Do trivial homology groups imply contractibility of a compact polyhedron? Is it true that a compact polyhedron X with trivial homology groups (except $H_{0}(X)$ of course) is necessarily contractible? If yes, what is the approach in proving it? If not, do you see a counter-example?
 A: (This is the deleted answer the comments refer to; it was missing the hypothesis about simple-connectedness)
Using the Hurewicz theorem, you deduce at once that such a polyhedron [Edit: if it is simply connected] has trivial homotopy groups, so that it is weakly homotopy equivalent to a point. Since it is a CW-complex, then Whitehead's theorem tells you that the polyhedron is in fact contractible.
A: The 2-skeleton of the Poincare homology sphere, also describable as the presentation complex of the binary icosahedral group, provides a counterexample to your original question.  The fundamental group is of order 120 and is perfect, which implies that that $H_1$ is trivial.  You can check from the group presentation $ <s,t | s^{-3}(st)^2, t^{-5}(st)^2> $ that the second homology group is trivial as well.
A: To sum up the comments: when Poincaré worked on the beginnings of algebraic topology, he originally thought that a space with trivial homology groups must be contractible. (More precisely, he thought that having the homology group of a 3-sphere implies being a 3-sphere.) However, he soon found a counterexample, the Poincaré homology sphere, which led him to the construction of the fundamental group.
When taking the fundamental group into account, the statement is indeed true: if a space has trivial fundamental group and trivial higher homology groups, then it must be contractible. This is a consequence of Whitehead's theorem and the Hurewicz map.
