Projective Resolution of Modules

The Hom Functor and Projective Modules

The cohomology groups of a space are not independent of its homology groups. The relationship is given by the Universal Coefficient Theorem:

Let be a chain complex of free abelian groups and be an abelian group. Then for each , there is a short exact sequence: This sequence splits, but the splitting is not natural.

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The issue here is that given a short exact sequence , applying the contravariant functor does not necessarily yield another short exact sequence. For example, consider the short exact sequence . Applying the functor gives the sequence . However, , , thus, the sequence is not exact at the second last position. To fix this problem, we will introduce the functor, which measures the failure of exactness when applying the functor. Before that, we first see when the functor preserves exactness, which leads us to the notion of projective modules.

Projective Module

An -module is projective if for any surjective homomorphism of -modules and any homomorphism , there exists a lift such that .

e.g. Free modules are projective, because we can choose such that for each basis element of , and define the lift on the basis and extend linearly.

Lemma

The following hold for -modules:

1. Let be a short exact sequence of -modules. Then for any -module , the induced sequence is exact at and .
2. A short exact sequence splits if is projective.
3. If is a split short exact sequence, then applying the functor results in another split short exact sequence.

Proof The proof of (1) requires showing exactness at each position.

• Exactness at (i.e., is injective): Suppose for some . By definition, . So, . Since the original sequence is exact, is surjective, so must be the zero map.
• Exactness at : First, we show . Since the original sequence is exact at , the composition is the zero map. Therefore, . Next, we show . Suppose , which means and . This implies that . Since the original sequence is exact, . Thus, . By the first isomorphism theorem, factors through the quotient . Since is surjective, we have an isomorphism . This gives a well-defined map such that . By definition, this means , so .

Now we prove (2). The following diagram illustrates the situation, so that the splitting lemma can be applied: (3) follows simply from the fact that .

Projective Resolutions

Projective Resolution and Ext Functor

A projective resolution of a module is an exact sequence where each is a projective module. We denote this by . Taking a projective resolution of , remove the term, and apply the functor . This yields a cochain complex: The cohomology of this complex is denoted . is defined as the -th cohomology group .

However, this is not well-defined so far. We need to check that different projective resolutions of yield isomorphic groups.

Fundamental Theorem of Homological Algebra

Let be a homomorphism of -modules. Let and be projective resolutions of and respectively. Then:

1. There exists a chain map that extends , meaning .
2. Any two such chain maps and that extend are chain homotopic.

Proof The proof is constructive and proceeds by induction on the degree .

• Base Case: By projectivity of , we can lift the map to a map such that :
• Inductive Step: Assuming have been constructed to form a chain map up to that degree, we construct as follows: Since the RHS diagram commutes, we can conclude that . So the projectivity implies the LHS diagram commutes, completing the inductive step.

Now suppose both and are chain maps extending . We show they are chain homotopic by constructing homotopy maps inductively:

Corollary

Let and be two projective resolutions of the same module . Then there is a chain homotopy equivalence between them. Consequently, for any module ,

Proof By the above theorem, we can find chain maps and , both extending the identity on : The composition is a chain map extending , must be chain homotopic to . Similarly, is chain homotopic to . Chain homotopic maps induce the same map on cohomology. Therefore, the induced maps on cohomology, and , are inverses of each other, establishing an isomorphism.

The Ext Functor

The previous result shows that the cohomology groups depend only on the modules and , not on the specific projective resolution chosen for . This allows us to make a well-defined definition of the Ext functor:

Proposition

We have the following properties of the Ext functor:

1. .
2. for .

Classification of Extensions

The Ext functor gets its name from its connection to extensions of modules. For , the group provides a classification of the collection of all short exact sequences of the form up to a natural equivalence. These sequences are called “extensions” of by . For the ring of integers , it is common to abbreviate to simply .