# Is there a filtered splitting of product labelling spaces?

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?

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 $$\tilde H_*(Z;\mathbb F)$$.
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.)
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.