Sufficient condition for a direct limit of abelian groups to be infinitely generated I have the following setup. The CW-complexes $\Gamma_n$ are equipped with maps $\gamma_n\colon\Gamma_{n+1}\rightarrow\Gamma_{n}$ and it is known that the rank of their first cohomology groups is finite and given by $\mbox{rk}( H^1(\Gamma_n;\mathbb{Z}))=s(n)$ where $s(n)\sim n^k$ for some fixed natural number $k\geq 2$.
I want to prove that the inverse limit $\Omega$ of the directed system $(\Gamma_n,\gamma_n)$ has infinitely generated first Cech cohomology. Now by standard results of Cech cohomology, $$\check{H}^\bullet(\Omega)=\check{H}^\bullet(\lim_{\leftarrow}(\Gamma_n,\gamma_n))\cong\lim_{\rightarrow}(\check{H}^\bullet(\Gamma_n),\gamma_n^*)$$ and $\Gamma_n$ is a CW complex so we can just look at simplicial cohomology giving $$\check{H}^\bullet(\Omega) \cong \lim_{\rightarrow}(H^\bullet(\Gamma_n),\gamma_n^*).$$
My question is, what conditions on the maps $\gamma_n$, or their induced maps in cohomology $\gamma_n^*$, would be sufficient to conclude that $\check{H}^1(\Omega)$ is infinitely generated, given that the rank of the first cohomology of the approximants grows like $n^k$? If it helps, we can also assume that the spaces $\Gamma_n$ are $1$-dimensional CW complexes (graphs) and so $H^1(\Gamma_n)$ is torsion-free.
 A: Well, if the group homomorphisms $\gamma_n^*$ are one-to-one, so you're essentially forming a directed union, then this will be infinitely generated, because any finitely many generators would be in one of the groups of your direct system and therefore can't generate the additional elements in the next group of the system.  
Presumably, you want sufficient conditions that are not quite so trivial, but this seems difficult to obtain.  Even if the homomorphisms are only "slightly" not one-to-one, you can have examples like the following (with torsion-free groups, as you requested).  Let the $n$-th group in your system be the direct sum of $n^k$ copies of $\mathbb Z$.  Let the homomorphism from the $n$-th to the $(n+1)$-th group send any tuple $(x_1,\dots,x_r)$, where $r=n^k$, to the tuple $(x_2,\dots,x_r,0,0,\dots,0)$, where there are $(n+1)^k-n^k+1$ zeros at the end.  Such a homomorphism has a kernel of rank $1$, which is as close as you can get to one-to-one without actually being one-to-one (for torsion-free groups).  Yet the direct limit of this system is the zero group.
