Statement If U\subseteq A \subseteq X are as above, we say that U can be
excised if the inclusion map of the pair (X \setminus U,A \setminus U ) into (X, A) induces an isomorphism on the relative homologies: The theorem states that if the
closure of U is contained in the
interior of A, then U can be excised. Often, subspaces that do not satisfy this containment criterion still can be excised—it suffices to be able to find a
deformation retract of the subspaces onto subspaces that do satisfy it.
Proof sketch The proof of the excision theorem is quite intuitive, though the details are rather involved. The idea is to subdivide the simplices in a relative cycle in (X, A) to get another chain consisting of "smaller" simplices (this can be done using
barycentric subdivision), and continuing the process until each simplex in the chain lies entirely in the interior of A or the interior of X \setminus U. Since these form an open cover for X and simplices are
compact, we can eventually do this in a finite number of steps. This process leaves the original homology class of the chain unchanged (this says the subdivision operator is
chain homotopic to the identity map on homology). In the relative homology H_n(X, A), then, this says all the terms contained entirely in the interior of U can be dropped without affecting the homology class of the cycle. This allows us to show that the inclusion map is an isomorphism, as each relative cycle is equivalent to one that avoids U entirely. == Applications ==