Timeline for Projective objects in the category of chain complexes

10 events

when toggle format	what		by	license	comment
Feb 21 at 16:40	comment	added	tj_		@locally trivial: $P(n)$ vanishes up to degrees $n,n-1$, thus $\big(\oplus_n P(n)\big)_m = \oplus_n P(n)_m=P(m)_m\oplus P(m+1)_m=P_m'' \oplus P_m'=P_m$. Similarly, you can show that the differentials in $\oplus_n P(n)$ and $P$ coincide.
Feb 21 at 16:21	comment	added	tj_		@Jolia: This is the special case that is shown in the question by the OP.
Feb 21 at 11:28	comment	converted from answer	Jolia		Why is $P(n)$ projective for $n\in\mathbb{Z}$?
Nov 9, 2023 at 6:44	comment	added	locally trivial		Is it immediate why P is a direct sum, and not a direct product here?
S Oct 1, 2015 at 13:42	history	suggested	Riccardo	CC BY-SA 3.0	there was a typo
Oct 1, 2015 at 13:33	review	Suggested edits
S Oct 1, 2015 at 13:42
Dec 6, 2012 at 0:30	vote	accept	user748
Dec 5, 2012 at 1:23	comment	added	Ralph		Hmm, I think their construction requires the complex to be bounded below. At least they obtain a decomposition $P=\oplus_{k \ge 0}D_k$ and $(D_k)_n=0$ for $n\neq k,k-1$ (also note that they only require $P$ to be acyclic and not contractible as Weibel does).
Dec 5, 2012 at 1:01	comment	added	David White		Indeed, this decomposition is mentioned in the link user49437 provided in the comments to the OP. A reference is Dwyer-Spalinski, Homotopy Theories and Model Categories, and in 7.10 of that paper they construct the decomposition, using the language of boundaries and cycles.
Dec 5, 2012 at 0:36	history	answered	Ralph	CC BY-SA 3.0