hcju/stacksqa huggingface dataset:
Suppose that $K^p = 0$ for $p < n$. Decompose $K^\bullet$ into short exact sequences as follows: Set $Z^p = \Ker(d^p)$, $B^p = \Im(d^{p - 1})$, $H^p = Z^p/B^p$, and consider $$ \begin{matrix} 0 \to Z^n \to K^n \to B^{n + 1} \to 0 \\ 0 \to B^{n + 1} \to Z^{n + 1} \to H^{n + 1} \to 0 \\ 0 \to Z^{n + 1} \to K^{n + 1} \to B^{n + 2} \to 0 \\ 0 \to B^{n + 2} \to Z^{n + 2} \to H^{n + 2} \to 0 \\ \ldots \end{matrix} $$ Set $I^{p, q} = 0$ for $p < n$. Inductively we choose injective resolutions as follows: \begin{enumerate} \item Choose an injective resolution $Z^n \to J_Z^{n, \bullet}$. \item Using Lemma \ref{lemma-injective-resolution-ses} choose injective resolutions $K^n \to I^{n, \bullet}$, $B^{n + 1} \to J_B^{n + 1, \bullet}$, and an exact sequence of complexes $0 \to J_Z^{n, \bullet} \to I^{n, \bullet} \to J_B^{n + 1, \bullet} \to 0$ compatible with the short exact sequence $0 \to Z^n \to K^n \to B^{n + 1} \to 0$. \item Using Lemma \ref{lemma-injective-resolution-ses} choose injective resolutions $Z^{n + 1} \to J_Z^{n + 1, \bullet}$, $H^{n + 1} \to J_H^{n + 1, \bullet}$, and an exact sequence of complexes $0 \to J_B^{n + 1, \bullet} \to J_Z^{n + 1, \bullet} \to J_H^{n + 1, \bullet} \to 0$ compatible with the short exact sequence $0 \to B^{n + 1} \to Z^{n + 1} \to H^{n + 1} \to 0$. \item Etc. \end{enumerate} Taking as maps $d_1^\bullet : I^{p, \bullet} \to I^{p + 1, \bullet}$ the compositions $I^{p, \bullet} \to J_B^{p + 1, \bullet} \to J_Z^{p + 1, \bullet} \to I^{p + 1, \bullet}$ everything is clear.
We will use the following remarks without further mention: \begin{enumerate} \item As $I^{p, \bullet}$ is an injective resolution of $K^p$ we see that $RF$ is defined at $K^p[0]$ with value $F(I^{p, \bullet})$. \item As $H^p_I(I^{\bullet, \bullet})$ is an injective resolution of $H^p(K^\bullet)$ the derived functor $RF$ is defined at $H^p(K^\bullet)[0]$ with value $F(H^p_I(I^{\bullet, \bullet}))$. \item By Homology, Lemma \ref{homology-lemma-double-complex-gives-resolution} the total complex $\text{Tot}(I^{\bullet, \bullet})$ is an injective resolution of $K^\bullet$. Hence $RF$ is defined at $K^\bullet$ with value $F(\text{Tot}(I^{\bullet, \bullet}))$. \end{enumerate} Consider the two spectral sequences associated to the double complex $L^{\bullet, \bullet} = F(I^{\bullet, \bullet})$, see Homology, Lemma \ref{homology-lemma-ss-double-complex}. These are both bounded, converge to $H^*(\text{Tot}(L^{\bullet, \bullet}))$, and induce finite filtrations on $H^n(\text{Tot}(L^{\bullet, \bullet}))$, see Homology, Lemma \ref{homology-lemma-first-quadrant-ss}. Since $\text{Tot}(L^{\bullet, \bullet}) = \text{Tot}(F(I^{\bullet, \bullet})) = F(\text{Tot}(I^{\bullet, \bullet}))$ computes $H^n(RF(K^\bullet))$ we find the final assertion of the lemma holds true. \medskip\noindent Computation of the first spectral sequence. We have ${}'E_1^{p, q} = H^q(L^{p, \bullet})$ in other words $$ {}'E_1^{p, q} = H^q(F(I^{p, \bullet})) = R^qF(K^p) $$ as desired. Observe for later use that the maps ${}'d_1^{p, q} : {}'E_1^{p, q} \to {}'E_1^{p + 1, q}$ are the maps $R^qF(K^p) \to R^qF(K^{p + 1})$ induced by $K^p \to K^{p + 1}$ and the fact that $R^qF$ is a functor. \medskip\noindent Computation of the second spectral sequence. We have ${}''E_1^{p, q} = H^q(L^{\bullet, p}) = H^q(F(I^{\bullet, p}))$. Note that the complex $I^{\bullet, p}$ is bounded below, consists of injectives, and moreover each kernel, image, and cohomology group of the differentials is an injective object of $\mathcal{A}$. Hence we can split the differentials, i.e., each differential is a split surjection onto a direct summand. It follows that the same is true after applying $F$. Hence ${}''E_1^{p, q} = F(H^q(I^{\bullet, p})) = F(H^q_I(I^{\bullet, p}))$. The differentials on this are $(-1)^q$ times $F$ applied to the differential of the complex $H^p_I(I^{\bullet, \bullet})$ which is an injective resolution of $H^p(K^\bullet)$. Hence the description of the $E_2$ terms.
If (2) holds, then (1) follows by evaluating the isomorphism $t$ on $RF(I) = F(I)$. Conversely, assume (1) holds. Let $A^\bullet$ be a bounded below complex of $\mathcal{A}$. Choose an injective resolution $A^\bullet \to I^\bullet$. The map $t$ is given (see proof of Lemma \ref{lemma-compose-derived-functors-general}) by the maps $$ R(G \circ F)(A^\bullet) = (G \circ F)(I^\bullet) = G(F(I^\bullet))) \to RG(F(I^\bullet)) = RG(RF(A^\bullet)) $$ where the arrow is an isomorphism by Lemma \ref{lemma-leray-acyclicity}.
We may represent $X$ by a bounded below complex $A^\bullet$. Choose an injective resolution $A^\bullet \to I^\bullet$. Choose a Cartan-Eilenberg resolution $F(I^\bullet) \to I^{\bullet, \bullet}$ using Lemma \ref{lemma-cartan-eilenberg}. Apply the second spectral sequence of Lemma \ref{lemma-two-ss-complex-functor}.
For every morphism $\alpha : K^\bullet \to L^\bullet$ of $K^{+}(\mathcal{A})$ there is a unique morphism $j(\alpha) : j(K^\bullet) \to j(L^\bullet)$ in $K^{+}(\mathcal{I})$ such that $$ \xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_{i_{K^\bullet}} & L^\bullet \ar[d]^{i_{L^\bullet}} \\ j(K^\bullet) \ar[r]^{j(\alpha)} & j(L^\bullet) } $$ is commutative in $K^{+}(\mathcal{A})$. To see this either use Lemmas \ref{lemma-morphisms-lift} and \ref{lemma-morphisms-equal-up-to-homotopy} or the equivalent Lemma \ref{lemma-morphisms-into-injective-complex}. The uniqueness implies that $j$ is a functor, and the commutativity of the diagram implies that $i$ gives a $2$-morphism which witnesses the $2$-commutativity of the diagram of categories in the statement of the lemma.
Consider the set of all objects $K^\bullet$ of $K^{+}(\mathcal{A})$. (Recall that by our conventions any category has a set of objects unless mentioned otherwise.) By Lemma \ref{lemma-injective-resolutions-exist} every object has an injective resolution. By the axiom of choice we can choose for each $K^\bullet$ an injective resolution $i_{K^\bullet} : K^\bullet \to j(K^\bullet)$.
Denote $i_{K^\bullet} : K^\bullet \to j(K^\bullet)$ the canonical maps of Definition \ref{definition-localization-functor}. First we discuss the existence of the functorial isomorphism $j(K^\bullet[1]) \to j(K^\bullet)[1]$. Consider the diagram $$ \xymatrix{ K^\bullet[1] \ar[d]^{i_{K^\bullet[1]}} \ar@{=}[rr] & & K^\bullet[1] \ar[d]^{i_{K^\bullet}[1]} \\ j(K^\bullet[1]) \ar@{..>}[rr]^{\xi_{K^\bullet}} & & j(K^\bullet)[1] } $$ By Lemmas \ref{lemma-morphisms-lift} and \ref{lemma-morphisms-equal-up-to-homotopy} there exists a unique dotted arrow $\xi_{K^\bullet}$ in $K^{+}(\mathcal{I})$ making the diagram commute in $K^{+}(\mathcal{A})$. We omit the verification that this gives a functorial isomorphism. (Hint: use Lemma \ref{lemma-morphisms-equal-up-to-homotopy} again.) \medskip\noindent Let $(K^\bullet, L^\bullet, M^\bullet, f, g, h)$ be a distinguished triangle of $K^{+}(\mathcal{A})$. We have to show that $(j(K^\bullet), j(L^\bullet), j(M^\bullet), j(f), j(g), \xi_{K^\bullet} \circ j(h))$ is a distinguished triangle of $K^{+}(\mathcal{I})$. Note that we have a commutative diagram $$ \xymatrix{ K^\bullet \ar[r]_f \ar[d] & L^\bullet \ar[r]_g \ar[d] & M^\bullet \ar[rr]_h \ar[d] & & K^\bullet[1] \ar[d] \\ j(K^\bullet) \ar[r]^{j(f)} & j(L^\bullet) \ar[r]^{j(g)} & j(M^\bullet) \ar[rr]^{\xi_{K^\bullet} \circ j(h)} & & j(K^\bullet)[1] } $$ in $K^{+}(\mathcal{A})$ whose vertical arrows are the quasi-isomorphisms $i_K, i_L, i_M$. Hence we see that the image of $(j(K^\bullet), j(L^\bullet), j(M^\bullet), j(f), j(g), \xi_{K^\bullet} \circ j(h))$ in $D^{+}(\mathcal{A})$ is isomorphic to a distinguished triangle and hence a distinguished triangle by TR1. Thus we see from Lemma \ref{lemma-exact-equivalence} that $(j(K^\bullet), j(L^\bullet), j(M^\bullet), j(f), j(g), \xi_{K^\bullet} \circ j(h))$ is a distinguished triangle in $K^{+}(\mathcal{I})$.
By Lemma \ref{lemma-bounded-derived} $Q$ is a localization functor. To prove the existence of $j'$ it suffices to show that any element of $\text{Qis}^{+}(\mathcal{A})$ is mapped to an isomorphism under the functor $j$, see Lemma \ref{lemma-universal-property-localization}. This is true by the remarks following Definition \ref{definition-localization-functor}.
Let $A \mapsto (A \to J(A))$ be a functorial injective embedding, see Homology, Definition \ref{homology-definition-functorial-injective-embedding}. We first note that we may assume $J(0) = 0$. Namely, if not then for any object $A$ we have $0 \to A \to 0$ which gives a direct sum decomposition $J(A) = J(0) \oplus \Ker(J(A) \to J(0))$. Note that the functorial morphism $A \to J(A)$ has to map into the second summand. Hence we can replace our functor by $J'(A) = \Ker(J(A) \to J(0))$ if needed. \medskip\noindent Let $K^\bullet$ be a bounded below complex of $\mathcal{A}$. Say $K^p = 0$ if $p < B$. We are going to construct a double complex $I^{\bullet, \bullet}$ of injectives, together with a map $\alpha : K^\bullet \to I^{\bullet, 0}$ such that $\alpha$ induces a quasi-isomorphism of $K^\bullet$ with the associated total complex of $I^{\bullet, \bullet}$. First we set $I^{p, q} = 0$ whenever $q < 0$. Next, we set $I^{p, 0} = J(K^p)$ and $\alpha^p : K^p \to I^{p, 0}$ the functorial embedding. Since $J$ is a functor we see that $I^{\bullet, 0}$ is a complex and that $\alpha$ is a morphism of complexes. Each $\alpha^p$ is injective. And $I^{p, 0} = 0$ for $p < B$ because $J(0) = 0$. Next, we set $I^{p, 1} = J(\Coker(K^p \to I^{p, 0}))$. Again by functoriality we see that $I^{\bullet, 1}$ is a complex. And again we get that $I^{p, 1} = 0$ for $p < B$. It is also clear that $K^p$ maps isomorphically onto $\Ker(I^{p, 0} \to I^{p, 1})$. As our third step we take $I^{p, 2} = J(\Coker(I^{p, 0} \to I^{p, 1}))$. And so on and so forth. \medskip\noindent At this point we can apply Homology, Lemma \ref{homology-lemma-double-complex-gives-resolution} to get that the map $$ \alpha : K^\bullet \longrightarrow \text{Tot}(I^{\bullet, \bullet}) $$ is a quasi-isomorphism. To prove we get a functor $inj$ it rests to show that the construction above is functorial. This verification is omitted. \medskip\noindent Suppose we have a functor $inj$ such that $s \circ inj = \text{id}$. For every object $K^\bullet$ of $\text{Comp}^{+}(\mathcal{A})$ we can write $$ inj(K^\bullet) = (i_{K^\bullet} : K^\bullet \to j(K^\bullet)) $$ This provides us with a resolution functor as in Definition \ref{definition-localization-functor}.
Let $p$ be the largest integer such that $F^pI \not = 0$. In particular $\text{gr}^p(I) = F^pI$. Let $I'$ be the object of $\text{Fil}^f(\mathcal{A})$ whose underlying object of $\mathcal{A}$ is $F^pI$ and with filtration given by $F^nI' = 0$ for $n > p$ and $F^nI' = I' = F^pI$ for $n \leq p$. Note that $I' \to I$ is a strict monomorphism too. The fact that $u$ is a strict monomorphism implies that $F^pI \to A/F^{p + 1}(A)$ is injective, see Homology, Lemma \ref{homology-lemma-characterize-strict}. Choose a splitting $s : A/F^{p + 1}A \to F^pI$ in $\mathcal{A}$. The induced morphism $s' : A \to I'$ is a strict morphism of filtered objects splitting the composition $I' \to I \to A$. Hence we can write $A = I' \oplus \Ker(s')$ and $I = I' \oplus \Ker(s'|_I)$. Note that $\Ker(s'|_I) \to \Ker(s')$ is a strict monomorphism and that $\Ker(s'|_I)$ is a filtered injective object. By induction on the length of the filtration on $I$ the map $\Ker(s'|_I) \to \Ker(s')$ is a split injection. Thus we win.
First choose a strict monomorphism $u_0 : A \to I^0$ of $A$ into a filtered injective object, see Lemma \ref{lemma-strict-monomorphism-into-filtered-injective}. Next, choose a strict monomorphism $u_1 : \Coker(u_0) \to I^1$ into a filtered injective object of $\mathcal{A}$. Denote $d^0$ the induced map $I^0 \to I^1$. Next, choose a strict monomorphism $u_2 : \Coker(u_1) \to I^2$ into a filtered injective object of $\mathcal{A}$. Denote $d^1$ the induced map $I^1 \to I^2$. And so on. This works because each of the sequences $$ 0 \to \Coker(u_n) \to I^{n + 1} \to \Coker(u_{n + 1}) \to 0 $$ is short exact, i.e., induces a short exact sequence on applying $\text{gr}$. To see this use Homology, Lemma \ref{homology-lemma-characterize-strict}.
As $A[0] \to I^\bullet$ and $C[0] \to J^\bullet$ are filtered quasi-isomorphisms we conclude that $a : A \to I^0$, $b : B \to J^0$ and all the morphisms $d_I^n$, $d_J^n$ are strict, see Homology, Lemma \ref{homology-lemma-filtered-acyclic}. We will inductively construct the maps $f^n$ in the following commutative diagram $$ \xymatrix{ A \ar[r]_a \ar[d]_f & I^0 \ar[r] \ar[d]^{f^0} & I^1 \ar[r] \ar[d]^{f^1} & I^2 \ar[r] \ar[d]^{f^2} & \ldots \\ B \ar[r]^b & J^0 \ar[r] & J^1 \ar[r] & J^2 \ar[r] & \ldots } $$ Because $A \to I^0$ is a strict monomorphism and because $J^0$ is filtered injective, we can find a morphism $f^0 : I^0 \to J^0$ such that $f^0 \circ a = b \circ f$, see Lemma \ref{lemma-injective-property-filtered-injective}. The composition $d_J^0 \circ b \circ f$ is zero, hence $d_J^0 \circ f^0 \circ a = 0$, hence $d_J^0 \circ f^0$ factors through a unique morphism $$ \Coker(a) = \Coim(d_I^0) = \Im(d_I^0) \longrightarrow J^1. $$ As $\Im(d_I^0) \to I^1$ is a strict monomorphism we can extend the displayed arrow to a morphism $f^1 : I^1 \to J^1$ by Lemma \ref{lemma-injective-property-filtered-injective} again. And so on.
As $A[0] \to I^\bullet$ and $C[0] \to J^\bullet$ are filtered quasi-isomorphisms we conclude that $a : A \to I^0$, $c : C \to J^0$ and all the morphisms $d_I^n$, $d_J^n$ are strict, see Homology, Lemma \ref{lemma-filtered-acyclic}. We are going to step by step construct the south-east and the south arrows in the following commutative diagram $$ \xymatrix{ B \ar[r]_\beta \ar[rd]^b & C \ar[r]_c \ar[rd]^{\overline{b}} & J^0 \ar[d]^{\delta^0} \ar[r] & J^1 \ar[d]^{\delta^1} \ar[r] & \ldots \\ A \ar[u]^\alpha \ar[r]^a & I^0 \ar[r] & I^1 \ar[r] & I^2 \ar[r] & \ldots } $$ As $A \to B$ is a strict monomorphism, we can find a morphism $b : B \to I^0$ such that $b \circ \alpha = a$, see Lemma \ref{lemma-injective-property-filtered-injective}. As $A$ is the kernel of the strict morphism $I^0 \to I^1$ and $\beta = \Coker(\alpha)$ we obtain a unique morphism $\overline{b} : C \to I^1$ fitting into the diagram. As $c$ is a strict monomorphism and $I^1$ is filtered injective we can find $\delta^0 : J^0 \to I^1$, see Lemma \ref{lemma-injective-property-filtered-injective}. Because $B \to C$ is a strict epimorphism and because $B \to I^0 \to I^1 \to I^2$ is zero, we see that $C \to I^1 \to I^2$ is zero. Hence $d_I^1 \circ \delta^0$ is zero on $C \cong \Im(c)$. Hence $d_I^1 \circ \delta^0$ factors through a unique morphism $$ \Coker(c) = \Coim(d_J^0) = \Im(d_J^0) \longrightarrow I^2. $$ As $I^2$ is filtered injective and $\Im(d_J^0) \to J^1$ is a strict monomorphism we can extend the displayed morphism to a morphism $\delta^1 : J^1 \to I^2$, see Lemma \ref{lemma-injective-property-filtered-injective}. And so on. We set $M^\bullet = I^\bullet \oplus J^\bullet$ with differential $$ d_M^n = \left( \begin{matrix} d_I^n & (-1)^{n + 1}\delta^n \\ 0 & d_J^n \end{matrix} \right) $$ Finally, the map $B[0] \to M^\bullet$ is given by $b \oplus c \circ \beta : M \to I^0 \oplus J^0$.
After replacing $K^\bullet$ by a shift (which is harmless for the proof) we may assume that $K^n = 0$ for $n < 0$. Consider the short exact sequences $$ \begin{matrix} 0 \to \Ker(d_K^0) \to K^0 \to \Coim(d_K^0) \to 0 \\ 0 \to \Ker(d_K^1) \to K^1 \to \Coim(d_K^1) \to 0 \\ 0 \to \Ker(d_K^2) \to K^2 \to \Coim(d_K^2) \to 0 \\ \ldots \end{matrix} $$ of the exact category $\text{Fil}^f(\mathcal{A})$ and the maps $u_i : \Coim(d_K^i) \to \Ker(d_K^{i + 1})$. For each $i \geq 0$ we may choose filtered quasi-isomorphisms $$ \begin{matrix} \Ker(d_K^i)[0] \to I_{ker, i}^\bullet \\ \Coim(d_K^i)[0] \to I_{coim, i}^\bullet \end{matrix} $$ with $I_{ker, i}^n, I_{coim, i}^n$ filtered injective and zero for $n < 0$, see Lemma \ref{lemma-filtered-injective-right-resolution-single-object}. By Lemma \ref{lemma-filtered-injective-right-resolution-map} we may lift $u_i$ to a morphism of complexes $u_i^\bullet : I_{coim, i}^\bullet \to I_{ker, i + 1}^\bullet$. Finally, for each $i \geq 0$ we may complete the diagrams $$ \xymatrix{ 0 \ar[r] & \Ker(d_K^i)[0] \ar[r] \ar[d] & K^i[0] \ar[r] \ar[d] & \Coim(d_K^i)[0] \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I_{ker, i}^\bullet \ar[r]^{\alpha_i} & I_i^\bullet \ar[r]^{\beta_i} & I_{coim, i}^\bullet \ar[r] & 0 } $$ with the lower sequence a termwise split exact sequence, see Lemma \ref{lemma-filtered-injective-right-resolution-ses}. For $i \geq 0$ set $d_i : I_i^\bullet \to I_{i + 1}^\bullet$ equal to $d_i =  \alpha_{i + 1} \circ u_i^\bullet \circ \beta_i$. Note that $d_i \circ d_{i - 1} = 0$ because $\beta_i \circ \alpha_i = 0$. Hence we have constructed a commutative diagram $$ \xymatrix{ I_0^\bullet \ar[r] & I_1^\bullet \ar[r] & I_2^\bullet \ar[r] & \ldots \\ K^0[0] \ar[r] \ar[u] & K^1[0] \ar[r] \ar[u] & K^2[0] \ar[r] \ar[u] & \ldots } $$ Here the vertical arrows are filtered quasi-isomorphisms. The upper row is a complex of complexes and each complex consists of filtered injective objects with no nonzero objects in degree $< 0$. Thus we obtain a double complex by setting $I^{a, b} = I_a^b$ and using $$ d_1^{a, b} : I^{a, b} = I_a^b \to I_{a + 1}^b = I^{a + 1, b} $$ the map $d_a^b$ and using for $$ d_2^{a, b} : I^{a, b} = I_a^b \to I_a^{b + 1} = I^{a, b + 1} $$ the map $d_{I_a}^b$. Denote $\text{Tot}(I^{\bullet, \bullet})$ the total complex associated to this double complex, see Homology, Definition \ref{homology-definition-associated-simple-complex}. Observe that the maps $K^n[0] \to I_n^\bullet$ come from maps $K^n \to I^{n, 0}$ which give rise to a map of complexes $$ K^\bullet \longrightarrow \text{Tot}(I^{\bullet, \bullet}) $$ We claim this is a filtered quasi-isomorphism. As $\text{gr}(-)$ is an additive functor, we see that $\text{gr}(\text{Tot}(I^{\bullet, \bullet})) = \text{Tot}(\text{gr}(I^{\bullet, \bullet}))$. Thus we can use Homology, Lemma \ref{homology-lemma-double-complex-gives-resolution} to conclude that $\text{gr}(K^\bullet) \to \text{gr}(\text{Tot}(I^{\bullet, \bullet}))$ is a quasi-isomorphism as desired.
Let $\alpha : K^\bullet \to I^\bullet$ be a morphism of complexes. Assume that $\alpha^j = 0$ for $j < n$. We will show that there exists a morphism $h : K^{n + 1} \to I^n$ such that $\alpha^n = h \circ d$. Thus $\alpha$ will be homotopic to the morphism of complexes $\beta$ defined by $$ \beta^j = \left\{ \begin{matrix} 0 & \text{if} & j \leq n \\ \alpha^{n + 1} - d \circ h & \text{if} & j = n + 1 \\ \alpha^j & \text{if} & j > n + 1 \end{matrix} \right. $$ This will clearly prove the lemma (by induction). To prove the existence of $h$ note that $\alpha^n \circ d_K^{n - 1} = 0$ since $\alpha^{n - 1} = 0$. Since $K^\bullet$ is filtered acyclic we see that $d_K^{n - 1}$ and $d_K^n$ are strict and that $$ 0 \to \Im(d_K^{n - 1}) \to K^n \to \Im(d_K^n) \to 0 $$ is an exact sequence of the exact category $\text{Fil}^f(\mathcal{A})$, see Homology, Lemma \ref{homology-lemma-filtered-acyclic}. Hence we can think of $\alpha^n$ as a map into $I^n$ defined on $\Im(d_K^n)$. Using that $\Im(d_K^n) \to K^{n + 1}$ is a strict monomorphism and that $I^n$ is filtered injective we may lift this map to a map $h : K^{n + 1} \to I^n$ as desired, see Lemma \ref{lemma-injective-property-filtered-injective}.
Proof of (1). Note that $$ (K^\bullet, L^\bullet, C(\alpha)^\bullet, \alpha, i, -p) $$ is a distinguished triangle in $K(\text{Fil}^f(\mathcal{A}))$ (Lemma \ref{lemma-the-same-up-to-isomorphisms}) and $C(\alpha)^\bullet$ is a filtered acyclic complex (Lemma \ref{lemma-filtered-acyclic}). Then $$ \xymatrix{ \Hom_{K(\text{Fil}^f(\mathcal{A}))}(C(\alpha)^\bullet, I^\bullet) \ar[r] & \Hom_{K(\text{Fil}^f(\mathcal{A}))}(L^\bullet, I^\bullet) \ar[r] & \Hom_{K(\text{Fil}^f(\mathcal{A}))}(K^\bullet, I^\bullet) \ar[lld] \\ \Hom_{K(\text{Fil}^f(\mathcal{A}))}(C(\alpha)^\bullet[-1], I^\bullet) } $$ is an exact sequence of abelian groups, see Lemma \ref{lemma-representable-homological}. At this point Lemma \ref{lemma-filtered-acyclic-is-zero} guarantees that the outer two groups are zero and hence $\Hom_{K(\mathcal{A})}(L^\bullet, I^\bullet) = \Hom_{K(\mathcal{A})}(K^\bullet, I^\bullet)$. \medskip\noindent Proof of (2). Let $a$ be an element of the right hand side. We may represent $a = \gamma\alpha^{-1}$ where $\alpha : K^\bullet \to L^\bullet$ is a filtered quasi-isomorphism and $\gamma : K^\bullet \to I^\bullet$ is a map of complexes. By part (1) we can find a morphism $\beta : L^\bullet \to I^\bullet$ such that $\beta \circ \alpha$ is homotopic to $\gamma$. This proves that the map is surjective. Let $b$ be an element of the left hand side which maps to zero in the right hand side. Then $b$ is the homotopy class of a morphism $\beta : L^\bullet \to I^\bullet$ such that there exists a filtered quasi-isomorphism $\alpha : K^\bullet \to L^\bullet$ with $\beta  \circ \alpha$ homotopic to zero. Then part (1) shows that $\beta$ is homotopic to zero also, i.e., $b = 0$.
The functor $K^{+}(\mathcal{I}^f) \to DF^{+}(\mathcal{A})$ is essentially surjective by Lemma \ref{lemma-right-resolution-by-filtered-injectives}. It is fully faithful by Lemma \ref{lemma-morphisms-into-filtered-injective-complex}. It is an exact functor by our definitions regarding distinguished triangles. The commutativity of the squares is immediate.
Choose a filtered quasi-isomorphism $K^\bullet \to I^\bullet$ with $I^\bullet$ a bounded below complex of filtered injective objects, see Lemma \ref{lemma-right-resolution-by-filtered-injectives}. Consider the complex $RT(K^\bullet) = T_{ext}(I^\bullet)$, see (\ref{equation-definition-filtered-derived-functor}). Thus we can consider the spectral sequence $(E_r, d_r)_{r \geq 0}$ associated to this as a filtered complex in $\mathcal{B}$, see Homology, Section \ref{homology-section-filtered-complex}. By Homology, Lemma \ref{homology-lemma-spectral-sequence-filtered-complex} we have $E_1^{p, q} = H^{p + q}(\text{gr}^p(T(I^\bullet)))$. By Equation (\ref{equation-decompose}) we have $E_1^{p, q} = H^{p + q}(T(\text{gr}^p(I^\bullet)))$, and by definition of a filtered injective resolution the map $\text{gr}^p(K^\bullet) \to \text{gr}^p(I^\bullet)$ is an injective resolution. Hence $E_1^{p, q} = R^{p + q}T(\text{gr}^p(K^\bullet))$. \medskip\noindent On the other hand, each $I^n$ has a finite filtration and hence each $T(I^n)$ has a finite filtration. Thus we may apply Homology, Lemma \ref{homology-lemma-biregular-ss-converges} to conclude that the spectral sequence is bounded, converges to $H^n(T(I^\bullet)) = R^nT(K^\bullet)$ moreover inducing finite filtrations on each of the terms. \medskip\noindent Suppose that $K^\bullet \to L^\bullet$ is a morphism of $\text{Comp}^{+}(\text{Fil}^f(\mathcal{A}))$. Choose a filtered quasi-isomorphism $L^\bullet \to J^\bullet$ with $J^\bullet$ a bounded below complex of filtered injective objects, see Lemma \ref{lemma-right-resolution-by-filtered-injectives}. By our results above, for example Lemma \ref{lemma-morphisms-into-filtered-injective-complex}, there exists a diagram $$ \xymatrix{ K^\bullet \ar[r] \ar[d] & L^\bullet \ar[d] \\ I^\bullet \ar[r] & J^\bullet } $$ which commutes up to homotopy. Hence we get a morphism of filtered complexes $T(I^\bullet) \to T(J^\bullet)$ which gives rise to the morphism of spectral sequences, see Homology, Lemma \ref{homology-lemma-spectral-sequence-filtered-complex-functorial}. The last statement follows from this.
For $i > p$ write any class $\xi$ as $\delta(E)$ where $E$ is a Yoneda extension $$ E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0 $$ This is possible by Lemma \ref{lemma-yoneda-extension}. Set $C = \Ker(Z_{p - 1} \to Z_p) = \Im(Z_p \to Z_{p - 1})$. Then $\delta(E)$ is the composition of $\delta(E')$ and $\delta(E'')$ where $$ E' : 0 \to C \to Z_{p - 1} \to \ldots \to Z_0 \to B \to 0 $$ and $$ E'' : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_p \to C \to 0 $$ Since $\delta(E') \in \Ext^p_\mathcal{A}(B, C) = 0$ we conclude.