TITLE: Does every complex have a quasi-isomorphic projective complex?
QUESTION [6 upvotes]: Let $C^{\textbf{.}}$ be a complex in some abelian category (edit: assuming it has enough projectives). I would like to know if there exist a complex $X^{\textbf{.}}$ consisting of projective objects and a quasi-isomorphism $f:C^{\textbf{.}}\to X^{\textbf{.}}$.
This question comes from "Cohomology of Number Fields", page 110 in the second edition. In the book it is assumed that the category is of modules over a dedekind domain, so mabye it is only true with this hypothesis. Another hypothesis about the special case in which the authors use this statement, is where $C^{\textbf{.}}$ is bounded above.

REPLY [6 votes]: $\require{AMScd}$
As noted above, the first argument provides an injection in homology, but not an isomorphism. I will set things up inductively so that we get a diagram 
$$\begin{CD}
  \\
&{}&{}& C_n  
@>>>
C_{n-1}   \\
{}&{}& {} @Af_n AA @A A f_{n-1} A 
  \\
{}&{}&{}&P_{n}
  @>>d_n>
P_{n-1} \\
\end{CD}$$
so that $f_{n-1}$ induces an isomorphism in homology and $f_n$ induces an epimorphism from $\ker d_n$ to $H_n(C_\cdot)$. When $n=0$ this is almost trivial: take $f_{n-1}=0$, $d_{-1}=0$ and $f_0$ induced from covering the zeroeth homology $\rho:P_0\to H_0$ by a projective and obtaining $f_0$ induced from the projection $\pi_0:C_0\to H_0$. This gives a commutative diagram 
$$\begin{CD}
  \\
&{}&{}& C_0 
@>\partial_{0}>>
0   \\
{}&{}& {} @Af_0 AA @AAA 
  \\
{}&{}&{}&P_0
  @>>d_{0}>
0\\
\end{CD}$$
Assume inductively that we have obtained a commutative diagram 
$$\begin{CD}
  \\
&{}&{}& C_n  
@>>>
C_{n-1}   \\
{}&{}& {} @Af_n AA @AA  f_{n-1} A 
  \\
{}&{}&{}&P_{n}
  @>>d_n>
P_{n-1} \\
\end{CD}$$
such that $f_{n-1}$ induces an isomorphism in homology and $f_n$ induces an epimorphism $\ker d_n\to H_n(C_\cdot )$. Take a cover $\rho_{n+1}:P_{n+1}' \to H_{n+1}$ with $P_{n+1}'$ projective and $f_{n+1}'$ induced from the projection $\ker{\partial_{n+1}}\to H_{n+1}$. Now consider the pullback diagram (at $Q_{n+1}$)
$$\begin{CD}
  \\
&{}&{}& {\rm im}\; \partial_{n+1}  
@>\iota>>
\ker \partial_{n+1}  @>\widetilde{j}>> 
C_n  \\
{}&{}& {} @A\psi_n AA @A\widetilde{f_n} AA @Af_nAA
  \\
P_{n+1}'' @>\Lambda_{n+1}>>
Q_{n+1}
  @>\varphi_n>>
\ker d_n  @>j>>
P_n  \\
\end{CD}$$
where we have taken $P_{n+1}''$ projective and $\Lambda_{n+1}$ an epimorphism. Define $d_{n+1}''= j\varphi_n\Lambda_{n+1}$. This also gives an arrow $\psi_n\Lambda_{n+1}:P_{n+1}''\longrightarrow {\rm im}\; \partial_{n+1}$ and we obtain $f_{n+1}'':P_{n+1}''\longrightarrow C_{n+1}$ from the epimorphism $ \tilde\partial_{n+1}:C_{n+1} \longrightarrow {\rm im}\; \partial_{n+1}$. Now let $d_{n+1}:P_{n+1}\longrightarrow  C_{n+1}$ where $P_{n+1}=P_{n+1}'\oplus P_{n+1}''$ and $(p',p'')\mapsto d_{n+1}''(p)$. Set $f_{n+1}=f_{n+1}'+f_{n+1}''$. Then
$(1)$ Since the image of $f_{n+1}'$ lies in the kernel of $\partial_{n+1}$, we have 
$$\begin{align} \partial_{n+1}f_{n+1}&= \partial_{n+1}f_{n+1}''\\&= \tilde j\iota \tilde \partial_{n+1} f_{n+1}''\\&=\tilde j\iota \psi_n\Lambda_{n+1}\\&=\tilde j\tilde{f_n}\varphi_n \Lambda_{n+1}\\&=f_n j\varphi_n \Lambda_{n+1}\\&=f_n d_{n+1}''=f_n d_{n+1}\end{align}$$
$(2)$ Evidently $d_nd_{n+1}=0$ since $d_{n+1}''$ factors through the kernel of $d_n$. Assume that $z\in\ker d_n''$ is such that $f_n(z)=\partial_{n+1}z ''$. Then $(\partial_{n+1}z'',z)$ lies in $Q_{n+1}$ so there is some $p\in P_{n+1}''$ so that $\Lambda_{n+1}(p)$ equals this. It then follows $d_{n+1}''(p)=z$ and since the images of $d$ and $d''$ coincide we obtain the desired isomorphism.
$(3)$ We have that $\ker d_{n+1}=P_{n+1}'\oplus \ker d_{n+1}''$ maps by $f_{n+1}$ onto $H_{n+1}(C_\cdot)$, since the first summand is already onto.