[proofplan] The quotient map coequalizes $f$ and $g$ because the congruence $\sim$ was chosen to identify each pair $f(a)$ and $g(a)$. To prove the universal property, take any homomorphism $h: B \to C$ with $h \circ f = h \circ g$ and show that its kernel congruence contains the generating pairs, hence contains all of $\sim$. This makes $h$ constant on $\sim$-equivalence classes, so it descends to a unique homomorphism $\bar h: B/{\sim} \to C$ satisfying $\bar h \circ q = h$. [/proofplan]

[step:Verify that the quotient map coequalizes the two homomorphisms] Let \begin{align*} q: B &\to B/{\sim} \\ b &\mapsto [b]_{\sim} \end{align*} be the canonical quotient homomorphism. For every $a \in A$, the defining property of $\sim$ gives $f(a) \sim g(a)$, and therefore \begin{align*} q(f(a)) = [f(a)]_{\sim} = [g(a)]_{\sim} = q(g(a)). \end{align*} Hence the two homomorphisms \begin{align*} q \circ f, q \circ g: A \to B/{\sim} \end{align*} agree on every element of $A$, so \begin{align*} q \circ f = q \circ g. \end{align*} [/step]

[step:Show that every coequalizing homomorphism is constant on $\sim$-classes] Let $C$ be an object of $\mathcal C$, and let \begin{align*} h: B \to C \end{align*} be a homomorphism satisfying \begin{align*} h \circ f = h \circ g. \end{align*} Define the kernel relation of $h$ by \begin{align*} \ker(h) := \{(b_1,b_2) \in B \times B : h(b_1) = h(b_2)\}. \end{align*} This relation is a congruence on $B$: it is an [equivalence relation](/page/Equivalence%20Relation) because equality in $C$ is an equivalence relation, and it is compatible with all basic operations because $h$ is a homomorphism. Indeed, if $\omega$ is an $n$-ary operation symbol and $(b_i,c_i) \in \ker(h)$ for $1 \le i \le n$, then \begin{align*} h(\omega_B(b_1,\dots,b_n)) &= \omega_C(h(b_1),\dots,h(b_n)) \\ &= \omega_C(h(c_1),\dots,h(c_n)) \\ &= h(\omega_B(c_1,\dots,c_n)), \end{align*} so \begin{align*} (\omega_B(b_1,\dots,b_n),\omega_B(c_1,\dots,c_n)) \in \ker(h). \end{align*} For each $a \in A$, the equality $h \circ f = h \circ g$ gives \begin{align*} h(f(a)) = h(g(a)), \end{align*} so $(f(a),g(a)) \in \ker(h)$. Thus $\ker(h)$ is a congruence on $B$ containing every generating pair $(f(a),g(a))$. Since $\sim$ is the smallest such congruence, we have \begin{align*} \sim \ \subseteq \ker(h). \end{align*} Consequently, whenever $b_1,b_2 \in B$ satisfy $b_1 \sim b_2$, we have $h(b_1)=h(b_2)$. [/step]

[step:Construct the induced homomorphism on the quotient] Define \begin{align*} \bar h: B/{\sim} &\to C \\ [b]_{\sim} &\mapsto h(b). \end{align*} This map is well-defined: if $[b_1]_{\sim} = [b_2]_{\sim}$, then $b_1 \sim b_2$, and the previous step gives $h(b_1)=h(b_2)$. The map $\bar h$ is a homomorphism. Let $\omega$ be an $n$-ary operation symbol and let $[b_1]_{\sim},\dots,[b_n]_{\sim} \in B/{\sim}$. Since the quotient algebra operations are defined by \begin{align*} \omega_{B/{\sim}}([b_1]_{\sim},\dots,[b_n]_{\sim}) = [\omega_B(b_1,\dots,b_n)]_{\sim}, \end{align*} we obtain \begin{align*} \bar h(\omega_{B/{\sim}}([b_1]_{\sim},\dots,[b_n]_{\sim})) &= \bar h([\omega_B(b_1,\dots,b_n)]_{\sim}) \\ &= h(\omega_B(b_1,\dots,b_n)) \\ &= \omega_C(h(b_1),\dots,h(b_n)) \\ &= \omega_C(\bar h([b_1]_{\sim}),\dots,\bar h([b_n]_{\sim})). \end{align*} Since the morphisms of $\mathcal C$ are exactly the homomorphisms between its objects, this homomorphism $\bar h$ is a morphism in $\mathcal C$. For every $b \in B$, \begin{align*} (\bar h \circ q)(b) = \bar h([b]_{\sim}) = h(b), \end{align*} so \begin{align*} \bar h \circ q = h. \end{align*} [/step]

[step:Prove uniqueness of the induced homomorphism] Let \begin{align*} k: B/{\sim} \to C \end{align*} be a homomorphism satisfying \begin{align*} k \circ q = h. \end{align*} For every equivalence class $[b]_{\sim} \in B/{\sim}$, we have \begin{align*} k([b]_{\sim}) = k(q(b)) = h(b) = \bar h([b]_{\sim}). \end{align*} Hence $k$ and $\bar h$ agree on every element of $B/{\sim}$, so $k=\bar h$. We have shown that $q \circ f = q \circ g$, and that every morphism $h: B \to C$ in $\mathcal C$ satisfying $h \circ f = h \circ g$ factors uniquely through the morphism $q$ by a morphism $\bar h: B/{\sim} \to C$ in $\mathcal C$. Therefore $q: B \to B/{\sim}$ is a coequalizer of $f$ and $g$ in $\mathcal C$. [/step]