[proofplan] We construct the comparison map from morphisms out of the pushout object $p$ to compatible pairs of morphisms out of $b$ and $c$. Commutativity of the pushout square makes this map well-defined. The existence part of the pushout universal property gives surjectivity, and the uniqueness part gives injectivity. [/proofplan] [step:Define the comparison map and verify compatibility] Fix an object $x$ of $\mathcal C$. Define the set of compatible pairs \begin{align*} P_x:=\{(r,s)\in \operatorname{Hom}_{\mathcal C}(b,x)\times \operatorname{Hom}_{\mathcal C}(c,x): r\circ f=s\circ g\}. \end{align*} Define the function \begin{align*} \Phi_x:\operatorname{Hom}_{\mathcal C}(p,x)&\to P_x\\ u&\mapsto (u\circ i,u\circ j). \end{align*} This is well-defined because, for every morphism $u:p\to x$, associativity of composition and the commutativity relation $i\circ f=j\circ g$ give \begin{align*} (u\circ i)\circ f &=u\circ (i\circ f)\\ &=u\circ (j\circ g)\\ &=(u\circ j)\circ g. \end{align*} Thus $(u\circ i,u\circ j)\in P_x$. [/step] [step:Use existence in the pushout property to prove surjectivity] Let $(r,s)\in P_x$. By definition of $P_x$, the morphisms $r:b\to x$ and $s:c\to x$ satisfy \begin{align*} r\circ f=s\circ g. \end{align*} Since the square with vertex $p$ is a pushout, the universal property gives a morphism $u:p\to x$ such that \begin{align*} u\circ i=r, \qquad u\circ j=s. \end{align*} Therefore $\Phi_x(u)=(r,s)$, so every element of $P_x$ lies in the image of $\Phi_x$. Hence $\Phi_x$ is surjective. [/step] [step:Use uniqueness in the pushout property to prove injectivity] Let $u,v:p\to x$ be morphisms such that $\Phi_x(u)=\Phi_x(v)$. Unpacking the definition of $\Phi_x$, this means \begin{align*} u\circ i=v\circ i, \qquad u\circ j=v\circ j. \end{align*} The pair $(u\circ i,u\circ j)$ is compatible by the first step. Since the square is a pushout, the morphism from $p$ to $x$ inducing a fixed compatible pair is unique. Both $u$ and $v$ induce the same compatible pair, so $u=v$. Hence $\Phi_x$ is injective. [/step] [step:Conclude the equivalence of data] The function $\Phi_x$ is both surjective and injective, hence bijective. Since $x$ was arbitrary, for every object $x$ of $\mathcal C$ a morphism $p\to x$ is equivalently a compatible pair of morphisms $r:b\to x$ and $s:c\to x$ satisfying $r\circ f=s\circ g$. [/step]