Let $X$ and $Y$ be connected CW complexes. By Kunneth's theorem we have a map of the form $H_i(X)\times H_j(Y) \rightarrow H_{i+j}(X\times Y)$. By Dold-Thom this translates to a map of homotopy groups of the following form (whenever $i,j>0$ since Dold-Thom is an isomorphism between the homotopy groups and reduced homology): $$\pi_i(SP(X))\times \pi_j(SP(Y))\rightarrow \pi_{i+j}(SP(X\times Y))$$
Here $SP$ denotes the infinite symmetric product. I was wondering whether there is an explicit map $SP(X)\wedge SP(Y)\rightarrow SP(X\wedge Y)$ such that passing to the homotopy groups the following map is the same as the Kunneth map?
$$\pi_i(SP(X))\times \pi_j(SP(Y))\rightarrow \pi_{i+j}(SP(X)\wedge SP(Y))\rightarrow \pi_{i+j}(SP(X\wedge Y)) \rightarrow \pi_{i+j}(SP(X\times Y))$$
Note that there is an obvious map $SP(X)\wedge SP(Y)\rightarrow SP(X\wedge Y)$ which seems should be the answer, I am looking for a proof. The obvious map is given by taking pairwise points and pairing them and considering them as a point in $X\times Y$. then looking at its image in $X\wedge Y$.