Why is even codimension necessary to apply excision for the Euler characteristic?

Question

In this answer on MathOverflow, it is claimed that $$\chi(X\setminus Z)=\chi(X)-\chi(Z)$$ holds for complex subvarieties $Z$ only because $Z$ has even codimension. It is implied that for $Z$ with odd codimension, this may fail. Why is this? Is this point discussed in some standard reference?

Answer 1

For a simple counterexample, take $X=S^1$ and $Z$ to be a point. Then $\chi(X\setminus Z)=1\neq \chi(X)-\chi(Z)=-1$. Morally, what's going wrong here is that you should consider $\chi(X\setminus Z)$ to be $-1$ rather than $1$, since it is made up of only a single (open) $1$-cell and no other cells. But if you take the standard homological definition of Euler characteristic, you get $1$, not $-1$.

So, why does $\chi(X\setminus Z)=\chi(X)-\chi(Z)$ if $Z$ has even codimension, at least in nice cases? Well, if $Z$ sits in $X$ nicely enough, then it has a tubular neighborhood $T\cong Z\times \mathbb{R}^d$ (where $d$ is even). Then Mayer-Vietoris tells us $\chi(X)=\chi(T)+\chi(X\setminus Z)-\chi(T\cap(X\setminus Z))$. We have $\chi(T)=\chi(Z)$ since $T$ deformation-retracts to $Z$ and $T\cap(X\setminus Z)\cong (\mathbb{R}^{d}\setminus\{0\})\times Z$ so $\chi(T\cap(X\setminus Z))=\chi((\mathbb{R}^{d}\setminus\{0\})\chi(Z)=0$ since $d$ is even (if $d$ were odd, the first factor would be $2$ instead of $0$). Thus $\chi(X)=\chi(Z)+\chi(X\setminus Z)$, which is what we wanted. In the odd codimension case, the problem is that the homological Euler characteristic of the open set $\chi(X\setminus Z)$ is "wrong" because it treats odd-dimensional open cells as if they could be contracted to a point (which makes them count as $1$ rather than $-1$ towards the Euler characteristic), as in the counterexample discussed above.