Lemma 59.40.3. Let X, Y be schemes. Any two morphisms a, b : X \to Y of schemes for which there exists a 2-isomorphism (a_{small}, a_{small}^\sharp ) \cong (b_{small}, b_{small}^\sharp ) in the 2-category of ringed topoi are equal.
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
t : a_{small}^{-1}h_ V = h_{a^{-1}(V)} \to b_{small}^{-1}h_ V = h_{b^{-1}(V)}
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
\xymatrix{ h_{b^{-1}(V)} \ar@{=}[r] & b_{small}^{-1}h_ V \ar[d]^{b_{small}^{-1}(f)} & & a_{small}^{-1}h_ V \ar[d]^{a_{small}^{-1}(f)} \ar[ll]^ t & h_{a^{-1}(V)} \ar@{=}[l] \\ & b_{small}^{-1}\mathcal{O}_ Y \ar[rd]_{b^\sharp } & & a_{small}^{-1}\mathcal{O}_ Y \ar[ll]^ t \ar[ld]^{a^\sharp } \\ & & \mathcal{O}_ X }
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
Comments (0)