Suppose that $f: M^{2n+1} \rightarrow \mathbb{R}$ is a self-indexing Morse function on a closed manifold (i.e. if $p$ is a critical point of index $k$, then $f(p) = k$). Then if $M$ is cut in half along the regular level set $f^{-1}(n+\frac{1}{2})$, are the two resulting submanifolds $M_0 = f^{-1}(\le n+\frac{1}{2})$ and $M_1 = f^{-1}(\ge n+\frac{1}{2})$ diffeomorphic?
More abstractly, suppose that $M_0, M_1$ are two $2n+1-$dimensional smooth, compact manifolds with the same boundary and admit proper Morse functions $f_i: M_i \rightarrow \mathbb{R}$ with all critical points of index at most $n$. Then I think that $M_0, M_1$ have the same homology since $H_k(M_0; \mathbb{Z}) \cong H_k(\partial M_0; \mathbb{Z})$ for $k < n$ by the LES on homology and $M_0, M_1$ have same boundary; furthermore, $H_k(M_0; \mathbb{Z}) = 0$ for $k > n$ since it doesn't have any index $k > n$ critical points.
Question Are $M_0, M_1$ diffeomorphic?
Remark: This is false if $M_0, M_1$ are $2n+1$-dimensional are allowed to have $n+1$ critical points. For example, $S^n \times S^{n+1} \backslash B^{2n+1}$ and $B^{2n+1}$ have the same boundary $S^{2n}$.