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?
Best Answer
The category of complexes in an abelian category $\mathcal{A}$ is a full subcategory of $\text{Fun}({\mathbb{Z}},\mathcal{A})$, where $\mathbb{Z}$ is partially ordered under reverse inequality. So if we know how these constructions are performed in the category of functors from $\mathbb{Z}$ to $\mathcal{A}$, we'll have the natural candidates to the category of complexes. The standard result is that (co)limits in $\text{Fun}({\mathcal{D}},\mathcal{C})$, where $\mathcal{D}$ is a small category and $\mathcal{C}$ is a category, are computed pointwise. Take a look at Borceux's 'Handbook of Categorical Algebra, Vol.$1$' section $2.15$. There he explains the precise meaning of being computed pointwise.
Since any abelian category is finitely (co)complete, we can compute any finite (co)limit in $\text{Fun}({\mathbb{Z}},\mathcal{A})$ pointwise. If we consider a (co)complete category, e.g., the category of modules over a ring, we can compute any (co)limit pointwise.
If you think Borceux's book is too terse, there is a similar discussion in Rotman's 'An Introduction to Homological Algebra' on page $317$.
Added: In order to remove the iotas you will need to prove that $\alpha_{n-1}\partial_n^\oplus\iota_n^B=\partial_n^C\alpha_n\iota_n^B$. Now use the fact that there is only one morphism $\varphi: A_n \oplus B_n \rightarrow C_{n-1}$ such that $\varphi \iota_n^A = \alpha_{n-1}\partial_n^\oplus\iota_n^A$ and $\varphi \iota_n^B = \alpha_{n-1}\partial_n^\oplus\iota_n^B $.