Local Homology for Graphs I'm trying to match the intuition of local homology to a simple computation. Consider the simplicial complex given by a graph with two vertices connected by an edge (a directed line segment).
My intuition tells me that the $0$th local homology group (with respect to an abelian group $G$) should be the same as the $0$th homology group, since if we look in a neighborhood of a point, the resulting complex is equivalent to the original. 
However, I am computing a trivial $0$th local homology group:
$$ C_0(X,G) = \{ g_1 v_1 + g_2 v_2 \Big| \ g_1, g_2 \in G \},  $$
$$ C_0(X \backslash \{v_1\},G) = \{ gv_2 \Big| \ g \in G \}. $$
$$ C_0(X,X \backslash\{v_1\},G):= C_0(X,G) / C_0(X \backslash \{v_1\},G) = \{ [g_1 v_1 + g_2 v_2]; \ \equiv [h_1 v_1 + h_2 v_2] \iff g_1=h_1 \} $$
Similarly,
$$ C_1(X,X \backslash\{v_1\},G) = \{[ge_1]\} \cong C_{1}(X,G),$$
so $$ \text{im} \ \partial_1^{'} = \partial_1^{'} [ge_1] = [\partial_1(ge_1)] = [g(v_2 - v_1)] = [g_1 v_1 + g_2 v_2] \cong C_0(X,X \backslash\{v_1\},G)$$
which can be seen to give a trivial $0$th homology group. Please excuse the lack of precise notation. I understand that I am denoting sets and omitting the group operations, but this is just for brevity. Any help here would be appreciated. I am also computing the first local homology to be $0$.
 A: Your computations seems fine, it is the intuition (that the local homology at the vertex should agree with the actual homology of the graph) that is incorrect. Recall that the local homology of any reasonable space $X$ at the point $x \in X$ is the relative homology of the pair $(X,X\setminus\{x\})$ with whatever coefficients. For most spaces that you'll care about (thanks to excision), the intuition you should have is that you're computing relative homology of a small closed neighborhood $K$ around $x$ in $X$ relative to its boundary $\partial K$.
So in your case, do the following: draw a tiny closed ball around $v_1$, and note that it intersects your graph in a small (closed) interval with $v_1$ as one endpoint. Now the boundary of the ball intersects that closed interval in the opposite endpoint, and you're simply computing the relative homology of a closed interval modulo one endpoint. The answer is, as your computation shows, trivial in all dimensions.
Here's a nice theorem: the relative homology of any vertex in a graph is trivial in all dimensions except one, where it is free with rank $G^{d-1}$, where $d$ is the degree of your vertex. The quotient space which computes this relative homology is always a wedge of circles modulo the common point.
