If $\Sigma$ is a smooth complex curve in a smooth projective surface $X$, then we can consider the homology class represented by $\Sigma^{[n]} \subset X^{[n]}$. $\ \ $ Where, $X^{[n]}, \Sigma^{[n]}$ stand for the Hilbert scheme of $n$-points on $X$ and $\Sigma$, respectively.  Is it possible to construct a homomorphism $\Phi_n: \rm{H}_2(X) \rightarrow H_w(X^{[n]})$, such that $[\Sigma] \mapsto [ \Sigma^{[n]} ]$? 

$\ \ \ $   One has the following at ones disposal:  we have the obvious quotient map $X^n \rightarrow S^nX$ (where $S^nX$ is the symmetric product of $X$).  Now, if $\beta \in H_2(X)$, then we can consider the image of $B := \beta \times \cdots \times \beta$ in $H_{2n}(S^nX)$.  If $\beta $ can be represented by an algebraic curve, we can take the proper transform of $B$ under the Chow map $X^{[n]} \rightarrow S^nX$.  If $\beta$ is not represented by such a curve, is there anything akin to proper transform that one can apply to $B$ to construct the desired homomorphism $\Phi_n$?   

 I am interested in studying the intersection theory between the classes $\Phi_n(\beta)$.  Nakajima in his book "Lectures  on  Hilbert  schemes  of points on surfaces" states the following nice result.  If $\Sigma$ and $\Sigma'$ are two smooth curves in $X$, then (page 102 of Nakajima's book):

$$\sum_n z^n \ [\Sigma^{[n]}] \cdot [\Sigma'^{[n]}] = (1+z)^{[\Sigma] \cdot [\Sigma']}$$

Does anyone know if there are related results for singular curves?  

As a side remark. the above formula is obvious if $\Sigma$ and $\Sigma'$ are two curves intersecting transversely.  All it says is that of the set of $m = [\Sigma]\cdot [\Sigma']$ points were it intersects, we choose $n$ of them (there are $\binom{m}{n}$ of these guys, which is what the formula is giving).  But the general proof of the formula is more intricate - one uses a representation  of the Heisenberg group on the space $\oplus_n H_*(X^{[n]})$ to derive it. This fancy shmancy  approach is more helpful when computing things like the self intersection of $\Sigma^{[n]}$ when $\Sigma$ is a $(-1)$-curve in $X$.  From it we get that $[\Sigma^{[n]}] \cdot[\Sigma^{[n]}] =  \binom{-1}{n} = (-1)^n$