$M$, $N$ are sets, $f: M \rightarrow N$ is a function, $R \subseteq M\times M$ is an equivalence relation on $M$, $\pi: M \rightarrow M/R$ is the canonical projection (i.e. $\pi(x)=[x]_R$), and $\ker f \subseteq M\times M$ is defined as $\{(x,y) \in M \times M: f(x)=f(y)\}$.
Show that $R \subseteq \ker f \Leftrightarrow (\exists F: M/R \rightarrow N)(f=F\circ \pi)$ and that there is exactly one $F$, if both statements in the logical equivalence are true.
Now, let's see what I've come up with so far:
"$\Rightarrow$": assume $R \subseteq \ker f$ and construct a function $F: M/R \rightarrow N$ which works like this: given a $c \in M/R$, choose an $x \in M$ with $c=[x]_R$ and let $F(c)=f(x)$. This works exactly because we know that $(\forall a,a' \in [x]_R)(f(a) = f(a'))$. Now let $x \in M$ and $y=\pi(x)=[x]_R$, then $F(y)=f(x)$ and $F(\pi(x))=(F\circ \pi)(x)=f(x)$ for all $x \in M$.
"$\Leftarrow$": assume $\exists F: M/R \rightarrow N$ with $f=F\circ \pi$: then $(\forall c \in \{[x]_R: x\in M\})(F(c)=f(x))$. Let $a \in M/R$ and $x,x' \in a$, then $\pi(x)=\pi(x')$. We can then apply $F$ to both sides and see that $f(x)=F(\pi(x))=F(\pi(x'))=f(x')$, i.e. $R\subseteq M\times M$ can be defined as $\{(x,y)\in M\times M: f(x)=f(y)\}$ and therefore $R\subseteq \ker f$.
Have I missed something important or is this correct? If so, how would I go about proving that there is exactly one F satisfying the given attributes?
As a side-note: I'm very new to mathematics (and I'm no native speaker either), so any hints regarding style/language are very welcome.