Weighted Lagrangian ribbon bordisms

The boundary inclusion kernel consists of pairs with $j_0{x}_0+j_1{x}_1=0$. It is Lagrangian by [F1] for $-{\omega }_{{\mathrm{\Sigma }}_0}\oplus {\omega }_{{\mathrm{\Sigma }}_1}$. The linear map $(x_0,x_1)\mapsto (x_0,-x_1)$ preserves that form and sends this kernel to ${\mathrm{\Lambda }}_M=\{j_0{x}_0=j_1{x}_1\}$. Thus each individual bordism defines a Lagrangian relation.

Transposing a Lagrangian relation reverses the sign of its symplectic form and preserves its Lagrangian property. Hence [F2] shows that $A({\lambda }_0)$ and $B^t({\lambda }_2)$ are Lagrangian; the third space is the specified ${\lambda }_1$. The Wall form in [F3] therefore defines exactly the middle-surface integer in the Definition, without any transversality or injectivity assumption.

Let $M=M_2\circ M_1$. Choose an open cover by extending each piece a short distance into the other through its product collar. In the collar coordinate, pushing that short extension linearly back to the seam is a deformation retraction onto the original piece; their intersection is a surface times an interval and retracts onto the common surface. By [F5], Mayer–Vietoris therefore gives exactness of $H_1({\mathrm{\Sigma }}_1)\to H_1(M_1)\oplus H_1(M_2)\to H_1(M)$ at the middle term, with first arrow $y\mapsto (j_{1}y,-j_1^{\mathrm{\prime }}y)$. If an incoming class $x_0$ and outgoing class $x_2$ become equal in $H_1(M)$, their difference $(j_0{x}_0,-j_2{x}_2)$ lies in the image of that arrow. Hence some $y$ satisfies $j_0{x}_0=j_{1}y$ inside $M_1$ and $j_1^{\mathrm{\prime }}y=j_2{x}_2$ inside $M_2$. Conversely these two equalities imply equality in the union. Thus ${\mathrm{\Lambda }}_M=BA$ as actual relations. Disconnected surfaces cause no exception: the proof uses exactness at the $H_1$ direct sum and makes no injectivity assertion involving $H_0$.

For three composable bordisms, their underlying relations compose by step 2.2. Apply [F4] with the four specified object Lagrangians. Its identity $m(B,A)+m(C,BA)=m(C,B)+m(CB,A)$ says precisely that the total correction, and therefore the total integer weight, agrees in both bracketings. The underlying geometric gluings agree up to the collar-respecting diffeomorphism that merely changes the placement of the two seams. At each seam the boundary parametrizations identify the same oriented arc germs and their product ribbons. Thus the glued ribbon germ is already smooth and fixed; changing collar lengths neither chooses an interpolation nor inserts a twist. The same two product seams occur in both bracketings. Thus weighted composition is associative on the stated diffeomorphism classes.

An identity cylinder has diagonal relation, so either one-sided correction has two equal Lagrangians and vanishes by [F4]. Its zero weight therefore gives the two identity laws. Surface homology, inclusion relations and Wall forms split as direct sums under disjoint union; [F4] makes corrections and weights add. The empty surface has zero homology and zero correction. Disjoint-union associators and symmetry cylinders permute these direct-sum data and have zero correction whenever their transported Lagrangians agree. Consequently the ordinary geometric symmetric-monoidal identifications respect all weights, units and composition laws.

For mapping cylinders the relations are graphs of $f_1,f_2$. The three middle Lagrangians are $f_1{\lambda }_0,{\lambda }_1,f_2^{-1}{\lambda }_2$. Symplectic invariance in [F4] transports them by $f_2$ and gives exactly $\mu (f_2{f}_1{\lambda }_0,f_2{\lambda }_1,{\lambda }_2)$. This proves the declared mapping-cylinder convention and finishes the category construction independently of the later surgery-defined RT maps.