TITLE: A $k$-component link defines a map $T^{k}\rightarrow \mathrm{Conf}_{k} S^{3}$. Does the homotopy type capture Milnor's invariants?
QUESTION [8 upvotes]: A $k$-component link defines a map $T^{k}\rightarrow \mathrm{Conf}_{k} S^{3}$. Does the homotopy  type of this map capture the Milnor invariants?
Some special cases:

$k=2$, no, it's null homologous, but you can look instead at the map $T^{2}\rightarrow \mathrm{Conf}_{2} R^{3}$, which captures  linking number.
$k=3$, Melvin et al. proved it does.

REPLY [8 votes]: It's been a while since I've thought about this but I think Koschorke answered much of your question back in 1997  "A generalization of Milnor's mu-invariants to higher-dimensional link maps" Topology 36 (1997), no 2. 301--324.  Scanning through the paper I see he recovers many of the mu invariants but not all.  He lists it as an open question (6.3) if the homotopy class of the map T^k --> C_k R^3 is a complete link homotopy invariant of the link.
Related: 
Brian Munson put these Koschorke "linking maps" into the context of the Goodwillie calculus in a recent arXiv paper.  I've wondered for a while if you could use these types of maps to create a direct construction of the Cohen-Wu correspondence between the homotopy groups of S^2 and their corresponding simplicial quotient object made from the brunnian braid groups.