Truncation of an injective resolution This piece comes from the proof of corollary 10.5.11 in Weibel's book on homological algebra. 
There we start with a cochain complex $X^{\bullet}$ and we are trying to construct a quasi-isomorphism $X^{\bullet} \to X'^{\bullet}$ with $X'^{\bullet}$ being complex of $F$-acyclic objects, where $F$ has cohomological dimension $n$.
We choose Cartan-Eilenberg resolution $X^{\bullet} \to I_{CE}^{\bullet, \bullet}$ and then Weibel suggests we take a (good) truncation $\tau_{\leq n}(I_{CE}^{p \bullet})$ to get a finite resolution by $F$-acyclic objects. 
So I cannon see why the last object we are truncating at is actually $F$ acyclic.  I initially wanted to splice the exact sequence $X^p \to I_{CE}^{p \bullet}$ and apply that with  $A'$ and $A$ $F$-acyclic in $$0 \to A' \to A \to A'' \to 0,$$ $A''$ is also $F$-acyclic. 
But than this doesn't work because $I^{p,0} \to I^{p,1}$ isn't injective. Can't quite see how cohomological dimension $n$ comes into play.  Any thoughts would be appreciated.

I do not have the book at my hand, and am happy to accept any possible corrections.
Let $X$ be a general object of the category and $X\to I^{\bullet}$ be an injective resolution of $X$. If $J$ is the image of the map $I^{n-1}\to I^n$, then $J\to I^n\to I^{n+1}\to\dots$ is an injective resolution of $J$. Also, because $F$ has cohomological dimension $n$, $F_*I^n\to F_*I^{n+1}\to\dots$ is exact. These two facts show that $J$ is $F$-acyclic. Therefore, $\tau_{\leq n}(I^{\bullet})$, ending with $I^{n-1}\to J\to 0\to \dots$, is an $F$-acyclic resolution of $X$.
