $i\circ r$ and the identity given by $h(t,y,z)=(ty,z)/\|(ty,z)\|$. Thus, in this case $S^{n}\setminus X$ is actually homotopy equivalent to $S^{n-k-1}$. However, that is not true in general. For example, let $X$ be a knotted circle in ${\mathbb{R}}^{3}$. We can identify ${\mathbb{R}}^{3}$ with $S^{3}\setminus\{\text{point}\}$ by stereographic projection, and thus think of $X$ as a subspace of $S^{3}$. The theorem tells us that in this case $S^{3}\setminus X$ has the same homology as $S^{1}$, but it can be shown that $S^{3}\setminus X$ is not homotopy equivalent to $S^{1}$.

We will also deduce the following result, for which the case $n=2$ is called the Jordan Curve Theorem:

###### Theorem 21.3.

Suppose that $n\geq 2$, and let $f\colon S^{n-1}\to{\mathbb{R}}^{n}$ be an injective continuous map. Then the complementary set ${\mathbb{R}}^{n}\setminus f(S^{n-1})$ has precisely two path components, one bounded and the other unbounded.

It is easy to see that the Jordan Curve Theorem is true for maps $u\colon S^{1}\to{\mathbb{R}}^{2}$ that are reasonably simple. However, it is hard to prove in the general case, where $u$ may wiggle in an extremely complicated way and can also be fractal or otherwise badly behaved.

We will work up to Theorem 21.1 by first considering $S^{n}\setminus X$ in cases where $X$ is homeomorphic to $B^{k}$ (or equivalently, $[0,1]^{k}$) rather than $S^{k}$.

Video (Definition 21.4 to Corollary 21.13)

###### Definition 21.4.

We say that a space $X$ is acyclic if $H_{0}(X)={\mathbb{Z}}$ and $H_{i}(X)=0$ for all $i>0$.

###### Remark 21.5.

We have seen previously that all contractible spaces are acyclic. Conversely, most commonly occurring acyclic spaces are contractible, but there are some exceptions.

###### Lemma 21.6.

Let $X$ be a space. Let $1=\{0\}$ denote the one-point space, so we have a constant map $p\colon X\to 1$. Then $X$ is acyclic iff the map $p_{*}\colon H_{*}(X)\to H_{*}(1)$ is an isomorphism.

###### Proof.

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 $p_{*}\colon H_{i}(X)\to H_{i}(1)$ 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 $p_{*}\colon H_{0}(X)\to H_{0}(1)$ is an isomorphism. ∎

###### Theorem 21.7.

Let $X$ be a subset of $S^{n}$ that is homeomorphic to $[0,1]^{k}$ for some $k\leq n$. Then $S^{n}\setminus X$ is acyclic.

Before proving the theorem we will prove a simpler lemma. This will not directly contribute to the theorem, but will introduce some relevant ideas.

###### Lemma 21.8.

Suppose that $Y$ and $Z$ are closed subsets of $S^{n}$ such that the sets $S^{n}\setminus Y$, $S^{n}\setminus Z$ and $S^{n}\setminus(Y\cap Z)$ are all acyclic. Then the set $S^{n}\setminus(Y\cup Z)$ is also acyclic.

###### Proof.

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

 $H_{m+1}(V\cup W)\xrightarrow{\delta}H_{m}(V\cap W)\xrightarrow{\alpha}H_{m}(V)% \oplus H_{m}(W)\xrightarrow{}H_{m}(V\cup W)$

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

 $0\to H_{0}(V\cap W)\to{\mathbb{Z}}\oplus{\mathbb{Z}}\to{\mathbb{Z}}\to 0$

and it is not hard to deduce that $H_{0}(V\cap W)\simeq{\mathbb{Z}}$. ∎

In Theorem 21.7, we assume that $X$ is homeomorphic to $[0,1]^{k}$. This means that we can choose an injective continuous map $f\colon[0,1]^{k}\to S^{n}$ with $f([0,1]^{k})=X$. We will make some constructions in that context.

###### Definition 21.9.

Let $f\colon[0,1]^{k}\to S^{n}$ be an injective continuous map. By a slice we mean a set of the form $f([0,1]^{k-1}\times[a,b])$ with $0\leq a\leq b\leq 1$. The width of such a slice is $b-a$. A coslice is the complement in $S^{n}$ of a slice; we define the width of a coslice to be the same as the width of the corresponding slice.

###### Remark 21.10.

We could attempt to prove Theorem 21.7 as follows. We argue by induction on $k$, the case $k=0$ being easy. For $k>0$, we divide $X$ into a large number of very thin slices. Any slice of width $0$ has acyclic complement, by the induction hypothesis. Our slices have very small width, so it seems reasonable to assume that they will also have acyclic complement. The intersection of two adjacent slices is a slice of width $0$, and so has acyclic complement. Using Lemma 21.8, we deduce that the union of two adjacent slices again has acyclic complement. By repeating this procedure, we see that the union of any block of adjacent slices has acyclic complement. In particular, the full set $X$ has acyclic complement, as required.

The problem with this approach is that our “reasonable assumption” is hard to justify directly. Our actual proof of Theorem 21.7 will use many of the same ideas, but arranged in a slightly different way.

###### Lemma 21.11.

Suppose that $Y$ and $Z$ are closed subsets of $S^{n}$ such that the set $S^{n}\setminus(Y\cap Z)$ is acyclic. Note that we have inclusions

 $S^{n}\setminus Y\xleftarrow{i}S^{n}\setminus(Y\cup Z)\xrightarrow{j}S^{n}% \setminus Z.$

Suppose that $u\in H_{m}(S^{n}\setminus(Y\cup Z))$ is nonzero; then either $i_{*}(u)$ is nonzero in $H_{m}(S^{n}\setminus Y)$, or $j_{*}(u)$ is nonzero in $H_{m}(S^{n}\setminus Z)$.

###### Proof.

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

 $H_{m+1}(V\cup W)\xrightarrow{\delta}H_{m}(V\cap W)\xrightarrow{\alpha}H_{m}(V)% \oplus H_{m}(W)\xrightarrow{}H_{m}(V\cup W)$

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

###### Lemma 21.12.

Let $U$ be a topological space, and let $U_{0},U_{1},U_{2},\dotsc$ be a sequence of open sets with $U_{0}\subseteq U_{1}\subseteq U_{2}\subseteq\dotsb$ and $U=\bigcup_{i=0}^{\infty}U_{i}$. Then any chain $w\in C_{p}(U)$ is contained in $C_{p}(U_{j})$ for some $j$.

###### Proof.

We can express $w$ as $m_{1}w_{1}+\dotsb+m_{r}w_{r}$ for some integers $m_{i}$ and continuous maps $w_{i}\colon\Delta_{p}\to U$. Note that the sets $w_{i}^{-1}(U_{j})$ are open in $\Delta_{p}$, and the union of all these sets is $w_{i}^{-1}(U)=\Delta_{p}$. As $\Delta_{p}$ 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 $w_{i}^{-1}(U_{j_{i}})=\Delta_{p}$. Equivalently, we have $w_{i}(\Delta_{p})\subseteq U_{j_{i}}$ or $w_{i}\in C_{p}(U_{j_{i}})$. Thus, if we put $j=\max(j_{1},\dotsc,j_{r})$ then we have $w_{i}\in C_{p}(U_{j})$ for all $i$, and so $w\in C_{p}(U_{j})$. ∎

###### Corollary 21.13.

In the context of Lemma 21.12, suppose we have an element $u\in H_{m}(U_{0})$ which maps to zero in $H_{m}(U)$. Then $u$ already maps to zero in $H_{m}(U_{j})$ for some $j$.

###### Proof.

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. ∎

###### Proof of Theorem 21.7.

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

Suppose (for a contradiction) that we have a nonzero element $u\in H_{m}(S^{n}\setminus X)$ for some $m\geq 0$. 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 $a_{\infty}$, and that the set $X(\infty)=\bigcap_{j}X(j)$ is just $f([0,1]^{k-1}\times\{a_{\infty}\})$. 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 $H_{m}(S^{n}\setminus X(\infty))$. (In the case $m=0$, we are using the assumption $p_{*}(u)=0$ here.) However, we can regard $S^{n}\setminus X(\infty)$ 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 $p_{*}\colon H_{0}(S^{n}\setminus X)\to{\mathbb{Z}}$ 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. ∎

###### Proof of Theorem 21.1.

Suppose that $X\subseteq S^{n}$ is homeomorphic to $S^{k}$ for some $k\leq n$. 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 ${\mathbb{R}}^{n}$, so removing two points gives ${\mathbb{R}}^{n}\setminus\{\text{point}\}$ 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 $V\cap W=\emptyset$ 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

 $H_{m+1}(V)\oplus H_{m+1}(W)\xrightarrow{}H_{m+1}(V\cup W)\xrightarrow{\delta}H% _{m}(V\cap W)\xrightarrow{}H_{m}(V)\oplus H_{m}(W)$

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

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

 $H_{1}(V\cup W)\xrightarrow{}H_{0}(V\cap W)\xrightarrow{}H_{0}(V)\oplus H_{0}(W% )={\mathbb{Z}}^{2}\xrightarrow{}H_{0}(V\cup W)={\mathbb{Z}}\xrightarrow{}0,$

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

 $H_{1}(V)\oplus H_{1}(W)\to H_{1}(V\cup W)\to H_{0}(V\cap W)\to H_{0}(V)\oplus H% _{0}(W)\to H_{0}(V\cup W)\to 0$

becomes

 $0\to{\mathbb{Z}}\xrightarrow{\delta}H_{0}(V\cap W)\xrightarrow{\alpha}{\mathbb% {Z}}^{2}\xrightarrow{\beta}{\mathbb{Z}}\to 0.$

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 $V\cap W=\emptyset$. The Mayer-Vietoris sequence

 $H_{1}(V\cup W)\to H_{0}(V\cap W)\to H_{0}(V)\oplus H_{0}(W)\to H_{0}(V\cup W)\to 0$

becomes

 $0\to H_{0}(V\cap W)\xrightarrow{\alpha}{\mathbb{Z}}^{2}\xrightarrow{\beta}{% \mathbb{Z}}^{2}\to 0.$

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 $\pi_{0}(V\cap W)=\emptyset$, so $V\cap W=\emptyset$ as claimed. ∎

###### Lemma 21.14.

If $U$ is an open subset of ${\mathbb{R}}^{n}$, then every path component of $U$ is also an open subset of ${\mathbb{R}}^{n}$. Similarly, if $U$ is an open subset of $S^{n}$, then every path component of $U$ is also an open subset of $S^{n}$.

###### Proof.

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 $\epsilon>0$ such that $OB(b,\epsilon)\cap S^{n}\subseteq U$. By reducing $\epsilon$ if necessary, we can assume that $\epsilon<1$. For any $c\in OB(b,\epsilon)\cap S^{n}$ we have a linear path $v$ from $b$ to $c$ in ${\mathbb{R}}^{n}$. As $\|b-c\|<\epsilon<1$ we see that this does not pass through $0$, so we can define $w(t)=v(t)/\|v(t)\|$; this gives a path from $b$ to $c$ in $S^{n}$. This stays within $OB(b,\epsilon)$ 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 $OB(b,\epsilon)\subseteq A$. As $b$ was arbitrary, this proves that $A$ is open as claimed.

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

###### Proof of Theorem 21.3.

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

We now recall a few more details about stereographic projection. We identify ${\mathbb{R}}^{n+1}$ with ${\mathbb{R}}^{n}\times{\mathbb{R}}$, so $S^{n}=\{(y,z)\;|\;\|y\|^{2}+z^{2}=1\}$. We put $a=(0,1)$ and $A_{+}=\{(y,z)\in S^{n}\;|\;z>0\}$. We have a homeomorphism $g\colon{\mathbb{R}}^{n}\to S^{n}\setminus\{a\}$ given by $g(u)=(u,\|u\|^{2}-1)/(\|u\|^{2}+1)$. This also gives a homeomorphism ${\mathbb{R}}^{n}\setminus f(S^{n-1})\to U\setminus\{a\}$, 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 $\|f(x)\|<1$ 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 ${\mathbb{R}}^{n}$ 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

 $H_{1}(A_{+}\cup A^{\prime})\to H_{0}(A_{+}\cap A^{\prime})\to H_{0}(A_{+})% \oplus H_{0}(A^{\prime})\to H_{0}(A_{+}\cup A^{\prime})\to 0.$

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 $n\geq 2$, 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

 $0\to{\mathbb{Z}}\to{\mathbb{Z}}\oplus H_{0}(A^{\prime})\to{\mathbb{Z}}\to 0.$

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. ∎

Video (Proposition 21.15 and Corollary 21.16)

###### Proposition 21.15.

Let $f\colon B^{n}\to S^{n}$ be continuous and injective. Then $f(OB^{n})$ is open in $S^{n}$.

###### Proof.

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 $V^{\prime}=f(OB^{n})$ 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 $x,y\in OB^{n}$ we have a path $t\mapsto f((1-t)x+ty)$ from $f(x)$ to $f(y)$ in the set $f(OB^{n})=V^{\prime}$; 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 $u\colon[0,1]\to U$ 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 $V=V^{\prime}=f(OB^{n})$. We have already remarked that $V$ is open, and it follows that $f(OB^{n})$ is open, as claimed. ∎

###### Corollary 21.16.

Let $U$ be an open subset of ${\mathbb{R}}^{n}$, and let $f\colon U\to{\mathbb{R}}^{n}$ be a continuous injective map. Then $f(U)$ is also open.

###### Proof.

As usual we can identify ${\mathbb{R}}^{n}$ 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 $\epsilon>0$ such that $OB(a,\epsilon)$ is contained in $U$. We then have a continuous injective map $g_{a}\colon B^{n}\to S^{n}$ given by $g_{a}(x)=f(a+\epsilon x/2)$. The proposition tells us that the set $V_{a}=g_{a}(OB^{n})$ is open, and it is clear that $f(a)\in V_{a}=f(OB(a,\epsilon/2))\subseteq f(U)$. It follows that $f(U)$ is the union of all these open sets $V_{a}$, and thus that $f(U)$ is open. ∎