[proofplan] We prove both implications directly from the categorical definition of an epimorphism. If $e$ is surjective, then equality after precomposition with $e$ forces two functions out of $B$ to agree at every element of $B$. Conversely, if $e$ misses an element $b_0 \in B$, we construct two distinct functions $B \to \{0,1\}$ that agree on the image of $e$, so their composites with $e$ are equal although the functions themselves are not. [/proofplan] [step:Use surjectivity to cancel precomposition by $e$] Assume that $e: A \to B$ is surjective. To prove that $e$ is an epimorphism in $\mathbf{Set}$, let $Y$ be a set and let \begin{align*} u: B &\to Y,\\ v: B &\to Y \end{align*} be functions such that \begin{align*} u \circ e = v \circ e. \end{align*} We must prove $u=v$. Let $b \in B$. Since $e$ is surjective, there exists $a \in A$ such that $e(a)=b$. Evaluating the equality $u \circ e = v \circ e$ at this element $a$ gives \begin{align*} u(b) = u(e(a)) = (u \circ e)(a) = (v \circ e)(a) = v(e(a)) = v(b). \end{align*} Since $b \in B$ was arbitrary, $u$ and $v$ agree on every element of $B$. Hence $u=v$ as functions $B \to Y$. Therefore $e$ is an epimorphism in $\mathbf{Set}$. [/step] [step:Construct two distinct maps when $e$ is not surjective] Assume that $e: A \to B$ is not surjective. Then there exists an element $b_0 \in B$ such that no element of $A$ maps to $b_0$: \begin{align*} \forall a \in A,\quad e(a) \ne b_0. \end{align*} Let $Y := \{0,1\}$. Define functions \begin{align*} u: B &\to Y,\\ b &\mapsto 0 \end{align*} and \begin{align*} v: B &\to Y,\\ b &\mapsto \begin{cases} 1, & b=b_0,\\ 0, & b \ne b_0. \end{cases} \end{align*} For every $a \in A$, the choice of $b_0$ gives $e(a) \ne b_0$, so \begin{align*} (u \circ e)(a) = u(e(a)) = 0 = v(e(a)) = (v \circ e)(a). \end{align*} Thus $u \circ e = v \circ e$ as functions $A \to Y$. However, \begin{align*} u(b_0)=0 \qquad\text{and}\qquad v(b_0)=1, \end{align*} so $u \ne v$ as functions $B \to Y$. Therefore precomposition with $e$ is not cancellative on maps out of $B$, and $e$ is not an epimorphism in $\mathbf{Set}$. [/step] [step:Combine the two implications] The first step proves that every surjective function $e: A \to B$ is an epimorphism in $\mathbf{Set}$. The second step proves the contrapositive of the reverse implication: if $e$ is not surjective, then $e$ is not an epimorphism in $\mathbf{Set}$. Hence $e$ is an epimorphism in $\mathbf{Set}$ if and only if $e$ is surjective. [/step]