Projective objects in the category of chain complexes  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.  
 A: 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. 
A: 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).
A: You can check out in Rotman's Book AIHA for a clear explanation, on the part of Cartan-Eilenberg resolutions.
A: 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 $\pi \circ g=f$.   
A: In the following [K] refers to the paper http://dml.cz/bitstream/handle/10338.dmlcz/120545/ActaOstrav_07-1999-1_3.pdf.  
That a split exact complex of projectives $(P,d)$ is a projective object can be seen as follows: 


*

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

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

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

*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.  
