Lemma 82.10.2. In Situation 82.2.1 let $X/B$ be good. Let $U \subset X$ be an open subspace. Let $Y$ be the reduced closed subspace of $X$ with $|Y| = |X| \setminus |U|$ and denote $i : Y \to X$ the inclusion morphism. For every $k \in \mathbf{Z}$ the sequence
is an exact complex of abelian groups.
Proof. Surjectivity of $j^*$ we saw above. First assume $X$ is quasi-compact. Then $Z_ k(X)$ is a free $\mathbf{Z}$-module with basis given by the elements $[Z]$ where $Z \subset X$ is integral closed of $\delta $-dimension $k$. Such a basis element maps either to the basis element $[Z \cap U]$ of $Z_ k(U)$ or to zero if $Z \subset Y$. Hence the lemma is clear in this case. The general case is similar and the proof is omitted. $\square$
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