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This turns out to be a remarkably interesting query. I shouldn't have a complete solution but right here is a start.

First, for any area $X$ I'll write $F_n(X)\subset X^n$ for the space distinct $n$-tuples. The question asks whether there exist $\Sigma_n$-equivariant maps $f:F_n(B^2)\to\Delta_n-1$ with the fixed point $b=(1/n,\dotsc,1/n)$ now not in the image. If there is such a map then we will push it away from $b$ to the boundary of $\Delta_n-1$, which is homeomorphic to $S^n-2$. On the other aspect, there is an obvious embedding $i:B\to\mathbbR^2$ and one can choose an embedding $j:\mathbbR^2\to B$ such that $ij$ and $ji$ are isotopic to the respective id maps; using this we see that $F_n(B^2)$ is equivariantly homotopy similar to the space $X=F_n(\mathbbR^2)$. This space is well-known: the following paper is one access point to the literature:

\bibMR1344842article author=Cohen, F. R., name=On configuration spaces, their homology, and Lie algebras, magazine=J. Pure Appl. Algebra, quantity=100, date=1995, number=1-3, pages=19--42, issn=0022-4049, overview=\MR1344842 (96d:55005), doi=10.1016/0022-4049(95)00054-Z,

In particular:

$\pi_1(X)$ is the pure braid staff $Br_n$ on $n$ strings. Moreover, the higher homotopy teams are trivial, so $X$ is the classifying house $BBr_n$.

$X$ has an equivariant deformation retract $X_0$ that is a finite simplicial complex of dimension $n-1$. This means that $H^k(X)=0$ for $okay>n-1$.

The cohomology of $X$ is utterly recognized, along side the action of $\Sigma_n$. In particular, the top workforce $H^n-1(X)$ is the module referred to as $\textLie(n)$ (or maybe the dual of that?). As a $\mathbbZ[\Sigma_n-1]$-module this is unfastened of rank one, however the $\Sigma_n$-action is tougher to describe. The usual description additionally implies that all Steenrod operations in $H^*(X_0;\mathbbZ/p)$ are trivial.

If we will show that there is no $\Sigma_n$-equivariant map from $X_0$ to $S^n-2$ then we will be done.

In the case $n=2$ we simply have $X_0=S^1$ and $S^n-2=S^0$ with $\Sigma_2$ appearing antipodally on each side: it is clear that there is no equivariant map, as required.

In the case $n=3$ we have $S^n-2=S^1=K(\mathbbZ,1)$, so the nonequivariant mapping set is $[X_0,S^1]=H^1(X_0)$, and one can check that this is simply $\mathbbZ^3$ with the motion given by means of permuting the coordinates and multiplying via the signature. The best fastened level for this action is 0, so any map $X_0\to S^1$ that is equivariant-up-to-homotopy is nonequivariantly homotopic to a constant map. Here the motion of $\Sigma_3$ on $S^1$ is generated via a mirrored image and a rotation thru

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\pi/3$, so there are not any mounted points. This implies that constant maps $X_0\to S^1$, even supposing equivariant-up-to-homotopy, cannot be equivariant on the nostril. I think that there are not any equivariant maps, and it will have to be imaginable to prove this by way of equivariant obstruction idea (ie operating up the skeleta of $X_0$) however I do not see the details these days.

For $n>3$ we nonetheless have an evident map $[X_0,S^n-2]\to H^n-2(X_0)$, but it don't need to be bijective. We can examine $S^n-2$ with the fibre of the map $Sq^2:Okay(\mathbbZ,n-2)\to K(\mathbbZ/2,n)$, recalling that $Sq^2$ acts trivially on $H^*(X_0)$, which will have to give an particular description of $[X_0,S^n-2]$. With somewhat of illustration theory we will have to be capable to calculate the crowd of equivariant-up-to-homotopy maps $X_0\to S^n-2$. We would then want some equivariant obstruction principle to toughen this to know whether there are any equivariant maps. Because $\Sigma_n$ acts freely on $X_0$ and $S^n-2$ is nonequivariantly $(n-3)$-connected and $X_0$ is $(n-1)$-dimensional, this obstruction idea will most effective contain the last two or 3 skeleta of $X_0$, so it should with a bit of luck be tractable.