[proofplan] We use the representing bijections explicitly, rather than appealing to a general uniqueness theorem. The representing object $P'$ gives a unique morphism $u:P\to P'$ whose associated pair of maps to $A$ and $B$ is $(\pi_A,\pi_B)$, and the representing object $P$ gives a unique morphism $v:P'\to P$ whose associated pair is $(\pi'_A,\pi'_B)$. Naturality identifies these associated pairs with composition by the projection maps, so $u$ and $v$ preserve the projections. Finally, the composites $v\circ u$ and $u\circ v$ preserve the same projection pairs as the identity maps, and uniqueness in the representing bijections forces both composites to be identities. [/proofplan] [step:Extract the projection-preserving morphisms from the representing bijections] For every object $X\in \operatorname{Ob}(\mathcal C)$, the component \begin{align*} \Phi_X:\operatorname{Hom}_{\mathcal C}(X,P)\to F(X) \end{align*} is a bijection, and the component \begin{align*} \Phi'_X:\operatorname{Hom}_{\mathcal C}(X,P')\to F(X) \end{align*} is a bijection. Apply the bijection $\Phi'_P:\operatorname{Hom}_{\mathcal C}(P,P')\to F(P)$ to the element $(\pi_A,\pi_B)\in F(P)$. There is a unique morphism $u:P\to P'$ such that \begin{align*} \Phi'_P(u)=(\pi_A,\pi_B). \end{align*} Similarly, applying the bijection $\Phi_{P'}:\operatorname{Hom}_{\mathcal C}(P',P)\to F(P')$ to $(\pi'_A,\pi'_B)\in F(P')$, there is a unique morphism $v:P'\to P$ such that \begin{align*} \Phi_{P'}(v)=(\pi'_A,\pi'_B). \end{align*} [guided] The representing property says that maps into the representing object are exactly the same data as pairs of maps into $A$ and $B$, naturally in the source object. Since $P'$ represents $F$, the component \begin{align*} \Phi'_P:\operatorname{Hom}_{\mathcal C}(P,P')\to F(P) \end{align*} is a bijection. The pair $(\pi_A,\pi_B)$ belongs to \begin{align*} F(P)=\operatorname{Hom}_{\mathcal C}(P,A)\times\operatorname{Hom}_{\mathcal C}(P,B), \end{align*} so there is a unique morphism $u:P\to P'$ satisfying \begin{align*} \Phi'_P(u)=(\pi_A,\pi_B). \end{align*} The same argument with the two representing objects interchanged gives a unique morphism $v:P'\to P$. Namely, since \begin{align*} \Phi_{P'}:\operatorname{Hom}_{\mathcal C}(P',P)\to F(P') \end{align*} is a bijection and $(\pi'_A,\pi'_B)\in F(P')$, there is a unique morphism $v:P'\to P$ such that \begin{align*} \Phi_{P'}(v)=(\pi'_A,\pi'_B). \end{align*} [/guided] [/step] [step:Use naturality to identify the representing bijections with composition by projections] We first compute $\Phi'_P(u)$. Naturality of $\Phi'$ for the morphism $u:P\to P'$ gives \begin{align*} \Phi'_P(u) &= \Phi'_P\bigl(\operatorname{id}_{P'}\circ u\bigr)\\ &= F(u)\bigl(\Phi'_{P'}(\operatorname{id}_{P'})\bigr)\\ &= F(u)(\pi'_A,\pi'_B)\\ &= (\pi'_A\circ u,\pi'_B\circ u). \end{align*} Since $\Phi'_P(u)=(\pi_A,\pi_B)$ by the definition of $u$, equality of ordered pairs gives \begin{align*} \pi'_A\circ u &= \pi_A,\\ \pi'_B\circ u &= \pi_B. \end{align*} The same naturality argument for $\Phi$ and the morphism $v:P'\to P$ gives \begin{align*} \Phi_{P'}(v) &= F(v)\bigl(\Phi_P(\operatorname{id}_P)\bigr)\\ &= F(v)(\pi_A,\pi_B)\\ &= (\pi_A\circ v,\pi_B\circ v). \end{align*} Since $\Phi_{P'}(v)=(\pi'_A,\pi'_B)$, we obtain \begin{align*} \pi_A\circ v &= \pi'_A,\\ \pi_B\circ v &= \pi'_B. \end{align*} [guided] We now translate the abstract equation $\Phi'_P(u)=(\pi_A,\pi_B)$ into the concrete projection equations. The point is that naturality tells us how the representing bijection behaves when we precompose a map into $P'$. Apply naturality of the natural transformation \begin{align*} \Phi':\operatorname{Hom}_{\mathcal C}(-,P')\Longrightarrow F \end{align*} to the morphism $u:P\to P'$. The [contravariant Hom functor](/theorems/3973) sends $u$ to the map \begin{align*} \operatorname{Hom}_{\mathcal C}(P',P')&\to \operatorname{Hom}_{\mathcal C}(P,P')\\ h&\mapsto h\circ u, \end{align*} and the functor $F$ sends $u$ to \begin{align*} F(P')&\to F(P)\\ (\alpha,\beta)&\mapsto (\alpha\circ u,\beta\circ u). \end{align*} Therefore naturality gives \begin{align*} \Phi'_P(\operatorname{id}_{P'}\circ u) &=F(u)\bigl(\Phi'_{P'}(\operatorname{id}_{P'})\bigr). \end{align*} Using $\operatorname{id}_{P'}\circ u=u$ and the definition \begin{align*} \Phi'_{P'}(\operatorname{id}_{P'})=(\pi'_A,\pi'_B), \end{align*} we get \begin{align*} \Phi'_P(u) &=F(u)(\pi'_A,\pi'_B)\\ &=(\pi'_A\circ u,\pi'_B\circ u). \end{align*} But $u$ was chosen so that $\Phi'_P(u)=(\pi_A,\pi_B)$, hence \begin{align*} \pi'_A\circ u &= \pi_A,\\ \pi'_B\circ u &= \pi_B. \end{align*} The same computation with $\Phi$ and the morphism $v:P'\to P$ yields \begin{align*} \Phi_{P'}(v) &=F(v)\bigl(\Phi_P(\operatorname{id}_P)\bigr)\\ &=F(v)(\pi_A,\pi_B)\\ &=(\pi_A\circ v,\pi_B\circ v). \end{align*} Since $v$ was chosen so that $\Phi_{P'}(v)=(\pi'_A,\pi'_B)$, we conclude \begin{align*} \pi_A\circ v &= \pi'_A,\\ \pi_B\circ v &= \pi'_B. \end{align*} [/guided] [/step] [step:Show that the two projection-preserving morphisms are inverse isomorphisms] Consider the composite $v\circ u:P\to P$. Using the projection identities already proved, \begin{align*} \pi_A\circ (v\circ u) &=(\pi_A\circ v)\circ u\\ &=\pi'_A\circ u\\ &=\pi_A, \end{align*} and similarly \begin{align*} \pi_B\circ (v\circ u) &=(\pi_B\circ v)\circ u\\ &=\pi'_B\circ u\\ &=\pi_B. \end{align*} By naturality of $\Phi$ applied to $v\circ u:P\to P$, \begin{align*} \Phi_P(v\circ u) &=F(v\circ u)\bigl(\Phi_P(\operatorname{id}_P)\bigr)\\ &=F(v\circ u)(\pi_A,\pi_B)\\ &=(\pi_A\circ v\circ u,\pi_B\circ v\circ u)\\ &=(\pi_A,\pi_B). \end{align*} Also $\Phi_P(\operatorname{id}_P)=(\pi_A,\pi_B)$. Since $\Phi_P$ is injective, it follows that \begin{align*} v\circ u=\operatorname{id}_P. \end{align*} The same argument gives \begin{align*} \pi'_A\circ (u\circ v)&=\pi'_A,\\ \pi'_B\circ (u\circ v)&=\pi'_B, \end{align*} and hence \begin{align*} \Phi'_{P'}(u\circ v)=\Phi'_{P'}(\operatorname{id}_{P'}). \end{align*} Since $\Phi'_{P'}$ is injective, \begin{align*} u\circ v=\operatorname{id}_{P'}. \end{align*} Thus $u$ is an isomorphism with inverse $v$. [/step] [step:Prove uniqueness of the isomorphism preserving the projections] Let $w:P\to P'$ be any morphism satisfying \begin{align*} \pi'_A\circ w &= \pi_A,\\ \pi'_B\circ w &= \pi_B. \end{align*} Naturality of $\Phi'$ for $w:P\to P'$ gives \begin{align*} \Phi'_P(w) &=F(w)\bigl(\Phi'_{P'}(\operatorname{id}_{P'})\bigr)\\ &=F(w)(\pi'_A,\pi'_B)\\ &=(\pi'_A\circ w,\pi'_B\circ w)\\ &=(\pi_A,\pi_B). \end{align*} The morphism $u$ was defined as the unique element of $\operatorname{Hom}_{\mathcal C}(P,P')$ mapped by $\Phi'_P$ to $(\pi_A,\pi_B)$. Therefore $w=u$. Hence the projection-preserving isomorphism $P\to P'$ is unique, completing the proof. [/step]