An old mathematical question

 Recently, I posted the following question

There is an old paper of mine ‘Homeomorphisms of compact 3-manifolds’ in Topology 1977 (accessible online). I was trying to extend Waldhausen theory to irreducible non-orientable 3-manifolds. So these contain two sided projective planes. The case when there are no such planes was done by W. Heil in his thesis. By decomposing further one obtains 3-manifolds with projective planes in the boundary but any two sided projective plane inside is parallel to the boundary. I was able to show that these were sufficiently large except in one case: when the boundary consists of exactly two projective planes. Now much more is known and it may be possible to handle this case.
Meanwhile, Jonathan Hillman sent me a preliminary draft in which he studies Poincaré complexes using Crisp’s thesis. One of the results he obtains is: If M is a closed irreducible manifold which has an element of finite order > 1 which has infinite centraliser then M is homotopy equivalent to projective plane cross a circle.
In the above, if N is an irreducible manifold with exactly two projective planes in the boundary, then we can glue the boundary components to obtain M as above and so the answer to the question is yes by Hillman’s result. Actually, it is a homeomrphism in Hillman’s result by an old theorem of G.R.Livesay.
Here is an argument which could have been done in 1977. Take M as above and let L be the cover corresponding to Z\cross Z_2. Take the Scott Core of L, which we may assume to have incompressible boundary. If L is not closed, the boundary has at least two projective planes. Take a map to the circle to split L along a projective plane. The result of the splitting is another compact 3- manifold K with at least four projective planes in the boundary and fundamental group Z_2. Now a famous theorem of D.B.A. Epstein from his thesis in 1960 asserts that any such K has to be projective plane times an interval. This contradiction shows that L is closed and thus equal to M itself. We are done.
If this is correct, I am still in the game but I find that I do not understand my 1977 paper. It is going to be hard work and I won’t even try.

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