Sum of semiartinians modules is preserved under morphism.

Question

Defining a semiartinian module M as module in which which every nonzero quotient of it contains a simple submodule and also define

$Sa(M):= \sum \lbrace N \leq M \: | \: \mbox{N is semiartinian} \rbrace$.

Given $\phi: M \to N$ module morphism. I want to prove $\phi (Sa(M)) \leq Sa(N)$. Im burned out trying this for a while, so far I have proved that $Sa(M)$ is semiartinian for every module $M$. Lets suppose an easier case where $Sa(M)=M'$ such $M'$ is semiartinian, so $M' / L$ has some simple submodule lets say $S_{1} /L$ but I need to prove $\phi(M') \in Sa(N)$ and this is where im stuck. Any help will be aprecciated, thanks!

Answer 1

Let $M'\le M$ be semiartinian.
Then, based on the isomorphism theorems, $\phi(M')$ is isomorphic to a quotient of $M'$ (namely $M'/\ker(\phi|_{M'})$), moreover each quotient of $\phi(M')$ is isomorphic to some quotient of $M'$.
Hence, all nonzero quotients of $\phi(M')$ have a simple submodule, as those are also (isomorphic to) quotients of the semiartinian $M'$.

Since $\phi(M')\le N$ is semiartinian, by definition of $Sa$, we have $\phi(M')\le Sa(N)$.
Finally, $$\phi(Sa(M))\ =\ \phi(\sum_{M'\text{ s.a.}}{M'})\ =\ \sum_{M'\text{ s.a.}}\phi(M')\ \le\ Sa(N)\,.$$