Yes, I see there are other Q&A's on this, for instance here: Projective objects in the category of chain complexes

I am wondering why a level-wise projective chain complex $P$ which is split exact can be written as a direct sum of the $P(n)$ from Ralph's first answer in the link above, rather than as a direct product. If $P$ is unbounded in both directions, for instance, then couldn't $P$ contain an element $(\dots, p_{-1},p_0, p_1, \dots )$ that is a priori not contained in $\bigoplus_{n\in \mathbb{Z}} P(n)$?

In the reference [DS95], Dwyer & Spalinski impose the condition that the chain complex be non-negatively graded [i.e., study projective objects in $Ch_{\geq 0}(\mathcal{A})$. I do not want to impose this condition: I am interested in projective objects in $Ch(\mathcal{A})$, that are unbounded. For that matter, even in the bounded below case $Ch_{\geq 0}(\mathcal{A})$, I don't understand why $P$ cannot have an element $p= (\dots, 0, \dots, 0, p_0, p_1, p_2, \dots )$ which could be zero in negative degrees and non-zero in all non-negative degrees, which is not an element of $\bigoplus_{n\in\mathbb{Z}}P(n)$.

[DS95] = Dwyer & Spalinski, Homotopy theories and model categories.


1 Answer 1


In this case, there's no difference between the direct sum and the direct product, because in every degree $n$, we are only taking the direct sum of two things. With reference to the answer you linked, it's important to note that, in the chain complex $P(n)$, the object $P_n''$ is in degree $n$ and $P_{n-1}'$ is in degree $n-1$. So, when we take the direct sum $\bigoplus_{n\in \mathbb{Z}} P(n)$, it's taken levelwise (see page 5 of Weibel) and at each level, it's just the direct sum of two objects of $\mathcal{A}$.

As a clarifying example, note that $P(3) \oplus P(2)$ is the chain complex $$ \cdots \to 0 \to P_{3}^{''} \to P_{2}^{'} \oplus P_2^{''} \to P_1^{'} \to \cdots$$

and $P(3) \oplus P(2) \oplus P(1)$ is the chain complex $$ \cdots \to 0 \to P_{3}^{''} \to P_{2}^{'} \oplus P_2^{''} \to P_1^{'} \oplus P_1^{''} \to P_0^{'} \to \cdots$$

This is why $\bigoplus _{n\in \mathbb{Z}} P(n) \cong P$, since $P_n' \oplus P_n'' \cong P_n$ by construction.

Also, the argument in the answer you linked does not require us to work with bounded chain complexes. In Weibel, "chain complex" means unbounded by default (see Definition 1.1.1). I think the comment I left in 2012 on the linked answer might have misled you, and if so, I'm sorry for that. Weibel's version is more general than the approach in Dwyer and Spalinski.

The difference between direct sum and direct product only appears when we combine infinitely many nonzero objects of $\mathcal{A}$ in some specific degree $n$. Whether or not the resulting chain complex has non-zero $P_k$ objects for $k < 0$ is entirely unrelated.

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    $\begingroup$ The answer to your first question is "yes." There's no finiteness restriction on a general chain complex $C_\bullet$. It's ok if it has all $C_i$ nonzero. $\endgroup$ Nov 10, 2023 at 0:40
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    $\begingroup$ Ah. So the direct sum $C_\bullet = \bigoplus_{n\in \mathbb{Z}} P(n)$ still contains the element $c = (..., c_{-1}, c_0, c_1, ... )$ which is non-zero in infinitely many degrees. Very good! Now we're clear! All good, thanks! $\endgroup$ Nov 10, 2023 at 1:25
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    $\begingroup$ I disagree with your previous two comments. This has nothing to do with Freyd-Mitchell. You wrote "If a chain complex 𝐶∙ is a $\mathbb{Z}$-graded 𝑅−module, then a typical element is non-zero only in finitely many degrees whether 𝐶∙ is unbounded or not" and I disagree. Consider the chain complex that is $C_n = \mathbb{Z}$ for all $n$, and $d_n = 0$ for all $n$. This is perfectly allowed by Weibel's Definition 1.1.1 $\endgroup$ Nov 11, 2023 at 13:46
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    $\begingroup$ It seems you are very set on the idea that every chain complex is equivalent to a graded $R$-module, and that $\bigoplus$ in the context of graded $R$-modules places a finiteness restriction. This is just not right. Show me where in Weibel the result you think is true is proven. If you can't, then go read Weibel's definition 1.1.1 and tell me why it rules out the example I gave you of a chain complex that has every $C_n$ nonzero. $\endgroup$ Nov 12, 2023 at 1:02
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    $\begingroup$ Lastly, you wrote to view $C_*$ as a graded $\mathbb{Z}$-module, and I don't know what you mean by this. Usually when we say M is a graded R module then R is a graded ring: en.wikipedia.org/wiki/Graded_ring $\endgroup$ Nov 12, 2023 at 1:02

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