Lemma 105.16.6. 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 $f_ i^* : \text{QCoh}(\mathcal{O}_\mathcal {X}) \to \text{Mod}_ B$. 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 99.6.2. 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 105.16.2 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. $\square$

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