Non-surjective epimorphism of algebras

Example

Work in the concrete category $\mathbf{Ring}$ of rings with identity and ring homomorphisms that preserve $1$. The inclusion $i \colon \Z \to \Q$ of the integers into the rationals is an epimorphism in $\mathbf{Ring}$, but it is not surjective.

Check: let $R$ be a ring and let $f, g \colon \Q \to R$ be ring homomorphisms with $f \circ i = g \circ i$, so $f(n) = g(n)$ for every integer $n$. For $n \ne 0$, $f(1/n) \, f(n) = f(1) = 1$ and $f(n) \, f(1/n) = 1$, so $f(n)$ is a unit of $R$ with inverse $f(1/n)$. In the same way $g(1/n)$ is the inverse of $g(n) = f(n)$. Inverses are unique, so $f(1/n) = g(1/n)$, and then $f(m/n) = f(m) f(1/n) = g(m) g(1/n) = g(m/n)$ for all $m, n$. Hence $f = g$, so $i$ is right-cancellable, that is, an epimorphism. It is not surjective, since $1/2 \notin \Z$.