TITLE: On spaces with finite homological dimension
QUESTION [5 upvotes]: Let $X$ be a connected $CW$-complex, such $\pi_1(X)$ is torsion-free and  $H_k(X,\mathbb Z) = 0$ for all $k \geq N$ and some $N \in \mathbb N$. Then 
$(1)$ Does it follow that $X$ is homotopy-equivalent to its $N$-skeleton, i.e $X \simeq X^{(N)}$ ?
$(2)$ If $(1)$ is false, does it follow that $X \simeq X^{(k)}$ for some $k \in \mathbb N$ ? 
$(3)$ If $(2)$ is also false, at the very least, does it follow that $X \simeq Y$ for some finite-dimensional $CW$-complex $Y$ ? 
It is clear to me that the question has a negative answer when $H_k$ is replaced by $\pi_k$ (take a $CW$-model of $BG$ for a group $G$ containing elements of finite order). I am also very certain that this question has been asked before and I just couldn't find the corresponding thread, so I have no problem with this question being marked as duplicate. 
Edit: One could then possibly find a finitely dominated $CW$-complex $X$, with $\pi_1(X)$ not torsion-free, such that its Wall finiteness obstruction element $w(X) \in \widetilde{K_0}(\mathbb Z[\pi_1(X)])$ is non-trivial. Any finitely dominated $CW$-complex has finite homological-dimension (so it satisfies the assumptions), but if its finiteness obstruction doesn't vanish, it doesn't have the homotopy type of a finite complex. 
Conversely, if $X$ is finitely-dominated and $w(X) = 0$, then it has the homotopy type of a finite-dimensional complex.
In particular, this must be the case if $\widetilde{K_0}(\mathbb Z[\pi_1(X)]) =0$. It is conjectured that $\widetilde{K_0}(\mathbb Z[G]) = 0$ if $G$ is torsion-free, hence the assumption in the statement of the question.
Edit 2: Mark has given an example of a $CW$-complex $X$ with finite homological dimension, but infinite cohomological dimension (and thus, has given a negative answer to my original questions) . It would be still interesting to know if there exist counter-examples $X$, such that for any $\pi_1(X)$-module $M$, both the homology $H_*(X,M)$ and the cohomology $H^*(X,M)$ have groups only in finitely many dimensions.

REPLY [8 votes]: You can take $X=BG$ where $G$ is a torsion-free, acyclic group of infinite cohomological dimension. Acyclic means $H_k(BG;\mathbb{Z})=0$ for $k>0$, and infinite cohomological dimension implies infinite geometric dimension, so $BG$ is not homotopy equivalent to any finite dimensional CW complex $Y$. Such a group therefore gives a (rather extreme) counter-example to (1), (2) and (3).
I believe an infinite direct product of copies of Higman's group $H$ gives such a $G$. It is obviously torsion-free (since $H$ is), and is acyclic since homology commutes with direct limits and a finite product of acyclic groups is acyclic. Since $H$ is of type $FP_\infty$ (iirc, the presentation complex is a model for $BH$), we can apply Benjamin Steinberg's answer here to conclude that $G$ has infinite cohomological dimension.

REPLY [3 votes]: For an elementary example, give $B^{n+1}$ the CW structure with a $0$-cell, a boundary $n$-cell, and an $n+1$-cell, then take $X=\bigvee_nB^{n+1}$.  Then $X$ is contractible but $X^{(n)}$ is homotopy equivalent to $S^n$ (from the $B^{n+1}$ term) so $X^{(n)}$ is never homotopy equivalent to $X$.  
It would be better to look for $Y$ with $X^{(n)}\subseteq Y\subseteq X^{(n+1)}$ such that the inclusion $Y\to X$ is a homotopy equivalence.  I suspect that that is not possible either, although it would be a bit more delicate to provide an example.  Better still, we can ask whether there is a CW complex $Y$ obtained from $X^{(n)}$ by attaching some $n+1$-cells, such that the inclusion $X^{(n)}\to X$ extends over $Y$, and the resulting map $Y\to X$ is a homotopy equivalence.  The attaching maps for $Y$ will lie in the relative homotopy group $\pi_{n+1}(X,X^{(n)})$, which maps to $H_{n+1}(X,X^{(n)})$.  If $X$ is simply connected then you can use the relative Hurewicz theorem and I think it works out that you can always find such a $Y$ provided that $H_{n+1}(X)$ is free abelian (perhaps zero) and $H_k(X)=0$ for $k\leq n$.  If $X$ is not simply connected then you need to worry about the Wall finiteness obstruction.