Trivial homology groups and fundamental group I was wondering if there are noncontractible spaces with trivial homology groups and trivial fundamental group. I can't think of any, but I suspect some must exist.
 A: The Warsaw circle $\mathbb{S}$ is such a space. To construct it start with the topologist's sine curve
$$X=\{(x,\sin(2\pi/x))\mid x\in(0,1]\}\cup(\{0\}\times[-1,1])$$
embedded in the euclidean plane. Recall that this space is often encountered as an example of a space which is connected, but not path-connected. Then the Warsaw circle is the union
$$\mathbb{S}=X\cup(\{0\}\times[-2,-1])\cup([0,1]\times\{-2\})\cup(\{1\}\times[-2,0]).$$
Thus $\mathbb{S}$ consists of the sine curve with its ends joined together by a simple arc in $\mathbb{R}^2$ which is disjoint from all other points of $X$. The result $\mathbb{S}$ is a compact metric space which is both connected and path-connected. However it is still highly pathological.
Basically, in terms of connectivity it thinks it is a circle, while in terms of path-connectivity is sees itself as a copy of the real line.
As a consequence all the homotopy groups and singular homology groups of $\mathbb{S}$ vanish. This is not difficult to see directly, but can be formally established as follows. There is an obvious bijective map $p:[0,\infty)\rightarrow\mathbb{S}$ that starts at $(0,1)$ and wraps itself anticlockwise around $\mathbb{S}$. For all intents and purposes this map acts like a covering projection with one-point fibres (though it is not locally-trivial - $\mathbb{S}$ is not locally connected). What this map is is the result of retopologising $\mathbb{S}$ by giving it the final topology with respect to all maps $M\rightarrow\mathbb{S}$ with locally path-connected domains. Thus in particular any map into $\mathbb{S}$ from a sphere $S^n$ lifts throught $p$ into $[0,\infty)$ and can be linearly null-homotoped. This gives us $\pi_*\mathbb{S}=0$, and working with simplices the same argument yields $\widetilde H_*\mathbb{S}=0$.
On the other hand, $\mathbb{S}$ is not contractible. In fact its first Cech cohomology group $\check H^1\mathbb{S}$ is nonzero. This can be computed as $\check H^1\mathbb{S}=[\mathbb{S},S^1]\cong\mathbb{Z}$, where $[\mathbb{S},S^1]$ is the group of homotopy classes of map $\mathbb{S}\rightarrow S^1$ (the group structure is induced by the multiplication on $S^1$). Thus you can very vividly see what is happening. For each $n\in\mathbb{Z}$ there is a map of $\mathbb{S}$ onto $S^1$ which wraps the 'pseudo-circle' $n$ times around the real circle. Because there are essential maps out of it, we see that $\mathbb{S}\not\simeq\ast$ .
For a formal calculation of the Cech group we can use Alexander duality, which states that if $A\subseteq\mathbb{R}^n$ is a compact subset, then there are isomorphisms
$$\check H^kA\cong \widetilde H_{n-k-1}(\mathbb{R}^n\setminus A),$$ where the group on the left is the Chech cohomology group, and the group on the right is ordinary (reduced) singular homology. Plugging in $\mathbb{S}\subseteq\mathbb{R}^2$ we get
$$\check H^1\mathbb{S}\cong \widetilde H_0(\mathbb{R}^2\setminus \mathbb{S}).$$
Now $\mathbb{S}$ cuts the plane into two connected components, and as such $H_0(\mathbb{R}^2\setminus \mathbb{S})\cong\mathbb{Z}^2$. In the reduced group we indeed retrieve
$$\check H^1\mathbb{S}\cong\mathbb{Z}.$$
A: Actually if a simply connected space $X$ has trivial homology then it must be weakly contractible (all homotopy groups vanish) by the Hurewicz theorem. If $X$ is a CW complex then it follows from Whitehead's theorem that $X$ is contractible. So a counterexample can't be homotopy equivalent to a CW complex; in other words it will only fail to be contractible for point-set reasons, not "purely homotopy-theoretic" reasons.
A more general version of this argument can be used to prove Whitehead's theorem for homology, which says that if a map $f : X \to Y$ between simply connected CW complexes induces an isomorphism on homology then it must be a homotopy equivalence (the simply connected hypothesis is crucial here).
