Relative tensor product gives associative composition of separable bimodules

We prove the projector construction directly under the present hypotheses, extending the construction of [F1]. For each composable pair $X:A\to B$, $Y:B\to D$, set $p=(r_X\otimes l_Y)(1_X\otimes e_B\otimes 1_Y)$ on $X\otimes Y$, where $e_B={\sigma }_B{\eta }_B$. Suppress coherent associators and write $e_B={\sum }_i{b}_i\otimes c_i$ only as notation for this morphism insertion. Centrality of $e_B$ gives $p(xb,y)=p(x,by)$. Any balanced $f$ satisfies $fp(x,y)={\sum }_{i}f(x{b}_i,c_{i}y)=f(x,({\sum }_i{b}_i{c}_i)y)=f(x,y)$ by $m_B{e}_B={\eta }_B$. These are equalities of action-map composites, independent of elements. Taking $f=p$ gives $p^2=p$. Split $p=iq$ through its image $P$ using [F3], with $qi=1_P$. The map $q$ is balanced because $i$ is monic and $p$ is balanced. Every balanced $f$ factors as $(fi)q=f$, uniquely since $q$ is split epic. Commutativity of the outer $A,D$ actions with the middle $B$ actions makes $p$ an outer bimodule endomorphism; the induced actions on its image satisfy their axioms after precomposition with the split epimorphism $q$ and its tensor translates. Thus $P=X{\otimes }_{B}Y$ is an $A$-$D$ bimodule with the required universal property, without rigidity, algebraic closedness or a simple tensor unit.

Let $X:A\to B$, $Y:B\to D$, and $Z:D\to E$. Both $(X{\otimes }_{B}Y){\otimes }_{D}Z$ and $X{\otimes }_B(Y{\otimes }_{D}Z)$ receive a map from $X\otimes Y\otimes Z$ that is balanced in the $B$- and $D$-slots. Conversely, a map from $X\otimes Y\otimes Z$ balanced in both slots factors uniquely through either parenthesization: first apply the universal property in step 1.1 to one slot, then to the other. Tensoring preserves its split epimorphic universal maps; the two balancing projectors commute because the two actions on $Y$ commute. Hence either iterated factorization retains balancing in the other slot. The two iterated products therefore represent the same functor of the target object.

The representing property in step 2.1 gives a unique isomorphism $a_{X,Y,Z}:(X{\otimes }_{B}Y){\otimes }_{D}Z⟶X{\otimes }_B(Y{\otimes }_{D}Z)$ preserving the universal three-balanced map; applying the same property in reverse gives its inverse. For four composable bimodules, both composites around the pentagon preserve the same universal four-balanced map. Uniqueness therefore makes the two composites equal, which is the pentagon equation.

The regular bimodule $A$ is a left and right unit. The action maps $A{\otimes }_{A}X\to X$ and $X{\otimes }_{A}A\to X$ are balanced. Their inverses are induced by the algebra unit ${\eta }_A$ and the relevant unitors. The action and unit laws show the two composites are identities; the triangle equation for these unitors follows because its two paths agree on the universal balanced map, so step 2.1 forces equality.

Bimodule maps $f:X\to X^{\mathrm{\prime }}$ and $g:Y\to Y^{\mathrm{\prime }}$ induce the unique map $f{\otimes }_{B}g$ preserving the universal balanced map. For composable triples of bimodule maps, either horizontal-then-vertical or vertical-then-horizontal composition has the same composite with that universal map, since each is the ordinary composite of $f,g$ on the tensor factors. Uniqueness from step 1.1 gives equality; this is interchange. Thus these data form the stated bicategory.