Let $f:X \to Y$ be a proper morphism between smooth algebraic varieties (say over $\mathbb{C}$), let me write $A\_X$ for the constant sheaf on $X$ with coefficients of the appropriate type. Then the decomposition theorem of [Beilinson-Bernstein-Deligne] tells me that $f\_\* A\_X$ splits as a direct sum of shifted semisimple perverse sheaves. 


So suppose I have collected some of the summands, say into $\mathcal{F}$, and want to know if I have found everything. Invoking proper base change, it is enough to ask, for every point $y \in Y$, whether 


$\dim H^\*(y,\mathcal{F}\_y) = \dim H^\*(X\_y)$, 


since any missing summands would have to contribute something somewhere. But computing the RHS requires actually thinking about the topology of $X\_y$. Something which is easier to compute are the virtual Betti numbers, defined for an arbitrary variety in terms of the weight filtration as 


$b^j\_{vir}(Z) = \sum\_i (-1)^{i+j} \dim Gr^j\_W H^i\_c(Z)$ 


At first glance these may not look easier, but the point is that they are motivic, i.e., additive under cutting into constructible pieces (and multiplicative under Zariski locally trivial fibrations); evidently they agree with the ordinary Betti numbers for smooth proper varieties. Thus for instance since a rational curve with a single node can be reassembled into $\mathbb{A}^1$, it has a single nonvanishing virtual Betti number, $b^2\_{vir} = 1$.



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> > To determine whether $f\_\* A\_X = \mathcal{F}$, the latter known to be a summand of the former, is it enough to compare virtual Betti numbers? 
> >  To be precise, does it suffice to know, for every $y \in Y$ and all $j$, that $b^j\_{vir}(X\_y) = \sum\_i (-1)^{i+j} \dim Gr^j\_W H^i(y,\mathcal{F}\_y)$?
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I sketch an argument in the affirmative which unfortunately due to my lack of knowledge about what weights are, I cannot seem to make precise. It is enough to show that for any nonzero summand $G$ of $f\_\* A\_X$, there is some $y \in Y$ where not all the virtual Betti numbers of $G\_y$ vanish. Let me take $y$ which has a neighborhood where $G$ is a sum of shifted local systems supported on some smooth subvariety $Y'$, and here it seems to me that restricting to $y$ must act as some global shift on the weights. Since $G$ was pure to begin with, being a summand of $f\_\* A\_X$, so is $G\_y$ (up to a shift) and hence its virtual Betti numbers are its actual Betti numbers, which see $G$. 


I have put the "reference request" tag in the hopes that if the answer to the question is yes, it is some well known thing, maybe somewhere in [BBD], to which someone can point me.



Decomposition theorem and virtual Poincare polynomial
(I'll just note beforehand that is this is a "I don't know the answer to your actual question, perhaps you'd rather know the answer to a slightly different one" answer). The answer is yes if you replace virtual Betti number with Frobenius trace; these are also motivic, if admittedly slightly harder to calculate, as they involve counting points. This is covered, for example, in Section 3.4 of [my second paper](https://arxiv.org/abs/0905.0486) with Geordie Williamson.


