Does every complex have a quasi-isomorphic projective complex? 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.
 A: The statement you've asked for is true provided some conditions on the abelian category: there are enough projectives, all colimits exist, and all filtered colimits are exact. 
Suppose first that a cochain complex $A^\bullet$ is bounded above. In this case, we can find a quasi-isomorphism $P^\bullet \to A^\bullet$ with $P^\bullet$ a complex of projectives as long as there are enough projectives in the abelian category. This is a classical result: you can do this by constructing a Cartan-Eilenberg resolution of $A^\bullet$ and taking its total complex, but you can also give an elementary proof by induction. Inductively, suppose that for all $i > n$, we have constructed projective objects $P^i$, differentials $d_P^i : P^i \to P^{i+1}$, and morphisms $f^i : P^i \to A^i$ which commute with the differentials and which induce isomorphisms $\ker(d_P^{i+1})/\mathrm{im}(d_P^i) \to H^{i+1}(A^\bullet)$. Since $A^\bullet$ is bounded above, the base case of the induction is satisfied since we can just take $P^i = 0$ for all big enough $i$. The fact that the map $f^{n+1} : P^{n+1} \to A^n$ commutes with the differentials tells us that there is an induced map $\ker(d_P^{n+1}) \to Z^{n+1}(A^\bullet)$. We know that $B^{n+1}(A^\bullet) \subset Z^{n+1}(A^\bullet)$. If we form the pullback $\ker(d_P^{n+1}) \times_{Z^{n+1}(A^\bullet)} B^{n+1}(A^\bullet)$, since there are enough projectives we can find an epimorphism 
$$ P^n \twoheadrightarrow \ker(d_P^{n+1}) \times_{Z^{n+1}(A^\bullet)} B^{n+1}(A^\bullet). $$
Now there is obviously a map $d_P^n : P^n \to P^{n+1}$ which is factors through $\ker(d_P^{n+1})$ so $d_P^{n+1} \circ d_P^n = 0$. There is also a map $P^n \to B^{n+1}(A^\bullet)$, and we have an epimorphism $A^n \twoheadrightarrow B^{n+1}(A^\bullet)$, so we can construct a map $f^n : P^n \to A^n$ using the fact that $P^n$ is projective. Now if you do a little diagram chasing, you can show that the inductive hypotheses continue to be satisfied, which completes the induction and produces us a quasi-isomorphism $f^\bullet : P^\bullet \to A^\bullet$ with $P^\bullet$ a complex of projectives. 
The difference between cochain complexes and chain complexes is notational, but sometimes people like their cochain complexes to be concentrated in non-negative degrees and their chain complexes to be concentrated in non-negative degrees. The above proof proves that we can find quasi-isomorphisms $P^\bullet \to A^\bullet$ when $A^\bullet$ is, for instance, concentrated in non-positive degrees, which, after a change of notation, proves that we can find quasi-isomorphisms $P_\bullet \to A_\bullet$ of chain complexes when $A_\bullet$ is concentrated in non-negative degrees. 
In any case, let's go back to thinking just about cochain complexes. Let $A^\bullet$ be an arbitrary (ie, possibly unbounded) complex. Fix some integer $n$. We know that the (good) truncation $\tau_{\leq n} A^\bullet$ is bounded above, so we can find a quasi-isomorphism $f_n^\bullet : P^\bullet_n \to \tau_{\leq n} A^\bullet$ where $P^\bullet_n$ is a complex of projectives. If we do a little mucking around, we can show that we can extend this to a quasi-isomorphism $f_{n+1}^\bullet : P^\bullet_{n+1} \to \tau_{\leq n+1}A^\bullet$ with $P_{n+1}^\bullet$ a complex of projectives. More precisely, by "extend" I mean that there is a map $P^\bullet_n \to P^\bullet_{n+1}$ which is the identity in all degrees strictly less than $n$, and such that the following diagram of complexes commutes.
$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
P_n^\bullet & \ra{} & P_{n+1}^\bullet \\
\da{f_n^\bullet} & & \da{f_{n+1}^\bullet} \\
\tau_{\leq n} A^\bullet & \ra{} & \tau_{\leq n+1} A^\bullet \\
\end{array} $$
Now since we're assuming that colimits exist, we can take the colimit of these maps
$$ P^\bullet := \mathrm{colim} P_n^\bullet \to \mathrm{colim} \tau_{\leq n} A^\bullet = A^\bullet. $$
It's clear that $P^\bullet$ will be a complex of projectives. Moreover, this colimit we're taking is a filtered colimit, and if we assume that filtered colimits in our abelian category are exact, then this map is guaranteed to be an quasi-isomorphism. 
This gives a proof of the statement you asked about, but I should probably point out that asking for a quasi-isomorphism $P^\bullet \to A^\bullet$ with $P^\bullet$ a complex of projectives is not a very useful thing to do unless $A^\bullet$ is bounded above. More specifically, I mean the following. A complex $P^\bullet$ of projective objects is called dg-projective if it satisfies the following property: whenever $C^\bullet$ is acyclic, any map $P^\bullet \to C^\bullet$ is null-homotopic. For bounded above complexes, being dg-projective is equivalent to being projective in all degrees, but this is not true for unbounded complexes. When $A^\bullet$ is bounded above, the proof we gave above shows that we can always find a quasi-isomorphism $P^\bullet \to A^\bullet$ with $P^\bullet$ a dg-projective complex, but for unbounded $A^\bullet$, proving that there is always a quasi-isomorphism $P^\bullet \to A^\bullet$ with $P^\bullet$ a dg-projective complex is more involved. 
A: $\require{AMScd}$
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.
