Natural transformations between tensor functors are bimodule maps

Define $f:={\rho }_{M^{\mathrm{\prime }}}\circ {\eta }_A\circ {\rho }_M^{-1}$, so $f(m)={\rho }_{M^{\mathrm{\prime }}}({\eta }_A(m\otimes 1))$ by [F1]. Then $f$ is $B$-linear, as a composite of the $B$-linear maps ${\rho }_M^{-1}$, ${\eta }_A$ (a morphism in $B\text{-}\mathbf{M}\mathbf{o}\mathbf{d}$ by [F2] and [F4]) and ${\rho }_{M^{\mathrm{\prime }}}$, which respect the outer structures by [F1]. Moreover $f(ma)=f(m)a$ for all $a\in A$: naturality at the left $A$-linear map $r_a:A\to A$, $r_a(x)=xa$, reads ${\eta }_A\circ (1_M\otimes r_a)=(1_{M^{\mathrm{\prime }}}\otimes r_a)\circ {\eta }_A$, and $ma\otimes 1=(1_M\otimes r_a)(m\otimes 1)$; evaluating there and using that ${\rho }_{M^{\mathrm{\prime }}}$ is $A$-linear, so that ${\rho }_{M^{\mathrm{\prime }}}((1_{M^{\mathrm{\prime }}}\otimes r_a)(y))={\rho }_{M^{\mathrm{\prime }}}(y)a$, gives $f(ma)={\rho }_{M^{\mathrm{\prime }}}({\eta }_A(m\otimes 1))a=f(m)a$. By [F3] the map $f$ is a $(B,A)$-bimodule map.

Conversely, let $f:M\to M^{\mathrm{\prime }}$ be a $(B,A)$-bimodule map and put ${\eta }_X:=f\otimes 1_X:M{\otimes }_{A}X\to M^{\mathrm{\prime }}{\otimes }_{A}X$ by [F5]. Each ${\eta }_X$ is $B$-linear, since ${\eta }_X(b(m\otimes x))=f(bm)\otimes x=b(f(m)\otimes x)$, and the family is natural: for $u:X\to Y$ functoriality in [F5] gives $(f\otimes 1_Y)\circ (1_M\otimes u)=f\otimes u=(1_{M^{\mathrm{\prime }}}\otimes u)\circ (f\otimes 1_X)$.

Let $\eta$ be natural with associated $f$ from step 1.1. Naturality at ${\mathrm{\ell }}_x:A\to X$, ${\mathrm{\ell }}_x(a)=ax$, gives ${\eta }_X\circ (1_M\otimes {\mathrm{\ell }}_x)=(1_{M^{\mathrm{\prime }}}\otimes {\mathrm{\ell }}_x)\circ {\eta }_A$; evaluated at $m\otimes 1$ the left side is ${\eta }_X(m\otimes x)$, while the right side is $(1_{M^{\mathrm{\prime }}}\otimes {\mathrm{\ell }}_x)(f(m)\otimes 1)=f(m)\otimes x$, using $f(m)={\rho }_{M^{\mathrm{\prime }}}({\eta }_A(m\otimes 1))$ and ${\rho }_{M^{\mathrm{\prime }}}^{-1}(f(m))=f(m)\otimes 1$ from [F1]. Both ${\eta }_X$ and $f\otimes 1_X$ are homomorphisms agreeing on every elementary tensor, so ${\eta }_X=f\otimes 1_X$; in particular $\eta$ is determined by $f$.

The assignments are inverse: starting from a bimodule map $f$, the transformation of step 1.2 has associated map $f^{\mathrm{\prime }}={\rho }_{M^{\mathrm{\prime }}}\circ (f\otimes 1_A)\circ {\rho }_M^{-1}$, and $f^{\mathrm{\prime }}(m)={\rho }_{M^{\mathrm{\prime }}}(f(m)\otimes 1)=f(m)$ by [F1] and [F5]; starting from $\eta$, its associated $f$ satisfies $f\otimes 1_X={\eta }_X$ for all $X$ by step 2.1. Hence $\eta \mapsto f$ is a bijection onto the set of $(B,A)$-bimodule maps.

Compatibility: sums of natural transformations, defined componentwise, are natural, and $f_{\eta +{\eta }^{\mathrm{\prime }}}=f_{\eta }+f_{{\eta }^{\mathrm{\prime }}}$ because ${\rho }_{M^{\mathrm{\prime }}}$ and $\eta \mapsto {\eta }_A$ are additive; conversely sums of bimodule maps are bimodule maps and $(f+f^{\mathrm{\prime }})\otimes 1_X=f\otimes 1_X+f^{\mathrm{\prime }}\otimes 1_X$ by [F5] and agreement on elementary tensors. The identity $1_{T_M}$ corresponds to $1_M$ in both directions, since ${\rho }_M(m\otimes 1)=m$ and $1_M\otimes 1_X=\mathrm{id}$ by [F5]. Vertical composition corresponds to composition: by [F6] and step 2.1, for $\xi :T_{M^{\mathrm{\prime }}}\Rightarrow T_{M^{\mathrm{\prime }\mathrm{\prime }}}$ with associated $g$ one has $(\xi \circ \eta {)}_A(m\otimes 1)={\xi }_A(f(m)\otimes 1)=g(f(m))\otimes 1$, so $\xi \circ \eta$ is associated with $g\circ f$, while $(g\otimes 1_X)\circ (f\otimes 1_X)=(g\circ f)\otimes 1_X$ by [F5]; by [F7] these operations are the additions and compositions on the two Hom-groups.

Steps 1.1-3.1 establish the bijection $\mathrm{Nat}(T_M,T_{M^{\mathrm{\prime }}})\cong {\mathrm{Hom}}_{B\text{-}A}(M,M^{\mathrm{\prime }})$ with ${\eta }_X=f\otimes 1_X$, and step 4.1 shows it is compatible with addition, identities and vertical composition. Nothing was chosen, and no commutativity was used.