MAS61015 Algebraic Topology

15. Homology of spheres

Consider a topological space X with open subsets U and V such that X=UV. There are many cases like this where we already understand three of the groups H*(U), H*(V), H*(UV) and H*(UV)=H*(X), and we want to determine the fourth one.

Video (Theorem 15.1 and Proposition 15.2)

We name the corresponding inclusion maps as shown on the left below; they induce homomorphisms of homology groups as shown on the right.

As the left hand diagram commutes and homology is functorial, we have k*i*=l*j*:H*(UV)H*(UV).

We can now combine the above maps to get maps as follows:


Here the notation [i*-j*] refers to the map sending a to (i*(a),-j*(a)), and the notation [k*l*] refers to the map sending (b,c) to k*(b)+l*(c). The composite of these two maps therefore sends a to k*i*(a)-l*j*(a), but this is zero because k*i*=l*j*. Thus, the above sequence has a chance to be exact. In fact, we have the following result:

Theorem 15.1 (Mayer-Vietoris).

In the above context, there are natural maps δ:Hn(UV)Hn-1(UV) such that the resulting sequence


is exact for all n.

In this section, we will explain some homology calculations based on the above theorem. The proof will be given in Section 19, using preliminary results from Sections 17 and 18.

We first note that the exact sequences in the theorem can be chained together to make an infinite sequence as follows:

The following slightly modified form is often convenient.

Proposition 15.2.

Suppose that UV is path connected (and therefore nonempty). Then we can modify the Mayer-Vietoris sequence by replacing all the H0 terms by zero, and the resulting sequence is still exact. Moreover, if U and V are also path connected then the same is true of UV, so all the H0 groups are isomorphic to .


Pick a point aUV, so H0(UV)=.[a]. Now H0(U) has a basis corresponding to the elements of π0(U), and i*[a] is one of these basis elements, so the map i*:H0(UV)H0(U) is injective. It follows that the map [i*-j*]:H0(UV)H0(U)H0(V) is also injective. Now consider the tail end of the Mayer-Vietoris sequence:


The map [i*-j*] is injective, and so the kernel is zero. As the sequence is exact, we see that the image of δ is also zero, so δ is the zero homomorphism. Using exactness again, we deduce that the map [k*l*] must be surjective. Using this together with the exactness of the original MV sequence, we see that the sequence


is exact. The rest of the modified MV sequences is the same as the original MV sequence, and so is also exact, as claimed.

Now suppose that U and V are also path connected. Any point in U can be joined to a by a path in U, and any point in V can be joined to a by a path in V, so any point in the space X=UV can be joined to a by a path in X. This shows that X is also path connected. We therefore see that the groups H0(X), H0(U), H0(V) and H0(UV) are all the same: a copy of , generated by [a]. ∎

We now want to calculate the homology groups of spheres Sn for n0. We will do this by induction, starting with n=0. Note that the point e0=(1,0,,0) always lies in Sn and so gives an element an=[e0]H0(Sn). Note also that


As in Remark 10.24, it follows that the elements a0=[e0] and [-e0] give a basis for H0(S0), and that Hk(S0)=0 for k0. However, it turns out to be more convenient to put b0=[-e0]-[e0]H0(S0); the elements a0 and b0 also form a basis.

Theorem 15.3.

For all n1 there is an element bnHn(Sn) such that

Hk(Sn)={an if k=0bn if k=n0 otherwise. 

Recall that


We will think of a point xn+1 as a pair (y,z) with yn and z, so x2=y2+z2, and x lies in Sn iff y2+z2=1. We put N=(0,1) (the “North pole”) and U=Sn{N}={(y,z)Sn|z1} and V=Sn{-N}={(y,z)Sn|z-1}. These are open sets with UV=Sn and


Just as in Example 4.13 we have a homeomorphism f:Un given by f(y,z)=y/(1-z) with f-1(y)=(2y,y2-1)/(y2+1). This proves that U is contractible. The map (y,z)(y,-z) gives a homeomorphism between U and V, so V is also contractible. It follows that H0(U)=H0(V)= and Hd(U)=Hd(V)=0 for d>0.

We also have a homeomorphism g:UVSn-1×(-1,1) given by g(y,z)=y/y and g-1(y,z)=(1-z2y,z). (It is valid to divide by y here because y0 whenever (y,z)UV.) As (-1,1) is contractible, it follows that UV is homotopy equivalent to Sn-1. More precisely, we can define p:Sn-1UV by p(y)=(y,0), and we find that p is a homotopy equivalence, so the map p*:H*(Sn-1)H*(UV) is an isomorphism.

Now consider the Mayer-Vietoris sequence

  • (a)

    For d2 we have Hd(U)=Hd(V)=Hd-1(U)=Hd-1(V)=0. Thus, Lemma 12.19 tells us that in these cases the map δ:Hd(Sn)Hd-1(UV)=Hd-1(Sn-1) is an isomorphism. This almost proves the induction step, apart from tweaks needed when d=0,1. We are assuming that n1 so Sn is path connected so H0(Sn) Thus, we just need to deal with the case d=1.

  • (b)

    Using the known values of Hd(U) and Hd(V) for d=0,1, the bottom end of the Mayer-Vietoris sequence is as follows:

  • (c)

    Consider the case where n=1. Point (a) tells us that for d2 we have Hd(S1)=Hd-1(S0)=0. Point (b) gives an exact sequence


    This means that δ gives an isomorphism from H1(S1) to the kernel of the map [k*l*].

    Here we have identified H0(U) with . More precisely, as U is connected and contains both e0 and -e0, we see that H0(U)=.[e0] and that [e0]=[-e0] in H0(U). It follows that k*(a0)=[e0] and k*(b0)=[-e0]-[e0]=0 in H0(U). For essentially the same reason we have l*(a0)=[e0] and l*(b0)=0 in H0(V). Thus, the kernel of [k*l*] is .b0. The exact sequence therefore means that there is a unique element b1H1(S1) such that δ(b1)=b0, and that H1(S1)=.b1. As S1 is path connected, we also know that H0(S1)=.a1. This proves that H*(S1) is as claimed.

  • (d)

    Now suppose instead that n2, and that we have already proved the claim for H*(Sn-1). Point (a) gives an isomorphism δ:Hn(Sn)Hn-1(Sn-1), so there is a unique element bnHn(Sn) with δ(bn)=bn-1, and we have Hn(Sn) Point (a) also shows that Hd(Sn)=0 for d2 with dn. As Sn is path connected, we have H0(Sn) This just leaves H1(Sn). The bottom end of the modified Mayer-Vietoris sequence has the form 0=H1(U)H1(V)H1(Sn)0, and exactness forces H1(Sn) to be zero as expected.

Remark 15.4.

If uHd(X) then we sometimes say that uH*(X) with |u|=d, and we call d the degree of u. We could then say that the elements an and bn form a basis for H*(Sn) with |an|=0 and |bn|=n. This formulation is valid for n=0 as well as for n>0.