Graded bimodule maps classify shift-compatible transformations

In the specified word-coded category of [L1], $T_M$ and $T_{M^{\mathrm{\prime }}}$ are represented by their canonical tensor generators. The equivalence $\mathrm{\Phi }$ gives the displayed $k$-linear bijection $\eta \mapsto {\eta }_A$ onto ${\mathrm{Hom}}_{B\text{-}A}(M,M^{\mathrm{\prime }})$ with inverse $f\mapsto f\otimes 1$. Its morphisms code exactly the coherent transformations between these two tensor functors, so this proves the asserted fixed-pair bijection. The unit isomorphisms of [L5] are degree-zero $(B,A)$-bimodule isomorphisms, and the inverse is the one recorded in [L3].

Direct direction check: for coherent $\eta$ the component ${\eta }_A$ is a morphism of ${\mathrm{GrMod}}_0(B)$, hence degree-zero $B$-linear [L6, L2]; writing $f$ for the map corresponding to ${\eta }_A$ through the unit isomorphism, one has $f(m)\otimes 1_A={\eta }_A(m\otimes 1_A)$. The equivariance square of [L2] at $X=A$ with parameter $d$ reads ${\theta }_{A,d}^{T_{M^{\mathrm{\prime }}}}\circ {\eta }_{A\{d\}}=({\eta }_A\{d\})\circ {\theta }_{A,d}^{T_M}$; both $\theta$'s are the identity on elementary tensors [L2] and the shift of the morphism ${\eta }_A$ is the same underlying map [L7], so ${\eta }_{A\{d\}}$ and ${\eta }_A$ are the same function. Naturality of $\eta$ at the degree-zero map $r_a:A\{d\}\to A$, $x\mapsto xa$, reads ${\eta }_A\circ (1\otimes r_a)=(1\otimes r_a)\circ {\eta }_{A\{d\}}$; evaluating at $m\otimes 1_A\in M{\otimes }_{A}A\{d\}$ and using ${\eta }_{A\{d\}}(m\otimes 1_A)={\eta }_A(m\otimes 1_A)=f(m)\otimes 1_A$ together with $r_a(1_A)=a$ gives ${\eta }_A(m\otimes a)=f(m)\otimes a$ for homogeneous $a\in A_d$. Then $f(ma)\otimes 1_A={\eta }_A(ma\otimes 1_A)={\eta }_A(m\otimes a)=f(m)\otimes a=f(m)a\otimes 1_A$, the middle equality using the balancing relation $ma\otimes 1_A=m\otimes a$ and its analogue for $f(m)$, so $f(ma)=f(m)a$ because $m^{\mathrm{\prime }}\mapsto m^{\mathrm{\prime }}\otimes 1_A$ is injective, being the inverse of the unit isomorphism [L5]; hence $f$ is a degree-zero $(B,A)$-bimodule map [L4], and conversely every such $f$ produces $f\otimes 1$ by [L3].

For $A=B=k$ concentrated in degree zero the unit isomorphism $u:T_k\Rightarrow \mathrm{i}\mathrm{d}$ has components the degree-zero isomorphisms $k{\otimes }_{k}X\to X$, $c\otimes x\mapsto cx$ [L5], and it is coherent for the canonical comparisons: ${\theta }^{T_k}$ is the identity on elementary tensors [L2] while ${\theta }_{X,r}^{\mathrm{i}\mathrm{d}}=1_{X\{r\}}$, so ${\theta }_{X,r}^{\mathrm{i}\mathrm{d}}{u}_{X\{r\}}=(u_X\{r\}){\theta }_{X,r}^{T_k}$; conjugating by the coherent isomorphism $u$ gives a $k$-linear bijection between the coherent endomorphisms of $\mathrm{i}\mathrm{d}$ and those of $T_k$, and by step 1.1 the latter correspond bijectively to ${\mathrm{Hom}}_{k\text{-}k}(k,k)\cong k$, since a $(k,k)$-bimodule map $k\to k$ is multiplication by the scalar $f(1)$; hence the coherent endomorphisms of the identity functor correspond bijectively to $k$ via $\eta \mapsto {\eta }_k$, and no family larger than the scalars satisfies the equivariance square.

Collecting steps 1.1, 2.1 and 2.2: the map $\eta \mapsto {\eta }_A$ is a $k$-linear bijection onto the degree-zero $(B,A)$-bimodule maps with inverse $f\mapsto f\otimes 1$, every coherent transformation is determined by its component at $A$, that component is automatically degree-zero, $B$-linear and right $A$-linear, and for $A=B=k$ the coherent endomorphisms of the identity functor are exactly the scalars; no selection of elements or bases was made, so no choice is used.