# 14. Excision and the homology of spheres

Roughly speaking, the homology functor $H_n$ measures the number of $(n+1)$-dimensional "holes" in a space. At the end of today's lecture we'll be able to prove this, but we can already see it in action with $H_1$.

Let $X$ denote the unit ball $B^2 \subset \mathbb{R}^2$. Then $X$ is contractible, so $H_n(X) = 0$ for all $n \ge 0$. Now let $Y$ denote the ball of radius $1/2$ in $ \mathbb{R}^2$. What happens to $H_1(X)$ when we gouge out a hole in $X$ (a.k.a remove $Y$)?

This is easy: $X \setminus Y$ is an annulus, and hence retracts onto $S^1$. Thus $H_1(X \setminus Y) \cong H_1(S^1) = \mathbb{Z}$ (apply the Hurewicz Theorem).

Thus when we carved out a hole in $X$, the homology functor $H_1$ noticed. We went from $H_1(X) = 0$ to $H_1(X \setminus Y) = \mathbb{Z}$, reflecting that $X \setminus Y$ has a two-dimensional hole sitting in the middle of it.

However, *relative* homology has a defect. Suppose we have topological spaces $X \subset Z$, and we lop out the *same *hole from both $X$ and $Z$. What happens? Again, we don't yet know how to prove this in general, but in dimension two we can see it directly. Take $Z$ to be the ball of radius two, and let $X$ be as before. Then since $H_1(X) = H_1(Z) =0$, from the long exact sequence of the pair $(Z,X)$, we also have $H_1(Z,X) = 0$.

Now suppose we scoop out $Y$ from both $Z$ and $X$. Then $Z \setminus Y$ is an annulus of radius $3/2$ and $X \setminus Y$ is an annulus of radius $1/2$. Both $Z \setminus X$ and $X \setminus Y$ retract onto $S^1$, and thus $H_1( Z \setminus Y, X \setminus Y) = H_1(S^1,S^1) = 0$. (For any space $A$, one always has $H_1(A,A) = 0$.)

Thus the homology *did not change*! This property: that relative homology is blind to casually mutilating a pair of topological spaces (provided we mutilate them both in the same way, as above), is known as *the excision axiom*. Its proof is the subject of today's lecture.

Comments and questions?