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

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 $H_*$.

Solutions Collecting From Web of "Does taking the direct limit of chain complexes commute with taking homology?"

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.