Does "maximal submodule <=> simple quotient module" generalize to abelian categories? Does the statement "If $A$, $B$ are modules over a commutative ring $R$, then $B$ is a maximal submodule of $A$ if and only if $A/B$ is a simple module" generalize to the setting of abelian categories?  That is, is it true to say "If C is an abelian category and $A, B \in$ C, then $B$ is a maximal subobject of $A$ if and only if $A/B$ is a simple object?"
My hunch is that this statement does generalize, since I've seen it stated that the Jordan-Holder theorem for abelian categories is a "straightforward generalization" of the version for modules (wherein this fact is used), but I've had a lot of trouble finding anything on maximal subobjects at all and I haven't figured out the details yet.
 A: An attempt (everything happens inside abelian categories here and sorry for not being able to draw commutative diagrams with named arrows):
$\textbf{Definition:}$ A monomorphism $\iota: B \rightarrow A$ is called a maximal subobject of $A$ if $\iota$ is not an isomorphism and the existence of two monomorphisms $B \rightarrow B'$ and $B' \rightarrow A$ sitting in a commutative diagram with $\iota$ implies that one of them is an isomorphism.
$\textbf{Definition:}$ An object $S$ is called simple if $0 \rightarrow S$ is a maximal subobbject of $S$.
$\textbf{Proposition:}$ Let $\iota: B \rightarrow A$ be a morphism with cokernel $\pi: A \rightarrow S$. If $\iota$ is a maximal subobject of $A$, then $S$ is simple. 
Proof. Note that since $\iota$ is not an isomorphism, we have $S \neq 0$ so we have the non-triviality condition. Let $\alpha: T \rightarrow S$ be a monomorphism. We want to show that either $\alpha = 0$ or $\alpha$ is an isomorphism. Form the pull-back along $\alpha$ and $\pi$, with parallel arrows to $\alpha$ and $\pi$ being $\alpha':P \rightarrow A$ and $\pi': P \rightarrow T$, respectively. 
First, $\alpha'$ is a monomorphism since pull-backs preserve kernels. 
Second, since $\pi$ is an epimorphism, the pull-back square is also a push-out; hence $\pi'$ is an epimorphism.
Third, since $\iota$ is a monomorphism, it is a kernel, hence a kernel of its cokernel $\pi$. Again since pull-backs preserve kernels, there is a morphism $\iota' : B \rightarrow P$ which is a kernel of $\pi'$ and makes the relevant triangle commute ($\alpha' \circ \iota' = \iota$).
Now, since $\iota$ is a maximal subobject, there are two possibilities: either $\iota'$ or $\alpha'$ is an isomorphism. The former yields $$\alpha \circ \pi' = \pi \circ \alpha' = \pi \circ \iota \circ (\iota')^{-1} = 0$$ hence $\alpha = 0$ as $\pi'$ is an epimorphism. And the latter yields that $\alpha$ is an isomorphism since $\alpha$ and $\alpha'$ are parallel in a push-out square.
I haven't tried but I believe the converse statement can be proved similarly.
