I want to prove the following characterization of an $R$-module homomorphism $g$ to be surjective:
"whenever the composition of $g:M→N$ and $k:N→Y$ is zero, then $k=0$".
It is easy to go one side: if $g$ is surjective then for any $n\in N$ we have $n=g(m)$ for some $m\in M$. Now, if the composition is zero we would have $k(n)=k(g(m))=0$. But, for the converse, I could not choose suitable $Y$ and $k$. Thanks, in advance, for any cooperation!
Let us assume that $g$ has the quoted property. Let $Y = N/\text{im}(g)$ and $ k : N \rightarrow Y$ the natural projection. Then $k \circ g = 0$, so $k = 0$ by assumption. But $k$ is surjective, so we have to have $Y = 0$, hence $N = \text{im}(g)$ and so $g$ is surjective.