Do unbounded chain complexes have enough injectives? I am not sure where one looks up this type of fact. Google was not very helpful.
 A: Yes, any Grothendieck abelian category has enough injectives. I believe this goes back to Grothendieck's Tohoku paper. The category in question is Grothendieck abelian since it is equivalent to a category of additive presheaves. The domain category has objects the integers, and morphisms generated by $d_n: n \rightarrow n+1$ (or the other way around depending on your grading conventions), subject to the relations $d_{n+1} \circ d_n=0$.
A: Appealing to Grothendieck's theorem is pretty much over the top. It's completely elementary. Suppose $\cal A$ is an abelian (or exact) category with enough injectives.

A complex is injective if and only if it is (a retract of a) split exact complex with injective components.

First note that the functor $A^\bullet \mapsto A^n$ that sends a complex to its $n$-th component is exact and has a right adjoint $D^n$. The right adjoint is given by the complex $D^n(A) = ( \cdots \to 0 \to A \stackrel{1}{\to} A \to 0 \to \cdots)$ where $A$ sits in degrees $n$ and $n+1$, so $D^n(I)$ is injective whenever $I$ is injective.
For a complex $A^\bullet$ let $IA^\bullet$ be the complex with components $(IA)^{n} =A^n \oplus A^{n+1}$ and differential $\begin{pmatrix} 0 & 1_{A^{n+1}} \\ 0 & 0\end{pmatrix}$, so $IA = \prod_{n \in \mathbb{Z}} D^n(A^n)$. Note that $IA$ is a split exact complex and the chain map $(1_{A^n} \, d^{n})^t \colon A^n \to A^{n} \oplus A^{n+1}$ is degreewise split monic. Now choose for each $A^n$ a monic $i^n \colon A^n \rightarrowtail I^n$ into an injective and put $J^n = I^n \oplus I^{n+1}$ to obtain the complex $J^\bullet$ with differential $\begin{pmatrix} 0 & 1_{I^{n+1}} \\ 0 & 0\end{pmatrix}$, so $J^\bullet = \prod_{n \in \mathbb{Z}} D^{n}(I^n)$. The complex $J^\bullet$ is injective as a product of injective complexes and the $i^n$ yield a chain map $i \colon IA \rightarrowtail J$ in an obvious way. The composite $A^\bullet \rightarrowtail IA^\bullet \rightarrowtail J^\bullet$ embeds the complex $A^\bullet$ into an injective complex.
To prove the emphasized statement, note that if we start with an injective complex $A^\bullet$, it is a retract of $J^\bullet$. Conversely, every retract of a split exact complex of injectives is a retract of a complex of the form $J^\bullet$, hence injective.
