Let $X$ be a complex algebraic variety, possibly singular and/or non-compact. It is well known that if $X$ is smooth then its Euler characteristic is equal to its Euler characteristic with compact support: $\chi(X)=\chi\_c(X)$.


**Questions.** (1) Is the same equality true if $X$ is singular? (I think it is still true for singular curves.)


(2) Is there an elementary proof of this fact at least in the smooth case?


**UPDATE:** As mentioned by Reladenine Vakalwe below the proof in the smooth case follows immediately from the Poincare duality.



Euler characteristics with and without compact support of algebraic varieties
Let $e(-)$ denote ordinary Euler characteristic and $\chi\_c(-)$ the compactly supported version.


Complex varieties admit Whitney stratifications. In particular, each closed stratum in such a stratification is a strong deformation retract of a tubular neighborhood. It follows (Mayer-Vietoris) that if $Y\subseteq X$ is such a closed strata, then $e(X) = e(Y) + e(X-Y)$.


**Added later**: Let me try and elucidate the above claim a bit more. For intuition let me make an informal remark: one of the reasons Whitney stratifications are so nice is that if one thinks of strata as spaces glued (or perhaps "linked" is a better word here) together, then in a Whitney stratification the gluing happens using locally trivial fibre bundles.


Now back to the claim above. Let $T$ be the tubular neighborhood of $Y$. This is a fibre bundle over $Y$, with fibre homeomeorphic to the mapping cone $cone(L)$ of a compact space $L$ (the link). Removing $Y$ from this neighborhood amounts to removing the vertex of $cone(L)$ (the "zero section"). So one is reduced to showing that the Euler characteristic of $L$ is $0$. Now the Whitney stratification of $X$ yields a Whitney stratification of $L$ by odd (real) dimensional oriented manifolds. By induction one sees that such a space must have zero Euler characteristic (I think this is originally an observation of Sullivan; the intuition is the same: at each step of your stratification you are attaching a compact space with the above properties to an oriented odd dimensional manifold using locally trivial maps whose fibre satisfies the same properties). 


Now, if $X = \bigsqcup\_i X\_i$ is a Whitney stratification with each $X\_i$ smooth, then by Poincare duality, $e(X\_i) = \chi\_c(X\_i)$ for each $i$. Let $Y$ be a stratum of minimal dimension. So $Y$ is closed. By induction on the number of strata we may assume $e(X-Y) = \chi\_c(X-Y)$. Now use the observation of the previous paragraph and you are done.


**Added even later:** I might be getting a bit carried away here (so do point out if you have a counterexample), but I think the following souped up version is true. Let $f\colon X\to Y$ be a morphism between complex varieties. Let $K\_0(X)$ and $K\_0(Y)$ denote the Grothendieck groups of the bounded derived categories of algebraically constructible sheaves on $X$ and $Y$. Then we have two maps $[f\_\*], [f\_!]\colon K\_0(X) \to K\_0(Y)$. I want to say that $[f\_\*] = [f\_!]$. I think I have a sketch of a proof of this using resolution of singularities (although maybe using a variant of the stratification method here might do it too). Anyway, food for thought.


