The Morita data satisfy the bicategory coherence axioms

(The associator is a natural bimodule isomorphism.) For a $(D,C)$-bimodule $H$, a $(C,B)$-bimodule $G$ and a $(B,A)$-bimodule $F$, the map ${\alpha }_{H,G,F}:(H{\otimes }_{C}G){\otimes }_{B}F\to H{\otimes }_C(G{\otimes }_{B}F)$ of [F2] is a group isomorphism sending $(h\otimes g)\otimes f$ to $h\otimes (g\otimes f)$. It respects the outer actions by [F2], and both sides are $(D,A)$-bimodules with the induced actions of [F1], so ${\alpha }_{H,G,F}$ is an isomorphism of $(D,A)$-bimodules, natural in each variable by [F2].

(The unitors are natural bimodule isomorphisms.) For a $(B,A)$-bimodule $F$, [F3] applied to the left $B$-module $F$ gives ${\lambda }_F:B{\otimes }_{B}F\to F$, $b\otimes f\mapsto bf$, and applied to the right $A$-module $F$ gives ${\rho }_F:F{\otimes }_{A}A\to F$, $f\otimes a\mapsto fa$; both are group isomorphisms, respect the outer actions by [F3] and are natural in $F$. These are the unitors of the Morita data at the identity 1-cells ${}_B{B}_B$ and ${}_A{A}_A$.

(Horizontal composition is a functor.) For rings $A,B,C$ the assignment $c$ sending a pair $(G,F)\in \mathbf{B}\mathbf{i}\mathbf{m}\mathbf{o}\mathbf{d}(B,C)\times \mathbf{B}\mathbf{i}\mathbf{m}\mathbf{o}\mathbf{d}(A,B)$ to $G{\otimes }_{B}F\in \mathbf{B}\mathbf{i}\mathbf{m}\mathbf{o}\mathbf{d}(A,C)$ and a pair of bimodule maps $(g,f)$ to the bimodule map $g\otimes f$ is well defined on objects and morphisms by [F1] and [F4]. It preserves identities and composition by the two laws of [F4], and it is functorial in each variable by the same formulas, hence a functor on the product of hom-categories.

(Pentagon.) Let $K,H,G,F$ be a $(E,D)$-, $(D,C)$-, $(C,B)$- and $(B,A)$-bimodule. Both sides of the pentagon identity are maps $((K{\otimes }_{D}H){\otimes }_{C}G){\otimes }_{B}F\to K{\otimes }_D(H{\otimes }_C(G{\otimes }_{B}F))$ of $(E,A)$-bimodules built from the associators of step 1.1 and whiskered identities, hence additive and action-preserving. On an elementary tensor $((k\otimes h)\otimes g)\otimes f$ a direct computation shows that both paths perform the same rebracketing: the left-hand path gives first $(k\otimes h)\otimes (g\otimes f)$ and then $k\otimes (h\otimes (g\otimes f))$, while the right-hand path gives successively $(k\otimes (h\otimes g))\otimes f$, then $k\otimes ((h\otimes g)\otimes f)$, then the same element $k\otimes (h\otimes (g\otimes f))$. Since the domain is generated additively by elementary tensors by [F5], the two maps are equal.

(Triangle.) Let $G$ be a $(C,B)$-bimodule and $F$ a $(B,A)$-bimodule. Both sides of the triangle identity are maps $(G{\otimes }_{B}B){\otimes }_{B}F\to G{\otimes }_{B}F$; on an elementary tensor $(x\otimes b)\otimes y$ the composite $(1_G\ast {\lambda }_F)\circ {\alpha }_{G,{}_B{B}_B,F}$ sends it to $x\otimes (by)$, while ${\rho }_G\ast 1_F$ sends it to $(xb)\otimes y$, and these agree by the balancing relation $xb\otimes y=x\otimes by$ of [F5]. By [F5] the two maps agree on the whole tensor product.

(Assembly.) Step 1.3 gives the composition functors with their functoriality, steps 1.1 and 1.2 give the associator and the two unitors as natural bimodule isomorphisms with the correct variances, and steps 2.1 and 2.2 verify the pentagon and triangle identities, every coherence equation being checked on elementary tensors and extended by additivity. All maps are the canonical tensor isomorphisms, no element outside the given modules is selected, and no commutativity of rings is assumed.