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Let $K_0$ be the Grothendieck group of complex algebraic varieties. This is the group generated by all complex algebraic varieties, subject to the relations:

(i) $[X]=[Y]$ if $X,Y$ are isomorphic,

(ii) $[X]= [Z]+[X-Z]$ if $Z \subset X$ is a closed subset.

(See the following notes: https://sites.math.northwestern.edu/~mpopa/571/chapter6.pdf)

Let $\chi_c(.)$ denote the Euler characteristic with compact support of a complex algebraic variety. It can be shown that it has the property that $\chi_c(X)= \chi_c(Z)+ \chi_c(X-Z)$. Hence, $\chi_c(.)$ gives a well-defined map $\phi: K_0 \to \mathbb{Z}$ taking a variety $V \mapsto \phi(V)= \chi_c(V)$.

Suppose that $C \subset X$ is a constructible set. In general, $C$ is not a variety. However an element $[C]\in K_0$ can still be defined as follows:

By definition of $C$ being constructible, we can write $C= V_1 \cup\dots\cup V_n$ where the $V_i$ are disjoint locally closed subvarieties of $X$. Let us now write $V_i = Y_i - Z_i$, with $Y_i, Z_i \subset X$ closed. Then $[C]$ can be viewed as an element in $K_0$ by setting $[C]:= \sum_i [Y_i]-[Z_i]$.

Question: is $\chi_c(C)= \phi([C])= \sum_i \chi_c(V_i)$?

Remark: if $C$ were a variety, then this would be true by definition the map $\phi$. However, $C$ is merely assumed here to be a constructible set.

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    $\begingroup$ It's obviously wrong as written, because your decomposition $C = V_1 \cup \ldots \cup V_n$ does not see points that are included multiple times. For example, you can add $V_{n+1} = \{p\}$ for any point $p \in C$, which doesn't change $C$ but it does change $[C]$. You should revise your definition. $\endgroup$ – R. van Dobben de Bruyn Dec 20 '18 at 22:38
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    $\begingroup$ @R.vanDobbendeBruyn Thank you for your comment. I added the additional assumption that the $V_i$ are assumed to be disjoint. $\endgroup$ – user142700 Dec 20 '18 at 22:47

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