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Gelfand and Manin give the definition of an internal $Hom^{\bullet}(A^{\bullet}, B^{\bullet})$ complex for two cochain complexes $A^{\bullet}, B^{\bullet}$:

$$Hom^n (A^{\bullet}, B^{\bullet})=\prod_{i} Hom (A^i, B^{i+n}).$$

Then in exercise IV.2.2.b as a hint for a construction of a spectral sequence they ask to take an internal Hom complex for two filtered complexes $(A^{\bullet} \supset ... \supset F^p A^{\bullet} \supset ... )$ and$(B^{\bullet} \supset ... \supset F^q B^{\bullet} \supset ...)$, with the induced filtration.

So how do I define $$(F^p Hom^{\bullet}(A^{\bullet}, B^{\bullet}))^n \ \ ?$$

I am struggling to set up the correct indexing for this natural filtration, and usually it helps to look at the easier tensor product case first, but I don't see how translate it to our case here (because of the contravariance in the first argument).

Edit: The following filtration on the level of the double complex seems to work: $$F^p Hom (A^n, B^m) = \{ f: A^n \to B^m | f(F^k A^n) \subset F^{k+p} B^m \text{ for every} \ k \}.$$ I'll leave the question in case someone finds it useful later on.

Bananeen
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  • Since $\operatorname{Hom}$ is left exact, my intuition would be to take cokernels of the inclusions for the contravariant argument. – jgon Mar 21 '19 at 12:40
  • @jgon, but since it should be a filtration, the resulting objects should be sitting inside $Hom(A^n, B^m)$, so I don't think cokernels would work. – Bananeen Mar 21 '19 at 12:48
  • But contravariant Hom turns surjective maps into inclusions. Thus my suggestion. – jgon Mar 21 '19 at 13:06
  • I didn't think too hard beyond that, and I'd be worried that the resulting subcomplexes don't actually form a filtration (i.e. that maybe they don't include into each other on the right way), but I didn't spend too much time checking it – jgon Mar 21 '19 at 13:09

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