Preservation of Socle Under Homomorphisms We know that if $f$ is an $R$-homomorphism from a, say, left $R$-module $M$ to a left $R$-module $N$ then $f(\operatorname{Soc}M)\leq \operatorname{Soc}N$, where $\operatorname{Soc}$ stands for "socle". Now, let $R$ and $S$ be two unital rings and $f :R\rightarrow S$ be a ring homomorphism. Is it necessarily true that $f(\operatorname{Soc}(R))\leq \operatorname{Soc}(S)$ ? (or, at least, when $f$ is onto.) I tried first to see if $x\in R$ lies in a minimal left ideal, does $f(x)\in S$ lies in a minimal left ideal, for then, the assertion follows, since $Soc $ is the sum of all the minimal left ideals. Is it a good start, or another attack should be done? Thanks, in advance, for any answer.

Let $I$ be a minimal left ideal of $R$. Then there are two cases: either $I\subseteq\ker f$ and so $f(I)=\\{0\\}$, or $I\cap\ker f=\\{0\\}$.

Assuming $f$ is surjective, we get that, in the second case, $f(I)$ is a minimal left ideal of $S$. It is a left ideal by surjectivity; if $y\in f(I)$, $y\
e0$, then $y=f(x)$ for some $x\in I$, so $I=Rx$ and therefore $Sy=f(I)$. Hence $f(I)$ is generated by each of its nonzero element, so it's a simple left $S$-module.

You need surjectivity (as rschwieb's example shows).