hcju/stacksqa huggingface dataset:
To check $\Picardfunctor_{X/B} \to B$ is quasi-separated we have to show that its diagonal is quasi-compact. This is immediate from Lemma \ref{lemma-pic-functor-diagonal-qc-immersion}. Since the morphism $\Picardstack_{X/B} \to \Picardfunctor_{X/B}$ is surjective, flat, and locally of finite presentation (by Lemma \ref{lemma-pic-gerbe-over-pic-functor} and Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-gerbe-fppf}) it suffices to prove that $\Picardstack_{X/B} \to B$ is locally of finite presentation, see Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-flat-finite-presentation-permanence}. This follows from Lemma \ref{lemma-pic-qs-lfp}.
Since $\Picardfunctor_{X/B} \to B$ is quasi-separated, it suffices to check the uniqueness part of the valuative criterion, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-valuative-criterion-separatedness}. This immediately reduces to the following problem: given \begin{enumerate} \item a valuation ring $R$ with fraction field $K$, \item an algebraic space $X$ proper and flat over $R$ with integral geometric fibre, \item an element $a \in \Picardfunctor_{X/R}(R)$ with $a|_{\Spec(K)} = 0$, \end{enumerate} then we have to prove $a = 0$. Applying Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-lift-valuation-ring-through-flat-morphism} to the surjective flat morphism $\Picardstack_{X/R} \to \Picardfunctor_{X/R}$ (surjective and flat by Lemma \ref{lemma-pic-gerbe-over-pic-functor} and Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-gerbe-fppf}) after replacing $R$ by an extension we may assume $a$ is given by an invertible $\mathcal{O}_X$-module $\mathcal{L}$. Since $a|_{\Spec(K)} = 0$ we find $\mathcal{L}_K \cong \mathcal{O}_{X_K}$ by Quot, Lemma \ref{quot-lemma-flat-geometrically-connected-fibres}. \medskip\noindent Denote $f : X \to \Spec(R)$ the structure morphism. Let $\eta, 0 \in \Spec(R)$ be the generic and closed point. Consider the perfect complexes $K = Rf_*\mathcal{L}$ and $M = Rf_*(\mathcal{L}^{\otimes -1})$ on $\Spec(R)$, see Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-flat-proper-perfect-direct-image-general}. Consider the functions $\beta_{K, i}, \beta_{M, i} : \Spec(R) \to \mathbf{Z}$ of Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-jump-loci} associated to $K$ and $M$. Since the formation of $K$ amd $M$ commutes with base change (see lemma cited above) we find $\beta_{K, 0}(\eta) = \beta_{M, 0}(\beta) = 1$ by Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-proper-geometrically-reduced-global-sections} and our assumption on the fibres of $f$. By upper semi-continuity we find $\beta_{K, 0}(0) \geq 1$ and $\beta_{M, 0} \geq 1$. By  Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-characterize-trivial-pic-integral} we conclude that the restriction of $\mathcal{L}$ to the special fibre $X_0$ is trivial. In turn this gives $\beta_{K, 0}(0) = \beta_{M, 0} = 1$ as above. Then by More on Algebra, Lemma \ref{more-algebra-lemma-lift-pseudo-coherent-from-residue-field} we can represent $K$ by a complex of the form $$ \ldots \to 0 \to R \to R^{\oplus \beta_{K, 1}(0)} \to R^{\oplus \beta_{K, 2}(0)} \to \ldots $$ Now $R \to R^{\oplus \beta_{K, 1}(0)}$ is zero because $\beta_{K, 0}(\eta) = 1$. In other words $K = R \oplus \tau_{\geq 1}(K)$ in $D(R)$ where $\tau_{\geq 1}(K)$ has tor amplitude in $[1, b]$ for some $b \in \mathbf{Z}$. Hence there is a global section $s \in H^0(X, \mathcal{L})$ whose restriction $s_0$ to $X_0$ is nonvanishing (again because formation of $K$ commutes with base change). Then $s : \mathcal{O}_X \to \mathcal{L}$ is a map of invertible sheaves whose restriction to $X_0$ is an isomorphism and hence is an isomorphism as desired.
By Lemma \ref{lemma-pic-curves-smooth} we know that $\Picardstack_{X/B} \to B$ is smooth. The morphism $\Picardstack_{X/B} \to \Picardfunctor_{X/B}$ is surjective and smooth by combining Lemma \ref{lemma-pic-gerbe-over-pic-functor} with Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-gerbe-smooth}. Thus if $U$ is a scheme and $U \to \Picardstack_{X/B}$ is surjective and smooth, then $U \to \Picardfunctor_{X/B}$ is surjective and smooth and $U \to B$ is surjective and smooth (because these properties are preserved by composition). Thus $\Picardfunctor_{X/B} \to B$ is smooth for example by Descent on Spaces, Lemma \ref{spaces-descent-lemma-syntomic-smooth-etale-permanence}.
By Lemma \ref{lemma-Mor-s-lfp} the morphism $\mathit{Mor}_B(Y, X) \to B$ is separated and locally of finite presentation. Thus it suffices to show that the open and closed subspace $\mathit{Mor}^{P, \mathcal{M}}_B(Y, X)$ of Remark \ref{remark-Mor-numerical} is quasi-compact over $B$. \medskip\noindent The question is \'etale local on $B$ (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-compact-local}). Thus we may assume $B$ is affine. \medskip\noindent Assume $B = \Spec(\Lambda)$. Note that $X$ and $Y$ are schemes and that $\mathcal{L}$ and $\mathcal{N}$ are ample invertible sheaves on $X$ and $Y$ (this follows immediately from the definitions). Write $\Lambda = \colim \Lambda_i$ as the colimit of its finite type $\mathbf{Z}$-subalgebras. Then we can find an $i$ and a system $X_i, Y_i, \mathcal{L}_i, \mathcal{N}_i$ as in the lemma over $B_i = \Spec(\Lambda_i)$ whose base change to $B$ gives $X, Y, \mathcal{L}, \mathcal{N}$. This follows from Limits, Lemmas \ref{limits-lemma-descend-finite-presentation} (to find $X_i$, $Y_i$), \ref{limits-lemma-descend-invertible-modules} (to find $\mathcal{L}_i$, $\mathcal{N}_i$), \ref{limits-lemma-descend-separated-finite-presentation} (to make $X_i \to B_i$ separated), \ref{limits-lemma-eventually-proper} (to make $Y_i \to B_i$ proper), and \ref{limits-lemma-limit-ample} (to make $\mathcal{L}_i$, $\mathcal{N}_i$ ample). Because $$ \mathit{Mor}_B(Y, X) = B \times_{B_i} \mathit{Mor}_{B_i}(Y_i, X_i) $$ and similarly for $\mathit{Mor}^P_B(Y, X)$ we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $B$ is a Noetherian affine scheme. By Properties, Lemma \ref{properties-lemma-ample-on-product} we see that $\mathcal{M}$ is ample. By Lemma \ref{lemma-hilb-qc-over-base} we see that $\Hilbfunctor^{P, \mathcal{M}}_{Y \times_B X/B}$ is of finite presentation over $B$ and hence Noetherian. By construction $$ \mathit{Mor}^{P, \mathcal{M}}_B(Y, X) = \mathit{Mor}_B(Y, X) \cap \Hilbfunctor^{P, \mathcal{M}}_{Y \times_B X/B} $$ is an open subspace of $\Hilbfunctor^{P, \mathcal{M}}_{Y \times_B X/B}$ and hence quasi-compact (as an open of a Noetherian algebraic space is quasi-compact).
To check $\Polarizedstack \to \Spec(\mathbf{Z})$ is quasi-separated we have to show that its diagonal is quasi-compact and quasi-separated. This is immediate from Lemma \ref{lemma-polarized-diagonal-separated-fp}. To prove that $\Polarizedstack \to \Spec(\mathbf{Z})$ is locally of finite presentation, it suffices to show that $\Polarizedstack$ is limit preserving, see Limits of Stacks, Proposition \ref{stacks-limits-proposition-characterize-locally-finite-presentation}. This is Quot, Lemma \ref{quot-lemma-polarized-limits}.
Observe that $|\Polarizedstack|$ is a locally Noetherian topological space, see Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-Noetherian-topology} (this also uses that $\Spec(\mathbf{Z})$ is Noetherian and hence $\Polarizedstack$ is a locally Noetherian algebraic stack by Lemma \ref{lemma-polarized-qs-lfp} and Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-locally-finite-type-locally-noetherian}). Thus any quasi-compact subset of $|\Polarizedstack|$ is a Noetherian topological space and any subset of such is also Noetherian, see Topology, Lemmas \ref{topology-lemma-finite-union-Noetherian} and \ref{topology-lemma-Noetherian}. Thus all we have to do is a find a quasi-compact subset containing $T$. \medskip\noindent By Lemma \ref{lemma-hilb-proper-over-base} the algebraic space $$ H = \Hilbfunctor^{P, \mathcal{O}(1)}_{\mathbf{P}^n_\mathbf{Z}/\Spec(\mathbf{Z})} $$ is proper over $\Spec(\mathbf{Z})$. By Quot, Lemma \ref{quot-lemma-extend-hilb-to-spaces}\footnote{We will see later (insert future reference here) that $H$ is a scheme and hence the use of this lemma and Quot, Lemma \ref{quot-lemma-extend-polarized-to-spaces} isn't necessary.} the identity morphism of $H$ corresponds to a closed subspace $$ Z \subset \mathbf{P}^n_H $$ which is proper, flat, and of finite presentation over $H$ and such that the restriction $\mathcal{N} = \mathcal{O}(1)|_Z$ is relatively ample on $Z/H$ and has Hilbert polynomial $P$ on the fibres of $Z \to H$. In particular, the pair $(Z \to H, \mathcal{N})$ defines a morphism $$ H \longrightarrow \Polarizedstack $$ which sends a morphism of schemes $U \to H$ to the classifying morphism of the family $(Z_U \to U, \mathcal{N}_U)$, see Quot, Lemma \ref{quot-lemma-extend-polarized-to-spaces}. Since $H$ is a Noetherian algebraic space (as it is proper over $\mathbf{Z})$) we see that $|H|$ is Noetherian and hence quasi-compact. The map $$ |H| \longrightarrow |\Polarizedstack| $$ is continuous, hence the image is quasi-compact. Thus it suffices to prove $T$ is contained in the image of $|H| \to |\Polarizedstack|$. However, assumptions (1) and (2) exactly express the fact that this is the case: any choice of a closed immersion $i : X \to \mathbf{P}^n_k$ with $i^*\mathcal{O}_{\mathbf{P}^n}(1) \cong \mathcal{L}$ we get a $k$-valued point of $H$ by the moduli interpretation of $H$. This finishes the proof of the lemma.
The representability of the diagonal (by algebraic spaces) was shown in Quot, Lemma \ref{quot-lemma-complexes-diagonal}. From the proof we find that we have to show: given a scheme $T$ over $B$ and objects $E, E' \in D(\mathcal{O}_{X_T})$ such that $(T, E)$ and $(T, E')$ are objects of the fibre category of $\Complexesstack_{X/B}$ over $T$, then $\mathit{Isom}(E, E') \to T$ is affine and of finite presentation. Here $\mathit{Isom}(E, E')$ is the functor $$ (\Sch/T)^{opp} \to \textit{Sets},\quad T' \mapsto \{\varphi : E_{T'} \to E'_{T'} \text{ isomorphism in }D(\mathcal{O}_{X_{T'}})\} $$ where $E_{T'}$ and $E'_{T'}$ are the derived pullbacks of $E$ and $E'$ to $X_{T'}$. Consider the functor $H = \SheafHom(E, E')$ defined by the rule $$ (\Sch/T)^{opp} \to \textit{Sets},\quad T' \mapsto \Hom_{\mathcal{O}_{X_{T'}}}(E_T, E'_T) $$ By Quot, Lemma \ref{quot-lemma-complexes-open-neg-exts-vanishing} this is an algebraic space affine and of finite presentation over $T$. The same is true for $H' = \SheafHom(E', E)$, $I = \SheafHom(E, E)$, and $I' = \SheafHom(E', E')$. Therefore we see that $$ \mathit{Isom}(E, E') = (H' \times_T H) \times_{c, I \times_T I', \sigma} T $$ where $c(\varphi', \varphi) = (\varphi \circ \varphi', \varphi' \circ \varphi)$ and $\sigma = (\text{id}, \text{id})$ (compare with the proof of Quot, Proposition \ref{quot-proposition-isom}). Thus $\mathit{Isom}(E, E')$ is affine over $T$ as a fibre product of schemes affine over $T$. Similarly, $\mathit{Isom}(E, E')$ is of finite presentation over $T$.
To check $\Complexesstack_{X/B} \to B$ is quasi-separated we have to show that its diagonal is quasi-compact and quasi-separated. This is immediate from Lemma \ref{lemma-complexes-diagonal-affine-fp}. To prove that $\Complexesstack_{X/B} \to B$ is locally of finite presentation, we have to show that $\Complexesstack_{X/B} \to B$ is limit preserving, see Limits of Stacks, Proposition \ref{stacks-limits-proposition-characterize-locally-finite-presentation}. This follows from Quot, Lemma \ref{quot-lemma-complexes-limits} (small detail omitted).
Consider a distinguished triangle $(X, Y, Z, f, g, h)$. We have already seen that $g \circ f = 0$, see Lemma \ref{lemma-composition-zero}. Suppose $a : W \to Y$ is a morphism such that $g \circ a = 0$. Then we get a commutative diagram $$ \xymatrix{ W \ar[r]_1 \ar@{..>}[d]^b & W \ar[r] \ar[d]^a & 0 \ar[r] \ar[d]^0 & W[1] \ar@{..>}[d]^{b[1]} \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] } $$ Both rows are distinguished triangles (use TR1 for the top row). Hence we can fill the dotted arrow $b$ (first rotate using TR2, then apply TR3, and then rotate back). This proves the lemma.
Assume that $a$ and $c$ are isomorphisms. For any object $W$ of $\mathcal{D}$ write $H_W( - ) = \Hom_\mathcal{D}(W, -)$. Then we get a commutative diagram of abelian groups $$ \xymatrix{ H_W(Z[-1]) \ar[r] \ar[d] & H_W(X) \ar[r] \ar[d] & H_W(Y) \ar[r] \ar[d] & H_W(Z) \ar[r] \ar[d] & H_W(X[1]) \ar[d] \\ H_W(Z'[-1]) \ar[r] & H_W(X') \ar[r] & H_W(Y') \ar[r] & H_W(Z') \ar[r] & H_W(X'[1]) } $$ By assumption the right two and left two vertical arrows are bijective. As $H_W$ is homological by Lemma \ref{lemma-representable-homological} and the five lemma (Homology, Lemma \ref{homology-lemma-five-lemma}) it follows that the middle vertical arrow is an isomorphism. Hence by Yoneda's lemma, see Categories, Lemma \ref{categories-lemma-yoneda} we see that $b$ is an isomorphism. This implies the other cases by rotating (using TR2).
Picture $$ \xymatrix{ X \ar[r] \ar[d]^0 & Y \ar[r] \ar[d]^{b, b'} \ar@{..>}[ld]^\alpha & Z \ar[r] \ar[d]^0 \ar@{..>}[ld]^\beta & X[1] \ar[d]^0 \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] } $$ Applying Lemma \ref{lemma-representable-homological} we find dotted arrows $\alpha$ and $\beta$ such that $b' = f \circ \alpha$ and $b = \beta \circ g$. Then $bb' = \beta \circ g \circ f \circ \alpha = 0$ as $g \circ f = 0$ by Lemma \ref{lemma-composition-zero}.
By TR3 there exists a morphism $b'$ such that $(a, b', c)$ is an endomorphism of $(X, Y, Z, f, g, h)$. Then $(0, (b')^2 - b', 0)$ is also an endomorphism. By Lemma \ref{lemma-third-map-square-zero} we see that $(b')^2 - b'$ has square zero. Set $b = b' - (2b' - 1)((b')^2 - b') = 3(b')^2 - 2(b')^3$. A computation shows that $(a, b, c)$ is an endomorphism and that $b^2 - b = (4(b')^2 - 4b' - 3)((b')^2 - b')^2 = 0$.
If we have a second morphism of triangles $(a, b', c)$ then $(0, b - b', 0)$ is a morphism of triangles. Hence we have to show: the only morphism $b : Y \to Y'$ such that the  $X \to Y \to Y'$ and $Y \to Y' \to Z'$ are zero is $0$. We will use Lemma \ref{lemma-representable-homological} without further mention. In particular, condition (3) implies (1). Given condition (1) if the composition $b : Y \to Y' \to Z'$ is zero, then $b$ lifts to a morphism $Y \to X'$ which has to be zero. This proves (1). \medskip\noindent The proof of (2) and (4) are dual to this argument. \medskip\noindent Assume (5). Consider the diagram $$ \xymatrix{ X \ar[r]_f \ar[d]^0 & Y \ar[r]_g \ar[d]^b & Z \ar[r]_h \ar[d]^0 \ar@{..>}[ld]^\epsilon & X[1] \ar[d]^0 \\ X' \ar[r]^{f'} & Y' \ar[r]^{g'} & Z' \ar[r]^{h'} & X'[1] } $$ We may choose $\epsilon$ such that $b = g \circ \epsilon$. Then $g' \circ \epsilon \circ g = 0$ which implies that $g' \circ \epsilon = \delta \circ h$ for some $\delta \in \Hom(X[1], Z')$. Since $\Hom(X[1], Z') = 0$ we conclude that $g' \circ \epsilon = 0$. Hence $\epsilon = f' \circ \gamma$ for some $\gamma \in \Hom(Z, X')$. Since $\Hom(Z, X') = 0$ we conclude that $\epsilon = 0$ and hence $b = 0$ as desired.
By TR1 the triangle $(X, X, 0, 1, 0, 0)$ is distinguished. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. By TR3 there is a map of distinguished triangles $(1, f, 0) : (X, X, 0) \to (X, Y, Z)$. If $f$ is an isomorphism, then $(1, f, 0)$ is an isomorphism of triangles by Lemma \ref{lemma-third-isomorphism-triangle} and $Z = 0$. Conversely, if $Z = 0$, then $(1, f, 0)$ is an isomorphism of triangles as well, hence $f$ is an isomorphism.
Assume (2). By TR1 we may choose a distinguished triangle $(X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$. By TR3 we can find morphisms of distinguished triangles $(X, Y, Z, f, g, h) \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$ and $(X', Y', Z', f', g', h') \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$. Taking the direct sum of these morphisms we obtain a morphism of triangles $$ \xymatrix{ (X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \ar[d]^{(1, 1, c)} \\ (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h''). } $$ In the terminology of Remark \ref{remark-special-triangles} this is a map of special triangles (because a direct sum of special triangles is special) and we conclude that $c$ is an isomorphism. Thus (1) holds. \medskip\noindent Assume (1). We will show that $(X, Y, Z, f, g, h)$ is a distinguished triangle. First observe that $(X, Y, Z, f, g, h)$ is a special triangle (terminology from Remark \ref{remark-special-triangles}) as a direct summand of the distinguished hence special triangle $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')$. Using TR1 let $(X, Y, Q, f, g'', h'')$ be a distinguished triangle. By TR3 there exists a morphism of distinguished triangles  $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \to (X, Y, Q, f, g'', h'')$. Composing this with the inclusion map we get a morphism of triangles $$ (1, 1, c) : (X, Y, Z, f, g, h) \longrightarrow (X, Y, Q, f, g'', h'') $$ By Remark \ref{remark-special-triangles} we find that $c$ is an isomorphism and we conclude that (2) holds.
To see (1) use that $\Hom_\mathcal{D}(Z, Y) \to \Hom_\mathcal{D}(Z, Z) \to \Hom_\mathcal{D}(Z, X[1])$ is exact by Lemma \ref{lemma-representable-homological}. By the same token, if $s$ is as in (2), then $h = 0$ and the sequence $$ 0 \to \Hom_\mathcal{D}(W, X) \to \Hom_\mathcal{D}(W, Y) \to \Hom_\mathcal{D}(W, Z) \to 0 $$ is split exact (split by $s : Z \to Y$). Hence by Yoneda's lemma we see that $X \oplus Z \to Y$ is an isomorphism. The last assertion follows from TR1 and Lemma \ref{lemma-direct-sum-triangles}.
Any morphism isomorphic to a map of the form $X' \oplus Z \to Z \oplus Y'$ has both a kernel and a cokernel. Hence (3) $\Rightarrow$ (1), (2). Next we prove (1) $\Rightarrow$ (3). Suppose first that $f : X \to Y$ is a monomorphism, i.e., its kernel is zero. By TR1 there exists a distinguished triangle $(X, Y, Z, f, g, h)$. By Lemma \ref{lemma-composition-zero} the composition $f \circ h[-1] = 0$. As $f$ is a monomorphism we see that $h[-1] = 0$ and hence $h = 0$. Then Lemma \ref{lemma-split} implies that $Y = X \oplus Z$, i.e., we see that (3) holds. Next, assume $f$ has a kernel $K$. As $K \to X$ is a monomorphism we conclude $X = K \oplus X'$ and $f|_{X'} : X' \to Y$ is a monomorphism. Hence $Y = X' \oplus Y'$ and we win. The implication (2) $\Rightarrow$ (3) is dual to this.
Part (1) is true because $[1]$ is an autoequivalence of $\mathcal{D}$ and because direct sums and products are defined in terms of the category structure. Let us prove (2)(a). Choose a distinguished triangle $\prod X_i \to \prod Y_i \to Z \to \prod X_i[1]$. For each $j$ we can use TR3 to choose a morphism $p_j : Z \to Z_j$ fitting into a morphism of distinguished triangles with the projection maps $\prod X_i \to X_j$ and $\prod Y_i \to Y_j$. Using the definition of products we obtain a map $\prod p_i : Z \to \prod Z_i$ fitting into a morphism of triangles from the distinguished triangle to the triangle made out of the products. Observe that the ``product'' triangle $\prod X_i \to \prod Y_i \to \prod Z_i \to \prod X_i[1]$ is special in the terminology of Remark \ref{remark-special-triangles} because products of exact sequences of abelian groups are exact. Hence Remark \ref{remark-special-triangles} shows that the morphism of triangles is an isomorphism and we conclude by TR1. The proof of (2)(b) is dual.
The replacement of $X, Y, Z$ by $X', Y', Z'$ is harmless by our definition of distinguished triangles and their isomorphisms. The lemma follows from the fact that the distinguished triangles $(X', Y', Q_1, f', p_1, d_1)$, $(X', Z', Q_2, g' \circ f', p_2, d_2)$ and $(Y', Z', Q_3, g', p_3, d_3)$ are unique up to isomorphism by Lemma \ref{lemma-cone-triangle-unique-isomorphism}.
Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of pre-triangulated categories. Since $(0, 0, 0, 1_0, 1_0, 0)$ is a distinguished triangle of $\mathcal{D}$ the triangle $$ (F(0), F(0), F(0), 1_{F(0)}, 1_{F(0)}, F(0)) $$ is distinguished in $\mathcal{D}'$. This implies that $1_{F(0)} \circ 1_{F(0)}$ is zero, see Lemma \ref{lemma-composition-zero}. Hence $F(0)$ is the zero object of $\mathcal{D}'$. This also implies that $F$ applied to any zero morphism is zero (since a morphism in an additive category is zero if and only if it factors through the zero object). Next, using that $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is a distinguished triangle, we see that $(F(X), F(X \oplus Y), F(Y), F(1, 0), F(0, 1), 0)$ is one too. This implies that the map $F(1, 0) \oplus F(0, 1) : F(X) \oplus F(Y) \to F(X \oplus Y)$ is an isomorphism, see Lemma \ref{lemma-split}. We omit the rest of the argument.
The ``only if'' part is clear. Assume $(F(X), F(Y), F(Z))$ is distinguished in $\mathcal{D}'$. Pick a distinguished triangle $(X, Y, Z', f, g', h')$ in $\mathcal{D}$. By Lemma \ref{lemma-cone-triangle-unique-isomorphism} there exists an isomorphism of triangles $$ (1, 1, c') : (F(X), F(Y), F(Z)) \longrightarrow (F(X), F(Y), F(Z')). $$ Since $F$ is fully faithful, there exists a morphism $c : Z \to Z'$ such that $F(c) = c'$. Then $(1, 1, c)$ is an isomorphism between $(X, Y, Z)$ and $(X, Y, Z')$. Hence $(X, Y, Z)$ is distinguished by TR1.
The notation signifies the following. If $0 \to A \xrightarrow{a} B \xrightarrow{b} C \to 0$ is a short exact sequence in $\mathcal{A}$, then $$ \delta = \delta_{A \to B \to C} : G(C) \to G(A)[1] $$ is a morphism in $\mathcal{D}$ such that $(G(A), G(B), G(C), a, b, \delta)$ is a distinguished triangle, see Definition \ref{definition-delta-functor}. Then $H^n(\delta) : H^n(G(C)) \to H^n(G(A)[1]) = H^{n + 1}(G(A))$ is clearly functorial in the short exact sequence. Finally, the long exact cohomology sequence (\ref{equation-long-exact-cohomology-sequence}) combined with the vanishing of $H^{-1}(G(C))$ gives a long exact sequence $$ 0 \to H^0(G(A)) \to H^0(G(B)) \to H^0(G(C)) \xrightarrow{H^0(\delta)} H^1(G(A)) \to \ldots $$ in $\mathcal{B}$ as desired.
We have to prove axioms MS1 -- MS6, see Categories, Definitions \ref{categories-definition-multiplicative-system} and \ref{categories-definition-saturated-multiplicative-system} and Definition \ref{definition-localization}. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and Lemma \ref{lemma-third-isomorphism-triangle}. By Lemma \ref{lemma-localization-conditions} we conclude that MS2 holds. To finish the proof we have to show that MS3 holds. To do this let $f, g : X \to Y$ be morphisms of $\mathcal{D}$, and let $t : Z \to X$ be an element of $S$ such that $f \circ t = g \circ t$. As $\mathcal{D}$ is additive this simply means that $a \circ t = 0$ with $a = f - g$. Choose a distinguished triangle $(Z, X, Q, t, d, h)$ using TR1. Since $a \circ t = 0$ we see by Lemma \ref{lemma-representable-homological} there exists a morphism $i : Q \to Y$ such that $i \circ d = a$. Finally, using TR1 again we can choose a triangle $(Q, Y, W, i, j, k)$. Here is a picture $$ \xymatrix{ Z \ar[r]_t & X \ar[r]_d \ar[d]^1 & Q \ar[r] \ar[d]^i & Z[1] \\ & X \ar[r]_a & Y \ar[d]^j \\ & & W } $$ OK, and now we apply the functor $F$ to this diagram. Since $t \in S$ we see that $F(Q) = 0$, see Lemma \ref{lemma-third-object-zero}. Hence $F(j)$ is an isomorphism by the same lemma, i.e., $j \in S$. Finally, $j \circ a = j \circ i \circ d = 0$ as $j \circ i = 0$. Thus $j \circ f = j \circ g$ and we see that LMS3 holds. The proof of RMS3 is dual.