Explicitly building Top. space with trivial homology group and non trivial fundamental group, w/o CW-complexes I know this is a pretty famous question here, but I was asked to show explicitly such space, during a bachelor lecture, without using any CW-complex result. I started working using some covering theory but I met several difficulties:
1) Before such request, th professor show us that $A_n$ with $n\geq 5$ is perfect. So I though that we have to use this group. Anyway I proved that the binary icosahedral group is perfect, and so I can use this one too
2) I think I have to work with some universal cover of my "target space", define a free action and conclude that the orbit space is the space I'm looking for.
3) the main problem is, I can't define a free action of $A_5$ on a sphere for example. The permutation action on $R^5$ for example is not free. I read that $A_5$ acts freely on $S^5 \times S^7$, but I cannot write down explicitly the action, and so I cannot proceed.
Any thoughts?
 A: Based on the professor's hint, I'll assume you're asking for an example of a space such that $H_1(X) = 0$ but such that $\pi_1(X) \neq 0$. Since $H_1(X) = \pi_1(X)/[\pi_1(X), \pi_1(X)]$ by Hurewicz, this means exhibiting a space whose $\pi_1(X)$ is perfect. As you say, a general construction involving CW complexes lets us exhibit a space with any fundamental group whatsoever, but the question is whether one can be more explicit than this. 
One option is to try to find a perfect group $G$ acting freely on a simply connected space, say a sphere, and then to take the quotient. This works, and leads to one of the most historically important examples: the binary icosahedral group $\widetilde{A_5}$ is naturally a subgroup of $\text{SU}(2) \cong S^3$, which acts on itself freely by left multiplication, hence $\widetilde{A_5}$ acts on $S^3$ freely as well. The resulting quotient is the Poincaré dodecahedral space, which in particular has the same homology as $S^3$, showing that to state the Poincaré conjecture it does not suffice to put conditions on homology.
More generally, any Lie group acts on itself freely by left multiplication, so if $G$ is a simply connected Lie group and $\Gamma$ is a discrete subgroup of it then $G/\Gamma$ is naturally a manifold of the same dimension as $G$ with fundamental group $\Gamma$. So to generalize the construction above we can, for example, try to embed finite groups as subgroups of $\text{SU}(n)$. It is of course straightforward to embed any finite group as a finite subgroup of $\text{U}(n)$ for some $n$; the resulting embedding automatically lands in $\text{SU}(n)$ if the group is perfect, since the composition of the embedding with the determinant map $\det : \text{U}(n) \to \text{U}(1)$ will vanish. This furnishes a large supply of manifolds with fundamental group any finite perfect group. 
A somewhat harder question is to ask for a space such that $\pi_1(X) \neq 0$ but such that all of its (reduced) homology groups vanish. Such spaces are called acyclic. Poincaré dodecahedral space minus a point is an example. 
