Does taking the direct limit of chain complexes commute with taking homology?

Question

Suppose I have a directed system $C_i$, $i\in\mathbb{N}$ of chain complexes over free abelian groups (bounded below degree $0$) $$C_i=0\rightarrow C^{0}_{(i)}\rightarrow C^{1}_{(i)}\rightarrow\cdots\rightarrow C^{n-1}_{(i)}\rightarrow C^n_{(i)}\rightarrow \cdots$$ with chain maps $f_i\colon C_i\rightarrow C_{i+1}$. Can I say that $$H_*\left(\lim_{\rightarrow}(C_i,f_i)\right)\cong\lim_{\rightarrow}\left(H_*(C_i),(f_i)_*\right),$$ where $\displaystyle\lim_{\rightarrow}$ is the direct limit (colimit) in the respective category, $H_*(C)$ is the *th homology of the the chain complex $C$, and $f_*$ is the induced homomorphism in homology of the chain map $f$?

I imagine the answer will involve some categorical property of the functor $H_*$.

Answer 1

Any exact functor between abelian categories will preserve homology, and colimits indexed by filtered or directed diagrams are exact in $\mathbf{Ab}$.

The first claim is straightforward, because homology is computed using kernels and cokernels. Indeed, given a chain complex $C_{\bullet}$, we form the object of cycles as a kernel, $$0 \longrightarrow Z_n \longrightarrow C_n \longrightarrow C_{n-1}$$ and we form the object of boundaries as a cokernel, $$0 \longrightarrow Z_{n+1} \longrightarrow C_{n+1} \longrightarrow B_n \longrightarrow 0$$ and then the homology object is also a cokernel: $$0 \longrightarrow B_n \longrightarrow Z_n \longrightarrow H_n \longrightarrow 0$$

The second claim is best checked by hand using the concrete description of filtered/directed colimits in $\mathbf{Ab}$. By general nonsense, colimits are additive and preserve cokernels, so it is enough to check that kernels are preserved by filtered/directed colimits.