Intuition for excision I am studying Algebraic Topology from Hatcher. In Chapter 2 Hatcher has a section on Excision which states that for triples $(X,U,A)$ where $A\subset U\subset X$ and $A\subset X$ is closed and $U\subset X$ is open we have that $H_n(X,U)\cong H_n(X\setminus A, U\setminus A)$.
I would like to know if there is any geometric intuition for the above. I have heard that $H_n(X,U)$ could be thought of as cycles away from the open set $U$, and since $A$ is contained in $U$, we might as well excise $A$ from $X$ and get the same result. Is there a better interpretation?
 A: I like to think of it this way.
First, rewrite the formula as $H_n(A\cup B,A) \approx H_n(B,A\cap B)$, which holds when $X = A\cup B = \text{int}(A)\cup\text{int}(B)$. (To relate this to Hatcher's form, let $B = X-Z$.)
The intuition is that stuff in $A$ gets ignored in $H_n(X,A)$, so it doesn't matter if we cut part of $A$ out. This works well for chains composed of "small" simplicies, which all belong either to $A$ or to $B$. Notation: $C_n(A+B)$ consists of chains made up of such simplicies. The diagram below shows a 
bunch of "small" simplicies.

But if a chain contains "large" simplicies, which don't lie entirely in $A$ or $B$, then we have a problem. Solution: use barycentric subdivision
to replace the "large" simplicies with "small" simplicies:

It may be necessary to repeat barycentric subdivision several times to obtain "small enough" vertices, as illustrated in this diagram:

The formal proof uses a chain homotopy, showing that $C_n(A+B)$ gives the same homology as $C_n(A\cup B)$.
A: In good cases, $H_*(X,A)$ is $\bar H_*(X/A)$, the (reduced) homology of the space obtained from X by contracting A to a point, so it does not really matter what's inside A.
Of course, this is not true always and so on, but it gives a very reasonable explanation of what relative homology is and why excision works.
