I want to find the cohomology with compact supports over $\mathbb Z$ of a) Figure 1, consisting of three rays emerging from a point, and b) Figure 3, the infinite trivalent tree.
Since both are noncompact and path connected, the zeroth cohomology with compact supports, $H^0_c(X)=H^0_c(Y)=0$, so the only nontrivial cohomology is in dimension $1$. We know that $H^1_c(X)=\varprojlim_K H^1(X,X\setminus K)$ over compact sets $K$ (in this case, they consist of cochains that are $0$ everywhere but a finite set of edges). Since every such compact set is contained in a compact set like in Figure 2, we need only take the colimit over such circles, but they are all homotopy equivalent, so all cohomology groups will be equal, and hence we need only compute one. But looking at the cohomology of the quotient, it is clear this is a graph with three loops, so the homology (and hence the cohomology by universal coefficient theorem) is $\mathbb Z^3$, so $H^1_c(X)=\mathbb Z^3$.
Let's now look at $Y$. We can apply a similar argument to show we need only take the colimit over circles in increasing radii. We could use some sort of combinatorial argument to find that at a circle of certain radius, the cohomology of the quotient is that of some (finite) bouquet of circles, but ultimately, that number will grow exponentially, and hence in the colimit, $H^1_c(Y)=\bigoplus_\mathbb{N}\mathbb Z$.
Is this true? The context of the question suggests more work might be needed but cannot see how this is an invalid argument.