As a well-known example of a ring homomorphism which is monic and epic, but not a ring isomorphism, serves the inclusion map $\iota:\mathbb Z\hookrightarrow\mathbb Q$. While the monocity follows immediately from $\mathbb Q$ being the quotient field of $\mathbb Z$, showing that $\iota$ is an epimorphism requires a little more (not much) work. One way is elementwise as done here.
Generalizing the aforementioned one arrives at the following fact
Let $R$ be an integral domain. The inclusion $\iota:R\hookrightarrow Q(R)$ into the quotient field $Q(R)$ is an epimorphism.
Proof$~~$Let $f,g:Q(R)\rightrightarrows F$ be field homomorphisms such that $f\circ \iota=g\circ\iota=:\varphi$. It follows that $\varphi:R\to F$, viewed as a ring homomorphism, is monic aswell. By the universal property of the quotient field this yields a unique field homomorphisms $\overline\varphi:Q(R)\to F$ such that $\varphi=\overline\varphi\circ\iota$. This gives a commutative diagram
From here it follows that $f=\overline\varphi=g$ as desired.
Is the given proof correct? If so, where can it be further improved; if not, where did I went wrong? Note that I am particularly interested in an elementfree proof using the universal property of the quotient field rather than an elementwise approach such as the one linked.
Thanks in advance!
The outline of your proof is fine, but you should give an argument as to why you can choose $F$ to be a field. At first sight you should allow $F$ to be any ring, since this is the category you are working in. Now since $Q(R)$ is a field, the image of any morphism $Q(R)\to A$ to a ring $A$ is a field, except when $A$ is the zero ring. Try to write an appropriate case distinction and it shoule be fine.