Suppose you have a module category, and an exact, contravariant endofunctor $F$ that preserves isomorphism classes of simple modules. If $M$ is a module of finite length, why is $F(M)$ also of finite length?
I figure there must be a way to construct a finite composition series of $F(M)$ from that of $M$, since $F$ preserves the isomorphism classes of the composition factors. But since a composition series is not exact and $F$ reverses the arrows, I don't see how you could do this.
(The motivation for my question is that dualization in category $\mathcal{O}$ preserves finitely generated modules.)
This follows immediately from the fact that length is additive in short exact sequences. So, by induction on $n$, we can show that if $M$ has length $n$ then $F(M)$ has length $n$. The case $n=0$ is trivial, since $M$ must be trivial. If $M$ has length $n>0$, there exists a short exact sequence $$0\to N\to M\to S\to 0$$ where $N$ has length $n-1$ and $S$ is simple. Applying $F$, we get a short exact sequence $$0\to F(S)\to F(M)\to F(N)\to 0.$$ By assumption $F(S)$ is simple and by induction $F(N)$ has length $n-1$. Thus $F(M)$ has length $n$.