Homological vs. cohomological dimension of a group/space I have several related questions regarding homological vs. cohomological dimension of a space/group (this is not a duplicate of this).
The standard definition of the cohomological dimension $cd(X)$ of a space $X$ is 
definition 1: $cd(X)=$ The minimum over $n\in \mathbb{N}$, such that $H^i(X,M)=0$ for all $i>n$ and all local coefficient systems $M$. 
One can similarly define the homological dimension $hd(X)$ (I am not sure it is a standard terminology)
definition 2: $hd(X)=$ The minimum over $n\in \mathbb{N}$, such that $H_i(X,M)=0$ for all $i>n$ and all local coefficient systems $M$. 
My main question is

Question1: Do we always have $cd(X)=hd(X)?$ What if one or both assumed to be finite? what if $X$ is a finite CW complex?

For $X = BG$, i.e a classifying space of a group $G$, the definitions reduce to the standard ones in group cohomology. I believe that the $cd(BG)$ coincides with the minimal length of a projective resolution of the trivial $G$-module $\mathbb{Z}$ over the group ring $\mathbb{Z}[G]$ (Namely, the projective dimension of $\mathbb{Z}$). 

Question 2: Is it true that the $hd(BG)$ coincides with the flat dimension of $\mathbb{Z}$ as a $\mathbb{Z}[G]$-module?

Finally, if $X$ is a connected CW complex with $cd(X)=0$, then $X$ is contractible. One can similarly ask

Question 3: If $X$ is a CW complex with $hd(X)=0$, is it true that $X$ is contractible?

(Edit: The following argument has a gap is wrong. It is not clear true that if $hd(X)=0$ then $X$ is a filtered colimit of finite subcomplexes $X'$ with $hd(X')=0$. An exmaple is given here)
Regardless of the answer to question 1, I think the answer to this is yes by the following argument. Since the universal cover is acyclic and simply connected, it is contractible, so this reduces to a question about group cohomology. If $G$ is finitely generated (if $X$ is a finite CW for example), then $\mathbb{Z}$ is a finitely presented $\mathbb{Z}[G]$-module so if it is flat then it is projective and hence the cohomological dimension is zero as well. Can one reduce the general case to this one since every CW complex is a filtered colimit of its finite subcomplexes. Is there a more direct argument?
 A: Q2) You mean $X=BG$, right? Then the answer is yes by standard homological algebra.
Q1) The answer to the first question is no, e.g. $X=M(\mathbb Q,n)=S^n_{\mathbb Q}$ the rational $n$-sphere has homological dimension $n$ but cohomological dimension $n+1$ (since $\mathbb Q$ is flat but not projective as an abelian group). For $X$ finite I don't know the answer in general, but if $X$ is in addition simply connected then yes, since the chain complex of $X$ is quasi-isomorphic to its homology and the projective and flat dimensions coincide on finitely generated abelian groups.
Q3) This is true if $X$ is simply connected (Whitehead's theorem) but false otherwise, e.g. $X=BG$ with $G$ acyclic. It is also positively answered in the question for $X=BG$ with $G$ finitely presented.
A: As noted in the question, "Question 3" reduces to asking whether, if $H_i(G,M)=0$ for all $i>0$ and all $\mathbb{Z}G$-modules $M$, then $G$ must be trivial.
But this is true, since if $G$ is non-trivial, we can choose a non-trivial abelian subgroup $A$, and then, by Shapiro's Lemma,
$$H_1\left(G,\mathbb{Z}[G/A]\right)\cong H_1\left(A,\mathbb{Z}\right)\cong A\neq0.$$
A: If $X$ is a finite CW complex then $hd(X) = cd(X)$. To prove this it is clearly enough to assume that $X$ is connected. Let $n$ be the dimension of $X$, let $G$ be the fundamental group of $X$ and let $\widetilde{X}$ be the universal cover of $X$. Let
$$ 0 \longrightarrow C_n \stackrel{\partial_n}{\longrightarrow} C_{n-1} \stackrel{\partial_{n-1}}{\longrightarrow} \dots \stackrel{\partial_{1}}{\longrightarrow} C_0 $$
be the cellular complex of $\widetilde{X}$ (i.e. $C_n = H_n(\widetilde{X}_n,\widetilde{X}_{n-1})$) considered as a complex of $G$-modules. The homological and cohomological dimensions of $X$ are the same as the flat and projective dimensions respectively of the complex of $G$-modules $C_\bullet$ (lets call these latter two guys $pd(C_\bullet)$ and $fd(C_\bullet)$). Since $X$ is finite each $C_n$ is a finitely generated free $G$-module. 
Given a complex of $G$-modules $D_\bullet$, indexed by non-negative integers, we will call the minimal $i$ such that $D_j = 0$ for $j > i$ the dimension of $D_\bullet$. Now let $D_\bullet$ be a complex of finitely generated projective $G$-modules which is quasi-isomorphic to $C_\bullet$ and has minimal dimension among all such complexes (since $C_\bullet$ is made of free finitely generated $G$-modules and has finite dimension such a $D_\bullet$ exists). Then $hd(X) = fd(C_\bullet) = fd(D_\bullet)$ and $cd(X) = pd(C_\bullet) = pd(D_\bullet)$. Let $m$ be the dimension of $D_\bullet$. We next claim that $fd(D_\bullet) = pd(D_\bullet) = m$. First it is clear that $fd(D_\bullet),pd(D_\bullet) \leq m$. Now if the map $d_m: D_m \longrightarrow D_{m-1}$ is not injective then $Tor^m(D_\bullet,\mathbb{Z}G) \neq 0$ and $Ext^m(D_\bullet,D_m) \neq 0$ and so $fd(D_\bullet)=pd(D_\bullet)=m$. Hence we may assume that the map $D_m \longrightarrow D_{m-1}$ is injective. Let $M$ be the cokernel of $d_m$. Then $M$ is finitely generated and by the minimality of $D_\bullet$ we see that $M$ cannot be projective. It follows that there exists a $G$-module $N$ such that the map $Hom_G(D_{m-1},N) \longrightarrow Hom_G(D_m,N)$ is not injective, and hence $Ext^m(D_\bullet,N) \neq 0$. This gives us $pd(D_\bullet) = m$. Now let $D'$ be a finitely generated projective $G$-module such that $D' \oplus D_{m-1}$ is a free $G$-module. Then $M$ is the quotient of $D' \oplus D_{m-1}$ by the finitely generated $G$-module $D' \oplus D_m$ and hence $M$ is finitely presented. It follows that $M$ cannot be flat, since any finitely presented flat $G$-module is projective. This means that there exists a $G$-module $N$ such that the map $N \otimes D_m \longrightarrow N \otimes D_{m-1}$ is not injective, and hence $Tor^m(D_\bullet,N) \neq 0$. We may now conclude that $fd(D_\bullet) = m$ as desired.
Regarding Q3, it seems very likely that $hd(X) = 0$ implies that $X$ is contractible, but the argument you present seems to have a problem. Indeed, $X$ is the filtered colimit of its finite sub complexes, but to use your argument we need to know that if $hd(X) = 0$ then $X$ can be written as a filtered colimit of finite complexes, each of whom has homological dimensional $0$.
