hcju/stacksqa huggingface dataset:
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 the long exact cohomology sequence (\ref{equation-long-exact-cohomology-sequence}). 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 $(Z, X, Q, t, g, h)$ using TR1 and TR2. Since $a \circ t = 0$ we see by Lemma \ref{lemma-representable-homological} there exists a morphism $i : Q \to Y$ such that $i \circ g = a$. Finally, using TR1 again we can choose a triangle $(Q, Y, W, i, j, k)$. Here is a picture $$ \xymatrix{ Z \ar[r]_t & X \ar[r]_g \ar[d]^1 & Q \ar[r] \ar[d]^i & Z[1] \\ & X \ar[r]_a & Y \ar[d]^j \\ & & W } $$ OK, and now we apply the functors $H^i$ to this diagram. Since $t \in S$ we see that $H^i(Q) = 0$ by the long exact cohomology sequence (\ref{equation-long-exact-cohomology-sequence}). Hence $H^i(j)$ is an isomorphism for all $i$ by the same argument, i.e., $j \in S$. Finally, $j \circ a = j \circ i \circ g = 0$ as $j \circ i = 0$. Thus $j \circ f = j \circ g$ and we see that LMS3 holds. The proof of RMS3 is dual.
The equivalence of (1), (2), and (3) is Homology, Lemma \ref{homology-lemma-kernel-localization}. If (2) holds, then $(Z'[-1], Z'[-1] \oplus Z, Z, (1, 0), (0, 1), 0)$ is a distinguished triangle (see Lemma \ref{lemma-split}) with ``$0 \in S$''. By rotating we conclude that (4) holds. If $(X, Y, Z \oplus Z', f, g, h)$ is a distinguished triangle with $f \in S$ then $Q(f)$ is an isomorphism hence $Q(Z \oplus Z') = 0$ hence $Q(Z) = 0$. Thus (1) -- (4) are all equivalent. \medskip\noindent Next, assume that $S$ is saturated. Note that each of (5), (6), (7) implies one of the equivalent conditions (1) -- (4). Suppose that $Q(Z) = 0$. Then $0 \to Z$ is a morphism of $\mathcal{D}$ which becomes an isomorphism in $S^{-1}\mathcal{D}$. According to Categories, Lemma \ref{categories-lemma-what-gets-inverted} the fact that $S$ is saturated implies that $0 \to Z$ is in $S$. Hence (1) $\Rightarrow$ (5). Dually (1) $\Rightarrow$ (6). Finally, if $0 \to Z$ is in $S$, then the triangle $(0, Z, Z, 0, \text{id}_Z, 0)$ is distinguished by TR1 and TR2 and is a triangle as in (4).
It is clear that $\mathcal{D}''$ is preserved under $[1]$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $F(X) = F(Y) = 0$, then also $F(Z) = 0$ as $(F(X), F(Y), F(Z), F(f), F(g), F(h))$ is distinguished. Hence we may apply Lemma \ref{lemma-triangulated-subcategory} to see that $\mathcal{D}''$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The final assertion of being saturated follows from $F(X) \oplus F(Y) = 0 \Rightarrow F(X) = F(Y) = 0$.
It is clear that $\mathcal{D}'$ is preserved under $[1]$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $H(X[n]) = H(Y[n]) = 0$ for all $n$, then also $H(Z[n]) = 0$ for all $n$ by the long exact sequence (\ref{equation-long-exact-cohomology-sequence}). Hence we may apply Lemma \ref{lemma-triangulated-subcategory} to see that $\mathcal{D}'$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The assertion of being saturated follows from \begin{align*} H((X \oplus Y)[n]) = 0 & \Rightarrow H(X[n] \oplus Y[n]) = 0 \\ & \Rightarrow H(X[n]) \oplus H(Y[n]) = 0 \\ & \Rightarrow H(X[n]) = H(Y[n]) = 0 \end{align*} for all $n \in \mathbf{Z}$.
Let us prove this for $\mathcal{D}_H^{+}$. It is clear that it is preserved under $[1]$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $H(X[n]) = H(Y[n]) = 0$ for all $n \ll 0$, then also $H(Z[n]) = 0$ for all $n \ll 0$ by the long exact sequence (\ref{equation-long-exact-cohomology-sequence}). Hence we may apply Lemma \ref{lemma-triangulated-subcategory} to see that $\mathcal{D}_H^{+}$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The assertion of being saturated follows from \begin{align*} H((X \oplus Y)[n]) = 0 & \Rightarrow H(X[n] \oplus Y[n]) = 0 \\ & \Rightarrow H(X[n]) \oplus H(Y[n]) = 0 \\ & \Rightarrow H(X[n]) = H(Y[n]) = 0 \end{align*} for all $n \in \mathbf{Z}$.
This lemma follows from Lemma \ref{lemma-universal-property-localization}. Namely, if $f : X \to Y$ is a morphism of $\mathcal{D}$ such that for some distinguished triangle $(X, Y, Z, f, g, h)$ the object $Z$ is isomorphic to an object of $\mathcal{B}$, then $H(f)$, resp.\ $F(f)$ is an isomorphism under the assumptions of (1), resp.\ (2). Details omitted.
First, let's start with a full triangulated subcategory $\mathcal{B}$. Then $\mathcal{B}(S(\mathcal{B})) = \Ker(Q : \mathcal{D} \to \mathcal{D}/\mathcal{B})$ and hence (2) is the content of Lemma \ref{lemma-kernel-quotient}. \medskip\noindent Next, suppose that $S$ is multiplicative system in $\mathcal{D}$ compatible with the triangulation on $\mathcal{D}$. Then $\mathcal{B}(S) = \Ker(Q : \mathcal{D} \to S^{-1}\mathcal{D})$. Hence (using Lemma \ref{lemma-third-object-zero} in the localized category) \begin{align*} S(\mathcal{B}(S)) & = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished}\\ \text{triangle }(X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with }Q(Z) = 0 \end{matrix} \right\} \\ & = \{f \in \text{Arrows}(\mathcal{D}) \mid Q(f)\text{ is an isomorphism}\} \\ & = \hat S = S' \end{align*} in the notation of Categories, Lemma \ref{categories-lemma-what-gets-inverted}. The final statement of that lemma finishes the proof.
The category $\Ker(H)$ is a strictly full saturated triangulated subcategory of $\mathcal{D}$ by Lemma \ref{lemma-homological-functor-kernel}. The set $S$ is a saturated multiplicative system compatible with the triangulated structure by Lemma \ref{lemma-homological-functor-localize}. Recall that the multiplicative system corresponding to $\Ker(H)$ is the set $$ \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle }\\ (X, Y, Z, f, g, h)\text{ with } H^i(Z) = 0 \text{ for all }i \end{matrix} \right\} $$ By the long exact cohomology sequence, see (\ref{equation-long-exact-cohomology-sequence}), it is clear that $f$ is an element of this set if and only if $f$ is an element of $S$. Finally, the factorization of $H$ through $Q$ is a consequence of Lemma \ref{lemma-universal-property-quotient}.
Let $X$ be an object of $\mathcal{D}$ and $A$ an object of $\mathcal{D}'$. Since $F$ is an exact functor we see that \begin{align*} \Mor_\mathcal{D}(X, G(A[1]) & = \Mor_{\mathcal{D}'}(F(X), A[1]) \\ & = \Mor_{\mathcal{D}'}(F(X)[-1], A) \\ & = \Mor_{\mathcal{D}'}(F(X[-1]), A) \\ & = \Mor_\mathcal{D}(X[-1], G(A)) \\ & = \Mor_\mathcal{D}(X, G(A)[1]) \end{align*} By Yoneda's lemma (Categories, Lemma \ref{categories-lemma-yoneda}) we obtain a canonical isomorphism $G(A)[1] = G(A[1])$. Let $A \to B \to C \to A[1]$ be a distinguished triangle in $\mathcal{D}'$. Choose a distinguished triangle $$ G(A) \to G(B) \to X \to G(A)[1] $$ in $\mathcal{D}$. Then $F(G(A)) \to F(G(B)) \to F(X) \to F(G(A))[1]$ is a distinguished triangle in $\mathcal{D}'$. By TR3 we can choose a morphism of distinguished triangles $$ \xymatrix{ F(G(A)) \ar[r] \ar[d] & F(G(B)) \ar[r] \ar[d] & F(X) \ar[r] \ar[d] & F(G(A))[1] \ar[d] \\ A \ar[r] & B \ar[r] & C \ar[r] & A[1] } $$ Since $G$ is the adjoint the new morphism determines a morphism $X \to G(C)$ such that the diagram $$ \xymatrix{ G(A) \ar[r] \ar[d] & G(B) \ar[r] \ar[d] & X \ar[r] \ar[d] & G(A)[1] \ar[d] \\ G(A) \ar[r] & G(B) \ar[r] & G(C) \ar[r] & G(A)[1] } $$ commutes. Applying the homological functor $\Hom_{\mathcal{D}'}(W, -)$ for an object $W$ of $\mathcal{D}'$ we deduce from the $5$ lemma that $$ \Hom_{\mathcal{D}'}(W, X) \to \Hom_{\mathcal{D}'}(W, G(C)) $$ is a bijection and using the Yoneda lemma once more we conclude that $X \to G(C)$ is an isomorphism. Hence we conclude that $G(A) \to G(B) \to G(C) \to G(A)[1]$ is a distinguished triangle which is what we wanted to show.
Since $F$ is fully faithful the adjunction map $\text{id} \to G \circ F$ is an isomorphism (Categories, Lemma \ref{categories-lemma-adjoint-fully-faithful}). Let $X$ be an object of $\mathcal{D}'$. Choose a distinguished triangle $$ F(G(X)) \to X \to Y \to F(G(X))[1] $$ in $\mathcal{D}'$. Applying $G$ and using that $G(F(G(X))) = G(X)$ we find a distinguished triangle $$ G(X) \to G(X) \to G(Y) \to G(X)[1] $$ Hence $G(Y) = 0$. Thus $Y = 0$. Thus $F(G(X)) \to X$ is an isomorphism.
The assumptions say that the diagram $$ \xymatrix{ K^\bullet \ar[r]_f \ar[d] & L^\bullet \ar[d]^g \\ 0 \ar[r] & M^\bullet } $$ commutes up to homotopy. Since the cone on $0 \to M^\bullet$ is $M^\bullet$ the map $C(f)^\bullet \to C(0 \to M^\bullet) = M^\bullet$ of Lemma \ref{lemma-functorial-cone} is the map in (1). The cone on $K^\bullet \to 0$ is $K^\bullet[1]$ and applying Lemma \ref{lemma-functorial-cone} gives a map $K^\bullet[1] \to C(g)^\bullet$. Applying $[-1]$ we obtain the map in (2).
Dual to Lemma \ref{lemma-make-injective}. Take $$ \tilde K^n = K^n \oplus L^{n - 1} \oplus L^n $$ and we define $$ d^n_{\tilde K} = \left( \begin{matrix} d^n_K & 0 & 0 \\ 0 & - d^{n - 1}_L & \text{id}_{L^n} \\ 0 & 0 & d^n_L \end{matrix} \right) $$ in other words $\tilde K^\bullet = K^\bullet \oplus C(1_{L^\bullet[-1]})$. Moreover, we set $$ \tilde \alpha = \left( \begin{matrix} \alpha & 0 & \text{id}_{L^n} \end{matrix} \right) $$ which is clearly a split surjection. It is also clear that it defines a morphism of complexes. We define $$ i = \left( \begin{matrix} \text{id}_{K^n} \\ 0 \\ 0 \end{matrix} \right) $$ so that clearly $\tilde \alpha \circ i = \alpha$. We set $$ s = \left( \begin{matrix} \text{id}_{K^n} & 0 & 0 \end{matrix} \right) $$ so that $s \circ i = \text{id}_{K^\bullet}$. Finally, let $h^n : \tilde K^n \to \tilde K^{n - 1}$ be the map which maps the summand $L^{n - 1}$ of $\tilde K^n$ via the identity morphism to the summand $L^{n - 1}$ of $\tilde K^{n - 1}$. Then it is a trivial matter to prove that $$ \text{id}_{\tilde K^\bullet} - i \circ s = d \circ h + h \circ d $$ which finishes the proof of the lemma.
By Lemma \ref{lemma-make-commute-map} we can replace $b$ and $b'$ by homotopic maps such that the right square of the left diagram commutes and the left square of the right diagram commutes. In other words, we have $\Im(b^n) \subset \Im(A_2^n \to B_2^n)$ and $\Ker((b')^n) \supset \Im(A_2^n \to B_2^n)$. Then $b \circ b' = 0$ as a map of complexes.
Let $a^{-1} : K_2^\bullet \to K_1^\bullet$ be a morphism of complexes which is inverse to $a$ in $K(\mathcal{A})$. Let $b^{-1} : L_2^\bullet \to L_1^\bullet$ be a morphism of complexes which is inverse to $b$ in $K(\mathcal{A})$. Let $c' : C(f_2)^\bullet \to C(f_1)^\bullet$ be the morphism from Lemma \ref{lemma-functorial-cone} applied to $f_1 \circ a^{-1} = b^{-1} \circ f_2$. If we can show that $c \circ c'$ and $c' \circ c$ are isomorphisms in $K(\mathcal{A})$ then we win. Hence it suffices to prove the following: Given a morphism of triangles $(1, 1, c) : (K^\bullet, L^\bullet, C(f)^\bullet, f, i, p)$ in $K(\mathcal{A})$ the morphism $c$ is an isomorphism in $K(\mathcal{A})$. By assumption the two squares in the diagram $$ \xymatrix{ L^\bullet \ar[r] \ar[d]_1 & C(f)^\bullet \ar[r] \ar[d]_c & K^\bullet[1] \ar[d]_1 \\ L^\bullet \ar[r] & C(f)^\bullet \ar[r] & K^\bullet[1] } $$ commute up to homotopy. By construction of $C(f)^\bullet$ the rows form termwise split sequences of complexes. Thus we see that $(c - 1)^2 = 0$ in $K(\mathcal{A})$ by Lemma \ref{lemma-nilpotent}. Hence $c$ is an isomorphism in $K(\mathcal{A})$ with inverse $2 - c$.
Proof of (1). We have $C(\alpha)^n = B^n \oplus A^{n + 1}$ and we simply define $C(\alpha)^n \to C^n$ via the projection onto $B^n$ followed by $\beta^n$. This defines a morphism of complexes because the compositions $A^{n + 1} \to B^{n + 1} \to C^{n + 1}$ are zero. To get a homotopy inverse we take $C^\bullet \to C(\alpha)^\bullet$ given by $(s^n , -\delta^n)$ in degree $n$. This is a morphism of complexes because the morphism $\delta^n$ can be characterized as the unique morphism $C^n \to A^{n + 1}$ such that $d \circ s^n - s^{n + 1} \circ d = \alpha \circ \delta^n$, see proof of Homology, Lemma \ref{homology-lemma-ses-termwise-split-cochain}. The composition $C^\bullet \to C(\alpha)^\bullet \to C^\bullet$ is the identity. The composition $C(\alpha)^\bullet \to C^\bullet \to C(\alpha)^\bullet$ is equal to the morphism $$ \left( \begin{matrix} s^n \circ \beta^n & 0 \\ -\delta^n \circ \beta^n & 0 \end{matrix} \right) $$ To see that this is homotopic to the identity map use the homotopy $h^n : C(\alpha)^n \to C(\alpha)^{n - 1}$ given by the matrix $$ \left( \begin{matrix} 0 & 0 \\ \pi^n & 0 \end{matrix} \right) : C(\alpha)^n = B^n \oplus A^{n + 1} \to B^{n - 1} \oplus A^n = C(\alpha)^{n - 1} $$ It is trivial to verify that $$ \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) - \left( \begin{matrix} s^n \\ -\delta^n \end{matrix} \right) \left( \begin{matrix} \beta^n & 0 \end{matrix} \right) = \left( \begin{matrix} d & \alpha^n \\ 0 & -d \end{matrix} \right) \left( \begin{matrix} 0 & 0 \\ \pi^n & 0 \end{matrix} \right) + \left( \begin{matrix} 0 & 0 \\ \pi^{n + 1} & 0 \end{matrix} \right) \left( \begin{matrix} d & \alpha^{n + 1} \\ 0 & -d \end{matrix} \right) $$ To finish the proof of (1) we have to show that the morphisms $-p : C(\alpha)^\bullet \to A^\bullet[1]$ (see Definition \ref{definition-cone}) and  $C(\alpha)^\bullet \to C^\bullet \to A^\bullet[1]$ agree up to homotopy. This is clear from the above. Namely, we can use the homotopy inverse $(s, -\delta) : C^\bullet \to C(\alpha)^\bullet$ and check instead that the two maps $C^\bullet \to A^\bullet[1]$ agree. And note that $p \circ (s, -\delta) = -\delta$ as desired. \medskip\noindent Proof of (2). We let $\tilde f : K^\bullet \to \tilde L^\bullet$, $s : L^\bullet \to \tilde L^\bullet$ and $\pi : \tilde L^\bullet \to L^\bullet$ be as in Lemma \ref{lemma-make-injective}. By Lemmas \ref{lemma-functorial-cone} and \ref{lemma-third-isomorphism} the triangles $(K^\bullet, L^\bullet, C(f), i, p)$ and $(K^\bullet, \tilde L^\bullet, C(\tilde f), \tilde i, \tilde p)$ are isomorphic. Note that we can compose isomorphisms of triangles. Thus we may replace $L^\bullet$ by $\tilde L^\bullet$ and $f$ by $\tilde f$. In other words we may assume that $f$ is a termwise split injection. In this case the result follows from part (1).
The case $n = 1$ is without content. Lemma \ref{lemma-make-injective} is the case $n = 2$. Suppose we have constructed the diagram except for $B_n^\bullet$. Apply Lemma \ref{lemma-make-injective} to the composition $B_{n - 1}^\bullet \to A_{n - 1}^\bullet \to A_n^\bullet$. The result is a factorization $B_{n - 1}^\bullet \to B_n^\bullet \to A_n^\bullet$ as desired.
Each of the categories mentioned is a full additive subcategory. We use the criterion of Lemma \ref{lemma-triangulated-subcategory} to show that they are triangulated subcategories. It is clear that each of the categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^b(\mathcal{A})$ is preserved under the shift functors $[1], [-1]$. Finally, suppose that $f : A^\bullet \to B^\bullet$ is a morphism in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^b(\mathcal{A})$. Then $(A^\bullet, B^\bullet, C(f)^\bullet, f, i, -p)$ is a distinguished triangle of $K(\mathcal{A})$ with $C(f)^\bullet \in K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^b(\mathcal{A})$ as is clear from the construction of the cone. Thus the lemma is proved. (Alternatively, $K^\bullet \to L^\bullet$ is isomorphic to an termwise split injection of complexes in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^b(\mathcal{A})$, see Lemma \ref{lemma-make-injective} and then one can directly take the associated distinguished triangle.)
Suppose $A^\bullet \to B^\bullet \to C^\bullet$ is a termwise split sequence of complexes of $\mathcal{A}$ with splittings $(s^n, \pi^n)$ and associated morphism $\delta : C^\bullet \to A^\bullet[1]$, see Definition \ref{definition-split-ses}. Then $F(A^\bullet) \to F(B^\bullet) \to F(C^\bullet)$ is a termwise split sequence of complexes with splittings $(F(s^n), F(\pi^n))$ and associated morphism $F(\delta) : F(C^\bullet) \to F(A^\bullet)[1]$. Thus $F$ transforms distinguished triangles into distinguished triangles.
We will use that $K(\mathcal{A})$ is a triangulated category by Proposition \ref{proposition-homotopy-category-triangulated}. Let $W^\bullet$ be the cone on $c : C^\bullet \to A^\bullet[1]$ with its maps $i : A^\bullet[1] \to W^\bullet$ and $p : W^\bullet \to C^\bullet[1]$. Then $(C^\bullet, A^\bullet[1], W^\bullet, c, i, -p)$ is a distinguished triangle by Lemma \ref{lemma-the-same-up-to-isomorphisms}. Rotating backwards twice we see that $(A^\bullet, W^\bullet[-1], C^\bullet, -i[-1], p[-1], c)$ is a distinguished triangle. By TR3 there is a morphism of distinguished triangles $(\text{id}, \beta, \text{id}) : (A^\bullet, B^\bullet, C^\bullet, a, b, c) \to (A^\bullet, W^\bullet[-1], C^\bullet, -i[-1], p[-1], c)$ which must be an isomorphism by Lemma \ref{lemma-third-isomorphism-triangle}. This finishes the proof because $0 \to A^\bullet \to W^\bullet[-1] \to C^\bullet \to 0$ is a termwise split short exact sequence of complexes by the very construction of cones in Section \ref{section-cones}.
Because $H^0$ is a functor, and by our definition of distinguished triangles it suffices to prove that given a termwise split short exact sequence of complexes $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ the sequence $H^0(A^\bullet) \to H^0(B^\bullet) \to H^0(C^\bullet)$ is exact. This follows from Homology, Lemma \ref{homology-lemma-long-exact-sequence-cochain}.
The initial statements follow from Lemma \ref{lemma-acyclic-general} by considering the restriction of the homological functor $H^0$. The statement on kernels in (1), (2), (3) is a consequence of the definitions in each case. Each of the functors is essentially surjective by Lemma \ref{lemma-complex-cohomology-bounded}. To finish the proof we have to show the functors are fully faithful. We first do this for the bounded below version. \medskip\noindent Suppose that $K^\bullet, L^\bullet$ are bounded above complexes. A morphism between these in $D(\mathcal{A})$ is of the form $s^{-1}f$ for a pair $f : K^\bullet \to (L')^\bullet$, $s : L^\bullet \to (L')^\bullet$ where $s$ is a quasi-isomorphism. This implies that $(L')^\bullet$ has cohomology bounded below. Hence by Lemma \ref{lemma-complex-cohomology-bounded} we can choose a quasi-isomorphism $s' : (L')^\bullet \to (L'')^\bullet$ with $(L'')^\bullet$ bounded below. Then the pair $(s' \circ f, s' \circ s)$ defines a morphism in $\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})$. Hence the functor is ``full''. Finally, suppose that the pair $f : K^\bullet \to (L')^\bullet$, $s : L^\bullet \to (L')^\bullet$ defines a morphism in $\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})$ which is zero in $D(\mathcal{A})$. This means that there exists a quasi-isomorphism $s' : (L')^\bullet \to (L'')^\bullet$ such that $s' \circ f = 0$. Using Lemma \ref{lemma-complex-cohomology-bounded} once more we obtain a quasi-isomorphism $s'' : (L'')^\bullet \to (L''')^\bullet$ with $(L''')^\bullet$ bounded below. Thus we see that $s'' \circ s' \circ f = 0$ which implies that $s^{-1}f$ is zero in $\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})$. This finishes the proof that the functor in (1) is an equivalence. \medskip\noindent The proof of (2) is dual to the proof of (1). To prove (3) we may use the result of (2). Hence it suffices to prove that the functor $\text{Qis}^b(\mathcal{A})^{-1}K^b(\mathcal{A}) \to \text{Qis}^{-}(\mathcal{A})^{-1}K^{-}(\mathcal{A})$ is fully faithful. The argument given in the previous paragraph applies directly to show this where we consistently work with complexes which are already bounded above.
We have already seen that this choice leads to a distinguished triangle whenever given a short exact sequence of complexes. We have to show that given a commutative diagram $$ \xymatrix{ 0 \ar[r] & A^\bullet \ar[r]_a \ar[d]_f & B^\bullet \ar[r]_b \ar[d]_g & C^\bullet \ar[r] \ar[d]_h & 0 \\ 0 \ar[r] & (A')^\bullet \ar[r]^{a'} & (B')^\bullet \ar[r]^{b'} & (C')^\bullet \ar[r] & 0 } $$ we get the desired commutative diagram of Definition \ref{definition-delta-functor} (2). By Lemma \ref{lemma-functorial-cone} the pair $(f, g)$ induces a canonical morphism $c : C(a)^\bullet \to C(a')^\bullet$. It is a simple computation to show that $q' \circ c = h \circ q$ and $f[1] \circ p = p' \circ c$. From this the result follows directly.
The case $n = 1$. Since $\tau_{\leq 0}K_0^\bullet = K_0^\bullet$ in $D(\mathcal{A})$ we can replace $K_0^\bullet$ by $\tau_{\leq 0}K_0^\bullet$ and $K_1^\bullet$ by $\tau_{\leq 0}K_1^\bullet$. Consider the distinguished triangle $$ \tau_{\leq -1}K_1^\bullet \to K_1^\bullet \to H^0(K_1^\bullet)[0] \to (\tau_{\leq -1}K_1^\bullet)[1] $$ (Remark \ref{remark-truncation-distinguished-triangle}). The composition $K_0^\bullet \to K_1^\bullet \to H^0(K_1^\bullet)[0]$ is zero as it is equal to $K_0^\bullet \to H^0(K_0^\bullet)[0] \to H^0(K_1^\bullet)[0]$ which is zero by assumption. The fact that $\Hom_{D(\mathcal{A})}(K_0^\bullet, -)$ is a homological functor (Lemma \ref{lemma-representable-homological}), allows us to find the desired factorization. For $n = 2$ we get a factorization $K_0^\bullet \to \tau_{\leq -1}K_1^\bullet$ by the case $n = 1$ and we can apply the case $n = 1$ to the map of complexes $\tau_{\leq -1}K_1^\bullet \to \tau_{\leq -1}K_2^\bullet$ to get a factorization $\tau_{\leq -1}K_1^\bullet \to \tau_{\leq -2}K_2^\bullet$. The general case is proved in exactly the same manner.
This follows from the fact that $H^0 : K(\mathcal{A}) \to \mathcal{A}$ is homological, see Lemma \ref{lemma-cohomology-homological} and the fact that the functors $\text{gr}, \text{gr}^p, (\text{forget }F)$ are exact functors of triangulated categories. See Lemma \ref{lemma-exact-compose-homological-functor}.
This follows from the universal property of localization, see Lemma \ref{lemma-universal-property-localization}, provided we can show that a filtered quasi-isomorphism is turned into a quasi-isomorphism by each of the functors $\text{gr}^p, \text{gr}, (\text{forget }F)$. This is true by definition for the first two. For the last one the statement we have to do a little bit of work. Let $f : K^\bullet \to L^\bullet$ be a filtered quasi-isomorphism in $K(\text{Fil}^f(\mathcal{A}))$. Choose a distinguished triangle $(K^\bullet, L^\bullet, M^\bullet, f, g, h)$ which contains $f$. Then $M^\bullet$ is filtered acyclic, see Lemma \ref{lemma-filtered-acyclic}. Hence by the corresponding lemma for $K(\mathcal{A})$ it suffices to show that a filtered acyclic complex is an acyclic complex if we forget the filtration. This follows from Homology, Lemma \ref{homology-lemma-filtered-acyclic}.
Suppose that $H^n(\text{gr}(K^\bullet)) = 0$ for all $n < a$. By Homology, Lemma \ref{homology-lemma-filtered-acyclic} the sequence $$ K^{a - 1} \xrightarrow{d^{a - 2}} K^{a - 1} \xrightarrow{d^{a - 1}} K^a $$ is an exact sequence of objects of $\mathcal{A}$ and the morphisms $d^{a - 2}$ and $d^{a - 1}$ are strict. Hence $\Coim(d^{a - 1}) = \Im(d^{a - 1})$ in $\text{Fil}^f(\mathcal{A})$ and the map $\text{gr}(\Im(d^{a - 1})) \to \text{gr}(K^a)$ is injective with image equal to the image of $\text{gr}(K^{a - 1}) \to \text{gr}(K^a)$, see Homology, Lemma \ref{homology-lemma-characterize-strict}. This means that the map $K^\bullet \to \tau_{\geq a}K^\bullet$ into the truncation $$ \tau_{\geq a}K^\bullet = (\ldots \to 0 \to K^a/\Im(d^{a - 1}) \to K^{a + 1} \to \ldots) $$ is a filtered quasi-isomorphism. This proves (1). The proof of (2) is dual to the proof of (1). Part (3) follows formally from (1) and (2).