I am struggling with the idea of local homology groups and would like to see an example of how to go about finding them in general.
I'm thinking of the most trivial case to apply the theory of local homology to try and understand how it can be applied to more complex topological spaces.
In the most trivial case, if I view a graph as a 1 dimensional delta complex, and take each vertex as a point $x \in X$, then defining the local homology as $H_n(X, X $ \ $ \{x\})$, how would I find the local homology of a graph as $x$ varies in $X$?
Thanks in advance for the help
Let $X$ be a graph. There are two types of points in $X$: the points $e$ interior to edges (I'll call them edge points) and the vertices $v$. Let's compute the local homology at each. To do this, we'll use the long exact sequence in homology: $$ \cdots\to H_{n+1}(X,A)\to H_n(A) \to H_n(X) \to H_n(X,A) \to H_n(A) \to \cdots $$ (The computations are made even easier by using the long exact sequence in reduced homology.)
Edge points. Let $e\in X$ be an edge point. By excision, the homology of the pair $(X,X-e)$ is the same as the homology of the pair $(U,U-e)$, where $U$ is any neighborhood of $e$. Taking $U$ to be the interior of the edge containing $e$, we find that $U$ is contractible and $U-e$ is homotopy equivalent to two points. Looking at the long exact sequence, we see immediately that $H_n(X,X-e)=0$ for $n\geq2$. For $n=1$, the long exact sequence looks like $$ H_1(U) \to H_1(U,U-e) \to H_0(U-e) \to H_0(U) \to H_0(U,U-e) \to 0 $$ Filling in the groups we know, this is $$ 0 \to H_1(U,U-e) \to \mathbb Z^2 \to \mathbb Z \to H_0(U,U-e) \to 0. $$ The map $H_0(U-e)\to H_0(U)$ is surjective (why?) and this forces $H_1(X,X-e)=\mathbb Z$ and $H_0(X,X-e)=0$.
Vertices. Let $v\in X$ be a vertex. Let $U$ denote the union of $v$ and the interiors of the edges incident to $v$. The space $U-v$ is homotopy equivalent to the discrete space with $d$ points, where $d=\deg v$ is the degree of $v$ (assume for simplicity that every vertex has finite degree.) The same argument as above, but with slightly different groups, shows that $H_n(X,X-v)$ equals $0$ for $n\neq1$ and equals $\mathbb Z^{d-1}$ for $n=1$.
I realized just now that, by subdividing $X$, we can think of any point of $X$ as a vertex; the edge points would then have degree $2$. With this convention, the local homology of $X$ is $$ H_n(X,X-x) = \begin{cases} \mathbb Z^{\deg x - 1} & n=1\\ 0 & n\neq1. \end{cases} $$
What does all of this mean?
For more practice with this concept, I recommend Exercise 28 of Chapter 2.1 of Hatcher's Algebraic Topology.