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 $H_1(A,A) = 0$.)

Homology didn't notice Y disappearing 😢

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?