Let $A, B$ be $k$-algebras, and let $M, M'$ be $A-B$-bimodules.
There is a bijection $Nat(Hom_A(M, −), Hom_A(M', −)) \cong Hom_{A⊗_kB^{op}} (M', M)$ sending a natural transformation $η : Hom_A(M, −) \to Hom_A(M', −)$ to $η(M)(Id_M)$.
In the proof, the author states that One needs to make use of the naturality of $η$ with respect to the right $B$-module structures on $M, M'$ to show that the map $η(M)(Id_M)$ is a bimodule homomorphism. This confused me as I cannot see the connection between naturality and being a bimodule homomorphism.
Any hints would be appreciated!
The key implicit information here is to consider both functors $\hom_A(M,-)$ and $\hom_A(M',-)$ as ${}_A{\bf Mod}\to {}_B{\bf Mod}$ (instead of the codomain being ${\bf Set}$ or ${}_k{\bf Vct}$), and thus assume that each component of $\eta$ is actually a $B$-module morphism.
If $M$ is an $A$-$B$-bimodule, we naturally get a left $B$-module structure on $\hom_A(M,N)$ for any ${}_AN$:
specifically we define $b\cdot f:=x\mapsto f(xb)$, i.e. we precompose $f$ by the map $(\cdot b):\underset{x\,\mapsto\, xb}{M\to M}\,$.
Now name $\,\varphi:=\eta_M({\rm id}_M)\in\hom_A(M',M)$. We need to show that it's also a right $B$-module morphism.
On the one hand, use the above assumption: $$\eta_M((\cdot b))\ =\ \eta_M(b\cdot {\rm id}_M)\ =\ b\cdot\eta_M({\rm id}_M)\ =\ b\cdot \varphi\ =\ m'\mapsto\varphi(m'b)$$ On the other hand, indeed use naturality of $\eta$ along the $A$-module morphism $(\cdot b):M\to M$, here we have postcomposition by it: $$\matrix{ \hom_A(M,M) & \overset{\eta_M}\longrightarrow & \hom_A(M',M) \\ {\scriptstyle (\cdot b)\circ}\Big\downarrow\phantom{\scriptstyle (\cdot b)\circ} && \phantom{\scriptstyle (\cdot b)\circ}\Big\downarrow {\scriptstyle (\cdot b)\circ} \\ \hom_A(M,M) & \underset{\eta_M}\longrightarrow & \hom_A(M',M) }$$ to obtain $\eta_M((\cdot b))\ =\ m'\mapsto \varphi(m')b$.