Excision theorem

Remarks

An equivalent form: if $A, B \subseteq X$ are subspaces whose interiors cover $X$, then the inclusion $(B, A \cap B) \to (X, A)$ induces isomorphisms on all $H_n$. The proof uses repeated barycentric subdivision to show that every singular chain is homologous to a sum of small simplices, each lying in the interior of $A$ or in $X \setminus Z$. The chains made of such small simplices compute the same homology as all chains, and on them the excision map is an isomorphism at chain level.

Example: if $A$ is a nonempty closed subspace that is a deformation retract of some open neighbourhood in $X$, then excision gives $H_n(X, A) \cong \tilde H_n(X / A)$, where $\tilde H$ is reduced homology. For $(D^n, S^{n-1})$ this gives $H_n(D^n, S^{n-1}) \cong \tilde H_n(S^n)$.

The closure hypothesis is needed. Take $X = [-1, 1]$, $A = [0, 1]$ and $Z = A$; the closure of $Z$ is not inside the interior $(0, 1]$ of $A$. Then $H_0(X \setminus Z, A \setminus Z) = H_0([-1, 0)) \cong \Z$, while $H_0(X, A) = 0$ because $X$ is path-connected and $A$ is nonempty.

Going deeper

For a student

Excision is one of the Eilenberg–Steenrod axioms, and it is the axiom that fails for homotopy groups. With the long exact sequence of a pair it gives the Mayer–Vietoris sequence and the suspension isomorphism $\tilde H_n(X) \cong \tilde H_{n+1}(\Sigma X)$. For manifolds it gives $H_n(M, M \setminus \set{x}) \cong H_n(\R^n, \R^n \setminus \set{0}) \cong \Z$, the starting point of orientations.