hcju/stacksqa huggingface dataset:
As $\mathcal{B}$ is abelian we see that all finite limits and colimits exist in $\mathcal{B}$ by Lemma \ref{lemma-colimit-abelian-category}. Since $b$ is a left adjoint we see that $b$ is also right exact and hence exact, see Categories, Lemma \ref{categories-lemma-exact-adjoint}. Let $\varphi : B_1 \to B_2$ be a morphism of $\mathcal{B}$. In particular, if $K = \Ker(B_1 \to B_2)$, then $K$ is the equalizer of $0$ and $\varphi$ and hence $bK$ is the equalizer of $0$ and $b\varphi$, hence $bK$ is the kernel of $b\varphi$. Similarly, if $Q = \Coker(B_1 \to B_2)$, then $Q$ is the coequalizer of $0$ and $\varphi$ and hence $bQ$ is the coequalizer of $0$ and $b\varphi$, hence $bQ$ is the cokernel of $b\varphi$. Thus we see that every morphism of the form $b\varphi$ in $\mathcal{A}$ has a kernel and a cokernel. However, since $ba \cong \text{id}$ we see that every morphism of $\mathcal{A}$ is of this form, and we conclude that kernels and cokernels exist in $\mathcal{A}$. In fact, the argument shows that if $\psi : A_1 \to A_2$ is a morphism then $$ \Ker(\psi) = b\Ker(a\psi), \quad\text{and}\quad \Coker(\psi) = b\Coker(a\psi). $$ Now we still have to show that $\Coim(\psi)= \Im(\psi)$. We do this as follows. First note that since $\mathcal{A}$ has kernels and cokernels it has all finite limits and colimits (see proof of Lemma \ref{lemma-colimit-abelian-category}). Hence we see by Categories, Lemma \ref{categories-lemma-exact-adjoint} that $a$ is left exact and hence transforms kernels (=equalizers) into kernels. \begin{align*} \Coim(\psi) & = \Coker(\Ker(\psi) \to A_1) & \text{by definition} \\ & = b\Coker(a(\Ker(\psi) \to A_1)) & \text{by formula above} \\ & = b\Coker(\Ker(a\psi) \to aA_1)) & a\text{ preserves kernels} \\ & = b\Coim(a\psi) & \text{by definition} \\ & = b\Im(a\psi) & \mathcal{B}\text{ is abelian} \\ & = b\Ker(aA_2 \to \Coker(a\psi)) & \text{by definition} \\ & = \Ker(baA_2 \to b\Coker(a\psi)) & b\text{ preserves kernels} \\ & = \Ker(A_2 \to b\Coker(a\psi)) & ba = \text{id}_\mathcal{A} \\ & = \Ker(A_2 \to \Coker(\psi)) & \text{by formula above} \\ & = \Im(\psi) & \text{by definition} \end{align*} Thus the lemma holds.
By Lemma \ref{lemma-localization-preadditive} we see that $S^{-1}\mathcal{C}$ is preadditive and that $Q$ is additive. Recall that the functor $Q$ commutes with finite colimits (resp.\ finite limits), see Categories, Lemmas \ref{categories-lemma-left-localization-limits} and \ref{categories-lemma-right-localization-limits}. We conclude that $S^{-1}\mathcal{C}$ has a zero object and direct sums, see Lemmas \ref{lemma-preadditive-zero} and \ref{lemma-preadditive-direct-sum}.
If (2) holds we see that $0 = Q(0) : Q(X) \to Q(Y)$ is an isomorphism. In the additive category $S^{-1}\mathcal{C}$ this implies that $Q(X) = 0$. Hence (2) $\Rightarrow$ (1). Similarly, (3) $\Rightarrow$ (1). Suppose that $Q(X) = 0$. This implies that the morphism $f : 0 \to X$ is transformed into an isomorphism in $S^{-1}\mathcal{C}$. Hence by Categories, Lemma \ref{categories-lemma-what-gets-inverted} there exists a morphism $g : Z \to 0$ such that $fg \in S$. This proves (1) $\Rightarrow$ (3). Similarly, (1) $\Rightarrow$ (2).
Assume $S$ is a left multiplicative system. Let $a : X \to Y$ be a morphism of $S^{-1}\mathcal{A}$. Then $a = s^{-1}f$ for some $s : Y \to Y'$ in $S$ and $f : X \to Y'$. Since $Q(s)$ is an isomorphism we see that the existence of $\Coker(a : X \to Y)$ is equivalent to the existence of $\Coker(Q(f) : X \to Y')$. Since $\Coker(Q(f))$ is the coequalizer of $0$ and $Q(f)$ we see that $\Coker(Q(f))$ is represented by $Q(\Coker(f))$ by Categories, Lemma \ref{categories-lemma-left-localization-limits}. This proves (1). \medskip\noindent Part (2) is dual to part (1). \medskip\noindent If $S$ is a multiplicative system, then $S$ is both a left and a right multiplicative system. Thus we see that $S^{-1}\mathcal{A}$ has kernels and cokernels and $Q$ commutes with kernels and cokernels. To finish the proof of (3) we have to show that $\Coim = \Im$ in $S^{-1}\mathcal{A}$. Again using that any arrow in $S^{-1}\mathcal{A}$ is isomorphic to an arrow $Q(f)$ we see that the result follows from the result for $\mathcal{A}$.
Assume (1). If $A$ is zero, then (2) holds. If $A$ is not zero, then there exists a smallest nonzero object $A_1 \subset A$ by the Artinian property. Of course $A_1$ is simple. If $A_1 = A$, then we are done. If not, then we can find $A_1 \subset A_2 \subset A$ minimal with $A_2 \not = A_1$. Then $A_2/A_1$ is simple. Continuing in this way, we can find a sequence $0 \subset A_1 \subset A_2 \subset \ldots $ of subobjects of $A$ such that $A_i/A_{i - 1}$ is simple. Since $A$ is Noetherian, we conclude that the process stops. Hence (2) follows. \medskip\noindent Assume (2). We will prove (1) by induction on $n$. If $n = 1$, then $A$ is simple and clearly Noetherian and Artinian. If the result holds for $n - 1$, then we use the short exact sequence $0 \to A_{n - 1} \to A_n \to A_n/A_{n - 1} \to 0$ and Lemmas \ref{lemma-ses-artinian} and \ref{lemma-ses-noetherian} to conclude for $n$.
Consider the set of arrows of $\mathcal{A}$ defined by the following formula $$ S = \{f \in \text{Arrows}(\mathcal{A}) \mid \Ker(f), \Coker(f) \in \Ob(\mathcal{C}) \}. $$ We claim that $S$ is a multiplicative system. To prove this we have to check MS1, MS2, MS3, see Categories, Definition \ref{categories-definition-multiplicative-system}. \medskip\noindent It is clear that identities are elements of $S$. Suppose that $f : A \to B$ and $g : B \to C$ are elements of $S$. There are exact sequences $$ \begin{matrix} 0 \to \Ker(f) \to \Ker(gf) \to \Ker(g) \\ \Coker(f) \to \Coker(gf) \to \Coker(g) \to 0 \end{matrix} $$ Hence it follows that $gf \in S$. This proves MS1. (In fact, a similar argument will show that $S$ is a saturated multiplicative system, see Categories, Definition \ref{categories-definition-saturated-multiplicative-system}.) \medskip\noindent Consider a solid diagram $$ \xymatrix{ A \ar[d]_t \ar[r]_g & B \ar@{..>}[d]^s \\ C \ar@{..>}[r]^f & C \amalg_A B } $$ with $t \in S$. Set $W = C \amalg_A B =  \Coker((t, -g) : A \to C \oplus B)$. Then $\Ker(t) \to \Ker(s)$ is surjective and $\Coker(t) \to \Coker(s)$ is an isomorphism. Hence $s$ is an element of $S$. This proves LMS2 and the proof of RMS2 is dual. \medskip\noindent Finally, consider morphisms $f, g : B \to C$ and a morphism $s : A \to B$ in $S$ such that $f \circ s = g \circ s$. This means that $(f - g) \circ s = 0$. In turn this means that $I = \Im(f - g) \subset C$ is a quotient of $\Coker(s)$ hence an object of $\mathcal{C}$. Thus $t : C \to C' = C/I$ is an element of $S$ such that $t \circ (f - g) = 0$, i.e., such that $t \circ f = t \circ g$. This proves LMS3 and the proof of RMS3 is dual. \medskip\noindent Having proved that $S$ is a multiplicative system we set $\mathcal{A}/\mathcal{C} = S^{-1}\mathcal{A}$, and we set $F$ equal to the localization functor $Q$. By Lemma \ref{lemma-localization-abelian} the category $\mathcal{A}/\mathcal{C}$ is abelian and $F$ is exact. If $X$ is in the kernel of $F = Q$, then by Lemma \ref{lemma-kernel-localization} we see that $0 : X \to Z$ is an element of $S$ and hence $X$ is an object of $\mathcal{C}$, i.e., the kernel of $F$ is $\mathcal{C}$. Finally, if $G$ is as in the statement of the lemma, then $G$ turns every element of $S$ into an isomorphism. Hence we obtain the functor $H : \mathcal{A}/\mathcal{C} \to \mathcal{B}$ from the universal property of localization, see Categories, Lemma \ref{categories-lemma-properties-left-localization}.
We will use the results of Lemma \ref{lemma-serre-subcategory-is-kernel} without further mention. The ``only if'' direction is true because the kernel of $\overline{F}$ is zero by construction. Namely, if $f : X \to Y$ is a morphism in $\mathcal{A}/\mathcal{C}$ such that $\overline{F}(f) = 0$, then $\overline{F}(\Im(f)) = \Im(\overline{F}(f)) = 0$, hence $\Im(f) = 0$ by the assumption on the kernel of $F$. Thus $f = 0$. \medskip\noindent For the ``if'' direction, let $X$ be an object of $\mathcal{A}$ such that $F(X) = 0$. Then $\overline{F}(\text{id}_X) = \text{id}_{\overline{F}(X)} = 0$, thus $\text{id}_X = 0$ in $\mathcal{A}/\mathcal{C}$ by faithfulness of $\overline{F}$. Hence $X = 0$ in $\mathcal{A}/\mathcal{C}$, that is $X \in \Ob(\mathcal{C})$.
Omitted. The maps come from the Snake Lemma \ref{lemma-snake} applied to the diagrams $$ \xymatrix{ & A_i/\Im(d_{A, i + 1}) \ar[r] \ar[d]^{d_{A, i}} & B_i/\Im(d_{B, i + 1}) \ar[r] \ar[d]^{d_{B, i}} & C_i/\Im(d_{C, i + 1}) \ar[r] \ar[d]^{d_{C, i}} & 0 \\ 0 \ar[r] & \Ker(d_{A, i - 1}) \ar[r] & \Ker(d_{B, i - 1}) \ar[r] & \Ker(d_{C, i - 1}) & } $$
Omitted. The maps come from the Snake Lemma \ref{lemma-snake} applied to the diagrams $$ \xymatrix{ & A^i/\Im(d_A^{i - 1}) \ar[r] \ar[d]^{d_A^i} & B^i/\Im(d_B^{i - 1}) \ar[r] \ar[d]^{d_B^i} & C^i/\Im(d_C^{i - 1}) \ar[r] \ar[d]^{d_C^i} & 0 \\ 0 \ar[r] & \Ker(d_A^{i + 1}) \ar[r] & \Ker(d_B^{i + 1}) \ar[r] & \Ker(d_C^{i + 1}) & } $$
We will use Categories, Lemma \ref{categories-lemma-left-dual} without further mention. Let $X$ be an object which has a left dual $Y$. We have $$ \Hom(X, X) = \Hom(\mathbf{1}, X \otimes Y) = \Hom(Y, Y) $$ If $a : X \to X$ corresponds to $b : Y \to Y$ then $b$ is the unique endomorphism of $Y$ such that precomposing by $a$ on $$ \Hom(Z' \otimes X, Z) = \Hom(Z', Z \otimes Y) $$ is the same as postcomposing by $1 \otimes b$. Hence the bijection $\Hom(X, X) \to \Hom(Y, Y)$, $a \mapsto b$ is an isomorphism of the opposite of the algebra $\Hom(X, X)$ with the algebra $\Hom(Y, Y)$. In particular, if $X = X_1 \oplus X_2$, then the corresponding projectors $e_1, e_2$ are mapped to idempotents in $\Hom(Y, Y)$. If $Y = Y_1 \oplus Y_2$ is the corresponding direct sum decomposition of $Y$ (Section \ref{section-karoubian}) then we see that under the bijection $\Hom(Z' \otimes X, Z) = \Hom(Z', Z \otimes Y)$ we have $\Hom(Z' \otimes X_i, Z) = \Hom(Z', Z \otimes Y_i)$ functorially as subgroups for $i = 1, 2$. It follows that $Y_i$ is the left dual of $X_i$ by the discussion in Categories, Remark \ref{categories-remark-left-dual-adjoint}.
As unit we take By Categories, Definition \ref{categories-definition-dual} we have maps $$ \eta : \mathbf{1} \to V \otimes W\quad \epsilon : W \otimes V \to \mathbf{1} $$ Since $\mathbf{1} = F$ placed in degree $0$, we may think of $\epsilon$ as a sequence of pairings $W^{-n} \times V^n \to F$ as in the statement of the lemma. Choose bases $\{e_{n, i}\}_{i \in I_n}$ for $V^n$ for all $n$. Write $$ \eta(1) = \sum e_{n, i} \otimes w_{-n, i} $$ for some elements $w_{-n, i} \in W^{-n}$ almost all of which are zero! The condition that $(\epsilon \otimes 1) \circ (1 \otimes \eta)$ is the identity on $W$ means that $$ \sum\nolimits_{n, i} \epsilon(w, e_{n, i})w_{-n, i} = w $$ Thus we see that $W$ is generated as a graded vector space by the finitely many nonzero vectors $w_{-n, i}$. The condition that $(1 \otimes \epsilon) \circ (\eta \otimes 1)$ is the identity of $V$ means that $$ \sum\nolimits_{n, i} e_{n, i}\ \epsilon(w_{-n, i}, v) = v $$ In particular, setting $v = e_{n, i}$ we conclude that $\epsilon(w_{-n, i}, e_{n, i'}) = \delta_{ii'}$. Thus we find that the statement of the lemma holds and that $\{w_{-n, i}\}_{i \in I_n}$ is the dual basis for $W^{-n}$ to the chosen basis for $V^n$.
Note that $\Coim(f) \to \Im(f)$ is an isomorphism of objects of $\mathcal{A}$, and that part (2) signifies that it is an isomorphism of filtered objects. By the description of kernels and cokernels in the proof of Lemma \ref{lemma-filtered} we see that the filtration on $\Coim(f)$ is the quotient filtration coming from $A \to \Coim(f)$. Similarly, the filtration on $\Im(f)$ is the induced filtration coming from the injection $\Im(f) \to B$. The definition of strict is exactly that the quotient filtration is the induced filtration.
Let $B$ a vector space over a field $k$ with basis $e_1, e_2$, with the filtration $F^nB = B$ for $n < 0$, with $F^0B = ke_1$, and $F^nB = 0$ for $n > 0$. Now take $A = k(e_1 + e_2)$ and $C = B/ke_2$ with filtrations induced by $B$, i.e., such that $A \to B$ and $B \to C$ are strict (Lemma \ref{lemma-induced-and-quotient-strict}). Then $F^n(A) = A$ for $n < 0$ and $F^n(A) = 0$ for $n \geq 0$.  Also $F^n(C) = C$ for $n \leq 0$ and $F^n(C) = 0$ for $n > 0$. So the (nonzero) composition $A \to C$ is not strict. \medskip\noindent Assume $g$ is injective. Then \begin{align*} g(f(F^pA)) & = g(f(A) \cap F^pB) \\ & = g(f(A)) \cap g(F^p(B)) \\ & = (g \circ f)(A) \cap (g(B) \cap F^pC) \\ & = (g \circ f)(A) \cap F^pC. \end{align*} The first equality as $f$ is strict, the second because $g$ is injective, the third because $g$ is strict, and the fourth because $(g \circ f)(A) \subset g(B)$. \medskip\noindent Assume $f$ is surjective. Then \begin{align*} (g \circ f)^{-1}(F^iC) & = f^{-1}(F^iB + \Ker(g)) \\ & = f^{-1}(F^iB) + f^{-1}(\Ker(g)) \\ & = F^iA + \Ker(f) + \Ker(g \circ f) \\ & = F^iA + \Ker(g \circ f) \end{align*} The first equality because $g$ is strict, the second because $f$ is surjective, the third because $f$ is strict, and the last because $\Ker(f) \subset \Ker(g \circ f)$.
Set $C \amalg_A B$ equal to $\Coker((1, -1) : A \to C \oplus B)$ in $\text{Fil}(\mathcal{A})$. This cokernel exists, by Lemma \ref{lemma-filtered}. It is a pushout, see Example \ref{example-fibre-product-pushouts}. Note that $F^p(C \amalg_A B)$ is the image of $F^pC \oplus F^pB$. Hence $$ (f')^{-1}(F^p(C \amalg_A B)) = g(f^{-1}(F^pB))) + F^pC $$ Whence the last statement.
The equivalence of (1) and (2) is Lemma \ref{lemma-characterize-strict-general}. By Lemma \ref{lemma-ses-gr} we see that (4), (5), (6) imply (3) and that (3) implies (4), (5), (6). Hence it suffices to show that (3) implies (2). Thus we have to show that if $f : A \to B$ is an injective and surjective map of finite filtered objects which induces and isomorphism $\text{gr}(A) \to \text{gr}(B)$, then $f$ induces an isomorphism of filtered objects. In other words, we have to show that $f(F^pA) = F^pB$ for all $p$. As the filtrations are finite we may prove this by descending induction on $p$. Suppose that $f(F^{p + 1}A) = F^{p + 1}B$. Then commutative diagram $$ \xymatrix{ 0 \ar[r] & F^{p + 1}A \ar[r] \ar[d]^f & F^pA \ar[r] \ar[d]^f & \text{gr}^p(A) \ar[r] \ar[d]^{\text{gr}^p(f)} & 0 \\ 0 \ar[r] & F^{p + 1}B \ar[r] & F^pB \ar[r] & \text{gr}^p(B) \ar[r] & 0 } $$ and the five lemma imply that $f(F^pA) = F^pB$.
Apply Lemma \ref{lemma-long-exact-sequence-cochain} to the short exact sequence of complexes $$ \begin{matrix} 0 & \to & A & \to & B & \to & C & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & A & \to & B & \to & C & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & A & \to & B & \to & C & \to & 0 \end{matrix} $$ where the vertical arrows are $d$.
If $\mathcal{A}$ has countable direct sums and if countable direct sums are exact, then this follows from the discussion above. In general, we proceed as follows; we strongly suggest the reader skip this proof. Consider the object $A = (F^{p + 1}K)$ of $\text{Gr}(\mathcal{A})$, i.e., we put $F^{p + 1}K$ in degree $p$ (the funny shift in numbering to get numbering correct later on). We endow it with a differential $d$ by using $d$ on each component. Then $(A, d)$ is a differential object of $\text{Gr}(\mathcal{A})$. Consider the map $$ \alpha : A \to A[-1] $$ which is given in degree $p$ by the inclusions $F^{p + 1}A \to F^pA$. This is clearly an injective morphism of differential objects $\alpha : (A, d) \to (A, d)[-1]$. Hence, we can apply Remark \ref{remark-differential-object-selfmap} with $S = \text{id}$ and $T = [1]$. The corresponding spectral sequence $(E_r, d_r)_{r \geq 0}$ in $\text{Gr}(\mathcal{A})$ is the spectral sequence we are looking for. Let us unwind the definitions a bit. First of all we have $E_r = (E_r^p)$ is an object of $\text{Gr}(\mathcal{A})$. Then, since $T^rS = [r]$ we have $d_r : E_r \to E_r[r]$ which means that $d_r^p : E_r^p \to E_r^{p + r}$. \medskip\noindent To see that the description of the graded pieces hold, we argue as above. Namely, first we have $E_0 = \Coker(\alpha : A \to A[-1])$ and by our choice of numbering above this gives $E_0^p = \text{gr}^pK$. The first differential is given by $d_0^p = \text{gr}^pd : E_0^p \to E_0^p$. Next, the description of the boundaries $B_r$ and the cocycles $Z_r$ in Remark \ref{remark-differential-object-selfmap} translates into a straightforward manner into the formulae for $Z_r^p$ and $B_r^p$ given above.
The objects $Z_\infty$, $B_\infty$, and the limit $E_\infty = Z_\infty/B_\infty$ of Definition \ref{definition-limit-spectral-sequence} are objects of $\text{Gr}(\mathcal{A})$ by our construction of the spectral sequence in the proof of Lemma \ref{lemma-spectral-sequence-filtered-differential}. Since $Z_r = \bigoplus Z_r^p$ and $B_r = \bigoplus B_r^p$, if we assume that $$ Z_\infty^p = \bigcap\nolimits_r Z_r^p = \frac{\bigcap_r (F^pK \cap d^{-1}(F^{p + r}K) + F^{p + 1}K)}{F^{p + 1}K} $$ and $$ B_\infty^p = \bigcup\nolimits_r B_r^p = \frac{\bigcup_r (F^pK \cap d(F^{p - r + 1}K) + F^{p + 1}K)}{F^{p + 1}K}. $$ exist, then $Z_\infty$ and $B_\infty$ exist with degree $p$ parts $Z_\infty^p$ and $B_\infty^p$ (follows from an elementary argument about unions and intersections of graded subobjects). Thus $$ E_\infty^p = \frac{\bigcap_r (F^pK \cap d^{-1}(F^{p + r}K) + F^{p + 1}K)} {\bigcup_r (F^pK \cap d(F^{p - r + 1}K) + F^{p + 1}K)}. $$ where the top and bottom exist. We have \begin{equation} \label{equation-on-top} \Ker(d) \cap F^pK + F^{p + 1}K \subset \bigcap\nolimits_r \left(F^pK \cap d^{-1}(F^{p + r}K) + F^{p + 1}K\right) \end{equation} and \begin{equation} \label{equation-at-bottom} \bigcup\nolimits_r \left(F^pK \cap d(F^{p - r + 1}K) + F^{p + 1}K\right) \subset \Im(d) \cap F^pK + F^{p + 1}K. \end{equation} Thus a subquotient of $E_\infty^p$ is $$ \frac{\Ker(d) \cap F^pK + F^{p + 1}K}{\Im(d) \cap F^pK + F^{p + 1}K} = \frac{\Ker(d) \cap F^pK}{\Im(d) \cap F^pK + \Ker(d) \cap F^{p + 1}K} $$ Comparing with the formula given for $\text{gr}^pH(K)$ in the discussion following Definition \ref{definition-filtration-cohomology-filtered-differential} we conclude.
If $\mathcal{A}$ has countable direct sums and if countable direct sums are exact, then this follows from the discussion above. In general, we proceed as follows; we strongly suggest the reader skip this proof. Consider the bigraded object $A = (F^{p + 1}K^{p + 1 + q})$ of $\mathcal{A}$, i.e., we put $F^{p + 1}K^{p + 1 + q}$ in degree $(p, q)$ (the funny shift in numbering to get numbering correct later on). We endow it with a differential $d : A \to A[0, 1]$ by using $d$ on each component. Then $(A, d)$ is a differential bigraded object. Consider the map $$ \alpha : A \to A[-1, 1] $$ which is given in degree $(p, q)$ by the inclusion $F^{p + 1}K^{p + 1 + q} \to F^pK^{p + 1 + q}$. This is an injective morphism of differential objects $\alpha : (A, d) \to (A, d)[-1, 1]$. Hence, we can apply Remark \ref{remark-differential-object-selfmap} with $S = [0, 1]$ and $T = [1, -1]$. The corresponding spectral sequence $(E_r, d_r)_{r \geq 0}$ of bigraded objects is the spectral sequence we are looking for. Let us unwind the definitions a bit. First of all we have $E_r = (E_r^{p, q})$. Then, since $T^rS = [r, -r + 1]$ we have $d_r : E_r \to E_r[r, -r + 1]$ which means that $d_r^{p, q} : E_r^{p, q} \to E_r^{p + r, q - r + 1}$. \medskip\noindent To see that the description of the graded pieces hold, we argue as above. Namely, first we have $$ E_0 = \Coker(\alpha : A \to A[-1, 1])[0, -1] = \Coker(\alpha[0, -1] : A[0, -1] \to A[-1, 0]) $$ and by our choice of numbering above this gives $$ E_0^{p, q} = \Coker(F^{p + 1}K^{p + q} \to F^pK^{p + q}) = \text{gr}^pK^{p + q} $$ The first differential is given by $d_0^{p, q} = \text{gr}^pd^{p + q} : E_0^{p, q} \to E_0^{p, q + 1}$. Next, the description of the boundaries $B_r$ and the cocycles $Z_r$ in Remark \ref{remark-differential-object-selfmap} translates into a straightforward manner into the formulae for $Z_r^{p, q}$ and $B_r^{p, q}$ given above.
The objects $Z_\infty$, $B_\infty$, and the limit $E_\infty = Z_\infty/B_\infty$ of Definition \ref{definition-limit-spectral-sequence} are bigraded objects of $\mathcal{A}$ by our construction of the spectral sequence in Lemma \ref{lemma-spectral-sequence-filtered-complex}. Since $Z_r = \bigoplus Z_r^{p, q}$ and $B_r = \bigoplus B_r^{p, q}$, if we assume that $$ Z_\infty^{p, q} = \bigcap\nolimits_r Z_r^{p, q} = \bigcap\nolimits_r \frac{F^pK^{p + q} \cap d^{-1}(F^{p + r}K^{p + q + 1}) + F^{p + 1}K^{p + q}} {F^{p + 1}K^{p + q}} $$ and $$ B_\infty^{p, q} = \bigcup\nolimits_r B_r^{p, q} = \bigcup\nolimits_r \frac{F^pK^{p + q} \cap d(F^{p - r + 1}K^{p + q - 1}) + F^{p + 1}K^{p + q}} {F^{p + 1}K^{p + q}} $$ exist, then $Z_\infty$ and $B_\infty$ exist with bidegree $(p, q)$ parts $Z_\infty^{p, q}$ and $B_\infty^{p, q}$ (follows from an elementary argument about unions and intersections of bigraded objects). Thus $$ E_\infty^{p, q} = \frac{\bigcap_r (F^pK^{p + q} \cap d^{-1}(F^{p + r}K^{p + q + 1}) + F^{p + 1}K^{p + q})} {\bigcup_r (F^pK^{p + q} \cap d(F^{p - r + 1}K^{p + q - 1}) + F^{p + 1}K^{p + q})}. $$ where the top and the bottom exist. With $n = p + q$ we have \begin{equation} \label{equation-on-top-bigraded} \Ker(d) \cap F^pK^{n} + F^{p + 1}K^{n} \subset \bigcap\nolimits_r \left( F^pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) + F^{p + 1}K^{n} \right) \end{equation} and \begin{equation} \label{equation-at-bottom-bigraded} \bigcup\nolimits_r \left( F^pK^{n} \cap d(F^{p - r + 1}K^{n - 1}) + F^{p + 1}K^{n} \right) \subset \Im(d) \cap F^pK^{n} + F^{p + 1}K^{n}. \end{equation} Thus a subquotient of $E_\infty^{p, q}$ is $$ \frac{\Ker(d) \cap F^pK^{n} + F^{p + 1}K^n} {\Im(d) \cap F^pK^{n} + F^{p + 1}K^{n}} = \frac{\Ker(d) \cap F^pK^n}{\Im(d) \cap F^pK^n + \Ker(d) \cap F^{p + 1}K^n} $$ Comparing with (\ref{equation-graded-cohomology}) we conclude.
Part (1) follows as $E_0^{p, n - p} = \text{gr}^p K^n$. Part (2) is clear from Equation (\ref{equation-filtration-cohomology}). We will use Lemma \ref{lemma-filtered-complex-ss-converges} to prove that the spectral sequence weakly converges. Fix $p, n \in \mathbf{Z}$. Looking at the right hand side of (\ref{equation-on-top-bigraded}) we see that we get $F^pK^n \cap \Ker(d) + F^{p + 1}K^n$ because $F^{p + r}K^n = 0$ for $r \gg 0$. Thus (\ref{equation-on-top-bigraded}) is an equality. Look at the left hand side of (\ref{equation-on-top-bigraded}). The expression is equal to the right hand side since $F^{p - r + 1}K^{n - 1} = K^{n - 1}$ for $r \gg 0$. Thus (\ref{equation-on-top-bigraded}) is an equality. Since the filtration on $H^n(K^\bullet)$ is finite by (2) we see that we have abutment. To prove we have convergence we have to show the spectral sequence is regular which follows as it is bounded (Lemma \ref{lemma-relate-boundedness}) and we have to show that $H^n(K^\bullet) = \lim_p H^n(K^\bullet)/F^pH^n(K^\bullet)$ which follows from the fact that the filtration on $H^*(K^\bullet)$ is finite proved in part (2). \medskip\noindent Proof of (4). Assume that for some $r \geq 0$ we have $E_r^{p, q} \in \mathcal{C}$ for some weak Serre subcategory $\mathcal{C}$ of $\mathcal{A}$. Then $E_{r + 1}^{p, q}$ is in $\mathcal{C}$ as well, see Lemma \ref{lemma-characterize-weak-serre-subcategory}. By boundedness proved above (which implies that the spectral sequence is both regular and coregular, see Lemma \ref{lemma-relate-boundedness}) we can find an $r' \geq r$ such that $E_\infty^{p, q} = E_{r'}^{p, q}$ for all $p, q$ with $p + q = n$. Thus $H^n(K^\bullet)$ is an object of $\mathcal{A}$ which has a finite filtration whose graded pieces are in $\mathcal{C}$. This implies that $H^n(K^\bullet)$ is in $\mathcal{C}$ by Lemma \ref{lemma-characterize-weak-serre-subcategory}.