Etymology of transpose of morphisms in an adjunction Let $F: \mathbf{C} \to \mathbf{D}$ be left adjoint to $G : \mathbf{D} \to \mathbf{C}$, witnessed by the family of bijections between hom-sets, natural in objects $X, Y$: $$\operatorname{Hom}_{\mathbf{C}}(X,GY) \overset{\psi_{X,Y}}{\longrightarrow} \operatorname{Hom}_{\mathbf{D}}(FX,Y).$$
I've noticed that among some authors, the image $\psi_{X,Y}(f)$ of a map $X \overset{f}{\to} GY$ is called the transpose of $f$. Does this have anything to do with a categorical realization of transpositions in linear algebra? Is there a canonical motivating example for adjoint functors involving transposes of linear operators, involving (say) adjoint representations of Lie groups?
 A: The etymology of the word "adjunction" is the characterization of the adjoint, or transpose, of a linear map between inner product spaces by the formula $\langle Av,w\rangle=\langle v,A^T w\rangle$. I guess that the use of "transpose" to indicate passing a morphisms across adjunction comes out of this history. But this seems to be only a psychological, not a technical, explanation. In particular, it's difficult to realize transposes in linear algebra as adjunctions in category theory.
A: Let $V$ and $W$ be finite free $R$-modules. Since the transpose reverses multiplication of matrices, evaluating the transpose $f^T$ of a map $f: V \to W$ at an element $(w_1, \dots, w_n)$ of $W$ amounts to precomposing the functional $[w_1, \dots, w_n]$ by $f$. So $f^T$ is just the image of $f$ under $\operatorname{Hom}_{R\text{-}\mathbf{Mod}}(-,R),$ which we'll call $(-)^{T}$.
With a little thought and a not-so-little amount of verification and diagram-chasing, one can see that $(-)^{T}$ is right adjoint to its opposite functor $\big((-)^{T} \big)^{\operatorname{op}}$ as functors $$\left((-)^{T}\right)^{\operatorname{op}} : R\text{-}\mathbf{Mod} \leftrightarrows R\text{-}\mathbf{Mod}^{\operatorname{op}} : (-)^{T}.$$
In particular, this restricts on finite free $R$-modules to an isomorphism of categories.
