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Theorem: Every epi is surjective.

Proof. Let $h : A \to B$ be an epimorphism. We define maps $f, g : B \to \mathcal{P}(B)$ by \begin{align*} f(b) &= \{b\} \cap \mathrm{im}(h)\\ g(b) &= \{b\} \end{align*} where we recall that $\mathrm{im}(h) = \{b \in B \mid \exists a \in A \,.\, h(a) = b\}$.

For every $a \in A$ we have $f(h(a)) = \{h(a)\} = g(h(a))$, therefore $f = g$ as $h$ is epi. Now, for every $y \in B$ we have $\{y\} = g(y) = f(y) = \{y\} \cap \mathrm{im}(h)$, therefore $y \in \mathrm{im}(h)$. QED.

Theorem: Every epi is surjective.

Proof. Let $h : A \to B$ be an epimorphism. We define maps $f, g : B \to \mathcal{P}(B)$ by \begin{align*} f(b) &= \{b\} \cap \mathrm{im}(h)\\ g(b) &= \{b\} \end{align*} where we recall that $\mathrm{im}(h) = \{b \in B \mid \exists a \in A \,.\, h(a) = b\}$.

For every $a \in A$ we have $f(h(a)) = \{h(a)\} = g(h(a))$, therefore $f = g$ as $h$ is epi. Now, for every $y \in B$ we have $\{y\} = g(y) = f(y) = \{y\} \cap \mathrm{im}(h)$, therefore $y \in \mathrm{im}(h)$. QED.

Supplemental 2022-01-09: Here is an improved version which uses a smaller codomain. We write $\Omega$ for the subobject classifier (the set of truth values).

Proof. Let $h : A \to B$ be an epimorphism and $b \in B$. We define maps $f, g : B \to \Omega$ by \begin{align*} f(b') &{}\mathbin{{:}{=}} (b = b' \land \exists a \in A . h(a) = b')\\ g(b') &{}\mathbin{{:}{=}} (b = b'). \end{align*}

For every $a \in A$ we have \begin{align*} f(h(a)) &\Leftrightarrow (b = h(a) \land \exists a' \in A . h(a') = h(a)) \\ &\Leftrightarrow (b = h(a)) \\ &\Leftrightarrow g(h(a)), \end{align*} therefore $f = g$ as $h$ is epi. Now \begin{align*} \top &\Leftrightarrow b = b \\ &\Leftrightarrow g(b) = f(b) \\ &\Leftrightarrow b = b \land \exists a \in A . h(a) = b \\ &\Leftrightarrow \exists a \in A . h(a) = b. \quad \Box \end{align*}

Theorem: Every epi is surjective.

Proof. Let $h : A \to B$ be an epimorphism. We define maps $f, g : B \to \mathcal{P}(B)$ by \begin{align*} f(b) &= \{b\} \cap \mathrm{im}(h)\\ g(b) &= \{b\} \end{align*} where we recall that $\mathrm{im}(h) = \{b \in B \mid \exists a \in A \,.\, h(a) = b\}$.

For every $x \in A$ we have $f(h(a)) = \{h(a)\} = g(h(a))$, therefore $f = g$ as $h$ is epi. Now, for every $y \in B$ we have $\{y\} = g(y) = f(y) = \{y\} \cap \mathrm{im}(h)$, therefore $y \in \mathrm{im}(h)$. QED.

Theorem: Every epi is surjective.

Proof. Let $h : A \to B$ be an epimorphism. We define maps $f, g : B \to \mathcal{P}(B)$ by \begin{align*} f(b) &= \{b\} \cap \mathrm{im}(h)\\ g(b) &= \{b\} \end{align*} where we recall that $\mathrm{im}(h) = \{b \in B \mid \exists a \in A \,.\, h(a) = b\}$.

For every $a \in A$ we have $f(h(a)) = \{h(a)\} = g(h(a))$, therefore $f = g$ as $h$ is epi. Now, for every $y \in B$ we have $\{y\} = g(y) = f(y) = \{y\} \cap \mathrm{im}(h)$, therefore $y \in \mathrm{im}(h)$. QED.

Theorem: Every epi is surjective.

Proof. Let $h : A \to B$ be an epimorphism and $y \in B$. We define maps $f, g : B \to \mathcal{P}(B)$ by \begin{align*} f(b) &= \{b\} \cap \mathrm{im}(h)\\ g(b) &= \{b\} \end{align*} where we recall that $\mathrm{im}(h) = \{b \in B \mid \exists a \in A \,.\, h(a) = b\}$.

For every $x \in A$ we have $f(h(a)) = \{h(a)\} = g(h(a))$, therefore $f = g$ as $h$ is epi. Now, for every $y \in B$ we have $\{y\} = g(y) = f(y) = \{y\} \cap \mathrm{im}(h)$, therefore $y \in \mathrm{im}(h)$. QED.

Theorem: Every epi is surjective.

Proof. Let $h : A \to B$ be an epimorphism. We define maps $f, g : B \to \mathcal{P}(B)$ by \begin{align*} f(b) &= \{b\} \cap \mathrm{im}(h)\\ g(b) &= \{b\} \end{align*} where we recall that $\mathrm{im}(h) = \{b \in B \mid \exists a \in A \,.\, h(a) = b\}$.

For every $x \in A$ we have $f(h(a)) = \{h(a)\} = g(h(a))$, therefore $f = g$ as $h$ is epi. Now, for every $y \in B$ we have $\{y\} = g(y) = f(y) = \{y\} \cap \mathrm{im}(h)$, therefore $y \in \mathrm{im}(h)$. QED.