Noncontractible manifold whose homology is zero Does there exist a noncontractible manifold whose integer reduced homology is zero? It is clear that, due to the Poincaré duality, this manifold is non-compact. Also, its fundamental group is nontrivial (otherwise it would be contractible by Whitehead's homology theorem).
Maybe a Poincaré homology sphere without a point will do? It is non-compact, so its $H_3$ is trivial. The fundamental group has not changed from cutting out the point, so $H_1$ too. What about $H_2$?
 A: Yes, take any finite, noncontractible CW complex with trivial homology and embed it in some $\mathbb{R}^n$, then take a regular neighborhood. This has the same homotopy type and is a manifold. In this way, all finite CW complexes have the homotopy type of a (not necessarily compact) manifold.
A: Let $M$ be a closed manifold of dimension $n\ge1$. Choose a point $p\in U\subseteq M$, where $(\overline{U},U)\cong(D^n,B^n)$. Consider the open cover $M=M\setminus\{p\}\cup U$, where $M\setminus\{p\}\cap U=U\setminus\{p\}\simeq S^{n-1}$. The reduced M-V-sequence of this open cover immediately implies that the map $\tilde{H}_k(M\setminus\{p\})\rightarrow\tilde{H}_k(M)$ induced by the inclusion is an isomorphism if $k\neq n-1,n$. As you've noted, $M\setminus\{p\}$ is non-compact (here, we use that $M$ is not zero-dimensional), so $\tilde{H}_n(M\setminus\{p\})=0$. The critical part of this sequence reads
$$0=\tilde{H}_n(M\setminus\{p\})\rightarrow\tilde{H}_n(M)\rightarrow\tilde{H}_{n-1}(S^{n-1})\rightarrow\tilde{H}_{n-1}(M\setminus\{p\})\rightarrow\tilde{H}_{n-1}(M)\rightarrow\tilde{H}_{n-2}(S^{n-1})=0.$$
I claim that if $M$ is orientable, the map $\tilde{H}_n(M)\rightarrow\tilde{H}_{n-1}(S^{n-1})$ is an isomorphism. From that, it follows that $\tilde{H}_{n-1}(M\setminus\{p\})\rightarrow\tilde{H}_{n-1}(M)$ is an isomorphism also.
To see this, consider the quotient $M/(M\setminus U)\cong S^n$ (here, we use that $(\overline{U},U)\cong(D^n,B^n)$ in full strength). The quotient map $M\rightarrow M/(M\setminus U)$ takes our open cover of $M$ to the standard cover of a sphere by the complements of two respective points (the points being $p$ and $\partial U/\partial U$). It induces an isomorphism on $\tilde{H}_n$ (the induced map is the composite of the isomorphism $\tilde{H}_n(M)\rightarrow H_n(M,M\setminus U)$, from the general theory of top homology, and the isomorphism $H_n(M,M\setminus U)\rightarrow H_n(M/(M\setminus U),(M\setminus U)/(M\setminus U))\cong\tilde{H}_n(M/(M\setminus U))$, from excision) and the boundary map of the M-V-sequence of the sphere is known to be an isomorphism, so by naturality the boundary map $\tilde{H}_n(M)\rightarrow\tilde{H}_{n-1}(S^{n-1})$ of our original M-V-sequence is an isomorphism as well.
(If $M$ is triangulable, here's an argument with more geometric flavor: the simplices of a triangulation with appropriate signs give a chain that represents a generator of $\tilde{H}_n(M)$. Assume WLOG that $p$ is an interior point of one (and necessarily only one) simplex of the triangulation and that this simplex is contained in $U$ (subdivide if necessary). The boundary map of the M-V-sequence then maps the fundamental class to the homology class represented by the boundary of that simplex, which is a generator of $\tilde{H}_{n-1}(S^{n-1})$.)
Furthermore, if $n\ge2$, the inclusion induces a surjection $\pi_1(M\setminus\{p\})\rightarrow\pi_1(M)$ (basepoints implicit) by the Seifert-van-Kampen theorem (applied to the same open cover $M=M\setminus\{p\}\cup U$; we need $n\ge2$ so that the intersection $M\setminus\{p\}\cap U=U\setminus\{p\}\simeq S^{n-1}$ is connected).
In total, if $M$ is a homology sphere, which is not a sphere, it has dmiension $n=\dim(M)\ge2$ by classification, is compact since it has non-vanishing top homology and is orientable, since $H_1(M)=0$ implies that the orientation character is trivial. Thus, $\tilde{H}_k(M\setminus\{p\})=\tilde{H}_k(M)=0$ for $k=0,\dotsc,n-1$ by the preceeding discussion and $\tilde{H}_k(M\setminus\{p\})=0$ for $k\ge n$ by general theory, so $M\setminus\{p\}$ is acyclic. Furthermore, $\pi_1(M)\neq1$ (otherwise $M$ would be a sphere), so $\pi_1(M\setminus\{p\})\neq1$, since it surjects onto $\pi_1(M)$. It follows that $M\setminus\{p\}$ is not contractible.
