A kind of uniqueness for the double of a manifold Given two smooth, connected manifolds, M, N, with the same boundary. If their doubles, D(M) and D(N) are diffeomorphic, does it follow that M and N are diffeomorphic ? The condition on the boundary is to exclude Mazur manifolds..
 A: The answer is still no. I mis-read your question the first time.  Finding a counter-example to your actual question was a little more difficult.    I will produce a manifold that has two fundamentally distinct ways of being represented as a double. 
In unreduced Seifert-fibre notation (click the link), the manifold I'm interested in will be
$$M[0,0; 1/3, -1/3, 1/5, -1/5]$$
This is a compact orientable 3-manifold which Seifert-fibres over $S^2$ with $4$ singular fibres, described by the four non-zero numbers in the above notation.
We can decompose this as a double by choosing a curve in the base $S^2$ disjoint from the singular fibres -- as long as a singular fibre is always on the opposite side of the curve from the singular fibre with the opposite sign. 
So for example, this manifold is the double of:
$$M[0,1; 1/3, 1/5]$$
but it is also the double of
$$M[0,1; 1/3, -1/5]$$
Both of the above manifold have a single torus as boundary.  By the classification of Seifert-fibred manifolds they are not diffeomorphic. 
So I think you can probably come up with quite a few examples of this sort. 
A: Here's an example in dimension 4: let $M$ and $N$ be the 4-dimensional 2-handlebodies associated to $-1$-surgery along the trefoil knot and $+1$-surgery along the figure-eight knot respectively.
They have the same boundary, as the two surgeries are diffeomorphic. On the other hand, Corollary 5.1.6 in Gompf and Stipsicz's 4-manifolds and Kirby calculus tells us that $D(M)$ and $D(N)$ are both diffeomorphic to $\mathbb{CP}^2\#\overline{\mathbb{CP}}^2$.
However, $M$ and $N$ can't even be homeomorphic, since the intersection form on $M$ is negative-definite and the intersection form on $N$ is positive-definite.
