Can the category of chain complexes be realized as a functor category?

As the title says, is there some sort of category $\mathsf{C}$ which can be thought of as the "walking chain complex", so that the category of chain complexes in some other category $\mathsf{A}$ can be realized as $\mathsf{A}^\mathsf{C}$ (or perhaps some subcategory thereof)?

Solution 1:

In the case of homological complexes bounded below by zero, C would be the preadditive category

• Objects are natural numbers
• $\hom(n+1, n) = \mathbb{Z}$ (call the generator $\partial$)
• $\hom(n, n) = \mathbb{Z}$ (the generator is $1_n$)
• All other homsets are zero

In other words, it's the preadditive category presented by

• Objects are natural numbers
• One generating arrow $\partial : n+ 1 \to n$ for each $n$
• All relations $\partial \circ \partial = 0$

Since an abelian group homomorphism $\mathbb{Z} \to H$ is the same thing as an element of $H$, an additive functor $F : \mathbf{C} \to \mathbf{A}$ is the same thing as:

• A choice $F_n$ of object of A for each natural number $n$
• A choice of morphism $\partial : F_{n+1} \to F_n$ for each natural number $n$
• Such that $\partial \circ \partial = 0$

IIRC, an amusing alternative way to construct C is the following procedure:

• Start with the opposite simplex category $\Delta^\circ$
• Linearize it (i.e. replace the homsets with the free abelian homgroups) to geta preadditive category $\mathbb{Z}\Delta^\circ$
• Split all idempotents

The additive category generated by this is equivalent to the additive category generated by C. (if you want C exactly, you can extract it as a full subcategory)

This is a sort of 'universal' form of the Dold-Kan correspondence, showing the walking simplicial abelian group is, in an appropriate sense, equivalent to the walking chain complex.