In the following short exact sequence,

Consider the functor . In particular, if we have , we can define by

If is a -algebra, i.e., a -vector space with (associative and bilinear) multiplication, then we can also take as the target of . Applying to the above short exact sequence,

Is this exact?

Take and consider the short exact sequence

Take . We have . For ,

So can only be or in , and . Applying , we have

For the first map, we have , and so we have an isomorphism. For the second map, we have , and so we have the zero map:

We find that the sequence fails to be exact in the last spot.

Let

be a short exact sequence. Then

is exact.

Suppose such that .

First, because the composition . Now, for ,

Let

be a short exact sequence. Then

is exact.

A functor is left exact if for all short exact sequences

the sequence

is exact. It is right exact if instead the sequence

is exact. It is exact if it is both left and right exact.

and are left exact.

A module is projective if is exact. Equivalently, for all surjective homomorphisms and , there exists such that

In other words, is surjective (think of as being the last map in short exact sequence). Here, we think of as a quotient of .

A module is injective if is exact. Equivalently, for all injective homomorphisms and , there exists such that

In other words, is injective (think of as being the last map first short exact sequence). Here, we think of as a submodule of .

An -module is free if (where is a possibly infinite indexing set). Equivalently, there exists such that every can be uniquely expressed in the form

for . is called a basis of .

Consider the diagram

For each basis element , surjectivity of means that there exists such that .

So for each , we choose such that , which is possible since is surjective. Now, the definition of is forced upon us by linearity:

works. This is well-defined because is a basis: the expression is unique.

is injective for silly reasons.
Is injective? No. Consider the diagram

Is injective? No. Consider the diagram

By the classification of finitely generated abelian groups, no non-zero finitely generated group is injective.

The problem in all the non-examples above is that we can’t divide. This turns out to be the only obstruction, and in fact is injective.

A module is projective if and only if it is a direct summand of a free module, i.e., there exists a module such that is free.

Mostly done already in the exampleexample above.

Fact: every module is the quotient of a free module, i.e., there exists a free module and a surjective homomorphism . So we consider the following diagram

Because is projective, there exists such that . By the fact belowfact below,

Given and such that ,

we have the decomposition