Lemma 101.6.2. Let f : \mathcal{X} \to \mathcal{Y} be a morphism of algebraic stacks. The following are equivalent:
the morphism f is representable by algebraic spaces,
the second diagonal of f is an isomorphism,
the group stack \mathcal{I}_{\mathcal{X}/\mathcal{Y}} is trivial over \mathcal X, and
for a scheme T and a morphism x : T \to \mathcal{X} the kernel of \mathit{Isom}_\mathcal {X}(x, x) \to \mathit{Isom}_\mathcal {Y}(f(x), f(x)) is trivial.
Proof. We first prove the equivalence of (1) and (2). Namely, f is representable by algebraic spaces if and only if f is faithful, see Algebraic Stacks, Lemma 94.15.2. On the other hand, f is faithful if and only if for every object x of \mathcal{X} over a scheme T the functor f induces an injection \mathit{Isom}_\mathcal {X}(x, x) \to \mathit{Isom}_\mathcal {Y}(f(x), f(x)), which happens if and only if the kernel K is trivial, which happens if and only if e : T \to K is an isomorphism for every x : T \to \mathcal{X}. Since K = T \times _{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}} as discussed above, this proves the equivalence of (1) and (2). To prove the equivalence of (2) and (3), by the discussion above, it suffices to note that a group stack is trivial if and only if its identity section is an isomorphism. Finally, the equivalence of (3) and (4) follows from the definitions: in the proof of Lemma 101.5.1 we have seen that the kernel in (4) corresponds to the fibre product T \times _{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}} over T. \square
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