Remark 2

If $\gamma :\mathscr{G}_1\rightarrow \mathscr{G}_2$ is a presheaf morphism on B, the presheaf morphisms $\beta_{\mathscr{G}_2}\circ \gamma$ and $u_*u^*(\gamma )\circ \beta_{\mathscr{G}_1}$ are equal. In fact, if V is an open set of B and $s\in \mathscr{G}_1(V),\beta_{\mathscr{G}_1}(s)$ is the section $t$ of $A\times_BE_{\mathscr{G}_1}$ over $\overset{-1}{u}(V)$, defined by $x\mapsto (x,[V, s, u(x)])$. The image of $t$ under $u^*(\gamma )$ is therefore the section of $A\times_BE_{\mathscr{G}_2}$ over $\overset{-1}{u}(V)$ given by $x\mapsto (x,[V, \gamma (s), u(x)])$. Hence indeed $u_*u^*(\gamma )\circ \beta_{\mathscr{G}_1}(s) =\beta_{\mathscr{G}_2}(\gamma (s))$.