Are acyclic subcomplexes of finite contractible 2-complexes contractible? Let $Y$ be a contractible finite simplicial 2-complex.
Let $X$ be an acyclic subcomplex of $Y$ (i.e. $X$ connected, $H_1(X)=0$, $H_2(X)=0$).
Is $X$ contractible?  (Equivalently, is $\pi_1(X)$ trivial?)
I asked this question among others here before, but it remained unanswered.
Apparently this question is equivalent to the following one.
If $m < n$, $G = \langle a_1,\dotsc,a_m \mid r_1 = \dotsb = r_m =1 \rangle$ and $J=\langle a_1,\dotsc,a_n \mid r_1 = \dotsb = r_n =1 \rangle$ are two groups given by (balanced) presentations, the presentation of $G$ is a sub-presentation of the presentation of $J$, and $J$ is trivial, is $G$ trivial? (It can be shown that the abelianization of $G$ is trivial.)
 A: One can show the following: 

Theorem: If the group given by the subpresentation is hyperlinear (in particular if it sofic), then it must be trivial.

The argument is based on an old Theorem of Gerstenhaber-Rothaus. I can only give a sketch here: Denote the group given by the subpresentation by $G$. We want to show that $G$ embeds into the group given by the full presentation to get a contradiction. Now, as $G$ is hyperlinear, one can satisfy the group multiplication on a finite subset of $G$ by unitaries up to any desired precision in the normalized Hilbert-Schmidt norm. Equivalently, $G$ embeds into a metric ultraproduct of unitary groups. Now, if unitaries $g_{m+1},\dots,g_n$ can be found which satisfy the relations $r_{m+1},\dots,r_n$, then the embedding of $G$ can be extended to homomorphism into the ultraproduct defined on $g_1,\dots,g_n$. In particular, $G$ embeds into the group given by the full presentation. Now, the fact that $g_{m+1},\dots,g_n$ can be found relies on the solvability of the equations $r_{m+1},\dots,r_n$ and can be proved along the lines of Gerstemhaber-Rothaus using degree theory and Hopf's computation of the cohomology of unitary groups.
Gerstenhaber-Rothaus proved some generalized form of the Kervaire-Laudenbach Conjecture for residually finite groups in
M. Gerstenhaber and O.S. Rothaus, The solution of sets of equations in groups, Proc. Nat. Acad. Sci. U.S.A. 48 (1962), 1531–1533.
The above argument is essentially only the observation that their argument extends without any problems to hyperlinear groups.
A: According to Sergei Ivanov, On balanced presentations of the trivial group, Invent. Math. 165, 525--549 (2006), for $n=m+1$, this is a particular case of Kervaire–Laudenbach conjecture which is also open, and according to one of Klyachko's results, the negative answer in this special case would imply the negative answer to Whitehead's asphericity conjecture.
It seems however that nothing is known in general, not even equivalence with the Kervaire–Laudenbach conjecture.  
