Consider a topological space with open subsets and such that . There are many cases like this where we already understand three of the groups , , and , and we want to determine the fourth one.
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 .
We can now combine the above maps to get maps as follows:
Here the notation refers to the map sending to , and the notation refers to the map sending to . The composite of these two maps therefore sends to , but this is zero because . Thus, the above sequence has a chance to be exact. In fact, we have the following result:
In the above context, there are natural maps such that the resulting sequence
is exact for all .
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.
Suppose that is path connected (and therefore nonempty). Then we can modify the Mayer-Vietoris sequence by replacing all the terms by zero, and the resulting sequence is still exact. Moreover, if and are also path connected then the same is true of , so all the groups are isomorphic to .
Pick a point , so . Now has a basis corresponding to the elements of , and is one of these basis elements, so the map is injective. It follows that the map is also injective. Now consider the tail end of the Mayer-Vietoris sequence:
The map is injective, and so the kernel is zero. As the sequence is exact, we see that the image of is also zero, so is the zero homomorphism. Using exactness again, we deduce that the map 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 and are also path connected. Any point in can be joined to by a path in , and any point in can be joined to by a path in , so any point in the space can be joined to by a path in . This shows that is also path connected. We therefore see that the groups , , and are all the same: a copy of , generated by . ∎
We now want to calculate the homology groups of spheres for . We will do this by induction, starting with . Note that the point always lies in and so gives an element . Note also that
As in Remark 10.24, it follows that the elements and give a basis for , and that for . However, it turns out to be more convenient to put ; the elements and also form a basis.
For all there is an element such that
Recall that
We will think of a point as a pair with and , so , and lies in iff . We put (the “North pole”) and and . These are open sets with and
Just as in Example 4.13 we have a homeomorphism given by with . This proves that is contractible. The map gives a homeomorphism between and , so is also contractible. It follows that and for .
We also have a homeomorphism given by and . (It is valid to divide by here because whenever .) As is contractible, it follows that is homotopy equivalent to . More precisely, we can define by , and we find that is a homotopy equivalence, so the map is an isomorphism.
Now consider the Mayer-Vietoris sequence
For we have . Thus, Lemma 12.19 tells us that in these cases the map is an isomorphism. This almost proves the induction step, apart from tweaks needed when . We are assuming that so is path connected so . Thus, we just need to deal with the case .
Using the known values of and for , the bottom end of the Mayer-Vietoris sequence is as follows:
Consider the case where . Point (a) tells us that for we have . Point (b) gives an exact sequence
This means that gives an isomorphism from to the kernel of the map .
Here we have identified with . More precisely, as is connected and contains both and , we see that and that in . It follows that and in . For essentially the same reason we have and in . Thus, the kernel of is . The exact sequence therefore means that there is a unique element such that , and that . As is path connected, we also know that . This proves that is as claimed.
Now suppose instead that , and that we have already proved the claim for . Point (a) gives an isomorphism , so there is a unique element with , and we have . Point (a) also shows that for with . As is path connected, we have . This just leaves . The bottom end of the modified Mayer-Vietoris sequence has the form , and exactness forces to be zero as expected.
∎
If then we sometimes say that with , and we call the degree of . We could then say that the elements and form a basis for with and . This formulation is valid for as well as for .