A monoidal category is

satisfies the interchange property

In a monoidal category, is commutative.

Let . Then

This is called the Eckmann-Hilton argument. This is also how one proves that is abelian for .

Let

Then is a graded algebra.

A -graded algebra is an algebra such that if and then .

A - or -graded algebra is graded commutative or super commutative if for all and then

Let and be complexes. Then there are isomorphisms

Now consider the diagram

This diagram does not commute, but anti-commutes:

This is the source of . There exists a derived version: a monoidal category the shift automorphisms implies that is graded commutative.

Suppose is “nice” and consider . Then is graded commutative.

Suppose we have a functor . Can we define

This works on the level of and . But does not (usually) give a derived functor.

Suppose we have and is a quasi-isomorphism in . So we have and thus . Then is a quasi-isomorphism in , i.e., it induces an isomorphism of cohomology etc. So we have problems if does not preserve exactness.

Let

be a short exact sequence in . The in . But now applying , we have

Is this in ? We need the cohomology to vanish, i.e., we need to be exact.

We need to replace by a projective/injective resolution (depending on which side is exact).

Let be an abelian category with enough injectives. Let be a complex such that for . Then there exists a complex with injective for all and a quasi-isomorphism .

We need to be left (right) exact. Then we replace by as in Theorem (Generalisation of existence of projective/injective resolutions)Theorem (Generalisation of existence of projective/injective resolutions) and apply term-wise to get a derived functor:

We inductively construct and , and .

We take for (in the range where ).

We now assume that the complex has been constructed for .

where is the pushout

We include this into an injective object (because we have enough injectives). This gives us the extension of the chain map. We need to show that this extends as a quasi-isomorphism.

Assume that so we can argue with elements. We are required to prove that is an isomorphism.

Injectivity: Suppose that . Then by the factorisation from the construction above,

Further, the map is surjective. So there exists (this should really be in the source ) such that in . Because is a pushout, there exists such that and . Therefore in .

Surjectivity: Let . Then

But also (still in )

So since is a pushout, there exists such that and

So in .

The Theorem (Generalisation of existence of projective/injective resolutions)Theorem (Generalisation of existence of projective/injective resolutions) implies that the obvious functor (going in the direction) is essentially surjective. Equality of spaces is from yesterdayyesterday.