Projective objects in chain complexes of an abelian category: Further question

Question

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.

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.