hcju/stacksqa huggingface dataset:
Choose $a, b$ such that $H^i(K) = 0$ for $i \not \in [a, b]$. We will prove the lemma by induction on $b - a$. If $b - a \leq 0$, then the result is clear. If $b - a > 0$, then we look at the distinguished triangle of truncations $$ \tau_{\leq b - 1}K \to K \to H^b(K)[-b] \to (\tau_{\leq b - 1}K)[1] $$ see Remark \ref{remark-truncation-distinguished-triangle}. By Lemma \ref{lemma-split} if the last arrow is zero, then $K \cong \tau_{\leq b - 1}K \oplus H^b(K)[-b]$ and we win by induction. Again using induction we see that $$ \Hom_{D(\mathcal{A})}(H^b(K)[-b], (\tau_{\leq b - 1}K)[1]) = \bigoplus\nolimits_{i < b} \Ext_\mathcal{A}^{b - i + 1}(H^b(K), H^i(K)) $$ Since $\Ext^i_\mathcal{A}(B, A) = 0$ for $i \geq 2$ and any pair of objects $A, B$ of $\mathcal{A}$ by our assumption and Lemma \ref{lemma-higher-ext-zero} we are done.
Since $RF$ is defined at $K$, we see that the rule which assigns to a $s : K \to I$ in $S$ the object $F(I)$ is essentially constant as an ind-object of $(S')^{-1}\mathcal{D}'$ with value $RF(K)$. Similarly, the rule which assigns to a $t : P \to M$ in $S'$ the object $G(P)$ is essentially constant as a pro-object of $S^{-1}\mathcal{D}$ with value $LG(M)$. Thus we have \begin{align*} \Hom_{(S')^{-1}\mathcal{D}'}(M, RF(K)) & = \colim_{s : K \to I} \Hom_{(S')^{-1}\mathcal{D}'}(M, F(I)) \\ & = \colim_{s : K \to I} \colim_{t : P \to M} \Hom_{\mathcal{D}'}(P, F(I)) \\ & = \colim_{t : P \to M} \colim_{s : K \to I} \Hom_{\mathcal{D}'}(P, F(I)) \\ & = \colim_{t : P \to M} \colim_{s : K \to I} \Hom_{\mathcal{D}}(G(P), I) \\ & = \colim_{t : P \to M} \Hom_{S^{-1}\mathcal{D}}(G(P), K) \\ & = \Hom_{S^{-1}\mathcal{D}}(LG(M), K) \end{align*} The first equality holds by Categories, Lemma \ref{categories-lemma-characterize-essentially-constant-ind}. The second equality holds by the definition of morphisms in $D(\mathcal{B})$, see Categories, Remark \ref{categories-remark-right-localization-morphisms-colimit}. The third equality holds by Categories, Lemma \ref{categories-lemma-colimits-commute}. The fourth equality holds because $F$ and $G$ are adjoint. The fifth equality holds by definition of morphism in $D(\mathcal{A})$, see Categories, Remark \ref{categories-remark-left-localization-morphisms-colimit}. The sixth equality holds by Categories, Lemma \ref{categories-lemma-characterize-essentially-constant-pro}. We omit the proof of functoriality.
Let $K^\bullet$ be an complex. Observe that the complex $$ C : \prod\nolimits_b \Hom(K^{-b}, I^{b - 1}) \to \prod\nolimits_b \Hom(K^{-b}, I^b) \to \prod\nolimits_b \Hom(K^{-b}, I^{b + 1}) $$ has cohomology $\Hom_{K(\mathcal{A})}(K^\bullet, I^\bullet)$ in the middle. Similarly, the complex $$ C_t : \prod\nolimits_b \Hom(K^{-b}, I_t^{b - 1}) \to \prod\nolimits_b \Hom(K^{-b}, I_t^b) \to \prod\nolimits_b \Hom(K^{-b}, I_t^{b + 1}) $$ computes $\Hom_{K(\mathcal{A})}(K^\bullet, I_t^\bullet)$. Next, observe that we have $$ C = \prod\nolimits_{t \in T} C_t $$ as complexes of abelian groups by our choice of $I$. Taking products is an exact functor on the category of abelian groups. Hence if $K^\bullet$ is acyclic, then $\Hom_{K(\mathcal{A})}(K^\bullet, I_t^\bullet) = 0$, hence $C_t$ is acyclic, hence $C$ is acyclic, hence we get $\Hom_{K(\mathcal{A})}(K^\bullet, I^\bullet) = 0$. Thus we find that $I^\bullet$ is K-injective. Having said this, we can use Lemma \ref{lemma-K-injective} to conclude that $$ \Hom_{D(\mathcal{A})}(K^\bullet, I^\bullet) = \prod\nolimits_{t \in T} \Hom_{D(\mathcal{A})}(K^\bullet, I_t^\bullet) $$ and indeed $I^\bullet$ represents the product in the derived category.
By Lemma \ref{lemma-derived-inverts} it suffices to show that $RF$ is defined at a K-injective complex, i.e., it suffices to show a K-injective complex $I^\bullet$ computes $RF$. Any quasi-isomorphism $I^\bullet \to N^\bullet$ is a homotopy equivalence as it has an inverse by Lemma \ref{lemma-K-injective}. Thus $I^\bullet \to I^\bullet$ is a final object of $I^\bullet/\text{Qis}(\mathcal{A})$ and we win.
To see this we apply Lemma \ref{lemma-find-existence-computes} with $\mathcal{I}$ the collection of K-injective complexes. Since (1) holds by assumption, it suffices to prove that if $I^\bullet \to J^\bullet$ is a quasi-isomorphism of K-injective complexes, then $F(I^\bullet) \to F(J^\bullet)$ is an isomorphism. This is clear because $I^\bullet \to J^\bullet$ is a homotopy equivalence, i.e., an isomorphism in $K(\mathcal{A})$, by Lemma \ref{lemma-K-injective}.
We urge the reader to skip the proof of this lemma. Let $M^\bullet$ be an acyclic complex. Let us abbreviate $H_n(a, b) = \Hom_\mathcal{A}(M^a, I_n^b)$. With this notation $\Hom_{K(\mathcal{A})}(M^\bullet, I^\bullet)$ is the cohomology of the complex $$ \prod_m \lim\limits_n H_n(m, m - 2) \to \prod_m \lim\limits_n H_n(m, m - 1) \to \prod_m \lim\limits_n H_n(m, m) \to \prod_m \lim\limits_n H_n(m, m + 1) $$ in the third spot from the left. We may exchange the order of $\prod$ and $\lim$ and each of the complexes $$ \prod_m H_n(m, m - 2) \to \prod_m H_n(m, m - 1) \to \prod_m H_n(m, m) \to \prod_m H_n(m, m + 1) $$ is exact by assumption (1). By assumption (2) the maps in the systems $$ \ldots \to \prod_m H_3(m, m - 2) \to \prod_m H_2(m, m - 2) \to \prod_m H_1(m, m - 2) $$ are surjective. Thus the lemma follows from Homology, Lemma \ref{homology-lemma-apply-Mittag-Leffler}.
By Lemma \ref{lemma-subcategory-right-resolution} we can find a quasi-isomorphism $\sigma_{\geq 0}K^\bullet \to M^\bullet$ with $M^n = 0$ for $n < 0$ and $d(M^n) = 0$ for $n \geq 0$. Then $K^\bullet$ is quasi-isomorphic to the complex $$ \ldots \to K^{-2} \to K^{-1} \to M^0 \to M^1 \to \ldots $$ Hence we may assume that $d(K^n) = 0$ for $n \gg 0$. Note that the condition $n + d(K^n) \to -\infty$ as $n \to -\infty$ is not violated by this replacement. \medskip\noindent We are going to improve $K^\bullet$ by an (infinite) sequence of elementary replacements. An {\it elementary replacement} is the following. Choose an index $n$ such that $d(K^n) > 0$. Choose an injection $K^n \to M$ where $d(M) = 0$. Set $M' = \Coker(K^n \to M \oplus K^{n + 1})$. Consider the map of complexes $$ \xymatrix{ K^\bullet : \ar[d] & K^{n - 1} \ar[d] \ar[r] & K^n \ar[d] \ar[r] & K^{n + 1} \ar[d] \ar[r] & K^{n + 2} \ar[d] \\ (K')^\bullet : & K^{n - 1} \ar[r] & M \ar[r] & M' \ar[r] & K^{n + 2} } $$ It is clear that $K^\bullet \to (K')^\bullet$ is a quasi-isomorphism. Moreover, it is clear that $d((K')^n) = 0$ and $$ d((K')^{n + 1}) \leq \max\{d(K^n) - 1, d(M \oplus K^{n + 1})\} \leq \max\{d(K^n) - 1, d(K^{n + 1})\} $$ and the other values are unchanged. \medskip\noindent To finish the proof we carefuly choose the order in which to do the elementary replacements so that for every integer $m$ the complex $\sigma_{\geq m}K^\bullet$ is changed only a finite number of times. To do this set $$ \xi(K^\bullet) = \max \{n + d(K^n) \mid d(K^n) > 0\} $$ and $$ I = \{n \in \mathbf{Z} \mid \xi(K^\bullet) = n + d(K^n) \text{ and }  d(K^n) > 0\} $$ Our assumption that $n + d(K^n)$ tends to $-\infty$ as $n \to -\infty$ and the fact that $d(K^n) = 0$ for $n >> 0$ implies $\xi(K^\bullet) < +\infty$ and that $I$ is a finite set. It is clear that $\xi((K')^\bullet) \leq \xi(K^\bullet)$ for an elementary transformation as above. An elementary transformation changes the complex in degrees $\leq \xi(K^\bullet) + 1$. Hence if we can find finite sequence of elementary transformations which decrease $\xi(K^\bullet)$, then we win. However, note that if we do an elementary transformation starting with the smallest element $n \in I$, then we either decrease the size of $I$, or we increase $\min I$. Since every element of $I$ is $\leq \xi(K^\bullet)$ we see that we win after a finite number of steps.
Let $g_i : K_{n_i} \to K_{n_{i + 1}}$ be the composition $f_{n_{i + 1} - 1} \circ \ldots \circ f_{n_i}$. We construct commutative diagrams $$ \vcenter{ \xymatrix{ \bigoplus\nolimits_i K_{n_i} \ar[r]_{1 - g_i} \ar[d]_b & \bigoplus\nolimits_i K_{n_i} \ar[d]^a \\ \bigoplus\nolimits_n K_n \ar[r]^{1 - f_n} & \bigoplus\nolimits_n K_n } } \quad\text{and}\quad \vcenter{ \xymatrix{ \bigoplus\nolimits_n K_n \ar[r]_{1 - f_n} \ar[d]_d & \bigoplus\nolimits_n K_n \ar[d]^c \\ \bigoplus\nolimits_i K_{n_i} \ar[r]^{1 - g_i} & \bigoplus\nolimits_i K_{n_i} } } $$ as follows. Let $a_i = a|_{K_{n_i}}$ be the inclusion of $K_{n_i}$ into the direct sum. In other words, $a$ is the natural inclusion. Let $b_i = b|_{K_{n_i}}$ be the map $$ K_{n_i} \xrightarrow{1,\ f_{n_i},\ f_{n_i + 1} \circ f_{n_i}, \ \ldots,\ f_{n_{i + 1} - 2} \circ \ldots \circ f_{n_i}} K_{n_i} \oplus K_{n_i + 1} \oplus \ldots \oplus K_{n_{i + 1} - 1} $$ If $n_{i - 1} < j \leq n_i$, then we let $c_j = c|_{K_j}$ be the map $$ K_j \xrightarrow{f_{n_i - 1} \circ \ldots \circ f_j} K_{n_i} $$ We let $d_j = d|_{K_j}$ be zero if $j \not = n_i$ for any $i$ and we let $d_{n_i}$ be the natural inclusion of $K_{n_i}$ into the direct sum. In other words, $d$ is the natural projection. By TR3 these diagrams define morphisms $$ \varphi : \text{hocolim} K_{n_i} \to \text{hocolim} K_n \quad\text{and}\quad \psi : \text{hocolim} K_n \to \text{hocolim} K_{n_i} $$ Since $c \circ a$ and $d \circ b$ are the identity maps we see that $\varphi \circ \psi$ is an isomorphism by Lemma \ref{lemma-third-isomorphism-triangle}. The other way around we get the morphisms $a \circ c$ and $b \circ d$. Consider the morphism $h = (h_j) : \bigoplus K_n \to \bigoplus K_n$ given by the rule: for $n_{i - 1} < j < n_i$ we set $$ h_j : K_j \xrightarrow{1,\ f_j,\ f_{j + 1} \circ f_j, \ \ldots,\ f_{n_i - 1} \circ \ldots \circ f_j} K_j \oplus \ldots \oplus K_{n_i} $$ Then the reader verifies that $(1 - f) \circ h = \text{id} - a \circ c$ and $h \circ (1 - f) = \text{id} - b \circ d$. This means that $\text{id} - \psi \circ \varphi$ has square zero by Lemma \ref{lemma-third-map-square-zero} (small argument omitted). In other words, $\psi \circ \varphi$ differs from the identity by a nilpotent endomorphism, hence is an isomorphism. Thus $\varphi$ and $\psi$ are isomorphisms as desired.
We have an exact sequence of complexes $$ 0 \to \bigoplus L_n^\bullet \to \bigoplus L_n^\bullet \to L^\bullet \to 0 $$ by Lemma \ref{lemma-compute-colimit}. The direct sums are direct sums in $D(\mathcal{A})$ by Lemma \ref{lemma-direct-sums}. Thus the result follows from the definition of derived colimits in Definition \ref{definition-derived-colimit} and the fact that a short exact sequence of complexes gives a distinguished triangle (Lemma \ref{lemma-derived-canonical-delta-functor}).
Write $K = \text{hocolim} K_n$. Apply $H$ to the defining distinguished triangle to get $$ \bigoplus H(K_n) \to \bigoplus H(K_n) \to H(K) \to \bigoplus H(K_n[1]) \to \bigoplus H(K_n[1]) $$ where the first map is given by $1 - H(f_n)$ and the last map is given by $1 - H(f_n[1])$. Apply Lemma \ref{lemma-compute-colimit} to see that this proves the lemma.
Consider the defining distinguished triangle $$ \bigoplus L_n \to \bigoplus L_n \to L \to \bigoplus L_n[1] $$ Apply the cohomological functor $\Hom_\mathcal{D}(K, -)$ (see Lemma \ref{lemma-representable-homological}). By elementary considerations concerning colimits of abelian groups we get the result.
It suffices to show that $\prod K_n$ exists in $D(\mathcal{A})$. For every $n$ we can represent $K_n$ by a bounded below complex $I_n^\bullet$ of injectives (Lemma \ref{lemma-injective-resolutions-exist}). Then $\prod K_n$ is represented by $\prod I_n^\bullet$, see Lemma \ref{lemma-product-K-injective}.
The statement of the lemma makes sense as $R\lim \tau_{\geq -n}K^\bullet$ exists by Lemma \ref{lemma-inverse-limit-bounded-below}. Each complex $I_n^\bullet$ is K-injective by Lemma \ref{lemma-bounded-below-injectives-K-injective}. Choose direct sum decompositions $I_{n + 1}^p = C_{n + 1}^p \oplus I_n^p$ for all $n \geq 1$. Set $C_1^p = I_1^p$. The complex $I^\bullet = \lim I_n^\bullet$ exists because we can take $I^p = \prod_{n \geq 1} C_n^p$. Fix $p \in \mathbf{Z}$. We claim there is a split short exact sequence $$ 0 \to I^p \to \prod I_n^p \to \prod I_n^p \to 0 $$ of objects of $\mathcal{A}$. Here the first map is given by the projection maps $I^p \to I_n^p$ and the second map by $(x_n) \mapsto (x_n - f^p_{n + 1}(x_{n + 1}))$ where $f^p_n : I_n^p \to I_{n - 1}^p$ are the transition maps. The splitting comes from the map $\prod I_n^p \to \prod C_n^p = I^p$. We obtain a termwise split short exact sequence of complexes $$ 0 \to I^\bullet \to \prod I_n^\bullet \to \prod I_n^\bullet \to 0 $$ Hence a corresponding distinguished triangle in $K(\mathcal{A})$ and $D(\mathcal{A})$. By Lemma \ref{lemma-product-K-injective} the products are K-injective and represent the corresponding products in $D(\mathcal{A})$. It follows that $I^\bullet$ represents $R\lim I_n^\bullet$ (Definition \ref{definition-derived-limit}). Moreover, it follows that $I^\bullet$ is K-injective by Lemma \ref{lemma-triangle-K-injective}. By the commutative diagram of Lemma \ref{lemma-special-inverse-system} we obtain a corresponding commutative diagram $$ \xymatrix{ K^\bullet \ar[r] \ar[d] & R\lim \tau_{\geq -n} K^\bullet \ar[d] \\ I^\bullet \ar[r] & R\lim I_n^\bullet } $$ in $D(\mathcal{A})$. Since the right vertical arrow is an isomorphism (as derived limits are defined on the level of the derived category and since $\tau_{\geq -n}K^\bullet \to I_n^\bullet$ is a quasi-isomorphism), the lemma follows.
By Lemma \ref{lemma-difficulty-K-injectives} it suffices to show that $K \to R\lim\tau_{\geq -n}K$ is an isomorphism for all $K$ in $D(\mathcal{A})$. Consider the defining distinguished triangle $$ R\lim\tau_{\geq -n}K \to \prod \tau_{\geq -n}K \to \prod \tau_{\geq -n}K \to (R\lim\tau_{\geq -n}K)[1] $$ By Lemma \ref{lemma-products} we have $$ H^p(\prod \tau_{\geq -n}K) = \prod\nolimits_{p \geq -n} H^p(K) $$ It follows in a straightforward manner from the long exact cohomology sequence of the displayed distinguished triangle that $H^p(R\lim \tau_{\geq -n}K) = H^p(K)$.
We will use Lemma \ref{lemma-cone-n} without further mention. We will prove (2) which trivially implies (1). We use induction on $b - a$. If $b - a = 0$, then $K$ is isomorphic to $H^i(K)[-a]$ in $\mathcal{D}$ and the result is immediate. If $b - a > 0$, then we consider the distinguished triangle $$ \tau_{\leq b - 1}K^\bullet \to K^\bullet \to K^b[-b] $$ and we conclude by induction on $b - a$. We omit the proof of (3) and (4).
The left equality in the displayed formula follows from Lemmas \ref{lemma-associativity-star} and \ref{lemma-smd-star} and induction. The middle equality is a matter of notation. Since $\langle E \rangle_1 = smd(add(E[-\infty, \infty])])$ and since $E[-\infty, \infty] = \bigcup_{m \geq 1} E[-m, m]$ we see from Remark \ref{remark-operations-unions} and Lemma \ref{lemma-smd-star} that we get the equality on the right. Then the final statement follows from the remark and the corresponding statement of Lemma \ref{lemma-cone-n}.
The equality on the right follows from Lemma \ref{lemma-generated-by-E-explicit}. It is clear that $\langle E \rangle = \bigcup \langle E \rangle_n$ contains $E$, is preserved under shifts, direct sums, direct summands. If $A \in \langle E \rangle_a$ and $B \in \langle E \rangle_b$ and if $A \to X \to B \to A[1]$ is a distinguished triangle, then $X \in \langle E \rangle_{a + b}$ by Lemma \ref{lemma-generated-by-E-explicit}. Hence $\bigcup \langle E \rangle_n$ is also preserved under extensions and it follows that it is a triangulated subcategory.  \medskip\noindent Finally, let $\mathcal{D}' \subset \mathcal{D}$ be a  strictly full, saturated, triangulated subcategory of $\mathcal{D}$ containing $E$. Then $\mathcal{D}'[-\infty, \infty] \subset \mathcal{D}'$, $add(\mathcal{D}) \subset \mathcal{D}'$, $smd(\mathcal{D}') \subset \mathcal{D}'$, and $\mathcal{D}' \star \mathcal{D}' \subset \mathcal{D}'$. In other words, all the operations we used to construct $\langle E \rangle$ out of $E$ preserve $\mathcal{D}'$. Hence $\langle E \rangle \subset \mathcal{D}'$ and this finishes the proof.
The implication (2) $\Rightarrow$ (1) is immediate. Conversely, assume (1). Then $\Hom(X, K) = 0$ for all $X$ in $\langle E \rangle_1$. Arguing by induction on $n$ and using Lemma \ref{lemma-representable-homological} we see that $\Hom(X, K) = 0$ for all $X$ in $\langle E \rangle_n$.
Assume $E$ is a classical generator. Let $K$ be an object of $\mathcal{D}$ such that $\Hom(E, K[i]) = 0$ for all $i \in \mathbf{Z}$. By Lemma \ref{lemma-right-orthogonal} $\Hom(E', K) = 0$ for all $E'$ in $\langle E \rangle$. However, since $\mathcal{D} = \langle E \rangle$ we conclude that $\text{id}_K = 0$, i.e., $K = 0$.
Let $E'$ be an object of $\mathcal{D}$ such that $\mathcal{D} = \langle E' \rangle_n$. Since $\mathcal{D} = \langle E \rangle$ we see that $E' \in \langle E \rangle_m$ for some $m \geq 1$ by Lemma \ref{lemma-find-smallest-containing-E}. Then $\langle E' \rangle_1 \subset \langle E \rangle_m$ hence $$ \mathcal{D} = \langle E' \rangle_n = smd( \langle E' \rangle_1 \star \ldots \star \langle E' \rangle_1) \subset smd( \langle E \rangle_m \star \ldots \star \langle E \rangle_m) = \langle E \rangle_{nm} $$ as desired. Here we used Lemma \ref{lemma-generated-by-E-explicit}.
Let $(X, Y, Z, f, g, h)$ be a distinguished triangle of $\mathcal{D}$ with $X$ and $Y$ compact. Then it follows from Lemma \ref{lemma-representable-homological} and the five lemma (Homology, Lemma \ref{homology-lemma-five-lemma}) that $Z$ is a compact object too. It is clear that if $X \oplus Y$ is compact, then $X$, $Y$ are compact objects too. Hence $\mathcal{D}_c$ is a saturated triangulated subcategory. Since $\mathcal{D}$ is Karoubian by Lemma \ref{lemma-projectors-have-images-triangulated} we conclude that the same is true for $\mathcal{D}_c$.
By Lemma \ref{lemma-representable-homological} the functor $\Hom(A, -)$ is homological and hence we get a long exact sequence as in (\ref{equation-long-exact-cohomology-sequence}). Assume (1) and let $A \in \Ob(\mathcal{A})$. Then we consider the exact sequence $$ \Hom(A[1], Z) \to \Hom(A, X) \to \Hom(A, Y) \to \Hom(A, Z) $$ Since $A[1] \in \Ob(\mathcal{A})$ we see that the first and last groups are zero. Thus we get (2). Assume (2) and let $A \in \Ob(\mathcal{A})$. Then we consider the exact sequence $$ \Hom(A, X) \to \Hom(A, Y) \to \Hom(A, Z) \to \Hom(A[-1], X) \to \Hom(A[-1], Y) $$ and we conclude that $\Hom(A, Z) = 0$ as desired.
It is immediate from the definitions that the orthogonals are preserved under taking shifts, direct sums, and direct summands. Consider a distinguished triangle $$ X \to Y \to Z \to X[1] $$ of $\mathcal{D}$. By Lemma \ref{lemma-triangulated-subcategory} it suffices to show that if $X$ and $Y$ are in $\mathcal{A}^\perp$, then $Z$ is in $\mathcal{A}^\perp$. This is immediate from Lemma \ref{lemma-pre-prepare-adjoint}.
Let $X_1 \to X_2 \to X_3 \to X_1[1]$ be a distinguished triangle and assume $P$ holds for $X_1$ and $X_2$. Choose distinguished triangles $$ A_1 \to X_1 \to B_1 \to A_1[1] \quad\text{and}\quad A_2 \to X_2 \to B_2 \to A_2[1] $$ as in condition $P$. Since $\Hom(A_1, A_2) = \Hom(A_1, X_2)$ by Lemma \ref{lemma-pre-prepare-adjoint} there is a unique morphism $A_1 \to A_2$ such that the diagram $$ \xymatrix{ A_1 \ar[d] \ar[r] & X_1 \ar[d] \\ A_2 \ar[r] & X_2 } $$ commutes. Choose an extension of this to a diagram $$ \xymatrix{ A_1 \ar[r] \ar[d] & X_1 \ar[r] \ar[d] & Q_1 \ar[r] \ar[d] & A_1[1] \ar[d] \\ A_2 \ar[r] \ar[d] & X_2 \ar[r] \ar[d] & Q_2 \ar[r] \ar[d] & A_2[1] \ar[d] \\ A_3 \ar[r] \ar[d] & X_3 \ar[r] \ar[d] & Q_3 \ar[r] \ar[d] & A_3[1] \ar[d] \\ A_1[1] \ar[r] & X_1[1] \ar[r] & Q_1[1] \ar[r] & A_1[2] } $$ as in Proposition \ref{proposition-9}. By TR3 we see that $Q_1 \cong B_1$ and $Q_2 \cong B_2$ and hence $Q_1, Q_2 \in \Ob(\mathcal{A}^\perp)$. As $Q_1 \to Q_2 \to Q_3 \to Q_1[1]$ is a distinguished triangle we see that $Q_3 \in \Ob(\mathcal{A}^\perp)$ by Lemma \ref{lemma-orthogonal-triangulated}. Since $\mathcal{A}$ is a full triangulated subcategory, we see that $A_3$ is isomorphic to an object of $\mathcal{A}$. Thus $X_3$ satisfies $P$. The other cases of (1) follow from this case by translation. Part (2) is a special case of (1) via Lemma \ref{lemma-split}.
Assume (1) and denote $v : \mathcal{D} \to \mathcal{A}$ the right adjoint. Let $X \in \Ob(\mathcal{D})$. Set $A = v(X)$. We may extend the adjunction mapping $A \to X$ to a distinguished triangle $A \to X \to B \to A[1]$. Since $$ \Hom_\mathcal{A}(A', A) = \Hom_\mathcal{A}(A', v(X)) = \Hom_\mathcal{D}(A', X) $$ for $A' \in \Ob(\mathcal{A})$, we conclude that $B \in \Ob(\mathcal{A}^\perp)$ by Lemma \ref{lemma-pre-prepare-adjoint}. \medskip\noindent Assume (2). We will contruct the adjoint $v$ explictly. Let $X \in \Ob(\mathcal{D})$. Choose $A \to X \to B \to A[1]$ as in (2). Set $v(X) = A$. Let $f : X \to Y$ be a morphism in $\mathcal{D}$. Choose $A' \to Y \to B' \to A'[1]$ as in (2). Since $\Hom(A, A') = \Hom(A, Y)$ by Lemma \ref{lemma-pre-prepare-adjoint} there is a unique morphism $f' : A \to A'$ such that the diagram $$ \xymatrix{ A \ar[d]_{f'} \ar[r] & X \ar[d]^f \\ A' \ar[r] & Y } $$ commutes. Hence we can set $v(f) = f'$ to get a functor. To see that $v$ is adjoint to the inclusion morphism use Lemma \ref{lemma-pre-prepare-adjoint} again. \medskip\noindent Proof of the final statement. In other to prove that $\mathcal{A}$ is saturated we may replace $\mathcal{A}$ by the strictly full subcategory having the same isomorphism classes as $\mathcal{A}$; details omitted. Assume $\mathcal{A}$ is strictly full. If we show that $\mathcal{A} = {}^\perp(\mathcal{A}^\perp)$, then $\mathcal{A}$ will be saturated by Lemma \ref{lemma-orthogonal-triangulated}. Since the incusion $\mathcal{A} \subset {}^\perp(\mathcal{A}^\perp)$ is clear it suffices to prove the other inclusion. Let $X$ be an object of ${}^\perp(\mathcal{A}^\perp)$. Choose a distinguished triangle $A \to X \to B \to A[1]$ as in (2). As $\Hom(X, B) = 0$ by assumption we see that $A \cong X \oplus B[-1]$ by Lemma \ref{lemma-split}. Since $\Hom(A, B[-1]) = 0$ as $B \in \mathcal{A}^\perp$ this implies $B[-1] = 0$ and $A \cong X$ as desired.
The case $n = 0$ is immediate as isomorphisms are invertible. The case $n = 1$ follows immediately from TR1 (existence of triangles) and TR3 (extending morphisms to triangles). For the case $n = 2$ we argue as follows. Set $Y_0 = X_0$. By the case $n = 1$ we can choose a Postnikov system $$ Y_1 \to X_1 \to Y_0 \to Y_1[1] $$ Since the composition $X_2 \to X_1 \to X_0$ is zero, we can factor $X_2 \to X_1$ (nonuniquely) as $X_2 \to Y_1 \to X_1$ by Lemma \ref{lemma-representable-homological}. Then we simply fit the morphism $X_2 \to Y_1$ into a distinguished triangle $$ Y_2 \to X_2 \to Y_1 \to Y_2[1] $$ to get the Postnikov system for $n = 2$. For $n > 2$ we cannot argue similarly, as we do not know whether the composition $X_n \to X_{n - 1} \to Y_{n - 1}$ is zero in $\mathcal{D}$.