Exact Sequences and Relative Homology

Long Exact Sequence of a Good Pair

Let’s recall the definition of exact sequences in the context of abelian groups:

A sequence of morphisms between objects in an abelian category is exact at if . The sequence is an exact sequence if it is exact at every object.

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Good Pair

A pair with is called a good pair if is a non-empty closed subset of that has an open neighborhood which deformation retracts onto .

Optionally, we can write the sequence in reduced homology as: But note that is always reduced.

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To prove this theorem we need the notion of relative homology and several lemmas:

Relative Chains and Homology

For any pair , the relative chain groups are defined as the quotient . The boundary map on induces a boundary map , turning it into a chain complex. The homology of this complex, denoted , is called the relative homology of the pair .

From Short to Long Exact Sequences

Lemma

For any pair , there is a long exact sequence relating their homology groups:

This sequence arises from a short exact sequence of chain complexes. Recall that the relative chain group is defined as the quotient . This gives a short exact sequence of chain complexes: where is induced by the inclusion and is the projection. The existence of the long exact sequence in homology is a general algebraic result stated as follows:

From Short to Long Exact Sequences

A short exact sequence of chain complexes induces a long exact sequence in homology:

Long Exact Sequence for a Triple

For a triple of spaces , there is a corresponding long exact sequence of relative homology groups: This arises from the short exact sequence of relative chain complexes: This follows from the isomorphism .

Proof We need to construct the connecting map explicitly, and show that it is well-defined, making the sequence exact. The construction of is a classic example of a diagram chase:

1. Take a homology class and choose a representative cycle for this class.
2. Since is surjective, there exists some such that .
3. Consider the boundary . By the commutativity of the diagram (), we have:
4. This implies . By exactness at , . Therefore, there exists a unique such that . Uniqueness follows from being injective.
5. This element is a cycle. To see this, note that is injective, and: Since is injective, .

Finally, we define . Notice that we made two choices in this construction: the choice of representative cycle for the homology class , and the choice of lift in . We need to show that the resulting homology class is independent of these choices, so that is well-defined.

• Independence of the representative : Suppose we choose another representative for the same homology class . Since is surjective, we can find such that . Let . Then . Now, we compute the boundary of : The new element we find is the same, so the result is unchanged.
• Independence of the lift : Suppose we choose a different element such that . Then , so . Thus, there exists some such that , so . Now consider the boundary of : The new element is defined by . Thus: Since is injective, . This means and are in the same homology class.

Thus, the connecting homomorphism is well-defined. To prove the sequence is exact, one must verify that and at each group , , and . We sketch two of the six required arguments.

1. Exactness at : We show .
   • : For any , . Since the sequence of chain complexes is exact, , so the induced map is also zero. Thus .
2. Exactness at : We show .
   • Let , represented by a cycle with . Then . To compute , we follow the construction:
     1. Representative cycle is .
     2. Lift to . We can choose itself.
     3. Compute its boundary: .
     4. This means the corresponding element is .
     5. Therefore, . So .
3. Exactness at : We show .
   • Let such that . We want to find a class such that .
     1. Let be a cycle representing .
     2. means that is a boundary in . So there is some with .
     3. Let .
     4. We check if is a cycle: . Since , we have .
     5. So, . By construction, is the class represented by the element such that . But , so .
     6. Thus , which shows .

Applications and Examples

e.g. Consider the pair . The long exact sequence in reduced homology is: Since the disk is contractible, its reduced homology is zero in all dimensions, i.e., for all . The long exact sequence breaks down into short exact sequences: This implies that the map between them is an isomorphism: Since is homeomorphic to the sphere , and we know is for and otherwise, we conclude:

e.g. Consider the pair where is a point in . The long exact sequence in reduced homology is: A single point has zero reduced homology in all dimensions, so for all . The sequence again breaks into short exact sequences: This gives an isomorphism: This shows that the relative homology of a space with respect to a point is precisely the reduced homology of that space.

Excision Theorem

For any such that , the inclusion induces an isomorphism .

Lemma

For a good pair , the relative homology is isomorphic to the reduced homology of the quotient space: