TITLE: Topological Spin manifolds in dimension 4
QUESTION [9 upvotes]: In his ICM Adress at Nice (Proceedings of the International Congress of Mathematicians Nice, September, 1970, Gauthier-Villars, editeur, Paris 6 e ,1971, Volume 2, pp. 133-163.), 
Robion Kirby  adresses  the  problem, whether  the  fourth topological Spin bordism group is  $\mathbb{Z}$ or $\mathbb{Z}\oplus \mathbb{Z}/2$.
Depending on whether  the  triangulation obstruction 
$$\Delta:\Omega_{4}^{\rm Spin Top}\to \mathbb{Z}/2 $$
is  zero  or  not. This being  equivalent  to  the  fact  that  Rokhlin's  divisibility property  of the  signature still  holds  modulo 2, in the  sense  that the  signature  is  eight times the  triangulation  obstruction  modulo  two. 
This  should  be  an  example  of  a  topological  spin  manifold  which  is  not  bordant  to  a  smooth spin manifold. 
¿Is  this  problem  solved? Are  there  newer  references for  this  problem I  am  not  aware  of?

REPLY [8 votes]: The map $\Omega_4^{\text{Spin}} \to \Omega_4^{\text{SpinTop}}$ is taken isomorphically by the signature to the inclusion $16\Bbb Z \hookrightarrow 8\Bbb Z$, so that the groups are abstractly isomorphic but that the natural map is not an isomorphism. This is obtained as Theorem 13.1 on page 325 of Kirby and Siebenmann's (1977) foundational essays.
A representative for the nonzero element of $\Omega_4^{\text{SpinTop}}/\Omega_4^\text{Spin} = \Bbb Z/2$ is given by Freedman's E8 manifold: take the E8 plumbing, a smooth manifold with intersection form E8 and boundary the Poincare homology sphere $\Sigma(2,3,5)$, and cap the boundary off with a contractible topological manifold. The result is topologically spinnable because the manifold is simply connected and the intersection form is even.