$(h_u \times_{\xi, G, a} F)(X) = \{(f, \xi) : x \xrightarrow{f} u, \xi' \in F(X), G(f)(\xi) = a_x(\xi')\}$ using definition of fibre product.

Question

I understand the fibre product definition now and have proved that for functors $F, G, H : C^{op} \to \text{Sets}, \ a : F \to G, \ b : H \to G$ that the fibre product can be defined by:

$$(F \times_{a, G, b} H)(x) = F(x) \times_{a_x, G(x), b_x} H(x) $$ for any object $x \in C$.

This is from the Stacks Project.

They next go on to give a formula for the fibre product of $a : F \to G$, $\xi: h_u \to G$ given by Yoneda and $u \in \text{Ob}(C), \xi' \in G(u)$:

$$ (h_u \times_{\xi, G, a}F)(x) = \{ (f, \xi') : x\xrightarrow{f} u, \xi'\in F(x), G(f)(\xi) = a_x(\xi')\}$$

Was wondering how they get $G(f)(\xi) = a_x(\xi')$. Shouldn't it be something like $\xi_x(f) = a_x(\xi')$? Confusion intusion!

Answer 1

Ah, I see. Looking again at the Yoneda lemma, we have $\xi_x(f) \equiv G(f)(\xi)$.