Textbook: Lectures on the topology of 3-manifolds by Nikolai Saveliev
Theorem. Any closed orientable 3-manifold admits a Heegaard splitting.
Proof. Let $T$ be a triangulation of a closed orientable $3$-manifold $M$. We associate with $T$ a Heegaard splitting of $M$ as follows. Let us replace each vertex of $T$ by a ball, each edge by a cylinder, each side of a tetrahedron by a plate, and each tetrahedron by a ball. The union $H(T)$ of the vertex balls and the cylinders is a handlebody, and so is the union $H'(T)$ of the tetrahedron balls and plates. Therefore, $M = H(T)\cup H'(T)$ is a Heegaard splitting of the manifold $M$.
If I understand correctly, $H(T)$ and $H'(T)$ are like dual graphs in planar graphs. I convinced myself $H(T)$ and $H'(T)$ are handlebodies (up to homeomorphism) but I can't see why two handlebodies have the same genus. Could you explan why?