# Hom cochain complex of two chain complexes

Does anybody know of a good reference (preferably online) where I can find a good, rigorous description of $Hom_R(C_\bullet,D_\bullet)_\bullet$, which is a cochain complex where the module is the nth degree is given as $\Pi_{i-j=n}Hom_R(C_i,D_j)$, where $C_\bullet$ and $D_\bullet$ are chain complexes? I would give what the coboundary maps are, but that is the main piece of information I am trying to find a definition of, as well as some other things about it. Thanks!

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The boundary in the complex $\hom(C_\bullet, D_\bullet)$ is simply the map $$\delta:\prod_{i-j=n}Hom_R(C_j,D_i) \to \prod_{i-j=n-1}Hom_R(C_j,D_i)$$ such that for each $f\in Hom_R(C_i,D_j)$ we have $$\delta(f)=f\circ d_C-d_D\circ f.$$ (You do have to figure out in which factor of the codomain of $\delta$ each of these two summands is; notice also that $0$-cycles are precisely the morphisms of complexes, and you should try to see what exactly does it mean for two such $0$-cocycles to be homologous)

This should be explained in pretty much every textbook on homological algebra. For example, this is one of the examples in section V.1 of P. J. Hilton, U. Stammbach's always useful Course on homological algebra.

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I have some difficulty with that definition. If $f:C_i\to D_j$ then when how can we compose $f$ with $d_C$, since after we apply $d_C$ we are in $C_{i-1}$? – Jon Beardsley Mar 11 '11 at 2:45
Well, $C$ has a lot of $d_C$s... use the one that has $C_i$ as codomain! – Mariano Suárez-Alvarez Mar 11 '11 at 2:48
Okay, this is fine. What I was failing to realize was that $f$ is actually an infinite collection of maps, not one map for one $i$, so when we talk about $d_Cf$ we mean applying the relevant coordinate of $f$ at the right spot. Thankyou! – Jon Beardsley Mar 11 '11 at 3:41