**Edit:** I put my answer into a broader perspective.

In the following [We] refers to Weibel's "An Introduction to Homological Algebra".

Recall that a chain complex $C$ is *split exact* if it is acyclic and $Z_n$ (the cycles) is a direct summand of $C_n$ for each $n$ [We, Def. 1.4.1, Ex. 1.4.2]. Moreover, let $Ch$ be the category of (unbounded) chain complexes over an abelian category with enough projectives. Then, by [We, Ex. 2.2.1] and my answer in

$\qquad$http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible

we have:

For a chain complex $P$ the following are equivalent:

- $P$ is a projective object in $Ch$
- $P$ is a split exact complex of projectives
- $P$ is a contractible complex of projectives

Also from the link we obtain the following **examples** where $Ch_R$ denotes the category of unbounded chain complexes of modules over the ring $R$:

If $R$ is hereditary (e.g. PID's, Dedekind domains), then the projective objects of $Ch_R$ are exactly the acyclic complexes of projective $R$-modules.

If $R$ is any ring with unit and the acyclic complex $P$ of projective $R$-modules is bounded below, then $P$ is a projective object in $Ch_R$.

However, not all acyclic complexes of projective or free modules are projective objects in $Ch_R$. A counter-example (due to Dold) is given in [We, Example 1.4.2]:

- Over $R=\mathbb{Z}/4$ the following complex is exact
$$\cdots \to \mathbb{Z}/4 \xrightarrow{2} \mathbb{Z}/4 \xrightarrow{2} \mathbb{Z}/4 \to \cdots $$
But it's no projective object in $Ch_R$ since $Z_n = \mathbb{Z}/2$ can't be a direct summand of $C_n = \mathbb{Z}/4$.

**Added:** Let $Ch_b \subseteq Ch$ be the subcategory of chain complexes that are bounded below. In contrast to $Ch$ the following holds in $Ch_b$:

The projective objects in $Ch_b$ are exactly the acyclic chain complexes (bounded below) of projectives.

Proof: Let $P$ be an acyclic chain complex of projectives that is bounded below. We have already seen in example 2 above (compare also [We, Ex. 1.4.1 2.]) that $P$ is projective in $Ch$. Since $P \in Ch_b$ it's a projective object in $Ch_b$

Now let $P \in Ch_b$ be a projective object. The same proof as in $Ch$ shows that each $P_i$ is projective (consider objects of the abelian category as chain complexes concentrated in a single degree). Also the same proof as in $Ch$ can be used to see that $P$ is acyclic: The mapping cone of $id: P[1] \to P[1]$ yields a short exact sequence
$$0 \to P[1] \to \operatorname{cone}(id_{P[1]}) \to P \to 0.$$
But $P, P[1] \in Ch_b$ and by definition $\operatorname{cone}(id)_i = P_i\oplus P_{i+1}$ whence it is also bunded below. So the short exact sequence is in $Ch_b$ and splits since $P$ is a projective object. Hence $id_{P[1]}$ is nullhomotopic. So in particular, $P[1]$ and thus $P$ is acyclic. q.e.d.