In this post we shall see a natural example that should give some motivation for derived functors of functors on the homotopy category of cochain complexes in an abelian category $ \mathcal{A}$. At first glance, the formalism of derived functors in this setting may seem less intuitive than the formalism of classical derived functors, partially because of the need to consider hypercohomology (cohomology of complexes), but hopefully in this and in future posts we will see some examples that show the elegance of the theory.

This material is basically contained in Weibel's book `An Introduction to Homological Algebra" in various places, but I thought it would be nice to have a condensed summary all in one place.`

Let $ A$ and $ B$ be cochain complexes whose objects are in an abelian category $ \mathcal{A}$ with arbitrary products. Recall that any cochain complex can be thought of as a chain complex by using the following reindexing: $ A_p := A^{-p}$.

Anyway, there is an interesting way that we can get a new cochain complex from $ A$ and $ B$. We start out with the Hom-double complex, which is a double complex that has in $ p,q$-degree $ \mathrm{Hom}(A,B)_{p,q} = \mathrm{Hom}(A_p,B^q)$.

Its differentials are defined by $ d^hf(p) = f(dp)$ and $ d^vf(p) = (-1)^{p+q+1}d(fp)$ where the domain of these differentials is $ \mathrm{Hom}(A,B)_{p,q}$. The sign in the vertical differential is used because we are using the sign convention that $ d^vd^h + d^hd^v = 0$; that is, the squares in our double complexes will always anticommute; this is convenient because then the total differential obtained in any total complex will be an actual differential.

We now form the total complex $ \mathrm{Tot}\mathrm{Hom}(A,B) := \mathrm{Tot}^{\prod_{}} \mathrm{Hom}(A,B)$, which has in degree $ n$ the object $ \prod_{p+q = n}\mathrm{Hom}(A,B)_{p,q}$; it has as a differential the product of all the horizontal and vertical differentials; that is, there are exactly two differentials coming into each component. Now, an element in $ \mathrm{Tot}\mathrm{Hom}(A,B)$ is just an infinite collection of maps $ f_p:A^p\to B^{n+p}$. Using $ f$ for all of these maps and ignoring indexing, that $ f$ is a $ n$-cycle means that

So if we modify $ f$ by $ (-1)$ in every second place in the product we see that the cohomology of the total complex $ H^n\mathrm{Tot}\mathrm{Hom}(A,B)$ is nothing other than equivalence classes of chain maps $ f:A\to B[-n]$ under the equivalence relation of homotopy equivalence. Recall that we say $ f,g:A\to B$ are homotopy equivalent if there are arbitrary $ \mathcal{A}$-maps $ s:A^n\to B^{n-1}$ such that $ f – g = ds + sd$. This notation and this last paragraph will probably confusing unless the reader writes this out in detail using diagrams; at least I always do this.

### Downstairs in the Triangulated Category

Now, let $ \mathrm{Ch}(\mathcal{A})$ denote the category of cochain complexes in $ \mathcal{A}$ and let $ \mathrm{K}(\mathcal{A})$ denote the triangulated category of cochain complexes, the morphisms $ \mathrm{Hom}_{\mathrm{K}(\mathcal{A})}(A,B)$ being equivalence classes of cochain maps under homotopy equivalence. What we have seen is that there is an isomorphism of abelian groups

where $ T^nB := B[-n]$ is the translation functor of the triangulated category $ \mathrm{K}(\mathcal{A})$. Now, we would like a similar statement for the derived category $ \mathrm{D}(\mathcal{A})$. Recall that this is the localization of $ \mathrm{K}(\mathcal{A})$ at the set of quasiisomorphisms; we assume here that the derived category $ \mathrm{D}^+$ exists (which it sometimes does not).

So if the derived category $ \mathrm{D}^+(\mathcal{A})$ does exist, we want a statement that looks like

where $ X$ is some object that depends on $ A$ and $ B$; since we are only assuming $ \mathrm{D}^+(\mathcal{A})$ exists, and additionally that derived functors will appear somewhere, we will have to assume that $ A$ and $ B$ are bounded below complexes.

What is this object you ask? Read on!

### Deep Down in the Derived Category

So, it's no surprise that the object $ X$ is constructed using a derived functor. To reiterate our assumptions, let us assume now that $ A$ and $ B$ are bounded below.

Let us also assume that $ \mathcal{A}$ has enough injectives. In this case, the derived functor $ \mathbf{R}^+\mathrm{Hom}(A,-):\mathbf{D}^+(\mathcal{A})\to\mathbf{D}(\mathrm{Ab})$ of $ \mathrm{Hom}(A,-):\mathrm{K}^+(\mathcal{A})\to \mathrm{K}(\mathrm{Ab})$ exists where $ \mathrm{Ab}$ is the category of abelian groups. We write $ \mathbf{R}\mathrm{Hom}(A,B)$ for $ \mathbf{R}^+\mathrm{Hom}(A,-)(B)$. The object $ \mathbf{R}\mathrm{Hom}(A,B)$ will be our $ X$.

Why is this? We shall sketch the idea of the proof and refer the reader to go through the proof in Weibel's book for more enlightenment. Basically, we first observe that if $ A$ is replaced by a quasiisomorphic complex $ A'$, then the new $ \mathbf{R}\mathrm{Hom}(A',B)$ is naturally isomorphic to the first. This allows us to view $ \mathbf{R}\mathrm{Hom}(A',B)$ as a bifunctor on derived categories, so we can compute the cohomology of the complex $ \mathbf{R}\mathrm{Hom}(A,B)$ by replacing $ B$ with a complex of injectives quasiisomorphic to $ B$ (why does this exist? hint: Cartan-Eilenberg resolutions); for notation's sake we will assume then that $ B$ is already a complex of injectives.

In this case though, $ H^n\mathbf{R}\mathrm{Hom}(A,B)$ is just the homology of the total complex of $ \mathrm{Hom}(A,B)$, which as we have seen is $ \mathrm{Hom}_{\mathrm{K}(\mathcal{A})}(A,B)$; since $ B$ is now a complex of injectives, this is the same thing as $ \mathrm{Hom}_{\mathrm{D}(\mathcal{A})}(A,B)$. Thus the object $ \mathbf{R}\mathrm{Hom}(A,B)$ that we got by applying the derived functor $ \mathbf{R}^+\mathrm{Hom}(A,-)$ to $ B$ is the $ X$ we were looking for; we get the isomorphism

### Conclusion

We have just seen one way which derived functors fit into the framework of derived categories. Of course, this is just a minor formalism; the true power of derived categories is in their applications to sheaves on manifolds and probably other things that I have yet to discover.