# A necessary and sufficient condition for contravariant auto-equivalence on module categories

I have a problem about the condition of contravariant auto-equivalence on module categories.

Let $R$ be a algebra over a field. Let $\mathcal{C}$ be a abelian subcategory of $R$-modules, and assume that every module in $\mathcal{C}$ has finite length. There is a fact that every contravariant auto-equivalence takes simple modules to simple modules (see, e.g. Images of simple modules under exact endofunctors).

$\bf My$ $\bf Question$: Is it true that the converse is true for exact endo-functors? More precisely, I really want to know that for give contravariant exact endo-functors $F,G:\mathcal{C} \rightarrow \mathcal{C}$, we have that $F$ is an contravariant auto-equivalence with inverse $G$ $\bf if$ $(F,G)$ forms a pair of adjoint functors and both $F,G$ take simple objects to simple objects. Thanks very much!

Yes, more generally, you have the following (which applies to your situation when setting ${\mathcal D} := {\mathcal C}^{\text{op}}$).
Let ${\mathcal C},{\mathcal D}$ be finite length abelian categories and let ${\mathscr F}\dashv{\mathscr G}$ be an adjunction between the exact functors ${\mathscr F}:{\mathcal C}\rightleftarrows {\mathcal D}: {\mathscr G}$ which both preserve simple objects. Then the adjunction is an adjoint equivalence.
Proof: If $X\in{\mathcal C}$ is simple, then so is ${\mathscr G}{\mathscr F}(X)$ by assumption. Further, the unit morphism $\eta_X: X\to {\mathscr G}{\mathscr F}(X)$ is nonzero, hence an isomorphism by Schur's lemma. Similarly, for $Y\in{\mathcal D}$ simple, the counit $\varepsilon_Y: {\mathscr F}{\mathscr G}(Y)\to Y$ is an isomorphism. Now consider the subcategories ${\mathcal C}^{\prime}$ resp. ${\mathcal D}^{\prime}$ of ${\mathcal C}$ resp. ${\mathcal D}$ on which the unit resp. counit is an isomorphism. Since ${\mathscr F}$ and ${\mathscr G}$ are exact, the Five Lemma shows that these are thick subcategories of ${\mathcal C}$ resp. ${\mathcal D}$ (i.e. if two terms in a short exact sequence belong to them, then so does the third), and by our previous considerations they contain all simples of ${\mathcal C}$ resp. ${\mathcal D}$ . Since ${\mathcal C}$ and ${\mathcal D}$ are finite length, we conclude ${\mathcal C}^{\prime}={\mathcal C}$ and ${\mathcal D}^{\prime}={\mathcal D}$, hence ${\mathscr F}\dashv {\mathscr G}$ is indeed an adjoint equivalence.