Lemma 106.16.6. Let \mathcal{X} be an algebraic stack with affine diagonal. Let B be a ring. Let f_ i : \mathop{\mathrm{Spec}}(B) \to \mathcal{X}, i = 1, 2 be two morphisms. Let t : f_1^* \to f_2^* be an isomorphism of the tensor functors f_ i^* : \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \text{Mod}_ B. Then there is a 2-arrow f_1 \to f_2 inducing t.
Proof. Choose an affine scheme U = \mathop{\mathrm{Spec}}(A) and a surjective smooth morphism g : U \to \mathcal{X}, see Properties of Stacks, Lemma 100.6.2. Since the diagonal of \mathcal{X} is affine, we see that U_ i = \mathop{\mathrm{Spec}}(B) \times _{f_ i, \mathcal{X}, g} U is affine. Say U_ i = \mathop{\mathrm{Spec}}(C_ i). Then C_ i is the B-algebra endowed with ring map A \to C_ i constructed in Lemma 106.16.2 using the functor F = f_ i^*. Therefore t induces an isomorphism C_1 \to C_2 of B-algebras, compatible with the ring maps A \to C_1 and A \to C_2. In other words, we have a commutative diagrams
This already shows that the objects f_1 and f_2 of \mathcal{X} over \mathop{\mathrm{Spec}}(B) become isomorphic after the smooth covering \{ U_1 \to \mathop{\mathrm{Spec}}(B)\} . To show that this descends to an isomorphism of f_1 and f_2 over \mathop{\mathrm{Spec}}(B), we have to show that our isomorphism (which comes from the commutative diagrams above) is compatible with the descent data over U_1 \times _{\mathop{\mathrm{Spec}}(B)} U_1. For this we observe that U \times _\mathcal {X} U is affine too, that we have the morphism g' : U \times _\mathcal {X} U \to \mathcal{X}, and that
It follows that the isomorphism C_1 \otimes _ B C_1 \to C_2 \otimes _ B C_2 coming from the isomorphism C_1 \to C_2 is compatible with the morphisms U_ i \times _{\mathop{\mathrm{Spec}}(B)} U_ i \to U \times _\mathcal {X} U. Some details omitted. \square
Comments (0)