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Excercise 2.2.1 in Weibel ("An Introduction to Homological Algebra") states that an object $P$ in the category of chain complexes over an abelian category is projective if and only it is a split exact complex of projectives.

I was able to solve the only-if-part but I have touble with the if-part and would be glad if someone can give me some help. This is no homework!

What have I a tried so far ? Given an epimorphism $\pi: X \to Y$ and a morphism $f: P \to Y$, it has to be shown that there is a morphism $g: P \to X$ s.t. $\pi \circ g=f$.

Weibel hints to consider the special case $0 \to P_1 \cong P_0 \to 0$. It's easy to construct $g$ in this case: $\pi$ epi means that each $\pi_i:X_i \to Y_i$ is epi. By projectivity of $P_1$ there is a hom. $g_1: P_1 \to X_1$ s.t. $\pi_1 \circ g_1 = f_1$. If $d^P$ resp. $d^X$ denotes the differential in $P$ resp. $X$, set $$g_0 := d^X_1 \circ g_1 \circ (d^P_1)^{-1}: P_0 \to X_0,\qquad g_i = 0: P_i \to X_i\; (i\neq 0,1)$$ Then $g=(g_i): P \to X$ is a morphisms s.t. $\pi \circ g=f$.

But I have no idea how to generalize this procedure to the general case where $d^P$ can not be expected to be an isomorphism.

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There is a paper on projective objects in the category of chain complexes:…. Maybe it's of help. – tj_ Dec 4 '12 at 23:02
See also: On the difference between a projective chain complex and a level-wise projective chain complex: – Martin Dec 4 '12 at 23:17
I seem to recall that the trick is to take an arbitrary split exact complex of projectives $Q$ and turn it into one like $0\to P_1\to P_0\to 0$ by setting $P_1 = P_0 = \bigoplus_i{Q_i}$, and choosing $d^P$ in an appropriate way – David White Dec 5 '12 at 0:21
You'll also have to use the fact that the identity map on your split exact complex of projectives is nullhomotopic. – David White Dec 5 '12 at 0:33
up vote 9 down vote accepted

The trick with Weibel's hint is to decompose $P$ as direct sum of complexes of type $$\cdots \to 0 \to P_1 \xrightarrow{\cong} P_0 \to 0 \to \cdots$$ Since $P$ is split exact, we can write $P_n=P_n^{'}\oplus P_n^{''}$ where $P_n^{'}=\text{ker}(d_n)$ and $d_n^{''} =d_n|P_n^{''}:P_n^{''} \to \text{im}(d_n)=P_{n-1}^{'}$ is an isomorphism. Note that since $P_n$ is projective, the direct summands $P_n^{'},P_n^{''}$ are projective as well. If we define a complex
$$P(n):\quad \cdots \to 0 \to P_{n}^{''} \xrightarrow{d_n^{''}} P_{n-1}^{'} \to 0 \to \cdots$$ then $P = \bigoplus_{n \in \mathbb{Z}}P(n)$. Now let's consider the extension problem $$\begin{array}{ccl} & & P \newline & & \;\downarrow f \newline X & \overset{\pi}{\twoheadrightarrow} & Y \end{array}$$ $f$ induces by restriction a morphism $f(n): P(n) \to Y$ with $f=\sum_n f(n)$ (the sum is finite in each degree). As already observed by the OP, there is a morphism $g(n): P(n) \to X$ with $\pi \circ g(n) = f(n)$. Hence $g := \sum_n g(n)\colon P \to X$ satisfies $\pi \circ g=f$.

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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. – David White Dec 5 '12 at 1:01
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). – Ralph Dec 5 '12 at 1:23

As another solution I want to offer a closed formula for the sought-after morphism $g=(g_i):P \to X$:

Since $P$ is split exact, it's contractible, i.e. there are maps $s_i : P_i \to P_{i+1}$ with $s_{i-1}d_i^P + s_i d_{i+1}^P=id_{P_i}$. Moreover, since each $P_i$ is projective we can choose $h_i: P_i \to X_i$ such that $\pi_i \circ h_i = f_i$. Now $$g_i := d_{i+1}^X h_{i+1}s_i + h_i s_{i-1}d_i^P: P_i \to X_i$$ does the trick.

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The question as asked has been answered, but to understand where bounded below enters the picture, it is helpful to think model categorically (as in Dwyer and Spalinsky or, more recently, chapter 18 of More Concise Algebraic Topology, by Kate Ponto and myself). With the usual model structure (there are others in the latter reference), a chain complex is acyclic and cofibrant if and only if it is a projective object. If it is cofibrant (not necessarily acyclic) then it is degreewise projective. If it is degreewise projective and bounded below, then it is cofibrant. However, it can be acyclic and degreewise projective and yet not cofibrant if it is not bounded below. There is a nice example in the paper [K] that TJ refers to: work over the ring $Z/4$ and take all $P_n$ to be free on one generator, with all differentials given by multiplication by $2$: $P$ is acyclic and degreewise free, but it is not cofibrant and not a projective object. Split exactness rules out such examples and is automatic when $P$ is exact, degreewise projective, and bounded below.

Incidentally, the role of $R$-split exactness becomes really interesting model theoretically when $R$ is commutative and not a field and one considers model structures on DG modules over a DG $R$-algebra. There are (at least) six different interesting projective type model structures, and the usual one is arguably not the most useful one (this is a shameless advertisement for a paper in the writing stage by Tobi Barthel, Emily Riehl, and myself).

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In the following [K] refers to the paper

That a split exact complex of projectives $(P,d)$ is a projective object can be seen as follows:

  1. $im(d_n)$ is projective since it is a direct summand of $P_{n-1}$

  2. By When is an acylic chain complex contractible a split exact complex is contractible, so $P$ is contractible.

  3. By [K], Lemma 4.4 a contractible complex (like $P$) is isomorphic to the mapping cone of the boundary subcomplex $$ \cdots \to im(d_{n+1}) \xrightarrow{0} im(d_n) \to \cdots$$

  4. By [K], Theorem 3.1, the mapping cone of a complex of projectives with zero differentials is a projective object. Hence $P$ is a projective object by 1. and 3.

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You can check out in Rotman's Book AIHA for a clear explanation, on the part of Cartan-Eilenberg resolutions.

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Could you give a summary here? – Dag Oskar Madsen Sep 29 '15 at 19:58

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