Our goal is the following theoremtheorem:

Let be an algebra and let be an -module. Then there exists an injective module which has as a submodule: . That is to say, the category has enough injectives.

We had a subalgebra previously. But all we need is an algebra homomorphism (not necessarily injective) such that . Then there is a restriction functor

where the -action on is for all , . is the right adjoint to (which we have done beforebefore, by thinking of as and as ).

Now, consider , a -bimodule with actions given by

Then we can write instead as as a tensor:

We now define the coinduction functor

The action of on is coming from the right -action on :

is the right adjoint to , i.e.,

By Tensor-Hom adjunction.

Let be a subgroup of where and are finite (possibly only need finite index). Let . Then , and we say that and are bi-adjoint.

Let be a homomorphism where and are finite. Let

If is invertible in , then .

sends injectives to injectives.

It has an exact left adjoint . The result follows by this propositionthis proposition.

Let be a field. Let be a -algebra. We have . We know that is an injective -module (-vector space). So

is an injective -module (the action of on is coming from the right action of on ). If is finite-dimensional over then is also finite-dimensional over (in fact it has the same dimension as ). If is infinite-dimensional over , then is massive.

Let be an algebra homomorphism. Let be an -module. If there exists an injective -module such that , then there exists an injective -module with .

Let . This is injective by the above Theorem ($ CoInd$ preserves injectives)Theorem ( preserves injectives). Further, we have by Theorem ($ CoInd$ and $ Res$)Theorem ( and ) that

so induces a map (not necessarily injective a priori, the abstract nonsense does not tell us such). But concretely, , and tracing through definitions, we have

Now if , then for all . In particular, taking , we have . Therefore is injective.

We already know that (by Baer’s Criterion)

where a -module is said to be divisible if for all , , , there exists such that .

Let . We can write as a quotient of a free module :

We can put into something divisible: we know that , and is divisible and therefore injective. The “fancy” way to write this is :

The dashed map exists because the composition

is , and therefore factors through the cokernel . We can show that is injective using the Snake Lemma or the -lemma or a diagram chase. We now claim that is injective. This is because any quotient of a divisible module is divisible: we can just go back up to to do the division and go back down. Therefore any quotient of an injective abelian group is injective. So is injective.

The property that any quotient of an injective abelian group is injective implies that

This is because if we try to take an injective resolution of ,

is already an injective resolution. Hence, can always be taken to be .

is an initial object in the category of unital algebras, i.e., for all algebras , there exists a unique (unital) homomorphism .