Is $Out(F_n)$ of type FP$_\infty$ Is $Out(F_n)$ of type $FP_\infty$?
$Out(F_n)$ acts on Outer space $X_n$ which is finite dimensional, locally finite and contractible. However the action is only proper (stabilizers of points correspond to groups of isometries of finite graphs), not free. Can this somehow be turned into $FP_\infty$?
 A: Yes, $Out(F_n)$ is of type $FP_\infty$.
Too see why, first, the group $Out(F_n)$ has a torsion free subgroup of finite index (a theorem of Serre), namely 
$$IA_n(\mathbb{Z}/3\mathbb{Z}) = \text{kernel} \bigl( Out(F_n) \to SL(n,\mathbb{Z}/3\mathbb{Z})\bigr)
$$
Second, $FP_\infty$ is invariant under passage to subgroups of finite index. So it suffices to prove that $IA_n(\mathbb{Z}/3\mathbb{Z})$ is $FP_\infty$.
Third, $Out(F_n)$ acts properly discontinuously on a contractible simplicial complex with finitely many orbits of cells and finite cell stabilizers, namely $K_n =$ the spine of outer space $X_n$ (see Vogtmann's survey).
Finally, the restricted action of $IA_n(\mathbb{Z}/3\mathbb{Z})$ on $K_n$ is not only properly discontinuous, and not only has finitely many orbits of cells, but its cell stabilizers are trivial (because it is torsion free and is the restriction of a properly discontinuous action with finite cell stabilizers). The action of $IA_n(\mathbb{Z}/3\mathbb{Z})$ on $K_n$ is therefore free. This is enough to conclude that $IA_n(\mathbb{Z}/3\mathbb{Z})$ is $FP_\infty$. 
