The Stacks project

Proof. Let $\{ y_ i \to y\}$ be a $\tau$-covering in $\mathcal{Y}$. By definition this simply means that $\{ q(y_ i) \to q(y)\}$ is a $\tau$-covering of schemes. By the final remark above the lemma we see that $\{ p(u(y_ i)) \to p(u(y))\}$ is the base change of the $\tau$-covering $\{ q(y_ i) \to q(y)\}$ by $p(u(y)) \to q(y)$, hence is itself a $\tau$-covering by the axioms of a site. Hence $\{ u(y_ i) \to u(y)\}$ is a $\tau$-covering of $\mathcal{X}$. This proves that $u$ is continuous.

Let's use the notation $u_ p, u_ s, u^ p, u^ s$ of Sites, Sections 7.5 and 7.13. If we can show the final assertion of the lemma, then we see that $f_* = u^ p = u^ s$ (by continuity of $u$ seen above) and hence by adjointness $f^{-1} = u_ s$ which will prove $u_ s$ is exact, hence that $u$ determines a morphism of sites, and the equality will be clear as well. To see that $f_*\mathcal{F}(y) = \mathcal{F}(u(y))$ note that by definition

Since $u(y)$ is a final object in the category the limit is taken over we conclude. $\square$