2
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This is really a nice proof! It help me a lot, thank you very much! $\endgroup$ – Khako Jan 19 '15 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.