Let $X$ be a topological space (a CW complex) and $\mathbb{P}^d$ here represents the complex projective space. Consider the space of base-point preserving maps from $X$ to $\text{Sym}^n(\mathbb{P}^d)$, denoted by $\text{Maps}(X, \text{Sym}^n(\mathbb{P}^d))$. Since there is a natural quotient map $(\mathbb{P}^d)^{\times n}\rightarrow \text{Sym}^n(\mathbb{P}^d)$. This induces a map of the form $\text{Maps}(X, (\mathbb{P}^d)^{\times n}) \rightarrow \text{Maps}(X, \text{Sym}^n(\mathbb{P}^d))$. Left side of this map is just $\text{Maps}(X, \mathbb{P}^d)^{\times n}$. So this map from $\text{Maps}(X, \mathbb{P}^d)^{\times n}$ to $\text{Maps}(X, \text{Sym}^n(\mathbb{P}^d))$ respects the action of the symmetric group so it factors through $\text{Sym}^n(\text{Maps}(X,\mathbb{P}^d))$. So now we have the following map $f_{n,d}:\text{Sym}^n(\text{Maps}(X,\mathbb{P}^d))\rightarrow \text{Maps}(X, \text{Sym}^n(\mathbb{P}^d))$. Now let $n$ and $d$ go to infinity and we denote the limiting map by $f$. So $f$ is the following maps: $$f: \text{Sym}^{\infty}(\text{Maps}(X,\mathbb{P}^{\infty}))\rightarrow \text{Maps}(X, \text{Sym}^{\infty}(\mathbb{P}^{\infty}))$$ Now note that $\mathbb{P}^{\infty}= \text{Sym}^{\infty}(\mathbb{P}^1)$ and by Dold-Thom this is $K(\mathbb{Z}, 2)$. Again by Dold-Thom $\text{Sym}^{\infty}(\mathbb{P}^{\infty})=\prod_{r\geq 1}K(\mathbb{Z}, 2r)$. Since the monoidal structure of $K(\mathbb{Z},2)\cong \text{Sym}^{\infty}(\mathbb{P}^1)$ induces a commutative $H$-space structure on $\text{Maps}(X, K(\mathbb{Z},2))$. The connected components of this $H$-space correspond to the elements in the group $H^2(X,\mathbb{Z} )$ so by focusing on the connected component of the identity (which is also a commutative $H$-space) we get the following map: $$f: \text{Sym}^{\infty}(\text{Maps}^0(X,\mathbb{P}^{\infty}))\rightarrow \text{Maps}(X, \text{Sym}^{\infty}(\mathbb{P}^{\infty}))$$
The superscript $0$ on the left indicates that it is the connected component of the zero.
Now let's look at the maps induced on the homotopy groups by $f$, denoted by $f_*$. We are considering the $i$-th homotopy group. On the left by Dold-Thom the $i$-th homotopy is $H_i(\text{Maps}^0(X, K(\mathbb{Z},2)))$. On the right we saw that the $i$-th homotopy should be the following: $$\bigoplus_{r\geq 1}H^{2r-i}(X, \mathbb{Z})$$ Now let's look at the map rationally (after tensoring with $\mathbb{Q}$) and using Milnor-Moore its rational homology is the universal enveloping algebra generated by its rational homotopy i.e. $H_i(\text{Maps}^0(X, K(\mathbb{Z},2)), \mathbb{Q})=U_{deg=i}(\bigoplus_{t}H^{2-t}(X, \mathbb{Q}))$$=U_{deg=i}(\mathbb{Q}\oplus H^1(X, \mathbb{Q})\oplus H^0(X, \mathbb{Q}))$.
The subscript $deg=i$ means the degree $i$ part of the universal enveloping algebra generated by rational homotopy groups. The $\mathbb{Q}$ has degree zero. The first cohomology group has degree one and the zero-th cohomology group has degree two.
Now somehow we have the following maps which is just $f_*\otimes \mathbb{Q}$ on the $i$-th homotopy group: $$U_{deg=i}(\mathbb{Q}\oplus H^1(X, \mathbb{Q})\oplus H^0(X, \mathbb{Q})) \rightarrow \bigoplus_{r\geq 1}H^{2r-i}(X, \mathbb{Q})$$
I was wondering what this map exactly is? What does the product structure of the universal enveloping algebra map to on the right side (is it cup product?)?