How to understand in context of category theory: combination of contravariant functor and natural transformation

Question

I have a category $\mathcal{C}$, a contravariant functor $D:\mathcal{C}\to\mathcal{C}$ (so that $\text{id}_{\mathcal{C}}$ and $D^2$ are covariant), and a natural transformation $\chi:\text{id}_{\mathcal{C}}\to D^2$. For each object $M$ of $\mathcal{C}$, I have shown that $D(\chi_M)\circ\chi_{DM}=\text{id}_{DM}$.

My question is this: how can the above identity be understood in the general context of category theory?

In other words, is there a name for this?

Extra context which may not be relevant: The category $\mathcal{C}$ in question is the category of modules over an arbitrary complete discrete valuation ring $A$. $D$ is the contravariant Hom functor $\text{Hom}(-,K/I_1)$ where $K$ is field of fractions of $A$ and $I_1$ is the unique maximal ideal of $A$, $\chi_M$ is the evaluation $m\mapsto(\alpha\mapsto\alpha(m)):M\to D^2M$.

Answer 1

This is a particular case of a functor being "adjoint to itself on the right", in Peter Freyd's terminology. Oswald Wyler calls this particular case being "self-adjoint on the right", but that's perhaps not suggestive enough of the special nature of the case.

Recall that a (covariant) adjunction $F\dashv G\colon C\to D$ consits of functors $F\colon C\to D$, $G\colon D\to C$, and natural transformations $\eta\colon\mathrm{id}_C\Rightarrow GF$, $\epsilon\colon FG\Rightarrow\mathrm{id}_D$ satisfying the triangle identities

1. $G\epsilon\circ \eta G\colon G\Rightarrow GFG\Rightarrow G\colon G\Rightarrow G$ is $\mathrm{id}_G$
2. $\epsilon F\circ F\eta\colon F\Rightarrow FGF\Rightarrow F$ is $\mathrm{id}_F$

The above data is equivalent to the existence of a natural isomorphism $D(F-,-)\cong C(-,G-)$, whence $F$ is called left adjoint to $G$ and $G$ is called right adjoint to $F$.

A contravariant adjunction of functors $F\colon C^{op}\to D$ and $G\colon D^{op}\to C$ would then be either a covariant adjunction $F\dashv G^{op}\colon C^{op}\to D$, equivalently a covariant adjunction $G\dashv F^{op}$, or a covariant adjunction $F^{op}\dashv G\colon C\to D^{op}$, equivalently $G^{op}\dashv F$. In the former case, one says that $F$ and $G$ are adjoint on the left, and in the latter that they are adjoint on the right.

For $F$ and $G$ adjoint on the right, say via $F^{op}\dashv G$, the natural transformations $\eta\colon\mathrm{id}_C\Rightarrow GF^{op}$ and $\epsilon\colon F^{op}G\Rightarrow\mathrm{id}_{D^{op}}$ correspond to natural transformations $\eta^{op}\colon GF^{op}\Rightarrow\mathrm{id_{C^{op}}}$ and $\epsilon^{op}\colon\mathrm{id}_D\Rightarrow FG^{op}$ whence the second triangle identity can be rewritten so that the two become

1. $G\epsilon\circ \eta G\colon G\Rightarrow GF^{op}G\Rightarrow G\colon G\Rightarrow G$ is $\mathrm{id}_G$
2. $F\eta^{op}\circ\epsilon^{op} F\colon F\Rightarrow FG^{op}F\Rightarrow F$ is $\mathrm{id}_F$

In the case of $F=G$, which implies $C=D$, it is possible to have $\eta=\epsilon^{op}\colon\mathrm{id}_C\Rightarrow GG$, equivalently $\eta^{op}=\epsilon\colon G^{op}G^{op}\Rightarrow\mathrm{id}_{C^{op}}$. This is the special case you have, and in it the two triangle identities coincide and become

• $G\eta^{op}\circ\eta G\colon G\Rightarrow GG^{op}G\Rightarrow G$ is $\mathrm{id}_G$.