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?



Boundary operator in the colimit of complexes.
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 $C$.


