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?
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$\begingroup$ I think you at least want to specify that $X$ is connected, or else you could take, say, two points. $\endgroup$ – Qiaochu Yuan Mar 30 '11 at 16:21
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$\begingroup$ The Poincare homology sphere is a counter-example which is a manifold. It's also called the Poincare Dodecahedral Space. There's a Wikipedia page for it. $\endgroup$ – Ryan Budney Mar 30 '11 at 16:22
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2$\begingroup$ @Ryan: But... Doesn't a homology $3$-sphere have $H_{3} \cong \mathbb{Z}$ by definition? $\endgroup$ – t.b. Mar 30 '11 at 16:28
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1$\begingroup$ @Ryan: In other words: what was wrong with Mariano's now deleted answer along the lines: If $X$ is connected and simply connected then it follows from Hurewicz and Whitehead that the inclusion of any point is a homotopy equivalence. What am I missing? $\endgroup$ – t.b. Mar 30 '11 at 16:36
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1$\begingroup$ @Theo: ah, right. I forgot about $H_3$. I'll give a proper example as an answer, if someone else doesn't beat me to it. $\endgroup$ – Ryan Budney Mar 30 '11 at 17:43
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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.
answered Mar 30 '11 at 17:41
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1$\begingroup$ I don't understand how the first two sentences are related. $\endgroup$ – Qiaochu Yuan Dec 31 '11 at 9:59
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$\begingroup$ @QiaochuYuan: If I were to conjecture that homology is a complete invariant, then this would imply both sentences. As to Poincaré's thinking, I am only aware of evidence that backs up the sentence in parentheses, however. $\endgroup$ – Greg Graviton Dec 31 '11 at 11:55
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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.
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2$\begingroup$ Very nice. This is the example I should have given! $\endgroup$ – Ryan Budney Mar 30 '11 at 17:51
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(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.
answered Mar 30 '11 at 16:23
