Let $\scr A$ be an abelian category with exact products and a cogenerator (e.g. $\scr A$ is a category of modules). Let ${\mathbf K}(\scr A)$ be the homotopy category of cochain complexes over $\scr A$. We call homotopcally injective a complex $X\in{\mathbf K}(\scr A)$ with the property that every map $N\to X$, starting from an acyclic complex $N$ is null-homotopic, that is the abelian group ${\mathbf K}({\scr A})(N,X)$ vanishes, for any acyclic $N$. Inductively it is easy to show that every left bounded complex of injectives (that is, with injective entries) is homotopically injective. It is also known that if $\scr A$ has finite injective dimension, then every complex of injectives is homotopically injective. I found this statement as an exercise for the reader in many places, but I didn't succeed to prove it. Clearly the statement is equivalent to another, namely: Every acyclic complex of injectives is contractible. Could someone help me?
2 Answers
Let $J^\bullet$ be an acyclic complex of injective objects in an abelian category $\mathcal A$. Consider its finite subquotient complexes of canonical truncation $0\to Z^m\to J^m\to J^{m+1}\to \dotsb\to J^{n-1}\to Z^n\to 0$, where $Z^i$ denotes the kernel of the differential $J^i\to J^{i+1}$. This finite complex is a right resolution of the object $Z^m$ whose terms are all injective objects in $\mathcal A$, with a possible exception of the rightmost term $Z^n$. Now if the category $\mathcal A$ has finite homological dimension and the number $n-m$ is chosen to be large enough, it follows that the object $Z^n$ is also injective. Finally, an acyclic complex of injectives $J^\bullet$ with injective objects of cocycles $Z^i$ is easily seen to be contractible.
Let $N$ be an acyclic complex of injectives. I will denote complexes with cohomological degre, i.e. degree $+1$ differentials. Notice that the complex $N$ is contractible iff it the kernel of each differential $d^k\colon N^k \rightarrow N^{k+1}$ is injective. Suppose that $\mathscr A$ has finite injective dimension $n$. For any $l<k$ we have an exact sequence
$$\ker d^l\hookrightarrow N^l\rightarrow N^{l+1}\rightarrow\cdots\rightarrow N^{k-1}\twoheadrightarrow \ker d^k$$
This is in particular true for $k-l>n$. By hypothesis, $\ker d^{l+n}$ is injective, hence the exact sequence $$\ker d^{l+n}\hookrightarrow N^{l+n}\rightarrow N^{l+n+1}\rightarrow\cdots\rightarrow N^{k-1}\twoheadrightarrow \ker d^k$$
shows that $\ker d^k$ must also be injective.
