We define a semi-artinian $R$ module $M$ as a $R$ module such
$$\operatorname{Soc}(M/U) \neq 0$$
for every $U \leq M$ submodule. Also we define semiartinianity of a $M$ as $R$ module is defined as
$$\operatorname{sa}(M):= \sum \lbrace M' \leq M \mid M' \: \mbox{is semi-artinian} \rbrace$$
What I want to prove is that for every $R$-module morphism $f:M \to N$ we have that $f(\operatorname{sa}(M)) \leq \operatorname{sa}(N)$. My idea is to visualize first that $f(M)$ is semiartinian if $M$ is semiartinian. Let suppose $M$ is semiartinian so $Soc(M/M') \neq 0$ for every $M' \leq M$ but $f(M)$ is not semiartinian this means $Soc(f(M)/N')=0$ for some $N' \leq f(M)$. So lets take preimage of $f^{-1}(N') \leq M$ so $Soc(M/f^{-1}(N'))\neq 0$ as $M$ is semiartinian. This way for $f \neq 0$ we have that $0 \neq f(Soc(M/f^{-1}(N'))) \leq Soc(f(M)/N')=0 $ by properties of socle which is a contradiction.