Let $Z$ be a simply connected, two dimensional simplicial complex.


Let $X\subset Z$ be a finite subcomplex with nontrivial $\pi\_{1}$.


Must there exist a finite, simply connected subcomplex $Y\subset Z$ such that $Y\supset X$?


(Motivation: the fact that Whitehead conjecture remains unproven indicates that there are probably some very weird things that can happen in two dimensional complexes. This question attempts to locate some of these pathologies.)



Simply connected simplicial complexes
The answer is negative unless you assume that $Z$ is aspherical (and even in this case I am not sure). Take Whitehead manifold $W$, which is a contractible open 3-manifold, not homeomorphic to ${\mathbb R}^3$ and let $Z$ be the 2-dimensional skeleton of a triangulation of $W$. Then $Z$ is a counter-example.


Proof. 


1. I claim that $W$ does not admit an exhaustion by simply-connected compact submanifolds with boundary. Indeed, if such an exhaustion $W\_i$ exists, then each $W\_i$ is bounded by some 2-spheres, all but one of which bounds a 3-ball in $W$. Filling in these spheres by 3-balls in $W$, we obtain an exhaustion $W\_i'$ of $W$ by closed 3-balls. This would imply that $W$ is homeomorphic to the 3-space. Contradiction.
2. Your question is equivalent to asking if $Z$ (provided that it is locally finite) is exhaustable by finite simply-connected subcomplexes. Note that my $Z$ is locally finite. Suppose that such an exhaustion $Z\_i$ does exist, we can then enlarge each $Z\_i$ by adding to it 3-simplices in $W$ whose boundary is contained in $Z\_i$. This does not change $\pi\_1$, of course. The resulting subcomplexes $V\_i$ will exhaust $W$ and will be simply-connected. Taking $W\_i$ to be a regular neighborhood of $V\_i$ we obtain an exhaustion of $W$ by simply-connected compact submanifolds with boundary. This contradicts Part 1.


