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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. –  Matt Dec 12 '10 at 17:37
I'm sorry, i was intended to write "I don't see it there, in Hilton Stammbach", in the chapter about these topics. –  tetrapharmakon Dec 12 '10 at 17:40
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. –  Vladimir Sotirov Dec 12 '10 at 19:00
What do you mean by terse? I added some more. –  tetrapharmakon Dec 12 '10 at 19:17
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2 Answers

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 $.

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I will check in Borceux tomorrow, thanks. I thought the same thing, but if it is true, then why it doesn't work with limits ( math.stackexchange.com/questions/7949/… )? Maybe the question is stupid, but I thought also that if it doesn't work with limits, I must be careful with products and coproducts too. –  tetrapharmakon Dec 13 '10 at 19:05
@tetrapharmakon: Well Borceux only proves that the category $\text{Fun}({\mathcal{D}},\mathcal{C})$ is (co)complete if $\mathcal{C}$ is and that (co)limits are computed pointwise. You are interested in the category of chain complexes which is a full subcategory of $\text{Fun}({\mathbb{Z}},\mathcal{A})$. So the (co)limits in $\text{Fun}({\mathbb{Z}},\mathcal{A})$ only are candidates for limits in the category of chain complexes (or in any full subcategory). But, if I'm not misremembering, direct limits and inverse limits are computed pointwise in the category of chain complexes... –  Nuno Dec 14 '10 at 15:33
... Also, remember that being exact is a much stronger condition, e.g., exact sequences are not preserved by additive functors. –  Nuno Dec 14 '10 at 15:34
I'm sorry but I don't see the difference between the exactness of $\bigoplus_{i\in I} A_i\hookrightarrow \bigoplus_{i\in I} B_i\twoheadrightarrow \bigoplus_{i\in I} C_i$ and the non-exactness of $\varinjlim_{i\in I} A_i\hookrightarrow \varinjlim_{i\in I} B_i\twoheadrightarrow \varinjlim_{i\in I} C_i$... I mean, I know the two are true, but I can't explain why, if I see the first as a paticular case of the second: what is missing in passing to the general concept of (co)limit? Cardinality of $I$? The structure in $I$ regarded as a category? Other? ... –  tetrapharmakon Dec 15 '10 at 17:23
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In a preadditive category (it's clear that chain complexes are preadditive), the idea of a direct sum can be described directly. Given two objects $A, B$, a direct sum $C$ has morphisms $i_1: A \to C, i_2: B \to C$ and projections $p_1: C \to A, p_2: C \to B$ that satisfy $i_1 p_1 + i_2 p_2 = 1_C$ and $p_1 i_1 = 1_A, p_2 i_2 = 1_B$. Indeed, it is easy to get such maps in a direct sum (inclusion and projection); conversely if we get such maps one can check that $C$ is a coproduct and a direct product by explicitly using these maps. Namely, one checks that such maps guarantee a direct sum in the case of abelian groups, and then applies hom-sets and Yoneda nonsense (if we have $\hom(X, C) = \hom(X, A) \oplus \hom(X, B)$ for all $X$ naturally, then we're done). I'm being a little vague here and only sketching it; if it's not clear, I can clarify.

This is clearly the case for the construction of chain complexes that you described above. The maps $p_1, p_2, i_1, i_2$ are the natural projections and inclusions piecewise. They are maps of complexes, as one can check, and they satisfy the required identities because they do for abelian groups.

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