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How can I figure out the classical construction (direct sum, product, pullbacks, and in general direct and inverse limits) in the category made by chain complexes and chain maps (of abelian groups or any abelian stuff)? Because of this category is abelian it must have (co)limits, isn't it?

In particular take $$ \begin{gather*} \dots\to A_n\to A_{n-1}\to\dots\\ \dots\to B_n\to B_{n-1}\to\dots \end{gather*} $$ I would like to say that $\mathcal A\oplus\mathcal B$ is "what I want it to be", $A_n\oplus B_n$ with obvious maps. But a standard argument doesn't allow me to conclude it: maybe it is false? A single word with the right reference will be enough to close the topic; I am now reading Hilton & Stammbach.

Edit: I would like to add what I've tried to do, but seems difficult not to invoke some diagrams, which I'm not able to draw without a suitable package here. However, first of all both complexes inject in the sum by maps which are part of a chian complex, $\iota_n^A,\iota_n^B$. Then, consider another complex $\{C_n,\partial_n^C\}$ and a couple of chain maps $\{a_n\colon A_n\to C_n\}$, $\{b_n\colon B_n\to C_n\}$. For each $n$ there exists a map $\alpha_n\colon A_n\oplus B_n\to C_n$ factoring the $a_n$ and the $b_n$s. Then I would like to show that the maps $\alpha_n$ are part of a chain map between the sum and the complex $\mathcal C$, but trying to prove it I can only conclude that in the diagram (vertical rows are the $\alpha_n$s) $$ \begin{array}{ccc} A_n\oplus B_n &\xrightarrow{\partial_n^\oplus}& A_{n-1}\oplus B_{n-1} \\ \downarrow && \downarrow\\ C_n &\xrightarrow[\partial_n^C]{}& C_{n-1} \end{array} $$ which I want to be commutative, aka $\alpha_{n-1}\partial_n^\oplus=\partial_n^C\alpha_n$, I have $\alpha_{n-1}\partial_n^\oplus\iota_n^A=\partial_n^C\alpha_n\iota_n^A$. How can I remove the iotas?

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    I hate to just blindly throw out a reference without checking whether it has it or not, but my guess is that Weibel talks about this.2010-12-12
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    I'm sorry, i was intended to write "I don't see it there, in Hilton Stammbach", in the chapter about these topics.2010-12-12
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    The direct sum is $A_n\oplus B_n$ with the 'obvious' maps; could you tell us what you've tried and why it fails so we could help you identify your error? For the record, showing $\mathbf{Ch}$ is abelian is exercise IV.1.3 in H&S and the discussion on p.5-7. If I were you, I'd start reading Weibel at this point since the latter chapters of H&S are pretty terse, imo.2010-12-12
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    What do you mean by terse? I added some more.2010-12-12

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