Constructing a Closed 4-manifold

Given a trisection diagram $\boldsymbol{(}\bf{F_g},\boldsymbol{\alpha, \beta, \gamma)},$ we can uniquely construct a smooth, connected, closed 4-manifold as follows. We first begin by constructing 3-dimensional handlebodies, $\bf{H}_{\boldsymbol{\alpha}},$ $\bf{H}_{\boldsymbol{\beta}},$ $\bf{H}_{\boldsymbol{\gamma}}$ by attaching 3-dimensional 2-handles to $\bf{F_g}$ along the $\bf{g}$-tuples $\boldsymbol{\alpha},$ $\boldsymbol{\beta},$ and $\boldsymbol{\gamma},$ respectively. We then attach $\bf{H}_{\boldsymbol{\alpha}} \boldsymbol{\times} \bf{I}$ to $\bf{F_g}\boldsymbol{\times} \bf{D^2}$ along $\bf{F_g }\boldsymbol{\times [-\varepsilon, \varepsilon]}$ (where we view $\boldsymbol{[-\varepsilon, \varepsilon]}$ as a subset of $\boldsymbol{\partial}\bf{D^2}$), and likewise for $\bf{H}_{\boldsymbol{\beta}} \boldsymbol{\times} \bf{I},$ $\bf{H}_{\boldsymbol{\gamma}}\boldsymbol{\times} \bf{I}.$ After smoothing corners, this gives us a smooth 4-manifold with boundary. A schematic of this intermediate step is given below.

By definition, we require that any two of $\boldsymbol{\alpha, \beta, \gamma}$ give a Heegaard diagram for$\bf{\boldsymbol{\#}^g\boldsymbol{(}S^1\boldsymbol{\times} S^2\boldsymbol{)}}.$ Thus we have constructed a 4-manifold with three disjoint boundary components, each of which is diffeomorphic to $\bf{\boldsymbol{\#}^g\boldsymbol{(}S^1\boldsymbol{\times} S^2\boldsymbol{)}}.$ Finally, we make use of a theorem of Laudenbach and Poénaru which tells us that there is a unique way to fill in this boundary components. This associates to the given diagram $\boldsymbol{(}\bf{F_g}, \boldsymbol{\alpha, \beta, \gamma)}$ a unique, closed 4-manifold $\bf{X},$ as desired.