Geometric vs cohomological dimension with families - on a proof of Lueck and Meintrup Let $G$ be a discrete group, and let $\mathcal{F}$ be a family of subgroups of $G$ (closed under conjugation and taking subgroups). Then we may define the geometric and cohomological dimensions of $G$ with respect to $\mathcal{F}$, denoted $\operatorname{gd}_\mathcal{F}(G)$ and $\operatorname{cd}_\mathcal{F}(G)$ respectively, with reference to classifying spaces for $G$-actions with isotropy in $\mathcal{F}$ and the homological algebra of modules over the $\mathcal{F}$-restricted orbit category. (If you've read this far, you probably already know the definitions.)
A well known result credited to Lueck and Meintrup says that $\operatorname{gd}_\mathcal{F}(G)\le \max\{3,\operatorname{cd}_\mathcal{F}(G)\}$. See Theorem 0.1(a) here.
I can understand the general framework of the proof given by Lueck and Meintrup, but there is a technical Lemma 2.7 about chain complexes which they do not prove (referring only to pp 279-280 of Lueck’s book Transformation Groups and Algebraic K-Theory, where he seems to be working in a more general framework to prove something about finite domination). Lemma 2.7 contains the statement "If $D_*$ is homotopic to a finite-dimensional free complex, then $C’_*$ can be chosen to be finite-dimensional" when it seems to me that what is needed is the stronger "If $D_*$ is homotopic to a $d$-dimensional free complex, then $C’_*$ can be chosen to be $d$-dimensional".
Does this stronger statement follow from the argument in Lueck’s book (which I therefore need to understand)? Is there any other source where this argument is written up in full?
 A: Note This is based on misreading the question as being about projective resolutions, not free ones. I had deleted it but the OP found it helpful so I have undeleted.
I have managed to cobble this out of the Brown references I gave in now deleted comments. In this answer all chain complexes are assumed to be bounded below.
Suppose that $D_\ast$ is a chain complex of projectives that is homotopy equivalent to a complex vanishing in dimensions $>n$.  Then $D_\ast$ is homotopy equivalent to a complex of projectives $C_\ast$ vanishing in dimensions $>n$.  Moreover, one can choose $C_\ast$ so that $C_q=D_q$ for $q<n$ and $C_n=D_n/B_n$ where $B_n$ is the $n$-boundaries of $D_\ast$.  In particular, if $D_\ast$ consists of finitely generated projectives then the same is true for $C$.
First of all we will use the standard fact that if you have a chain map between (bounded below) chain complexes which are degreewise projective, then it induces a homotopy equivalence if and only if it is an isomorphism on homology.
Define $C_\ast$ as proposed above.  Clearly the quotient map $D_\ast\to C_\ast$ which is $0$ for in degree $q>n$, the identity for $q<n$ and the projection $D_n\to D_n/B_n$ gives a homology isomorphism since $D_\ast$ is homotopy equivalent to a complex vanishing in dimension $>n$.  It remains to show that $C_n$ is projective.  Consider the cochain complex $K^\ast=\mathrm{Hom}(D_\ast,B_n)$.  Since $D_\ast$ is homotopy equivalent to a chain complex vanishing in degree $n+1$, we have $H^{n+1}(K^\ast)=0$.  Clearly $d_{n+1}\colon D_{n+1}\to B_n$ is an $(n+1)$-cocycle in $K^{n+1}$, and hence there is $g\colon D_n\to B_n$ with $g\circ d_{n+1}=d_{n+1}$. Then $g$ is a retraction onto $B_n$ and so $D_n= B_n\oplus \ker g$.  Thus $\ker g\cong D_n/B_n=C_n$ is projective, as required.
