TITLE: Realizing braid group by homeomorphisms
QUESTION [18 upvotes]: Markovich and Saric proved the following remarkable theorem.  Let $S$ be a compact surface of genus at least $2$ and let $MCG(S)=\pi_0(Homeo^{+}(S))$ be the mapping class group of $S$.  There is then no right inverse to the natural projection $Homeo^{+}(S) \rightarrow MCG(S)$.  This should be contrasted with the solution to the Nielsen realization problem by Kerckhoff, who proved that any finite subgroup of $MCG(S)$ can be realized by homeomorphisms of the surface.
My question is whether this is known for braid groups.  Let me make this a little more precise.  Let $X_n$ be a $2$-dimensional disc with $n$ punctures.  By $Homeo^{+}(X_n)$, we mean homeomorphisms of $X_n$ that are the identity on the boundary and extend over the punctures (but are allowed to permute the punctures).  The group $\pi_0(Homeo^{+}(X_n))$ is then the $n$-strand braid group $B_n$.  My question is if it is known whether or not there is a right inverse to the natural projection $Homeo^{+}(X_n) \rightarrow B_n$.
Since $B_1=1$ and $B_2=\mathbb{Z}$, the first nontrivial case is $n=3$.
I would find it shocking if this is not known for $n=3$.  The braid group then has two generators $a$ and $b$ and only one relation $aba=bab$.  The problem is then asking whether or not we can find homeomorphisms $f,g \in Homeo^{+}(X_n)$ in the homotopy classes of $a$ and $b$ such that $fgf=gfg$.  Surely this cannot be open!

REPLY [8 votes]: Hot off the presses: a new preprint by Nick Salter and Bena Tshishiku proves that the braid groups cannot be realized by diffeomorphisms for $n \geq 5$.