The following is taken from: $\textit{Groups, Rings, Modules}$ by: Auslander and Buchsbaum
$\color{Green}{Background:}$
$\textbf{Proposition:}$ Let $M$ be an $R-$module.
Supppose $f:M\to N$ is a morphism of $R-$modules. Then:
$(a)$ if $M'$ is a submodule of $M,$ then $M'$ is the kernel of the canonical surjective morphism $k_{M/M'}:M\to M/M'.$
$(b)$ $f$ is the zero morphism if and only if $\text{Ker} f=M.$ $\text{Im} f$ is the kernel of $\text{Coker }f,$ that is, $\text{Im }f$ is the kernel of the surjective morphism $N\to \text{Coker }f.$
$\color{Red}{Questions:}$
For the proposition quoted above, what does it to say in more math notation, in part $(a)$ that $M'$ is the kernel of the canonical surjective morphism $k_{M/M'}:M\to M/M'?$ Does it mean that $\text{ker }k_{M/M'}=\{m\in M':k_{M/M'}(m)=0\}$
and also for part $(b)$ where it says: $\text{Im} f$ is the kernel of $\text{Coker }f.$ To translate their "that is..." part. Let the morphism $N\to \text{Coker }f.$ be denoted by $b,$ then $b:N\to \text{Coker }f,$ and $\text{Im }f=\text{ker }b=\{n\in N:b(n)=0\}.$
Thank you in advance