When does $U\subset V$ induce maps on homology of finite rank? 
Question 1. Let $X$ be a topological manifold, $U\subset V\subset X$ open subsets. The inclusion $\iota:U\to V$ induces corresponding maps on homology $\iota_*:H_\bullet(U)\to H_\bullet(V)$. We take the coefficients to lie in some fixed field. Under what conditions can we conclude that $\iota_*$ has finite rank (as a linear map)? In particular, I am interested in whether the following is sufficient: $\overline{U}\subset V$ and $\overline{U}$ is compact.

I encountered this problem when reading Gromov's paper Curvature, diameter and Betti numbers. The setting is on a complete Riemannian manifold $(M,g)$ where we may consider balls $B(p,r):=\{x\in M:d(p,x)<r\}$. The content of such a ball is defined by $$\operatorname{cont}B(p,r):=\operatorname{rank}(B(p,r/5)\subset B(p,r)),$$ where the rank refers to the rank of the induced map on homology. I need to know that this number is always finite.
I encountered the second question when reading the proof of the results in the aforementioned paper of Gromov in the book Riemannian Geometry by Peter Petersen. The author remarks without proof the following:

Question 2. Notation as above. For $r$ large enough, $$\operatorname{cont}B(p,r)=\sum b_i(M)$$ where $b_i:=\dim H_i(M)$ is the Betti number of $M$.

Now this is obvious if $M$ is compact. But why is it true for noncompact complete manifolds?
I'm a bit rusty on algebraic topology, but I really don't see why these should hold. Any help is highly appreciated!
 A: For the first question, it is sufficient for $U$ to be precompact in $V$ when $X$ is smooth.  In that case, you can triangulate $V$.  Since $\overline{U}$ is compact, it is covered by finitely many cells of the triangulation, and so there is a finite simplicial complex $K$ such that $\overline{U}\subseteq K\subseteq V$.  Then the map $H_*(U)\to H_*(V)$ factors through the finite-dimensional space $H_*(K)$, so it has finite rank.
(I suspect it is also true for topological manifolds but don't see a proof at the moment.  It would suffice, for instance, to show there is a compact manifold with boundary $K$ such that $\overline{U}\subseteq K\subseteq V$, since compact manifolds with boundary also have finitely generated homology.)
For the second question, the statement is just wrong without some additional assumption about $M$.  Indeed, every manifold admits a complete metric, so if you take a manifold $M$ with infinitely generated homology, the statement cannot be true.
It can even be false for manifolds with finitely generated homology.  For instance, take a smooth map $f:\mathbb{R}^2\to\mathbb{R}$ which is $0$ except that it has spikes near the points $(n,0)$ for each $n\in\mathbb{N}$ which get higher and higher very fast as $n\to\infty$.  Let $M\subset\mathbb{R}^3$ be the graph of $f$ with the induced metric.  Then $M\cong\mathbb{R}^2$ is contractible, but the content of arbitrarily large balls will be nontrivial, since they (and the subball with $1/5$ the radius) will contain the bottom of the spikes in the graph but not the top so there will be noncontractible loops in them.
