Lemma 42.14.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $U \subset X$ be an open subscheme, and denote $i : Y = X \setminus U \to X$ as a reduced closed subscheme of $X$. For every $k \in \mathbf{Z}$ the sequence

$\xymatrix{ Z_ k(Y) \ar[r]^{i_*} & Z_ k(X) \ar[r]^{j^*} & Z_ k(U) \ar[r] & 0 }$

is an exact complex of abelian groups.

Proof. 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 $[Z \cap U]$ 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. $\square$

Above the second arrow in the diagram, it should be j^?42? instead of lower star.

Comment #2297 by Daniel on

Why cannot there be to different cycles $[Z_1]$ and $[Z_2]$ in $Z_k(X)$ both mapping to $[Z_1\cap U]=[Z_2\cap U]$??

Comment #2323 by on

@Daniel: If $Z_1$ and $Z_2$ are integral closed subschemes of $X$ which both meet $U$ and if $Z_1 \cap U = Z_2 \cap U$, then we have $Z_1 = Z_2$. One argument to see this is if $\eta_i \in Z_i$ is the generic point, then $Z_1 \cap U = Z_2 \cap U$ nonempty implies $\eta_1 = \eta_2$, hence $Z_1 = \overline{\{\eta_1\}} = \overline{\{\eta_2\}} = Z_2$.

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