MAS61015 Algebraic Topology

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 XSn and that X is homeomorphic to Sk for some kn.

  • (a)

    If k=n then X is just equal to Sn and so SnX is empty and H*(SnX)=0.

  • (b)

    If k=n-1 then H0(SnX)2 and Hm(SnX)=0 for all m>0.

  • (c)

    If k<n-1 then H0(SnX)Hn-k-1(SnX) 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*(SnX)H*(Sn-1-k). For the most obvious case of the theorem, we can express n+1 as k+1n-k, so


The space X={(y,0)|y=1}Sn is then homeomorphic to Sk, with


We can define maps Sn-k-1𝑖SnX𝑟Sn-k-1 by i(z)=(0,z) and r(y,z)=z/z. We then have ri=id, and we have a homotopy between ir and the identity given by h(t,y,z)=(ty,z)/(ty,z). Thus, in this case SnX is actually homotopy equivalent to Sn-k-1. However, that is not true in general. For example, let X be a knotted circle in 3. We can identify 3 with S3{point} by stereographic projection, and thus think of X as a subspace of S3. The theorem tells us that in this case S3X has the same homology as S1, but it can be shown that S3X is not homotopy equivalent to S1.

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

Theorem 21.3.

Suppose that n2, and let f:Sn-1n be an injective continuous map. Then the complementary set nf(Sn-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:S12 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 SnX in cases where X is homeomorphic to Bk (or equivalently, [0,1]k) rather than Sk.

Video (Definition 21.4 to Corollary 21.13)

Definition 21.4.

We say that a space X is acyclic if H0(X)= and Hi(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:X1. Then X is acyclic iff the map p*:H*(X)H*(1) is an isomorphism.


If p* is an isomorphism then H*(X) is isomorphic to H*(1) i.e. H0(X) and Hi(X)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 Hi(X) and Hi(1) are both zero, so the map p*:Hi(X)Hi(1) is automatically an isomorphism. Next, we know that H0(X){π0(X)}. As X is acyclic we have H0(X) so |π0(X)|=1 so X is nonempty and path connected. We can therefore choose aX and we have H0(X)=.[a]. We also have p(a)=0 and H0(1)=.[0] so p*:H0(X)H0(1) is an isomorphism. ∎

Theorem 21.7.

Let X be a subset of Sn that is homeomorphic to [0,1]k for some kn. Then SnX 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 Sn such that the sets SnY, SnZ and Sn(YZ) are all acyclic. Then the set Sn(YZ) is also acyclic.


Put V=SnY and W=SnZ. These are open subsets of Sn with VW=Sn(YZ) and VW=Sn(YZ). Thus, our assumptions are that V, W and VW are acyclic, and we need to prove that VW is acyclic. We have a Mayer-Vietoris sequence


If m>0 then the first and third terms are zero so Hm(VW)=0 as expected. If m=0 we instead have an exact sequence


and it is not hard to deduce that H0(VW). ∎

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:[0,1]kSn with f([0,1]k)=X. We will make some constructions in that context.

Definition 21.9.

Let f:[0,1]kSn be an injective continuous map. By a slice we mean a set of the form f([0,1]k-1×[a,b]) with 0ab1. The width of such a slice is b-a. A coslice is the complement in Sn 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 Sn such that the set Sn(YZ) is acyclic. Note that we have inclusions


Suppose that uHm(Sn(YZ)) is nonzero; then either i*(u) is nonzero in Hm(SnY), or j*(u) is nonzero in Hm(SnZ).


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


We again have Hm+1(VW)=0, so α is injective, so α(u)0. However, α(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 U0,U1,U2, be a sequence of open sets with U0U1U2 and U=i=0Ui. Then any chain wCp(U) is contained in Cp(Uj) for some j.


We can express w as m1w1++mrwr for some integers mi and continuous maps wi:ΔpU. Note that the sets wi-1(Uj) are open in Δp, and the union of all these sets is wi-1(U)=Δp. As Δp is compact, it must be covered by a finite subcollection of the sets wi-1(Uj). As these sets are nested inside each other, this just means that there exists ji with wi-1(Uji)=Δp. Equivalently, we have wi(Δp)Uji or wiCp(Uji). Thus, if we put j=max(j1,,jr) then we have wiCp(Uj) for all i, and so wCp(Uj). ∎

Corollary 21.13.

In the context of Lemma 21.12, suppose we have an element uHm(U0) which maps to zero in Hm(U). Then u already maps to zero in Hm(Uj) for some j.


Choose a cycle zZm(U0) such that u=[z]. As u maps to zero in Hm(U), there must be a chain wCm+1(U) with z=(w) in Cm(U). By the lemma, the element w lies in Cm(Uj) for some j. We can therefore interpret the equation (w)=z as an equation in Cm(Uj), showing that [z]=0 in Hm(Uj) as required. ∎

Proof of Theorem 21.7.

If k=0 then X is just a single point, so U is homeomorphic to 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]kX.

Suppose (for a contradiction) that we have a nonzero element uHm(SnX) for some m0. If m=0 we also assume that p*(u)=0 in H0(1)=. We will define a sequence of slices X(j) of width 2-j such that u has nonzero image in Hm(SnX(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)+X(j)-, where X(j)+ and X(j)- are slices of width 2-j-1, and the set X(j)0=X(j)+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 Hm(SnX(j)+) or in Hm(SnX(j)-). We choose X(j+1)=X(j)+ or X(j+1)=X(j)- as appropriate to ensure that u has nonzero image in Hm(SnX(j+1)).

By construction we have X(j)=f([0,1]k-1×[aj,aj+2-j]) for some sequence (aj) with aj+1{aj,aj+2-j-1}. It follows that the numbers aj converge to a limit a, and that the set X()=jX(j) is just f([0,1]k-1×{a}). 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 Hm(SnX()). (In the case m=0, we are using the assumption p*(u)=0 here.) However, we can regard SnX() as the union of the nested open sets SnX(j), so u must map to zero in Hm(SnX(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 Hm(SnX)=0, and also the kernel of the map p*:H0(SnX) is zero, so p* is injective. On the other hand, X is contractible but Sn is not, so X cannot be equal to Sn, so we can choose a point aSnX, 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 XSn is homeomorphic to Sk for some kn. If k=0 then this just means that X consists of two points. If n=0 this clearly means that X=S0={1,-1} as claimed. If n>0 then we recall that Sn{point} is homeomorphic to n, so removing two points gives n{point} which is homotopy equivalent to Sn-1 and so has the same homology as Sn-1, as claimed.

We now suppose that k>0, and argue by induction on k. We can write Sk as the union of two hemispheres, with intersection Sk-1. Correspondingly, we can write X as YZ, where Y and Z are homeomorphic to Bk (or [0,1]k) and YZ is homeomorphic to Sk-1. Theorem 21.7 tells us that the sets V=SnY and W=SnZ are acyclic. The induction hypothesis tells us that the space VW=Sn(YZ) has the same homology as Sn-(k-1)-1=Sn-k. We need to show that the space VW=SnX has the same homology as Sn-1-k (or that VW= if k=n). For m>0 we note that Hm(V)=Hm(W)=Hm+1(V)=Hm+1(W)=0 so the Mayer-Vietoris sequence


shows that δ:Hm+1(VW)Hm(VW) is an isomorphism. This means that Hm(VW)=0 for all m>0 with mn-1-k, but that if n-1-k>0 (or equivalently k<n-1) then Hn-1-k(VW).

This just leaves a few exceptional cases to consider. First suppose that k<n-1, so n-k>1, so H1(VW)=H1(Sn-k)=0. We then have a Mayer-Vietoris sequence


and it follows easily that H0(VW) as required.

Now suppose instead that k=n-1. In this case we must show that VW has the same homology as S0, or in other words that H0(VW)2 and Hm(VW)=0 for m>0. The case m>0 is covered by our main discussion above. The induction hypothesis tells us that VW has the same homology as S1, so the Mayer-Vietoris sequence




We can describe H0 as the free abelian group generated by π0. Using this we see that β is essentially the addition map (n,m)n+m, with kernel generated by (1,-1). We can choose an element uH0(VW) with α(u)=(1,-1), then it is not hard to deduce that {δ(1),u} is a basis for H0(VW). This means that H0(VW)2, or in other words that VW has precisely two path components.

Finally suppose that k=n. Here the induction hypothesis shows that VW has the same homology as S0, so in particular it has two path components. We must show that VW=. The Mayer-Vietoris sequence




Choose aV and bW, so H0(V)=.[a] and H0(W)=.[b]. The sequence shows that β is surjective, but that is only possible if a lies in one path component of VW and b lies in the other. That implies that β is actually an isomorphism, and then exactness shows that H0(VW)=0. However, we know that H0(VW) is the free abelian group on π0(VW), so π0(VW)=, so VW= as claimed. ∎

Lemma 21.14.

If U is an open subset of n, then every path component of U is also an open subset of n. Similarly, if U is an open subset of Sn, then every path component of U is also an open subset of Sn.


First let U be open in Sn, and suppose we have a point aU, with path component A say. Suppose that bA, 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)SnU. By reducing ϵ if necessary, we can assume that ϵ<1. For any cOB(b,ϵ)Sn we have a linear path v from b to c in n. As b-c<ϵ<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 Sn. 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 cA. This proves that OB(b,ϵ)A. As b was arbitrary, this proves that A is open as claimed.

The argument for n is similar but easier. ∎

Proof of Theorem 21.3.

Let f:Sn-1n be an injective continuous map, where n2. First note that f(Sn-1) is compact, so it is bounded and closed in n. It will be harmless to multiply f by a small positive constant, so we can assume that f(x)<1 for all xSn-1.

We now recall a few more details about stereographic projection. We identify n+1 with n×, so Sn={(y,z)|y2+z2=1}. We put a=(0,1) and A+={(y,z)Sn|z>0}. We have a homeomorphism g:nSn{a} given by g(u)=(u,u2-1)/(u2+1). This also gives a homeomorphism nf(Sn-1)U{a}, where U=Sng(f(Sn-1)). Theorem 21.1 tells us that U has the same homology as S0, so in particular H0(U)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=Hm(U)=Hm(A)Hm(B), so Hm(A)=Hm(B)=0. This shows that both A and B are acyclic.

Because f(x)<1 for all xSn-1 we see that the last coordinate of g(f(x)) is always negative. Thus, the whole upper hemisphere A+ of Sn is contained in Sng(f(Sn)). Moreover, A+ is clearly path connected and contains a so A+A. It follows that B is contained in the lower hemisphere, and so the corresponding subset of n is bounded.

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


Here A+A is just A, and A is acyclic. Thus, the first and fourth groups above are 0 and . The space A+A is the same as A+{a}, which is homeomorphic to (0,1)×Sn-1 and so homotopy equivalent to Sn-1. In particular, as we are assuming that n2, we know that A+A is connected. Thus, the second group in our sequence is . It is also clear that A+ is contractible and so H0(A+)=. We therefore have an exact sequence


This is only consistent if H0(A), which means that A is path connected.

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

Video (Proposition 21.15 and Corollary 21.16)

Proposition 21.15.

Let f:BnSn be continuous and injective. Then f(OBn) is open in Sn.


Put U=Snf(Sn-1). As f(Sn-1) is compact, it is closed in Sn, so U is open. Theorem 21.1 tells us that H*(U)H*(S0), 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 Sn by Lemma 21.14. Now put V=f(OBn) and W=Snf(Bn). Using the injectivity of f we see that V and W are disjoint and U=VW. Given x,yOBn we have a path tf((1-t)x+ty) from f(x) to f(y) in the set f(OBn)=V; this shows that V is path connected. As f(0)V we see that VV. Next, Theorem 21.7 tells us that W is acyclic and therefore also path connected. If there was a path in U joining some point in V to some point in W, then we could conclude that the whole space U=VW was path connected. However, we know that U has two path components, so no such path can exist.

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

Corollary 21.16.

Let U be an open subset of n, and let f:Un be a continuous injective map. Then f(U) is also open.


As usual we can identify n with the complement of a point in Sn, which is an open subset of Sn. It will therefore be enough to show that f(U) is open in Sn. Consider a point aU. 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 ga:BnSn given by ga(x)=f(a+ϵx/2). The proposition tells us that the set Va=ga(OBn) is open, and it is clear that f(a)Va=f(OB(a,ϵ/2))f(U). It follows that f(U) is the union of all these open sets Va, and thus that f(U) is open. ∎