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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 $H_*$.

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  • $\begingroup$ Do we know that the direct limit of chain complexes is a chain complex? I believe so. $\endgroup$ – user38268 May 24 '13 at 16:16
  • $\begingroup$ Related: math.stackexchange.com/questions/121122/… $\endgroup$ – Grigory M May 24 '13 at 16:23
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    $\begingroup$ Yes taking homology commutes with direct limits, by a pretty direct argument. Just write down the map and check it is 1-1 and onto. $\endgroup$ – Cheerful Parsnip May 24 '13 at 16:25
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    $\begingroup$ It is an exercise in these notes: win.tue.nl/~aeb/at/algtop-6.html $\endgroup$ – Cheerful Parsnip May 24 '13 at 16:26
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    $\begingroup$ One should probably point out that this is a special property of module categories, namely Grothendieck's axiom AB5: filtered colimits of exact sequences are exact). The dual property fails in the category of modules and the failure of $\varprojlim$ to preserve cokernels is measured by $\varprojlim\nolimits^1$, the derived functor of the (inverse) limit. $\endgroup$ – Martin May 24 '13 at 19:54
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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.

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    $\begingroup$ I'm sorry to hijack the answer and the whole question, but I was wondering how I could contact you privately. I'm just insterested in some advice you could give me. I'm also sorry if this breaks the meta and therefore ask for some meta-complying way of asking this question. Thanks, Andy. $\endgroup$ – Andy May 24 '13 at 17:38
  • $\begingroup$ Thanks, this is a great answer. $\endgroup$ – Dan Rust May 24 '13 at 18:59

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