Let $A$ be an abelian category (you can assume additional conditions for its goodness). Let $\mathrm{Seq}(A) = \mathrm{Func}(\mathbb{Z}, A)$, where $\mathbb{Z}$ is the standard order category on integers. Let $\mathrm{Chain}(A)$ be a subcategory of complexes in it.



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> Is $\mathrm{Chain}(A)$ a reflexive or coreflexive subcategory of $\mathrm{Seq}(A)$?
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As far as I can see, the inclusion functor $\mathrm{Chain}(A) \to \mathrm{Seq}(A)$ preserves all limits and colimits. I'm also interested in answers for variations like complexes bounded on one side.



Is the category of chain complexes a reflexive or coreflexive subcategory of the category of functors?
Yes, it's reflective and coreflective, under mild assumptions on the codomain category $\mathcal A.$ The adjoints are given, by definition, by Kan extension along the quotient from the abelian group-enriched category freely generated by $\mathbb Z$ to the abelian group-enriched category $\mathbb Z\_\partial$, Ab-functors out of which define chain complexes. (This map imposes the relations $d\_{i-1}d\_i=0$ for every $i.$) So it's sufficient to assume $\mathcal A$ is complete and cocomplete.


Explicitly, I believe the reflection $B\_\bullet$ of a sequence $\cdots A\_i \stackrel{f\_i}{\to} A\_{i-1}\cdots$ is given by $B\_{i-1}=A\_{i-1}/f\_if\_{i+1}(A\_{i+1})$ and the coreflection $C\_\bullet$, by $C\_{i+1}=f\_{i+1}^{-1}(\mathrm{ker}f\_i),$ both with the differential induced by the $f\_i.$ So any abelian $\mathcal A$ should be fine.


