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 function $\Phi_n: \rm{H}_2(X) \rightarrow H_w(X^{[n]})$, 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 function $\Phi_n$?

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 $[\Sigma^{[n]}] \cdot[\Sigma^{[n]}] = \binom{-1}{n} = (-1)^n$

EDITED: In view of Nakajima's comment below, please replace function for homomorphism when reading the above question. Notice that, as stated in my comment below, the extension of the map $[\Sigma] \rightarrow [\Sigma^{[n]}]$ should be a "nice" one.