Let be a bounded below complex of injectives. Let be a quasi-isomorphism. Then there exists such that in (homotopy equivalent in ).

Consider the cone , which comes equipped with a map to . Now, because is a quasi-isomorphism, then is acyclic, i.e., it is a long exact sequence: this is because we have a long exact sequence

so must be .

Now, is a bounded below complex of injectives. Recall that given two injective resolutions and , we can construct unique up to homotopy:

However, the construction only requires that we are mapping from an acyclic complex to a complex of injectives. In our case, we apply this to : is unique up to homotopy. But since works, then is homotopic to . That is, there exists such that

Now, we have

Claim: This (the composition of inclusion and ) works. Exercise: Check that is a chain map and that

Suppose we have a morphism in . This is the data of

where is a quasi-isomorphism.
Claim: This morphism is equal to (in ) (an actual chain map).
To verify this, we need the commutative diagram (by definitiondefinition)

The key here is that the red is a quasi-isomorphism.

Let be a bounded below complex of injectives in an abelian category . Then for all ,

Similarly, if is a bounded above complex of projectives, then

The natural map is in the direction.

We just discussed this: given , we have .

If , when does in ? We need a diagram

for some object and a quasi-isomorphism. By the Key lemmaKey lemma, there exists such that in . But so . So .

If , there are functors

where we “put in degree ”, i.e., . What is if . Take an injective resolution

This means that in , we have

So now,

What does a chain map look like? We have a diagram

So the chain maps are

What about homotopy? We have , with in blue in the above diagram. But now is forced. So there is no homotopy. Therefore . Therefore is fully faithful if has enough projectives or has enough injectives (because we needed to take a resolution).

Now we have

The chain maps are

What about homotopy? When is ? We need such that . That is, the existence of

To get the in the homotopy category, we need to take the chain maps and mod out by homotopy. But we have

and therefore we have precisely

Therefore .

There is a product on :

which is just

The name “product” is appropriate because when , we can then turn into an algebra.

The derived version is

called the Yoneda product. But in the derived category, this is just composition again. We can identify

Note that the middle blue part of the computation works in any triangulated category (or even more generality, where the category has an autoequivalence).

We didn’t check associativity. It holds.

is an algebra, called .

If we take where is “nice”. Then this algebra is .

One answer: Forced by a tensor structure (e.g. on for a group).

Let be a category with and , where

(where the isomorphisms are part of the data), then we get the concept of a monoidal category.

In a monoidal category, is commutative.

In -mod for a group, the unit is , the -dimensional trivial module, and (where we take tensor over ).

Let be a commutative ring. In -mod, .