If $h_u \times_G F$ is representable for every $u \in \text{Ob}(C)$ and $G$ is representable then so is $F$.

Question

This is from the Stacks Project.

I have so far the following pullback squares:

$$ \require{AMScd} \begin{CD} D @>{Q}>> C^{op}\\ @V{P}VV @V{H}VV \\ C^{op} @>{h_u \times_G F}>> \text{Sets} \end{CD} \ \ \ \ \ \ \begin{CD} E @>{Q'}>> C^{op} \\ @V{P'}VV @V{H'}VV \\ C^{op} @>{G}>> \text{Sets} \end{CD} \ \ \ \ \ \ \begin{CD} (h_u \times_G F)(x) @>{q_x}>> F(x) \\ @V{p_x}VV @V{g_x}VV \\ h_u(x) @>{f_x}>> G(x) \end{CD} \forall x \in \text{Ob}(C) $$

where $h_u \times_G F$ means $ h_u \times_{\xi, G, a} F$ for every $\xi \in G(u)$.

Answer 1

Let $u$ be a representing object for $G$ and $v$ be a representing object for $h_u \times_G F$. Then,

$$ \begin{align} F &\cong G \times_{G} F \cong h_u \times_G F \cong h_v \end{align}$$

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Note that for any morphism $f$ in any category, the following square is a pullback: $$\require{AMScd} \begin{CD} X @>1_x>> X \\ @VfVV @VVf V \\ Y @>>1_Y> Y \end{CD} $$ More generally, for any isomorphisms $u,v$, the following square is a pullback: $$\require{AMScd} \begin{CD} W @>u>> X \\ @Vv^{-1}fuVV @VVf V \\ Y @>>v> Z \end{CD} $$