Boundary operator in the colimit of complexes.

Question

Let $\mathcal{A}$ be an abelian category and let $\mathcal{C}$ be the (abelian) category of complexes of objects in $\mathcal{A}$. Suppose we have a small indexing category $\mathcal{I}$ and a functor $F:\mathcal{I}\to \mathcal{C}$, which defines a diagram indexed by $\mathcal{I}$ which gives a complex $C_i$ for each $i\in\cal{I}$.

Suppose then we form the colimit $\mathrm{colim}_{i\in \cal I}C_i$ of this diagram (so $C=\mathrm{colim}_{i\in\cal{I}}$ is a complex along with a family of chain maps $f_i:C_i\to C$ such that if $m:j\to k$ is a morphism in $\cal I$ then $f_k\circ F(m)=f_j$, and $C$ is universal with respect to this property).

My question is: given that we have boundary operators $\partial_i$ in each complex $i$, how is the boundary operator in the colimit complex $C$ defined, or forced to be equal to?

Answer 1

For each integer $n$, the $n$th term $C[n]$ in the colimit complex $C$ is the colimit of the $n$th terms $C_i[n]$. For each $i \in \mathcal{I}$ and each integer $n$, the $n$th boundary map in $C_i$ is $\partial: C_i[n] \to C_i[n-1]$, so you can compose with your chain map $f_i$ (restricted to $C_i[n-1]$) to get $C_i[n] \to C_i[n-1] \to C[n-1]$. The resulting map $C_i[n] \to C[n-1]$ is compatible with the morphisms $m$ in the category $\mathcal{I}$, so there is an induced map $C[n] \to C[n-1]$. This is the $n$th boundary map in $C$.