Associativity of Brauer and partition diagram composition

Regard the three diagrams as disjoint finite graphs (for Brauer) or disjoint finite vertex sets carrying partitions (for partition diagrams). Glue both middle interfaces at once, and let $\sim$ be the least equivalence relation generated by all diagram edges or blocks and both sets of interface identifications. The relation $\sim$ is independent of the order in which the two interfaces are glued, since the least equivalence relation generated by a fixed set of pairs is unique. Thus both bracketings retain exactly the same equivalence classes meeting the outer boundary.

The classes meeting no outer vertex are also the same for both bracketings. Any class removed when the first two diagrams are composed meets neither of their exposed boundaries, so it contains no vertex of the third diagram; it can never merge with another class in the second composition. Conversely, every final class meeting no outer vertex is either removed at the first composition or is removed at the second. Hence the sum of the two loop counts in either bracketing equals the number of classes of $\sim$ meeting no outer vertex.

By [F1], each bracketing therefore has the same outer diagram and coefficient ${\delta }^{\mathrm{\ell }}$, with the same integer $\mathrm{\ell }$. The two bracketings agree on basis diagrams, so bilinearity gives associativity on the free modules. The identity diagrams add no new equivalence identifications and remove no class; they act as left and right units.