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. 



Projective objects in the category of chain complexes 
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 $\pi \circ g=f$. 


