The Stacks project

Proof. Let us argue this carefully since it is a bit confusing. Let $t : a_{small}^{-1} \to b_{small}^{-1}$ be the $2$-isomorphism. Consider any open $V \subset Y$. Note that $h_ V$ is a subsheaf of the final sheaf $*$. Thus both $a_{small}^{-1}h_ V = h_{a^{-1}(V)}$ and $b_{small}^{-1}h_ V = h_{b^{-1}(V)}$ are subsheaves of the final sheaf. Thus the isomorphism

has to be the identity, and $a^{-1}(V) = b^{-1}(V)$. It follows that $a$ and $b$ are equal on underlying topological spaces. Next, take a section $f \in \mathcal{O}_ Y(V)$. This determines and is determined by a map of sheaves of sets $f : h_ V \to \mathcal{O}_ Y$. Pull this back and apply $t$ to get a commutative diagram

where the triangle is commutative by definition of a $2$-isomorphism in Modules on Sites, Section 18.8. Above we have seen that the composition of the top horizontal arrows comes from the identity $a^{-1}(V) = b^{-1}(V)$. Thus the commutativity of the diagram tells us that $a_{small}^\sharp (f) = b_{small}^\sharp (f)$ in $\mathcal{O}_ X(a^{-1}(V)) = \mathcal{O}_ X(b^{-1}(V))$. Since this holds for every open $V$ and every $f \in \mathcal{O}_ Y(V)$ we conclude that $a = b$ as morphisms of schemes. $\square$