Let $X$ be a projective variety over $\mathbb{C}$. Is there a way to define some number $\tilde{\chi}(X)\in \mathbb{Z}$ satisfying both of the following two properties?



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> $\boldsymbol{(1)} \;$ When $X$ is smooth, $\tilde{\chi}(X)=\chi(X)$, where $\chi(X)$ is the usual Euler number of $X$ (as a topological space).
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> $\boldsymbol{(2)} \;$ $\tilde{\chi}(X)$ is invariant under deformation, i.e. if we have a family of variety over a connected base $B$, then any two fibers have the same $\tilde{\chi}$.
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If I consider an affine variety (still over $\mathbb{C}$) instead of a projective variety, then what is the answer to the corresponding question?



Topological Euler number of a singular variety
The following example shows that the answer to abx's down-to-earth question 



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> Is the Euler number of the general fiber independent of the smoothing?
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is in general *no*.


Take $X \subset \mathbb{P}^5$, the cone over a rational normal curve $C\_4 \subset \mathbb{P}^4$. It is well-known that $X$ admits two differents smoothings: a $\mathbb{Q}$-Gorenstein smoothing whose general fibre isomorphic to $\mathbb{P}^2$ and a non-$\mathbb{Q}$-Gorenstein smoothing whose general fibre is isomorophic to $\mathbb{P}^1 \times \mathbb{P}^1$, see
for instance [this](https://mathoverflow.net/questions/244562/some-examples-of-mathbb-q-gorenstein-smoothing?rq=1) MathOverflow question and the corresponding answers.


In the first case, the topological Euler number of the smooth fibre is $3$, whereas in the second case it is $4$. 


So, at least for varieties that are not $\mathbb{Q}$-Gorenstein, the topological Euler number of the general fibre actually *depends* on the smoothing we are choosing. 


