In the paper Configuration-Spaces and Iterated Loop-Spaces. Graeme Segal, page 213-214, it is obtained that the labelled configuration space $C_n$ is homotopy equivalent to a topological monoid $C'_n$. Let a homotopy equivalent map be $$ \phi: C_n\to C'_n.$$
On the other hand, we have the path-connected components decomposition $$ C_n=\bigsqcup_{k\geq 0} F(\mathbb{R}^n,k)/\Sigma_k $$ as the disjoint union of $k$-th unordered configuration spaces. Hence $$ C'_n=\phi (C_n)=\bigsqcup_{k\geq 0}\phi(F(\mathbb{R}^n,k)/\Sigma_k)). $$ Question: Does the multiplication $\mu$ of $C'_n$ given in line 1-2, page 214 of Configuration-Spaces and Iterated Loop-Spaces. Graeme Segal satisfy $$ \mu: \phi(F(\mathbb{R}^n,k)/\Sigma_k))\times \phi(F(\mathbb{R}^n,l)/\Sigma_l))\to \phi(F(\mathbb{R}^n,{k+l})/\Sigma_{k+l}))? $$ How to verify it?
It's easier to think about this the other way around. By definition, $C'_n$ is the set of pairs $(c,t)$ where $c \in C_n$ is a configuration, $t \ge 0$ is a real number, and the configuration $c$ is in $(0,t) \times \mathbb{R}^{n-1}$. Then there is a homotopy equivalence $\psi : C'_n \to C_n$ that simply forgets the parameter $t$ (ie. $\psi(c,t) = c$). This is easily seen to be a homotopy equivalence (it's a fibration and the fiber is contractible).
Now the product of $C'_n$ is concatenation on configurations, and addition on the parameter. So in particular, yes, the concatenation of two configurations $c \in C_{n,k}$ and $\bar{c} \in C_{n,l}$ is in $C_{n, k+l}$, directly by definition. In other words, the concatenation of a configuration with $k$ points and a configuration with $l$ points has $(k+l)$ points. It follows that the product maps $\psi^{-1}(C_{n,k}) \times \psi^{-1}(C_{n,l})$ to $\psi^{-1}(C_{n,k+l})$.