The Stacks project

Proof. The maps $[f]$ and $[g]$ come from an application of Lemma 78.21.1 and the other two maps come from Lemma 78.20.2 (and the fact that $(U'', R'', s'', t'', c'')$ lives over $U'$). To show the $2$-fibre product property, it suffices to prove the lemma for the diagram

of categories fibred in groupoids, see Stacks, Lemma 8.9.3. In other words, it suffices to show that an object of the $2$-fibre product $\mathcal{S}_ U \times _{[U'/_{\! p}R']} [U/_{\! p}R]$ over $T$ corresponds to a $T$-valued point of $U''$ and similarly for morphisms. And of course this is exactly how we constructed $U''$ and $R''$ in the first place.

In detail, an object of $\mathcal{S}_ U \times _{[U'/_{\! p}R']} [U/_{\! p}R]$ over $T$ is a triple $(u', u, r')$ where $u'$ is a $T$-valued point of $U'$, $u$ is a $T$-valued point of $U$, and $r'$ is a morphism from $u'$ to $f(u)$ in $[U'/R']_ T$, i.e., $r'$ is a $T$-valued point of $R$ with $s'(r') = u'$ and $t'(r') = f(u)$. Clearly we can forget about $u'$ without losing information and we see that these objects are in one-to-one correspondence with $T$-valued points of $R''$.

Similarly for morphisms: Let $(u'_1, u_1, r'_1)$ and $(u'_2, u_2, r'_2)$ be two objects of the fibre product over $T$. Then a morphism from $(u'_2, u_2, r'_2)$ to $(u'_1, u_1, r'_1)$ is given by $(1, r)$ where $1 : u'_1 \to u'_2$ means simply $u'_1 = u'_2$ (this is so because $\mathcal{S}_ U$ is fibred in sets), and $r$ is a $T$-valued point of $R$ with $s(r) = u_2$, $t(r) = u_1$ and moreover $c'(f(r), r'_2) = r'_1$. Hence the arrow

is completely determined by knowing the pair $(r, r'_2)$. Thus the functor of arrows is represented by $R''$, and moreover the morphisms $s''$, $t''$, and $c''$ clearly correspond to source, target and composition in the $2$-fibre product $\mathcal{S}_ U \times _{[U'/_{\! p}R']} [U/_{\! p}R]$. $\square$