Trying to understand Heegaard diagrams I have been looking through Rolfsen's "Knots and Links" and I have come across some questions that I am confused about regarding Heegaard diagrams.  Let $H_1$ and $H_2$ be genus $g$ handlebodies and let $\phi :\partial H_1 \to \partial H_2$ be a homeomorphism of there boundaries.  Then we can form the closed orientable 3-manifold $M = H_1 \cup_\phi H_2$.  
Often Heegaard diagrams are drawn that just display the images of the curves that are the centers of the $1$-handles on $H_1$ on $\partial H_2$ which Rolfsen calls the "characteristic curves".  I am confused about the following questions that seem foundational to the idea of Heegaard diagrams:
1) Why does it suffice to just specify the characteristic curves?  That is, given two maps $\phi_1$ and $\phi_2$ that both have the same characteristic curves, why is it that $H_1 \cup_{\phi_1} H_2 \cong H_1 \cup_{\phi_2} H_2$?  In fact Rolfsen asks for more: if there is a homeomorphism of $H_2$ that takes the characteristic curves of $\phi_1$ to the characteristic curves of $\phi_2$ (in any order), then $H_1 \cup_{\phi_1} H_2 \cong H_1 \cup_{\phi_2} H_2$.  
2) Give necessary and sufficient conditions for a collection of curves on the boundary of a handlebody to be the characteristic curves of a Heegaard diagram.  
Thank you!
 A: Once the characteristic curves are given on $\partial H_2$, the construction of the 3-manifold is automatic, by the following procedure:


*

*For each characteristic curve $c$, attach a 2-handle $D^2 \times [0,1]$ to $H_2$ by identifying $\partial D^2 \times [0,1]$ to a regular neighborhood of $c$. Doing this for each characteristic curve, let $N$ be the resulting 3-manifold with boundary.

*Each component of $\partial N$ is homeomorphic to a 2-sphere.

*Attach a 3-handle $D^3$ to each component of $\partial N$. The result is homeomorphic to the 3-manifold $H_1 \cup_\phi H_2$.


It follows if two Heegard diagrams have the same characteristic curves on $\partial H_2$, then both of those two 3-manifolds are homeomorphic to the construction above, and hence those two 3-manifolds are homeomorphic to each other. Similarly, if there is a homeomorphism of $H_2$ taking characteristic curve set #1 to characteristic curve set #2, then the homeomorphism extends over the 2-handles and then over the 3-handles, thus giving a homeomorphism betwen the two 3-manifolds.
As for your question 2), ask yourself: What is the homeomorphism type of the surface obtained from $\partial H_1$ by cutting open along the "centers of the 1-handles on $H_1$? Whatever that homeomorphism type is, it must be homeomorphic to the surface obtained from $\partial H_2$ by cutting open along the characteristic curves. So for a set of curves on $\partial H_2$ to be a valid set of characteristic curves, it is necessary and sufficient that their complement be a surface of that appropriate homeomorphism type. 
Added: To answer the question in the comment, let me make a few definitions. Given a handlebody $H$, a "meridian curve" $c \subset \partial H$ is a circle having the property that there is an embedding $D^2 \times [0,1] \to H_c \subset H$ intersecting $\partial H$ in $\partial D^2 \times [0,1]$ such that $\partial D^2 \times \frac{1}{2} \to c$. A collection of merician curves $c_1,...,c_g \subset \partial H$ forms what you call "centers of 1-handles" if $B = H - (H_{c_1} \cup \cdots \cup H_{c_g})$ is a 3-ball. So when you attach $H$ to another handlebody by matching the curves $c_1,...,c_g$ with characteristic curves on that other handlebody, its equivalent to doing the attachment in pieces: first attach $H_{c_1} \cup ... \cup H_{c_g}$ (which is a bunch of 2-handles from the point of iew of that other handlebody); then attach $B$ (which is a 3-handle).  
