Let $M$ be a smooth, closed, connected, oriented 3-manifold and let $f: M \rightarrow \mathbb{R}$ be a self-indexing Morse function. Since $\frac{3}{2}$ is a regular value of $f$ it follows from Morse theory that there is a Heegaard splitting of $M$ given by $M = f^{-1}(\infty, \frac{3}{2}]\cup_{f^{-1}(\frac{3}{2})} f^{-1}[\frac{3}{2}, \infty)$.
When we glue the submanifolds $f^{-1}(\infty, \frac{3}{2}] \subset M$ and $f^{-1}[\frac{3}{2}, \infty) \subset M$ do we glue by the identity map of $f^{-1}(\frac{3}{2})$ or not?
Suppose that we know that the genus of this Heegaard splitting is $g$ and we identify $f^{-1}(\frac{3}{2})$ with $\#^g S^1 \times S^1$ i.e. the standard closed, orientable surface of genus $g$ and we identify $f^{-1}(\infty, \frac{3}{2}]$ with $\#^g S^1 \times D^2$ and $f^{-1}[\frac{3}{2},\infty)$ with $\#^g S^1 \times D^2$ i.e. the standard 3-manifolds with boundary $\#^g S^1 \times S^1$. Then there is a homeomorphism $M \cong (\#^g S^1 \times D^2) \cup_{\#^g S^1 \times S^1} (\#^g D^2 \times S^1)$.
Is union $(\#^g S^1 \times D^2) \cup_{\#^g S^1 \times S^1} (\#^g D^2 \times S^1)$ twisted, that is do we have to glue over a homeomorphism of $\#^g S^1 \times S^1$ and not just the identity map? Is the twisting simply determined the choice of identifications we made of $f^{-1}(\infty, \frac{3}{2}]$ with $(\#^g S^1 \times D^2)$ and $f^{-1}[\frac{3}{2},\infty)$ with $\#^g S^1 \times D^2$?