Projective object in the category of chain complexes I have the following sequence of projective $\mathbb{Z}$-modules:
$\cdots \rightarrow  0 \rightarrow \mathbb{Z} \overset{\times 2}\rightarrow  \mathbb{Z} \rightarrow 0 \rightarrow  \cdots $
This is clearly a chain comlex. I wanna show that it's not projective in the category of chain complexes... I started with the formal definition of projective objects and it got me nowhere!
Help please?

Let's say these two $\mathbb{Z}$ live in degrees $0,1$. Then a map from this chain complex $P$ to another chain complex $C$ is the same as two elements $a \in C_0$, $b \in C_1$ such that $d(a)=2 b$. So we have to find suitable $C,D$ and an epimorphism $D \to C$ and two elements $a,b$ as above, which does not lift to elements $a',b'$ with $d(a')=2 b'$. Of course they lift to elements $a' \in D_0, b' \in D_1$, but the problem is that we cannot really ensure that $d(a')=2 b'$ is satisfied. In fact, this has a homological obstruction. Can you now find an example on your own?
