Can an acyclic space be determined by its fundamental group

Question

Let $X$ be any acyclic CW complex (i.e. reduced homology in all degree is zero). Clearly its fundamental group $\pi_{1}(X)$ is perfect group. Is it true that if $Y$ is any acyclic CW complex with same fundamental group as $X$ then $Y$ is homotopy equivalent to $X$?

Answer 1

The comment is correct but does not give a very strong result. If $H \subset \pi_1 X$ is a subgroup and $X'$ is the associated cover, the homology $H_*(X; A[\pi_1 X/H]) \cong H_*(X'; A)$. In particular if $\pi_1 X$ is non-trivial it has a non-trivial cyclic (so Abelian) subgroup $H$, so that $H_1(X'; \Bbb Z) = H$. So any space with $\pi_1 X$ non-trivial has a local coefficient system with non-trivial homology.

The answer to your question itself is no. See here for a counterexample.