Injective resolutions of a complex Let $\mathcal{A}$ be an abelian category, $M\in\mathcal{A}$. An injective resolution of $M$ is a quasi-isomorphism $M\longrightarrow I$, where $I$ is a complex of injective objects. This can be made more explicit: It is the same thing as an exact sequence $0\longrightarrow M\longrightarrow I_0\longrightarrow I_1\longrightarrow\dots$, where the $I_i$ are injective.

Is it possible to make the notion of an injective resolution of a bounded-below complex $C$, i.e. a quasi-isomorphism $C\longrightarrow I$ "more explicit" in the same spirit?

If $\mathcal{A}$ has enough Injectives, it can be shown that to any object $M$ there exists an injective resolution. The proof is a ping-pong of taking Cokernels and embedding into injectives, since taking colimits extends the sequence in an exact way and an embedding does not change the kernel. The same should be true for any bounded-below complex.

How can I see that if $\mathcal{A}$ has enough injectives, then every bounded-below complex has an injective resolution.

Thank you very much in advance.
 A: I found this.
Let $X$ be an Abelian category with enough injectives. Let $\mathcal A^{\bullet} \in C(X)$, the category of complexes in $X$.
Definition 1.9 A resolution of $\mathcal A^{\bullet}$
is defined as a quasi-isomorphism
$\mathcal A^{\bullet} \rightarrow \mathcal F^{\bullet}$.
Assume $\mathcal A^{\bullet}$ is bounded below. Now construct an injective resolution $\mathcal A^{\bullet} \rightarrow \mathcal I^{\bullet}$ of $\mathcal A^{\bullet}$.


*

*(Proposition 2.3) Find a double complex $\mathcal I^{\bullet,\bullet}$, such that for all $q$, $\mathcal I^{\bullet,q}$ is an injective resolution of $\mathcal A^q$.

*(Definition 2.1) Then let $\mathcal I^{\bullet}$ be the (simple) complex associated to the double complex $\mathcal I^{\bullet,\bullet}$.  ($\mathcal I^{\bullet}$ is defined as: $\mathcal I^i = \otimes_{p+q=i}\mathcal I^{p,q}$, moreover, for $d':\mathcal I^{p,q} \rightarrow \mathcal I^{p+1,q}$, $d'':\mathcal I^{p,q} \rightarrow \mathcal I^{p,q+1}$, set $d=d'+d''$.)

*The morphism $\mathcal A^{\bullet} \rightarrow \mathcal I^{\bullet}$ can be defined in a natural way.
Please check the hyperlink and read the pdf file. Although I really hope I've made no mistakes in understanding and typing.
A: I was looking for a slick proof of the same fact, avoiding taking the Cartan-Eilenberg resolutions, so I came across this question. 
I think the proof with Cartan-Eilenberg resolutions is an overkill; it is useful for different reasons, namely for the spectral sequences associated to the corresponding total complex.
Let me write down a direct proof by induction. Even though the question is old, it might be useful to someone.

Let $A^\bullet$ be a (cohomological) complex such that $A^n = 0$ for $n \ll 0$. Then we claim that there exists a complex $I^\bullet$ consisting of injective objects, together with a quasi-isomorphism of complexes
  $$f^\bullet\colon A^\bullet \to I^\bullet.$$


First of all, note that

$f^\bullet$ is a quasi-isomorphism if and only if its cone $C (f)$ is acyclic (i.e. $H^n (C (f)) = 0$ for all $n$).

Namely, one defines $C (f)$ to be the complex consisting of objects
$$C (f)^n = I^n \oplus A^{n+1}$$
and differentials
$$d^n_{C (f)} = \begin{pmatrix}
d^n_I & f^{n+1} \\
0 & -d_A^{n+1}
\end{pmatrix}\colon I^n \oplus A^{n+1}\to I^{n+1} \oplus A^{n+2}.$$
This complex sits in the short exact sequence
$$0 \to I^\bullet \to C (f) \to A^\bullet [1] \to 0$$
and in the corresponding long exact sequence in cohomology
$$\cdots \to H^{n-1} (C (f)) \to H^n (A^\bullet) \to H^n (I^\bullet) \to H^n (C (f)) \to \cdots$$
the connecting morphism $H^n (A^\bullet) \to H^n (I^\bullet)$ is actually $H^n (f)$. This shows that acyclicity of $C (f)$ indeed corresponds to all $H^n (f)$ being isomorphisms.

Now we may construct the complex $I^\bullet$ by induction, and at each step our goal is to keep $C (f)$ acyclic.
For the induction base, if $A^k = 0$ for all $k \le N$, we may put $I^k = 0$ for all $k \le N$.
For the induction step, let's assume we have defined in degrees $k \le n$ injective objects $I^k$, differetials $d_I^{k-1}\colon I^{k-1}\to I^k$, and morphisms $f^k\colon A^k \to I^k$ such that $d_I^{k-1}\circ d_I^{k-2} = 0$, the differentials $d_A$ and $d_I$ commute with $f$, and the cone $C (f)$ is acyclic in degrees $k \le n-1$.
$$\require{AMScd}\begin{CD}
\cdots @>>> A^{n-1} @>d^{n-1}_A>> A^n @>d^n_A>> A^{n+1} @>>> \cdots \\
@. @VVf^{n-1}V @VVf^nV \\
\cdots @>>> I^{n-1} @>d^{n-1}_I>> I^n
\end{CD}$$
We are going to define $I^{n+1}$ and morphisms $f^{n+1}\colon A^{n+1}\to I^{n+1}$ and $d^n_I\colon I^n \to I^{n+1}$, such that


*

*$d^n_I\circ d^{n-1}_I = 0$,

*$d^n_I\circ f^n = f^{n+1}\circ d^n_A$,

*$C (f)$ is acyclic in degree $n$.


The last condition suggests that we should do the following thing: take the cokernel of the differential $d^{n-1}_{C(f)}$:
$$I^{n-1}\oplus A^n \xrightarrow{d^{n-1}_{C(f)}} I^n\oplus A^{n+1} \twoheadrightarrow C^{n+1}$$
Sadly, $C^{n+1}$ doesn't have to be injective, but we can always pick a monomorphism $C^{n+1} \rightarrowtail I^{n+1}$ to some injective object (using the hypothesis on enough injectives).
Now the morphism
$$I^n\oplus A^{n+1} \twoheadrightarrow C^{n+1} \rightarrowtail I^{n+1}$$
is uniquely determined by a pair of arrows $I^n \to I^{n+1}$ and $A^{n+1} \to I^{n+1}$, and it's tempting to call them $d^n_I$ and $f^{n+1}$.
The composition
$$I^{n-1}\oplus A^n \xrightarrow{d^{n-1}_{C (f)}} I^n\oplus A^{n+1} \xrightarrow{(d^n_I, f^{n+1})} I^{n+1}$$
is zero by our construction, so
$$(d^n_I, f^{n+1}) \circ \begin{pmatrix}
d^{n-1}_I & f^n \\
0 & -d_A^n
\end{pmatrix} = (d^n_I\circ d^{n-1}_I, d_I^n\circ f^n - f^{n+1}\circ d^n_A) = (0,0).$$
We obtained the desired identities $d^n_I\circ d^{n-1}_I = 0$ and $d_I^n\circ f^n = f^{n+1}\circ d^n_A$. The complex $C (f)$ is acyclic in degree $n$ by our construction.
We are done.

Note that this argument generalizes the proof that one normally uses when the complex $A^\bullet$ consists of a single object: we repeatedly take certain cokernels and then embed them in injective objects.

The very same argument dualizes to show that

if $A_\bullet$ is a (homological) complex such that $A_n = 0$ for $n \ll 0$, then there exists a complex of projective objects $P_\bullet$ together with a quasi-isomorphism $P_\bullet \to A_\bullet$ (assuming the category has enough projectives).

Namely, for the induction step you have to take the kernel of the differential of $C (f)$, and then pick an epimorphism from some projective object:
$$P_n \twoheadrightarrow K_n \rightarrowtail A_n \oplus P_{n-1} \xrightarrow{d_n^{C (f)}} A_{n-1}\oplus P_{n-2}$$
(The numbering is homological now.)
