hcju/stacksqa huggingface dataset:
By Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation} we may assume that $\mathcal{X} = [U/R]$ for some smooth groupoid in algebraic spaces. By Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-locally-finite-type} it suffices to check the property fppf locally on $T$. Thus we may assume that $x, y$ come from morphisms $x', y' : T \to U$. By Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-morphisms} we see that in this case $\mathit{Isom}_\mathcal{X}(x, y) = T \times_{(y', x'), U \times_S U} R$. Hence it suffices to prove that $R \to U \times_S U$ is locally of finite type. This follows from the fact that the composition $s : R \to U \times_S U \to U$ is smooth (hence locally of finite type, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-smooth-locally-finite-presentation} and \ref{spaces-morphisms-lemma-finite-presentation-finite-type}) and Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-finite-type}.
Let $T$ be a scheme and let $a : T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ be a morphism. By definition of the fibre product and the $2$-Yoneda lemma the morphism $a$ is given by a triple $a = (x, x', \alpha)$ where $x, x'$ are objects of $\mathcal{X}$ over $T$, and $\alpha : f(x) \to f(x')$ is a morphism in the fibre category of $\mathcal{Y}$ over $T$. By definition of an algebraic stack the sheaves $\mathit{Isom}_\mathcal{X}(x, x')$ and $\mathit{Isom}_\mathcal{Y}(f(x), f(x'))$ are algebraic spaces over $T$. In this language $\alpha$ defines a section of the morphism $\mathit{Isom}_\mathcal{Y}(f(x), f(x')) \to T$. A $T'$-valued point of $\mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}, a} T$ for $T' \to T$ a scheme over $T$ is the same thing as an isomorphism $x|_{T'} \to x'|_{T'}$ whose image under $f$ is $\alpha|_{T'}$. Thus we see that \begin{equation} \label{equation-diagonal} \vcenter{ \xymatrix{ \mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}, a} T \ar[d] \ar[r] & \mathit{Isom}_\mathcal{X}(x, x') \ar[d] \\ T\ar[r]^-\alpha & \mathit{Isom}_\mathcal{Y}(f(x), f(x')) } } \end{equation} is a fibre square of sheaves over $T$. In particular we see that $\mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}, a} T$ is an algebraic space which proves part (1) of the lemma. \medskip\noindent To prove the second statement we have to show that the left vertical arrow of Diagram (\ref{equation-diagonal}) is locally of finite type. By Lemma \ref{lemma-isom-locally-finite-type} the algebraic space $\mathit{Isom}_\mathcal{X}(x, x')$ and is locally of finite type over $T$. Hence the right vertical arrow of Diagram (\ref{equation-diagonal}) is locally of finite type, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-finite-type}. We conclude by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-type}.
We have already seen in Lemma \ref{lemma-properties-diagonal} that $\Delta_f$ is representable by algebraic spaces. Hence the statements (2) -- (5) make sense, see Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. Also Lemma \ref{lemma-properties-diagonal} guarantees (2) holds. Let $T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ be a morphism and contemplate Diagram (\ref{equation-diagonal}). By Algebraic Stacks, Lemma \ref{algebraic-lemma-criterion-map-representable-spaces-fibred-in-groupoids} the right vertical arrow is injective as a map of sheaves, i.e., a monomorphism of algebraic spaces. Hence also the morphism $T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X} \to T$ is a monomorphism. Thus (3) holds. We already know that $T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X} \to T$ is locally of finite type. Thus Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-loc-finite-type-loc-quasi-finite} allows us to conclude that $T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X} \to T$ is locally quasi-finite and separated. This proves (4) and (5). Finally, Morphisms of Spaces, Proposition \ref{spaces-morphisms-proposition-locally-quasi-finite-separated-over-scheme} implies that $T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X}$ is a scheme which proves (1).
The statements ``$f$ is separated'', ``$\Delta_f$ is a closed immersion'', ``$\Delta_f$ is universally closed'', and ``$\Delta_f$ is proper'' refer to the notions defined in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = \mathcal{X} \times_\mathcal{Y} V$ which is an algebraic space by assumption, and the morphism $U \to \mathcal{X}$ is surjective and smooth. By Categories, Lemma \ref{categories-lemma-base-change-diagonal} and Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering} we see that for any property $P$ (as in that lemma) we have: $\Delta_f$ has $P$ if and only if $\Delta_{U/V} : U \to  U \times_V U$ has $P$. Hence the equivalence of (2), (3) and (4) follows from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-separated-diagonal-proper} applied to $U \to V$. Moreover, if (1) holds, then $U \to V$ is separated and we see that $\Delta_{U/V}$ is a closed immersion, i.e., (2) holds. Finally, assume (2) holds. Let $T$ be a scheme, and $a : T \to \mathcal{Y}$ a morphism. Set $T' = \mathcal{X} \times_\mathcal{Y} T$. To prove (1) we have to show that the morphism of algebraic spaces $T' \to T$ is separated. Using Categories, Lemma \ref{categories-lemma-base-change-diagonal} once more we see that $\Delta_{T'/T}$ is the base change of $\Delta_f$. Hence our assumption (2) implies that $\Delta_{T'/T}$ is a closed immersion, hence $T' \to T$ is separated as desired.
The statements ``$f$ is quasi-separated'', ``$\Delta_f$ is quasi-compact'', and ``$\Delta_f$ is of finite type'' refer to the notions defined in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. Note that (2) and (3) are equivalent in view of the fact that $\Delta_f$ is locally of finite type by Lemma \ref{lemma-properties-diagonal-representable} (and Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-transformations-property-implication}). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = \mathcal{X} \times_\mathcal{Y} V$ which is an algebraic space by assumption, and the morphism $U \to \mathcal{X}$ is surjective and smooth. By Categories, Lemma \ref{categories-lemma-base-change-diagonal} and Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering} we see that we have: $\Delta_f$ is quasi-compact if and only if $\Delta_{U/V} : U \to  U \times_V U$ is quasi-compact. If (1) holds, then $U \to V$ is quasi-separated and we see that $\Delta_{U/V}$ is quasi-compact, i.e., (2) holds. Assume (2) holds. Let $T$ be a scheme, and $a : T \to \mathcal{Y}$ a morphism. Set $T' = \mathcal{X} \times_\mathcal{Y} T$. To prove (1) we have to show that the morphism of algebraic spaces $T' \to T$ is quasi-separated. Using Categories, Lemma \ref{categories-lemma-base-change-diagonal} once more we see that $\Delta_{T'/T}$ is the base change of $\Delta_f$. Hence our assumption (2) implies that $\Delta_{T'/T}$ is quasi-compact, hence $T' \to T$ is quasi-separated as desired.
The statements ``$f$ is locally separated'', and ``$\Delta_f$ is an immersion'' refer to the notions defined in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. Proof omitted. Hint: Argue as in the proofs of Lemmas \ref{lemma-representable-separated-diagonal-closed} and \ref{lemma-representable-quasi-separated-diagonal-quasi-compact}.
To see (1) note that a proper morphism of algebraic spaces is quasi-compact and quasi-separated, see Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-proper}. To see (2) note that an unramified morphism of algebraic spaces is locally quasi-finite, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-unramified-quasi-finite}. Finally (3) follows from Lemma \ref{lemma-properties-diagonal-representable}.
Let $f : \mathcal{X} \to \mathcal{Y}$ and $\mathcal{Y}' \to \mathcal{Y}$ be morphisms of algebraic stacks. Let $f' : \mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'$ be the base change of $f$ by $\mathcal{Y}' \to \mathcal{Y}$. Then $\Delta_{f'}$ is the base change of $\Delta_f$ by the morphism $\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}' \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$, see Categories, Lemma \ref{categories-lemma-base-change-diagonal}. By the results of Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} each of the properties of the diagonal used in Definition \ref{definition-separated} is stable under base change. Hence the lemma is true.
Denote $g : W \times_\mathcal{Y} \mathcal{X} \to W$ the base change. Then $\Delta_g$ is the base change of $\Delta_f$ by the morphism $q : W \times_\mathcal{Y} (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$. Since $q$ is the base change of $W \to \mathcal{Y}$ we see that $q$ is representable by algebraic spaces, surjective, flat, and locally of finite presentation. Hence the result follows from Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-weak-covering}.
The following diagram $$ \xymatrix{ \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \ar[r]_i \ar[d] & \mathcal{X} \times_\mathcal{T} \mathcal{Y} \ar[d] \\ \mathcal{Z} \ar[r]^-{\Delta_{\mathcal{Z}/\mathcal{T}}} \ar[r] & \mathcal{Z} \times_\mathcal{T} \mathcal{Z} } $$ is a $2$-fibre product diagram, see Categories, Lemma \ref{categories-lemma-fibre-product-after-map}. Hence $i$ is the base change of the diagonal morphism $\Delta_{\mathcal{Z}/\mathcal{T}}$. Thus the lemma follows from Lemma \ref{lemma-properties-diagonal}, and the material in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}.
Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks to which the axiom in question applies. The diagonal $\Delta_{\mathcal{X}/\mathcal{Z}}$ is the composition $$ \mathcal{X} \longrightarrow \mathcal{X} \times_\mathcal{Y} \mathcal{X} \longrightarrow \mathcal{X} \times_\mathcal{Z} \mathcal{X}. $$ Our separation axiom is defined by requiring the diagonal to have some property $\mathcal{P}$. By Lemma \ref{lemma-fibre-product-after-map} above we see that the second arrow also has this property. Hence the lemma follows since the composition of morphisms which are representable by algebraic spaces with property $\mathcal{P}$ also is a morphism with property $\mathcal{P}$, see our general discussion in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} and Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-unramified}, \ref{spaces-morphisms-lemma-composition-quasi-finite}, \ref{spaces-morphisms-lemma-composition-proper}, \ref{spaces-morphisms-lemma-composition-quasi-compact}, and \ref{spaces-morphisms-lemma-composition-separated}.
Parts (1), (2), (3), and (4) follow immediately from Lemma \ref{lemma-composition-separated} and Definition \ref{definition-absolute-separated}. For (5), (6), (7), and (8) think of $\mathcal{X}$ and $\mathcal{Y}$ as algebraic stacks over $\Spec(\mathbf{Z})$ and apply Lemma \ref{lemma-composition-separated}. Details omitted.
Parts (5), (6), (7), and (8) follow immediately from Lemma \ref{lemma-compose-after-separated} and Spaces, Definition \ref{spaces-definition-separated}. To prove (3) and (4) think of $X$ and $Y$ as algebraic stacks over $\Spec(\mathbf{Z})$ and apply Lemma \ref{lemma-compose-after-separated}. Similarly, to prove (1) and (2), think of $\mathcal{X}$ as an algebraic stack over $\Spec(\mathbf{Z})$ consider the morphisms $$ \mathcal{X} \longrightarrow \mathcal{X} \times_S \mathcal{X} \longrightarrow \mathcal{X} \times_{\Spec(\mathbf{Z})} \mathcal{X} $$ Both arrows are representable by algebraic spaces. The second arrow is unramified and locally quasi-finite as the base change of the immersion $\Delta_{S/\mathbf{Z}}$. Hence the composition is unramified (resp.\ locally quasi-finite) if and only if the first arrow is unramified (resp.\ locally quasi-finite), see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-unramified} and \ref{spaces-morphisms-lemma-permanence-unramified} (resp.\ Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-quasi-finite} and \ref{spaces-morphisms-lemma-permanence-quasi-finite}).
Note that if $f$ is unramified, then it is \'etale by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-unramified-flat-lfp-etale}. This explains the parenthetical remark in (1). Assume $f$ is unramified (resp.\ locally quasi-finite). We have to show that $\Delta_\mathcal{X} : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is unramified (resp.\ locally quasi-finite). Note that $W \times W \to \mathcal{X} \times \mathcal{X}$ is also surjective, flat, and locally of finite presentation. Hence it suffices to show that $$ W \times_{\mathcal{X} \times \mathcal{X}, \Delta_\mathcal{X}} \mathcal{X} = W \times_\mathcal{X} W \longrightarrow W \times W $$ is unramified (resp.\ locally quasi-finite), see Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering}. By assumption the morphism $\text{pr}_i : W \times_\mathcal{X} W \to W$ is unramified (resp.\ locally quasi-finite). Hence the displayed arrow is unramified (resp.\ locally quasi-finite) by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-unramified} (resp.\ Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-quasi-finite}).
If $f : \mathcal{X} \to \mathcal{Y}$ is a monomorphism of algebraic stacks, then $\Delta_f$ is an isomorphism, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-monomorphism}. Since an isomorphism of algebraic spaces is proper and unramified we see that $f$ is separated and DM. The second assertion follows from the first as an immersion is a monomorphism, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-immersion-monomorphism}.
By Categories, Lemma \ref{categories-lemma-inertia-fibred-category} there are equivalences $$ \mathcal{I}_\mathcal{X} \to \mathcal{X} \times_{\Delta, \mathcal{X} \times_S \mathcal{X}, \Delta} \mathcal{X} \quad\text{and}\quad \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X} \times_{\Delta, \mathcal{X} \times_\mathcal{Y} \mathcal{X}, \Delta} \mathcal{X} $$ which shows that the inertia stacks are algebraic stacks. Let $T \to \mathcal{X}$ be a morphism given by the object $x$ of the fibre category of $\mathcal{X}$ over $T$. Then we get a $2$-fibre product square $$ \xymatrix{ \mathit{Isom}_\mathcal{X}(x, x) \ar[d] \ar[r] & \mathcal{I}_\mathcal{X} \ar[d] \\ T \ar[r]^x & \mathcal{X} } $$ This follows immediately from the definition of $\mathcal{I}_\mathcal{X}$. Since $\mathit{Isom}_\mathcal{X}(x, x)$ is always an algebraic space locally of finite type over $T$ (see Lemma \ref{lemma-isom-locally-finite-type}) we conclude that $\mathcal{I}_\mathcal{X} \to \mathcal{X}$ is representable by algebraic spaces and locally of finite type. Finally, for the relative inertia we get $$ \vcenter{ \xymatrix{ \mathit{Isom}_\mathcal{X}(x, x) \ar[d] & K \ar[l] \ar[d] \ar[r] & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] \\ \mathit{Isom}_\mathcal{Y}(f(x), f(x)) & T \ar[l]_-e \ar[r]^x & \mathcal{X} } } $$ with both squares $2$-fibre products. This follows from Categories, Lemma \ref{categories-lemma-relative-inertia-as-fibre-product}. The left vertical arrow is a morphism of algebraic spaces locally of finite type over $T$, and hence is locally of finite type, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-finite-type}. Thus $K$ is an algebraic space and $K \to T$ is locally of finite type. This proves the assertion on the relative inertia.
Part (1) follows from Definition \ref{definition-isom}. Part (2) comes from the exact sequence (\ref{equation-exact-sequence-isom}) \'etale locally on $Z$. Part (3) can be seen by unwinding the definitions. Locally on $Z$ in the \'etale topology part (4) reduces to part (2) of Lemma \ref{lemma-isom-pseudo-torsor-aut}.
The inertia stack $\mathcal{I}_{\mathcal{X}'/\mathcal{Y}'}$ is defined as the category of pairs $(x', \alpha')$ where $x'$ is an object of $\mathcal{X}'$ and $\alpha'$ is an automorphism of $x'$ with $\pi'(\alpha') = \text{id}$, see Categories, Section \ref{categories-section-inertia}. Suppose that $x'$ lies over the scheme $U$ and maps to the object $x$ of $\mathcal{X}$. By the construction of the $2$-fibre product in Categories, Lemma \ref{categories-lemma-2-product-categories-over-C} we see that $x' = (U, x, y', \beta)$ where $y'$ is an object of $\mathcal{Y}'$ over $U$ and $\beta$ is an isomorphism $\beta : \pi(x) \to f(y')$ in the fibre category of $\mathcal{Y}$ over $U$. By the very construction of the $2$-fibre product the automorphism $\alpha'$ is a pair $(\alpha, \gamma)$ where $\alpha$ is an automorphism of $x$ over $U$ and $\gamma$ is an automorphism of $y'$ over $U$ such that $\alpha$ and $\gamma$ are compatible via $\beta$. The condition $\pi'(\alpha') = \text{id}$ signifies that $\gamma = \text{id}$ whereupon the condition that $\alpha, \beta, \gamma$ are compatible is exactly the condition $\pi(\alpha) = \text{id}$, i.e., means exactly that $(x, \alpha)$ is an object of $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$. In this way we see that the left square is a fibre product square (some details omitted).
This follows immediately from the fact that $f$ is fully faithful (see Properties of Stacks, Lemma \ref{stacks-properties-lemma-monomorphism}) and the definition of the inertia in Categories, Section \ref{categories-section-inertia}. Namely, an object of $\mathcal{I}_\mathcal{X}$ over a scheme $T$ is the same thing as a pair $(x, \alpha)$ consisting of an object $x$ of $\mathcal{X}$ over $T$ and a morphism $\alpha : x \to x$ in the fibre category of $\mathcal{X}$ over $T$. As $f$ is fully faithful we see that $\alpha$ is the same thing as a morphism $\beta : f(x) \to f(x)$ in the fibre category of $\mathcal{Y}$ over $T$. Hence we can think of objects of $\mathcal{I}_\mathcal{X}$ over $T$ as triples $((y, \beta), x, \gamma)$ where $y$ is an object of $\mathcal{Y}$ over $T$, $\beta : y \to y$ in $\mathcal{Y}_T$ and $\gamma : y \to f(x)$ is an isomorphism over $T$, i.e., an object of $\mathcal{I}_\mathcal{Y} \times_\mathcal{Y} \mathcal{X}$ over $T$.
Proof of (1), (2), and (3). Choose an algebraic space $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then $G = U \times_\mathcal{X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an algebraic space over $U$ (Lemma \ref{lemma-inertia}). In fact, $G$ is a group algebraic space over $U$ by the group law on relative inertia constructed in Remark \ref{remark-inertia-is-group-in-spaces}. Moreover, $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is surjective and smooth as a base change of $U \to \mathcal{X}$. Finally, the base change of $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ by $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is the identity $U \to G$ of $G/U$. Thus the equivalence of (a) and (c) follows from Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-group-scheme-separated}. Since $\Delta_{f, 2}$ is the diagonal of $\Delta_f$ we have (b) $\Leftrightarrow$ (c) by definition. \medskip\noindent Proof of (4) and (5). Recall that (4)(b) means $\Delta_f$ is unramified and (5)(b) means that $\Delta_f$ is locally quasi-finite. Choose a scheme $Z$ and a morphism $a : Z \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$. Then $a = (x_1, x_2, \alpha)$ where $x_i : Z \to \mathcal{X}$ and $\alpha : f \circ x_1  \to f \circ x_2$ is a $2$-morphism. Recall that $$ \vcenter{ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2) \ar[d] \ar[r] & Z \ar[d] \\ \mathcal{X} \ar[r]^{\Delta_f} & \mathcal{X} \times_\mathcal{Y} \mathcal{X} } } \quad\text{and}\quad \vcenter{ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2) \ar[d] \ar[r] & Z \ar[d]^{x_2} \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} } } $$ are cartesian squares. By Lemma \ref{lemma-isom-pseudo-torsor-aut-over-space} the algebraic space $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2)$ is a pseudo torsor for $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)$ over $Z$. Thus the equivalences in (4) and (5) follow from Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-pseudo-torsor-implications}.
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 \ref{algebraic-lemma-characterize-representable-by-algebraic-spaces}. 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 \ref{lemma-inertia} 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 $T$.
Recall from Properties of Stacks, Lemma \ref{stacks-properties-lemma-monomorphism} that a morphism of algebraic stacks is a monomorphism if and only if its diagonal is an isomorphism of stacks. Thus Lemma \ref{lemma-second-diagonal} can be rephrased as saying that a morphism is representable by algebraic spaces if the diagonal is a monomorphism. In particular, it shows that condition (3) of Lemma \ref{lemma-properties-diagonal-representable} is actually an if and only if, i.e., a morphism of algebraic stacks is representable by algebraic spaces if and only if its diagonal is a monomorphism.
Proof of (1). Assume that $\Delta_{f, 2}$ and $\Delta_{f, 1}$ are universally closed. Then $\Delta_{f, 1}$ is separated and universally closed by Lemma \ref{lemma-first-diagonal-separated-second-diagonal-closed}. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-universally-closed-quasi-compact} and Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-transformations-property-implication} we see that $\Delta_{f, 1}$ is quasi-compact. Hence it is quasi-compact, separated, universally closed and locally of finite type (by Lemma \ref{lemma-properties-diagonal}) so proper. This proves ``$\Leftarrow$'' of (1). The proof of the implication in the other direction is omitted. \medskip\noindent Proof of (2). This follows immediately from Lemma \ref{lemma-first-diagonal-separated-second-diagonal-closed}. \medskip\noindent Proof of (3). This follows from the fact that $\Delta_{f, 2}$ is always locally quasi-finite by Lemma \ref{lemma-properties-diagonal-representable} applied to $\Delta_f = \Delta_{f, 1}$. \medskip\noindent Proof of (4). This follows from the fact that $\Delta_{f, 2}$ is always unramified as Lemma \ref{lemma-properties-diagonal-representable} applied to $\Delta_f = \Delta_{f, 1}$ shows that $\Delta_{f, 2}$ is locally of finite type and a monomorphism. See More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-universally-injective-unramified}.