If $X$ is a space, then let us say that $\pi_0^\infty(X)$ is the set of equivalence classes of proper maps $[0,\infty) \to X$, modulo proper homotopy (the map $[0,1] \times [0,\infty) \to X$ should be proper). Call an element of this set an "end of X".
If $X$ is compact, $X$ has no ends. If $X = \Bbb R$, then $X$ has two ends, corresponding to the identity and negation maps $[0, \infty) \to \Bbb R$. If $X = \Bbb R^n$ for $n > 1$, then $X$ has one end.
In fact, generalizing this, if $M^\circ$ is the interior of some compact manifold with boundary $M$, then we may identify $\pi_0^\infty(X) = \pi_0(\partial M)$.
Suppose $M$ is connected and thus, if $M^\circ$ has more than one end, then $H_0(\partial M;\Bbb Z/2)$ is larger than 1-dimensional (all (co)homology groups will have $\Bbb Z/2$ coefficients from now on); the relative long exact sequence then implies that $H_1(M, \partial M)$ is non-trivial ; applying Poincare-Lefschetz duality we find that $H^{n-1}(M)$ is nonzero.
So we conclude: if $M$ is a connected compact manifold with boundary whose interior has more than on end, then $H^{n-1}(M) \neq 0$. This fits with situations we see in practice: the easiest way to construct $M$ with two ends is to take the interior of $N \times [0,1]$ for $N$ a closed connected manifold.
After all this, a new definition. An $n$-dimensional homology manifold is a locally compact separable Hausdorff space so that at each point $x \in M$, we have $H_k(M, M -x) = H_k(\Bbb R^n, \Bbb R^n – 0)$. Theorems like Alexander duality and Poincare duality continue to hold in this context.
Is it still true that a connected $n$-dimensional homology manifold with more than one end has $H^{n-1}(M) \neq 0$?
The name of the game seems to be to figure out if one can find a proof just using homological duality theorems, and not using some sort of compactification to a manifold with boundary.
This question came.up in the course of answering this question, where I needed to show that certain contractible homology manifolds have 1 end. I ended up restricting to the 2-dimensional case, where a contractible homology manifold must be $\Bbb R^2$.
Best Answer
First of all, let $X$ be a reasonably nice space, say, metrizable and locally compact. Define $$ H^i(Ends(X))=\lim_K H^i(X-K), $$ where the direct limit is taken over compact subsets $K$ in $X$. (Similarly, one defines $H_i(Ends(X))$ by taking the inverse limit.) In fact, these groups are the Chech cohomology groups of the space of ends of $X$ but I will not need this.
The space $X$ has more than one end if and only if $$ \tilde{H}^0(Ends(X))\ne 0, $$ where I am using the reduced cohomology. On the other hand, the cohomology with local support of $X$ satisfies $$ H^1_c(X)\cong \lim_K H^1(X, X-K). $$ Assume now that $X$ is acyclic as in your case. Then by the long exact sequence of a pair, $$ \lim_K H^1(X, X-K) \cong \lim_K \tilde{H}^0(X-K)\cong \tilde{H}^0(Ends(X)). $$ By the Alexander duality, assuming that $X$ is an $n$-dimensional homology manifold, $$ H^1_c(X)\cong H_{n-1}(X). $$ Hence, since $X$ is acyclic, $$\tilde{H}^0(Ends(X))\cong H^1_c(X)\cong H_{n-1}(X)=0,$$ i.e. $X$ has exactly one end.