I have a module $M$ and $M_1$, $M_2$ submodules of $M$ such that $M_1\le M_2$. I have to proof that the mapping $f: M/M_1\to M/M_2$ given by $f(\overline{m})=\overline{m}$ is a surjective homomorfism. I just wanted to know if my proof is correct.
First, to show that $f$ is an homomorfism let $\overline{x},\overline{y}\in M/M_1$. So,
\begin{align*} f(\overline{x}+\overline{y})=f(\overline{x+y}) =\overline{x+y} =\overline{x}+\overline{y} =f(\overline{x})+f(\overline{y}). \end{align*}
And if $a\in R$ and $\overline{x}\in M/M_1$, then \begin{align*} f(a\overline{x})=f(\overline{ax}) =\overline{ax} =a\overline{x} =af(\overline{x}). \end{align*}
Now for the surjective, let $y\in M/M_2$ then exists $m\in M$ such that $y=m+M_2$. Let $x=m+M_1$. Then, $x\in M/M_1$ and $f(x)=f(m+M_1)=f(\overline{m})=\overline{m}=m+M_2$. Is that right?