MAS61015 Algebraic Topology

19. Construction of the Mayer-Vietoris sequence

Video (All of Section 19)

This section will constitute the proof of Theorem 15.1. As before, let X be a topological space, and let U and V be open sets with X=UV. We name the inclusion maps as shown below:

This gives a sequence of chain maps as follows:


If this was a short exact sequence, then we could apply Theorem 17.2 to get the Mayer-Vietoris sequence. Unfortunately, however, this is not quite a short exact sequence: we will see that the first map is injective and the image of the first map is the kernel of the second one, but the second map is not surjective. We will need an extra step involving subdivision to deal with this issue.

Definition 19.1.


Sn(U) ={u:ΔnU|u(Δn)UV}
Sn(V) ={u:ΔnV|u(Δn)UV}
Sn(X) ={u:ΔnX|u(Δn)U and u(Δn)V}
Sn(U,V) ={u:ΔnX|u(Δn)U or u(Δn)V}.

We also define Cn(U)={Sn(U)} and similarly for Cn(V), Cn(X) and Cn(U,V).

Remark 19.2.

If uSn(U) then u does not send the whole of Δn into UV, but it may send some faces of Δn into UV. Because of this, C*(U) need not be closed under , so it need not be a subcomplex of C*(U). However, this will not matter for our immediate purposes.

We now note that

Sn(U) =Sn(UV)Sn(U)
Sn(V) =Sn(UV)Sn(V)
Sn(U,V) =Sn(UV)Sn(U)Sn(V)
Sn(UV) =Sn(UV)Sn(U)Sn(V)Sn(X),

and all these unions involve disjoint sets. It follows that

Cn(U) =Cn(UV)Cn(U)
Cn(V) =Cn(UV)Cn(V)
Cn(U,V) =Cn(UV)Cn(U)Cn(V)
Cn(UV) =Cn(UV)Cn(U)Cn(V)Cn(X),

Thus, in our earlier sequence, the map [i*-j*]:Cn(UV)Cn(U)Cn(V) becomes the map


given by a(a,0,-a,0). Similarly, the map [k*l*]:Cn(U)Cn(V)Cn(UV) becomes the map


given by (a,b,a,c)(a+a,b,c,0). From this it is clear that we have a short exact sequence


and an inclusion C*(U,V)C*(X) of chain complexes. By applying Theorem 17.2 to the short exact sequence, we get something which is essentially the Mayer-Vietoris sequence except that it involves H*(C*(U,V)) instead of H*(UV)=H*(X). To complete the construction of the Mayer-Vietoris sequence, we need to show that H*(C*(U,V)) is actually the same as H*(X).

Lemma 19.3.

For any uCn(X) there exists k0 such that sdk(u)Cn(U,V).


We can easily reduce to the case where u is a single map ΔnX. Put U=u-1(U) and V=u-1(V), so U and V are open in Δn with UV=Δn. This means that {U,V} is an open cover of Δn, so Proposition 8.31 tells us that there is a Lebesgue number, say ϵ>0. The identity chain ιnCn(Δn) has diameter 2, so sdk(ιn) has diameter at most (n/(n+1))k2. If we choose k large enough, then this diameter will be less than ϵ, and it will follow that every simplex involved in sdk(ιn) is either contained in U or contained in V. It follows that every simplex involved in the chain sdk(u)=u*(sdk(ιn)) is either contained in U or contained in V, so sdk(u)Cn(U,V) as claimed. ∎

Corollary 19.4.

The homology of the quotient complex Q*=C*(X)/C*(U,V) is zero.


First, note that the subdivision map sd:C*(X)C*(X) sends C*(U) to C*(U) and C*(V) to C*(V) so it also sends the subcomplex C*(U,V)=C*(U)+C*(V) to itself. We therefore have an induced map sd:Q*Q* given by sd(u+Cn(U,V))=sd(u)+Cn(U,V). Similarly, the chain homotopy σ:Cn(X)Cn+1(X) induces a chain homotopy σn:QnQn+1. We showed previously that σ+σ=id-sd on C*(X), and it follows that we have the same relation on Q*. We therefore deduce from Proposition 14.7 that the map sd*:Hn(Q*)Hn(Q*) is the identity. Consider an element qHn(Q*). This has the form q=z+Bn(Q*) for some zZn(Q*)Qn=Cn(X)/Cn(U,V). This in turn has the form z=u+Cn(U,V) for some uCn(X). For sufficiently large k we have sdk(u)Cn(U,V), so sdk(z)=0, so sd*k(q)=0. As sd* is the identity this means that q=0. Thus, we have Hn(Q*)=0 as claimed. ∎

Corollary 19.5.

The inclusion C*(U,V)C*(X) induces an isomorphism H*(C*(U,V))H*(X).


We have a short exact sequence of chain complexes C*(U,V)C*(X)Q*. Theorem 17.2 therefore gives us exact sequences


The first and last groups are zero by Corollary 19.5, so exactness forces the middle map to be an isomorphism. ∎

This completes the construction of the Mayer-Vietoris sequence.