For a well-based space $X$ denote by $C(\mathbb{R};X)$ the unordered configuration space of points on the real line with labels in $X$, and a point can vanish if its label reaches the basepoint. (Alternatively, you can think about the free $E\_1$-algebra over the based space $X$.) Note that $C(\mathbb{R};X)$ is filtered by subspaces $C\_{\le r}(\mathbb{R};X)$ which contain configurations of at most $r$ labelled points.


Now let $X$ and $Y$ be well-based and path-connected spaces. According to Segal, we have homotopy equivalences
$$C(\mathbb{R};X\times Y)\to \Omega\Sigma(X\times Y),$$
and we use that $\Sigma(X\times Y)$ splits up to homotopy into $\Sigma X\vee \Sigma Y\vee \Sigma(X\wedge Y)$. We also have a homotopy equivalence
$$C(\mathbb{R};X\vee Y\vee (X\wedge Y))\to \Omega \Sigma(X\vee Y\vee (X\wedge Y)).$$
Now my question is: Can we invert the homotopy equivalences in such a way that the resulting equivalence
$$C(\mathbb{R};X\vee Y\vee (X\wedge Y)) \to C(\mathbb{R};X\times Y)$$
is filtration-preserving? Or is there even an explicit, geometric description of such a map?



Is there a filtered splitting of product labelling spaces?
The answer to your first question is no. And this can be seen by homology considerations. Note that this equivalence induces an isomorphism of Hopf algebras
$$ H\_\*(C(\mathbb R; X \vee Y \vee (X\wedge Y)); \mathbb F) \simeq H\_\*(C(\mathbb R; X \times Y); \mathbb F)$$
for any field coefficients $\mathbb F$. Also, one knows that $H\_\*(C(\mathbb R; Z); \mathbb F)$ is isomorphic to the free tensor algebra generated by $\tilde H\_\*(Z;\mathbb F)$.


The isomorphism must send primitives to primitives.


Examples show that this will not preserve the filtration on homology induced by the filtration you are asking about. Here is a simple example: let $X=Y=S^2$ and $\mathbb F = \mathbb Q$. Let $x \in H\_2(X;\mathbb Q)$ and $y \in H\_2(Y;\mathbb Q)$ be generators. Then the filtration 1 primitive one might write as $x \bar{\times} y \in H\_4(X \wedge Y;\mathbb Q)$ must map to the primitive $x \times y + x \* y \in H\_4(X \times Y;\mathbb Q)$. But this element has filtration 2. (Here $\*$ is the multiplication induced by the H-space structure.)


(Of course, even more simply, filtration 1 on the one side certainly can't be mapped as an equivalence to filtration 1 on the other side!)


As to your second question, just add the three maps
$$X \times Y \rightarrow X \hookrightarrow C(\mathbb R; X \vee Y \vee (X\wedge Y)),$$
$$X \times Y \rightarrow Y \hookrightarrow C(\mathbb R; X \vee Y \vee (X\wedge Y)),$$
$$X \times Y \rightarrow X \wedge Y \hookrightarrow C(\mathbb R; X \vee Y \vee (X\wedge Y)),$$
and then extend to a map of $\mathbb E\_1$-algebras.


