hcju/stacksqa huggingface dataset:
Denote $E_r^{even}$ and $E_r^{odd}$ the even and odd part of $E_r$ defined as the direct sum of the $(p, q)$ components with $p + q$ even and odd. The differential $d_r$ defines maps $\varphi : E_r^{even} \to E_r^{odd}$ and $\psi : E_r^{odd} \to E_r^{even}$ whose compositions either way give zero. Then we see that \begin{align*} [E_r^{even}] - [E_r^{odd}] & = [\Ker(\varphi)] + [\Im(\varphi)] - [\Ker(\psi)] - [\Im(\psi)] \\ & = [\Ker(\varphi)/\Im(\psi)] - [\Ker(\psi)/\Im(\varphi)] \\ & = [E_{r + 1}^{even}] - [E_{r + 1}^{odd}] \end{align*} Note that all the intervening objects are in the smallest Serre subcategory containing the objects $E_r^{p, q}$. Continuing in this manner we see that we can increase $r$ at will. Since there are only a finite number of pairs $(p, q)$ for which $E_r^{p, q}$ is nonzero, a property which is inherited by $E_{r + 1}, E_{r + 2}, \ldots$, we see that we may assume that $d_r = 0$. At this stage we see that $H^n(K^\bullet)$ has a finite filtration (Lemma \ref{lemma-biregular-ss-converges}) whose graded pieces are exactly the $E_r^{p, n - p}$ and the result is clear.
Fix $n$. Using the long exact cohomology sequence associated to the short exact sequence of complexes $$ 0 \to F^{p + 1}K^\bullet \to F^pK^\bullet \to \text{gr}^pK^\bullet \to 0 $$ we find that $E_1^{p, n - p} = 0$ for $p \geq \max(p_0(n), p_0(n + 1))$ and $p < \min(p_1(n), p_1(n + 1))$. Hence the spectral sequence is bounded (Definition \ref{definition-bounded-ss}). This proves (1). \medskip\noindent It is clear from the assumptions and Definition \ref{definition-filtration-cohomology-filtered-complex} that the filtration on $H^n(K^\bullet)$ is finite. This proves (2). \medskip\noindent Next we prove that the spectral sequence weakly converges to $H^*(K^\bullet)$ using Lemma \ref{lemma-filtered-complex-ss-converges}. Let us show that we have equality in (\ref{equation-on-top-bigraded}). Namely, for $p + r > p_0(n + 1)$ the map $$ d : F^pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) \to F^{p + r}K^{n + 1} $$ ends up in the image of $d : F^{p + r}K^n \to F^{p + r}K^{n + 1}$ because the complex $F^{p + r}K^\bullet$ is exact in degree $n + 1$. We conclude that $F^pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) = d(F^{p + r}K^n) + \Ker(d) \cap F^pK^n$. Hence for such $r$ we have $$ \Ker(d) \cap F^pK^{n} + F^{p + 1}K^{n} = F^pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) + F^{p + 1}K^{n} $$ which proves the desired equality. To show that we have equality in (\ref{equation-at-bottom-bigraded}) we use that for $p - r + 1 < p_1(n - 1)$ we have $$ d(F^{p - r + 1}K^{n - 1}) = \Im(d) \cap F^{p - r + 1}K^n $$ because the map $F^{p - r + 1}K^\bullet \to K^\bullet$ induces an isomorphism on cohomology in degree $n - 1$. This shows that we have $$ F^pK^{n} \cap d(F^{p - r + 1}K^{n - 1}) + F^{p + 1}K^{n} = \Im(d) \cap F^pK^{n} + F^{p + 1}K^{n} $$ for such $r$ which proves the desired equality. \medskip\noindent To see that the spectral sequence abuts to $H^*(K^\bullet)$ using Lemma \ref{lemma-filtered-complex-ss-converges} we have to show that $\bigcap_p (\Ker(d) \cap F^pK^n + \Im(d) \cap K^n) = \Im(d) \cap K^n$ and $\bigcup_p (\Ker(d) \cap F^pK^n + \Im(d) \cap K^n) = \Ker(d) \cap K^n$. For $p \geq p_0(n)$ we have $\Ker(d) \cap F^pK^n + \Im(d) \cap K^n = \Im(d) \cap K^n$ and for $p \leq p_1(n)$ we have $\Ker(d) \cap F^pK^n + \Im(d) \cap K^n = \Ker(d) \cap K^n$. Combining weak convergence, abutment, and boundedness we see that (2) and (3) are true.
The map is simply the map given by the morphisms $K^n \to A^{n, 0} \to \text{Tot}^n(A^{\bullet, \bullet})$, which are easily seen to define a morphism of complexes. Consider the spectral sequence $({}'E_r, {}'d_r)_{r \geq 0}$ associated to the double complex $A^{\bullet, \bullet}$. By Lemma \ref{lemma-first-quadrant-ss} this spectral sequence converges and the induced filtration on $H^n(\text{Tot}(A^{\bullet, \bullet}))$ is finite for each $n$. By Lemma \ref{lemma-ss-double-complex} and assumption (4) we have ${}'E_1^{p, q} = 0$ unless $q = 0$ and $'E_1^{p, 0} = K^p$ with differential ${}'d_1^{p, 0}$ identified with $d_K^p$. Hence ${}'E_2^{p, 0} = H^p(K^\bullet)$ and zero otherwise. This clearly implies $d_2^{p, q} = d_3^{p, q} = \ldots = 0$ for degree reasons. Hence we conclude that $H^n(\text{Tot}(A^{\bullet, \bullet})) = H^n(K^\bullet)$. We omit the verification that this identification is given by the morphism of complexes $K^\bullet \to \text{Tot}(A^{\bullet, \bullet})$ introduced above.
If there exists a $t \in \mathbf{Z}$ such that $A_0^q = 0$ for $q < t$, then this follows immediately from Lemma \ref{lemma-double-complex-gives-resolution} (with $p$ and $q$ swapped as in the final statement of that lemma). OK, but for every $t \in \mathbf{Z}$ we have a complex $$ 0 \to \sigma_{\geq t}M^\bullet \to \sigma_{\geq t}A_0^\bullet \to \sigma_{\geq t}A_1^\bullet \to \sigma_{\geq t}A_2^\bullet \to \ldots $$ of stupid truncations. Denote $A(t)^{\bullet, \bullet}$ the corresponding double complex. Every element $\xi$ of $H^n(\text{Tot}(A^{\bullet, \bullet}))$ is the image of an element of $H^n(\text{Tot}(A(t)^{\bullet, \bullet}))$ for some $t$ (look at explicit representatives of cohomology classes). Hence $\xi$ is in the image of $H^n(\sigma_{\geq t}M^\bullet)$. Thus the map $H^n(M^\bullet) \to H^n(\text{Tot}(A^{\bullet, \bullet}))$ is surjective. It is injective because for all $t$ the map $H^n(\sigma_{\geq t}M^\bullet) \to H^n(\text{Tot}(A(t)^{\bullet, \bullet}))$ is injective and similar arguments.
Abbreviating $T^\bullet = \text{Tot}_\pi(A^{\bullet, \bullet})$ we define $$ T^n = \prod\nolimits_{p + q = n} A^{p, q} = \prod\nolimits_{p + q = n} A_p^q \quad\text{with}\quad \text{d}_{T^\bullet}^n = \prod\nolimits_{n = p + q} (f_p^q + (-1)^pd_{A_p^\bullet}^q) $$ where $f_p^\bullet : A_p^\bullet \to A_{p + 1}^\bullet$ are the maps of complexes in the lemma. \medskip\noindent We will show that $H^0(M^\bullet) \to H^0(T^\bullet)$ is an isomorphism. The same argument works for other degrees. Let $x \in \Ker(\text{d}_{T^\bullet}^0)$ represent $\xi \in H^0(T^\bullet)$. Write $x = (x_i)$ with $x_i \in A_i^{-i}$. Note that $x_0$ maps to zero in $\Coker(A_1^{-1} \to A_1^0)$. Hence we see that $x_0 = m_0 + d_{A_0^\bullet}^{-1}(y)$ for some $m_0 \in M^0$ and $y \in A_0^{-1}$. Then $d_{M^\bullet}(m_0) = 0$ because $\text{d}_{A_0^\bullet}(x_0) = 0$ as $\text{d}_{T^\bullet}(x) = 0$. Thus, replacing $\xi$ by something in the image of $H^0(M^\bullet) \to H^0(T^\bullet)$ we may assume that $x_0$ is in $\Im(d^{-1}_{A_0^\bullet})$. \medskip\noindent Assume $x_0 \in \Im(d^{-1}_{A_0^\bullet})$. We claim that in this case $\xi = 0$. To prove this we find, by induction on $n$ elements $y_0, y_1, \ldots, y_n$ with $y_i \in A_i^{-i - 1}$ such that $x_0 = \text{d}_{A_0}^{-1}(y_0)$ and $x_j = f_{j - 1}^{-j}(y_{j - 1}) + (-1)^j d^{-j - 1}_{A_{-j}^\bullet}(y_j)$ for $j = 1, \ldots, n$. This is clear for $n = 0$. Proof of induction step: suppose we have found $y_0, \ldots, y_{n - 1}$. Then $w_n = x_n - f_{n - 1}^{-n}(y_{n - 1})$ is in the kernel of $d^{-n}_{A_n^\bullet}$ and maps to zero in $H^n(A_{n + 1}^\bullet)$ (because it maps to an element which is a boundary the boundary of $x_{n + 1}$ up to sign). Exactly as in the proof of Lemma \ref{lemma-good-resolution-gives-qis} the assumptions of the lemma imply that $$ 0 \to H^p(M^\bullet) \to H^p(A_0^\bullet) \to H^p(A_1^\bullet) \to H^p(A_2^\bullet) \to \ldots $$ is exact for all $p \in \mathbf{Z}$. Thus after changing $y_{n - 1}$ by an element in $\Ker(d^{n - 1}_{A_{n - 1}^\bullet})$ we may assume that $w_n$ maps to zero in $H^{-n}(A_n^\bullet)$. This means we can find $y_n$ as desired. Observe that this procedure does not change $y_0, \ldots, y_{n - 2}$. Hence continuing ad infinitum we find an element $y = (y_i)$ in $T^{n - 1}$ with $d_{T^\bullet}(y) = \xi$. This shows that $H^0(M^\bullet) \to H^0(T^\bullet)$ is surjective. \medskip\noindent Suppose that $m_0 \in \Ker(d^0_{M^\bullet})$ maps to zero in $H^0(T^\bullet)$. Say it maps to the differential applied to $y = (y_i) \in T^{-1}$ . Then $y_0 \in A_0^{-1}$ maps to zero in $\Coker(d^{-2}_{A_1^\bullet})$. By assumption this means that $y_0 \bmod \Im(d^{-2}_{A_0^\bullet})$ is the image of some $z \in M^{-1}$. It follows that $m_0 = d^{-1}_{M^\bullet}(z)$. This proves injectivity and the proof is complete.
By the Yoneda lemma (Categories, Lemma \ref{categories-lemma-yoneda}) the object $Q$ of $\mathcal{A}$ corresponding to $P$ is defined up to unique isomorphism by the formula $\Hom_\mathcal{A}(Q, A) = \Hom_\mathcal{B}(P, u(A))$. Let us write $Q = v(P)$. Denote $i_P : P \to u(v(P))$ the map corresponding to $\text{id}_{v(P)}$ in $\Hom_\mathcal{A}(v(P), v(P))$. Functoriality in (2) implies that the bijection is given by $$ \Hom_\mathcal{A}(v(P), A) \to \Hom_\mathcal{B}(P, u(A)),\quad \varphi \mapsto u(\varphi) \circ i_P $$ For any pair of elements $P_1, P_2 \in \mathcal{P}$ there is a canonical map $$ \Hom_\mathcal{B}(P_2, P_1) \to \Hom_\mathcal{A}(v(P_2), v(P_1)),\quad \varphi \mapsto v(\varphi) $$ which is characterized by the rule $u(v(\varphi)) \circ i_{P_2} = i_{P_1} \circ \varphi$ in $\Hom_\mathcal{B}(P_2, u(v(P_1)))$. Note that $\varphi \mapsto v(\varphi)$ is compatible with composition; this can be seen directly from the characterization. Hence $P \mapsto v(P)$ is a functor from the full subcategory of $\mathcal{B}$ whose objects are the elements of $\mathcal{P}$. \medskip\noindent Given an arbitrary object $B$ of $\mathcal{B}$ choose an exact sequence $$ P_2 \to P_1 \to B \to 0 $$ which is possible by assumption (1). Define $v(B)$ to be the object of $\mathcal{A}$ fitting into the exact sequence $$ v(P_2) \to v(P_1) \to v(B) \to 0 $$ Then \begin{align*} \Hom_\mathcal{A}(v(B), A) & = \Ker(\Hom_\mathcal{A}(v(P_1), A) \to \Hom_\mathcal{A}(v(P_2), A)) \\ & = \Ker(\Hom_\mathcal{B}(P_1, u(A)) \to \Hom_\mathcal{B}(P_2, u(A))) \\ & = \Hom_\mathcal{B}(B, u(A)) \end{align*} Hence we see that we may take $\mathcal{P} = \Ob(\mathcal{B})$, i.e., we see that $v$ is everywhere defined.
Assume (1)(a), i.e., $\mathcal{I}$ is filtered and $M$ is essentially constant. Let $X = \colim M_i$. Choose $i$ and $X \to M_i$ as in Categories, Definition \ref{categories-definition-essentially-constant-diagram}. Let $\mathcal{I}'$ be the full subcategory consisting of objects which are the target of a morphism with source $i$. Suppose $i' \in \Ob(\mathcal{I}')$ and choose a morphism $i \to i'$. Then $X \to M_i \to M_{i'}$ composed with $M_{i'} \to X$ is the identity on $X$. As $\mathcal{A}$ is Karoubian, we find a direct summand decomposition $M_{i'} = X_{i'} \oplus Z_{i'}$, where $Z_{i'} = \Ker(M_{i'} \to X)$ and $X_{i'}$ maps isomorphically to $X$. Pick $i \to k$ and $i' \to k$ such that $M_{i'} \to X \to M_i \to M_k$ equals $M_{i'} \to M_k$ as in Categories, Definition \ref{categories-definition-essentially-constant-diagram}. Then we see that $M_{i'} \to M_k$ annihilates $Z_{i'}$. Thus (1)(b) holds. \medskip\noindent Assume (1)(b), i.e., $\mathcal{I}$ is filtered and we have $\mathcal{I}' \subset \mathcal{I}$ and for $i' \in \Ob(\mathcal{I}')$ a direct sum decomposition $M_{i'} = X_{i'} \oplus Z_{i'}$ as stated in the lemma. To see that $M$ is essentially constant we can replace $\mathcal{I}$ by $\mathcal{I}'$, see Categories, Lemma \ref{categories-lemma-cofinal-essentially-constant}. Pick any $i \in \Ob(\mathcal{I})$ and denote $X \to M_i$ the inverse of the isomorphism $X_i \to X$ followed by the inclusion map $X_i \to M_i$. If $j$ is a second object, then choose $j \to k$ such that $Z_j \to M_k$ is zero. Since $\mathcal{I}$ is filtered we may also assume there is a morphism $i \to k$ (after possibly increasing $k$). Then $M_j \to X \to M_i \to M_k$ and $M_j \to M_k$ both annihilate $Z_j$. Thus after postcomposing by a morphism $M_k \to M_l$ which annihilates the summand $Z_k$, we find that $M_j \to X \to M_i \to M_l$ and $M_j \to M_l$ are equal, i.e., $M$ is essentially constant. \medskip\noindent The proof of (2) is dual.
Assume (1) holds. Set $W = \colim_\mathcal{I} F \oplus G$. Note that the projection onto $F$ defines natural transformation $F \oplus G \to F \oplus G$ which is idempotent. Hence we obtain an idempotent endomorphism $W \to W$ by Categories, Lemma \ref{categories-lemma-functorial-colimit}. Since $\mathcal{A}$ is Karoubian we get a corresponding direct sum decomposition $W = X \oplus Y$, see Lemma \ref{lemma-karoubian}. A straightforward argument (omitted) shows that $X = \colim_\mathcal{I} F$ and $Y = \colim_\mathcal{I} G$. Thus (2) holds. We omit the proof that (2) implies (1).
Assume (1) holds. In particular $W = \colim_\mathcal{I} F \oplus G$ exists and hence by Lemma \ref{lemma-direct-sum-from-product-colimit} we have $W = X \oplus Y$ with $X = \colim_\mathcal{I} F$ and $Y = \colim_\mathcal{I} G$. A straightforward argument (omitted) using for example the characterization of Categories, Lemma \ref{categories-lemma-characterize-essentially-constant-ind} shows that $F$ is essentially constant with value $X$ and $G$ is essentially constant with value $Y$. Thus (2) holds. The proof that (2) implies (1) is omitted.
Let $Z_i = \Ker(C_i \to D_i)$ and $I_i = \Im(A_i \to B_i)$. Then $\lim Z_i = \Ker(\lim C_i \to \lim D_i)$ and we get a short exact sequence of systems $$ 0 \to (I_i) \to (B_i) \to (Z_i) \to 0 $$ Moreover, by Lemma \ref{lemma-Mittag-Leffler} we see that $(I_i)$ has (ML), thus another application of Lemma \ref{lemma-Mittag-Leffler} shows that $\lim B_i \to \lim Z_i$ is surjective which proves the lemma.
After renumbering we may assume that $C_i = C \oplus Z_i$ compatible with transition maps and that for all $i$ there exists an $i' \geq i$ such that $Z_{i'} \to Z_i$ is zero, see Lemma \ref{lemma-essentially-constant}. Pick $i$. Let $c \geq i$ by an integer such that $\Im(A_c \to A) = \Im(A_{i'} \to A_i)$ for all $i' \geq c$. Let $c' \geq c$ be an integer such that $Z_{c'} \to Z_c$ is zero. For $i' \geq c'$ consider the maps $$ \xymatrix{ 0 \ar[r] & A_{i'} \ar[d] \ar[r] & B_{i'} \ar[d] \ar[r] & C \oplus Z_{i'} \ar[d] \ar[r] & 0 \\ 0 \ar[r] & A_{c'} \ar[d] \ar[r] & B_{c'} \ar[d] \ar[r] & C \oplus Z_{c'} \ar[d]\ar[r] & 0  \\ 0 \ar[r] & A_c \ar[d] \ar[r] & B_c \ar[d] \ar[r] & C \oplus Z_c \ar[d]\ar[r] & 0  \\ 0 \ar[r] & A_i \ar[r] & B_i \ar[r] & C \oplus Z_i \ar[r] & 0 } $$ Because $Z_{c'} \to Z_c$ is zero the image $\Im(B_{c'} \to B_c)$ is an extension $C$ by a subgroup $A' \subset A_c$ which contains the image of $A_{c'} \to A_c$. Hence $\Im(B_{c'} \to B_i)$ is an extension of $C$ by the image of $A'$ which is the image of $A_c \to A_i$ by our choice of $c$. In exactly the same way one shows that $\Im(B_{i'} \to B_i)$ is an extension of $C$ by the image of $A_c \to A_i$. Hence $\Im(B_{c'} \to B_i) = \Im(B_{i'} \to B_i)$ and we win.
Let $Z^j_i = \Ker(A^j_i \to A^{j + 1}_i)$ and $I^j_i = \Im(A^{j - 1}_i \to A^j_i)$. Note that $\lim Z^0_i = \Ker(\lim A^0_i \to \lim A^1_i)$ as taking kernels commutes with limits. The systems $(I^{-1}_i)$ and $(I^0_i)$ have ML as quotients of the systems $(A^{-2}_i)$ and $(A^{-1}_i)$, see Lemma \ref{lemma-Mittag-Leffler}. Thus an exact sequence $$ 0 \to (I^{-1}_i) \to (Z^{-1}_i) \to (H^{-1}_i) \to 0 $$ of inverse systems where $(I^{-1}_i)$ has ML and where $(H^{-1}_i)$ is essentially constant by assumption. Hence $(Z^{-1}_i)$ has ML by Lemma \ref{lemma-exact-sequence-ML}. The exact sequence $$ 0 \to (Z^{-1}_i) \to (A^{-1}_i) \to (I^0_i) \to 0 $$ and an application of Lemma \ref{lemma-Mittag-Leffler} shows that $\lim A^{-1}_i \to \lim I^0_i$ is surjective. Finally, the exact sequence $$ 0 \to (I^0_i) \to (Z^0_i) \to (H^0_i) \to 0 $$ and Lemma \ref{lemma-Mittag-Leffler} show that $\lim I^0_i \to \lim Z^0_i \to \lim H^0_i \to 0$ is exact. Putting everything together we win.
The assumptions imply that for every object $y$ of $\mathcal{D}$ the functor $x \mapsto \Mor_\mathcal{D}(y, G(x))$ satisfies the assumptions of Lemma \ref{lemma-a-version-of-brown}. Thus it is representable by an object, let's call it $F(y)$. An application of Yoneda's lemma (Lemma \ref{lemma-yoneda}) turns the rule $y \mapsto F(y)$ into a functor which by construction is an adjoint to $G$. We omit the details.
Let $F : \mathcal{A} \to \mathcal{B}$ be essentially surjective and fully faithful. As by convention all categories are small and as $F$ is essentially surjective we can, using the axiom of choice, choose for every $X \in \Ob(\mathcal{B})$ an object $j(X)$ of $\mathcal{A}$ and an isomorphism $i_X : X \to F(j(X))$. Then we apply Lemma \ref{lemma-construct-quasi-inverse} using that $F$ is fully faithful.
We will continue to use the Yoneda lemma to identify $F(U)$ with transformations $h_U \to F$ of functors. \medskip\noindent Equivalence of (2) and (3). Let $U, \xi, V, \xi'$ be as in (3). Both (2) and (3) tell us exactly that $h_U \times_{\xi, F, \xi'} h_V$ is representable; the only difference is that the statement (3) is symmetric in $U$ and $V$ whereas (2) is not. \medskip\noindent Assume condition (1). Let $U, \xi, V, \xi'$ be as in (3). Note that $h_U \times h_V = h_{U \times V}$ is representable. Denote $\eta : h_{U \times V} \to F \times F$ the map corresponding to the product $\xi \times \xi' : h_U \times h_V \to F \times F$. Then the fibre product $F \times_{\Delta, F \times F, \eta} h_{U \times V}$ is representable by assumption. This means there exists $W \in \Ob(\mathcal{C})$, morphisms $W \to U$, $W \to V$ and $h_W \to F$ such that $$ \xymatrix{ h_W \ar[d] \ar[r] & h_U \times h_V \ar[d]^{\xi \times \xi'} \\ F \ar[r] & F \times F } $$ is cartesian. Using the explicit description of fibre products in Lemma \ref{lemma-fibre-product-presheaves} the reader sees that this implies that $h_W = h_U \times_{\xi, F, \xi'} h_V$ as desired. \medskip\noindent Assume the equivalent conditions (2) and (3). Let $U$ be an object of $\mathcal{C}$ and let $(\xi, \xi') \in (F \times F)(U)$. By (3) the fibre product $h_U \times_{\xi, F, \xi'} h_U$ is representable. Choose an object $W$ and an isomorphism $h_W \to h_U \times_{\xi, F, \xi'} h_U$. The two projections $\text{pr}_i : h_U \times_{\xi, F, \xi'} h_U \to h_U$ correspond to morphisms $p_i : W \to U$ by Yoneda. Consider $W' = W \times_{(p_1, p_2), U \times U} U$. It is formal to show that $W'$ represents $F \times_{\Delta, F \times F} h_U$ because $$ h_{W'} =  h_W \times_{h_U \times h_U} h_U = (h_U \times_{\xi, F, \xi'} h_U) \times_{h_U \times h_U} h_U = F \times_{F \times F} h_U. $$ Thus $\Delta$ is representable and this finishes the proof.
One can prove this by showing that $\mathcal{I}$ is cofinal in $\mathcal{I}'$ and applying Lemma \ref{lemma-cofinal}. But we can also prove it directly as follows. It suffices to show that for any object $T$ of $\mathcal{C}$ we have $$ \lim_{\mathcal{I}^{opp}} \Mor_\mathcal{C}(M_{F(i)}, T) = \lim_{(\mathcal{I}')^{opp}} \Mor_\mathcal{C}(M_{i'}, T) $$ If $(g_{i'})_{i' \in \Ob(\mathcal{I}')}$ is an element of the right hand side, then setting $f_i = g_{F(i)}$ we obtain an element $(f_i)_{i \in \Ob(\mathcal{I})}$ of the left hand side. Conversely, let $(f_i)_{i \in \Ob(\mathcal{I})}$ be an element of the left hand side. Note that on each (connected) fibre category $\mathcal{I}_{i'}$ the functor $M \circ F$ is constant with value $M_{i'}$. Hence the morphisms $f_i$ for $i \in \Ob(\mathcal{I})$ with $F(i) = i'$ are all the same and determine a well defined morphism $g_{i'} : M_{i'} \to T$. By assumption (2) the collection $(g_{i'})_{i' \in \Ob(\mathcal{I}')}$ defines an element of the right hand side.
Say $\Ob(\mathcal{I}) = \{x_1, \ldots, x_n\}$. Denote $s, t : \{1, \ldots, m\} \to \{1, \ldots, n\}$ the functions such that $f_j : x_{s(j)} \to x_{t(j)}$. We set $\Ob(\mathcal{J}) = \{y_1, \ldots, y_n, z_1, \ldots, z_n\}$ Besides the identity morphisms we introduce morphisms $g_j : y_{s(j)} \to z_{t(j)}$, $j = 1, \ldots, m$ and morphisms $h_i : y_i \to z_i$, $i = 1, \ldots, n$. Since all of the nonidentity morphisms in $\mathcal{J}$ go from a $y$ to a $z$ there are no compositions to define and no associativities to check. Set $F(y_i) = F(z_i) = x_i$. Set $F(g_j) = f_j$ and $F(h_i) = \text{id}_{x_i}$. It is clear that $F$ is a functor. It is clear that $\mathcal{J}$ is finite. It is clear that $\mathcal{J}$ is connected, resp.\ nonempty if and only if $\mathcal{I}$ is so. \medskip\noindent Let $M : \mathcal{I} \to \mathcal{C}$ be a diagram. Consider an object $W$ of $\mathcal{C}$ and morphisms $q_i : W \to M(x_i)$ as in Definition \ref{definition-limit}. Then by taking $q_i : W \to M(F(y_i)) = M(F(z_i)) = M(x_i)$ we obtain a family of maps as in Definition \ref{definition-limit} for the diagram $M \circ F$. Conversely, suppose we are given maps $qy_i : W \to M(F(y_i))$ and $qz_i : W \to M(F(z_i))$ as in Definition \ref{definition-limit} for the diagram $M \circ F$. Since $$ M(F(h_i)) = \text{id} : M(F(y_i)) = M(x_i) \longrightarrow M(x_i) = M(F(z_i)) $$ we conclude that $qy_i = qz_i$ for all $i$. Set $q_i$ equal to this common value. The compatibility of $q_{s(j)} = qy_{s(j)}$ and $q_{t(j)} = qz_{t(j)}$ with the morphism $M(f_j)$ guarantees that the family $q_i$ is compatible with all morphisms in $\mathcal{I}$ as by assumption every such morphism is a composition of the morphisms $f_j$. Thus we have found a canonical bijection $$ \lim_{B \in \Ob(\mathcal{J})} \Mor_\mathcal{C}(W, M(F(B))) = \lim_{A \in \Ob(\mathcal{I})} \Mor_\mathcal{C}(W, M(A)) $$ which implies the statement on limits in the lemma. The statement on colimits is proved in the same way (proof omitted).
Since equalizers and fibre products are finite connected limits we see that (1) implies (2). For the converse, let $\mathcal{I}$ be a finite connected diagram category. Let $F : \mathcal{J} \to \mathcal{I}$ be the functor of diagram categories constructed in the proof of Lemma \ref{lemma-finite-diagram-category}. Then we see that we may replace $\mathcal{I}$ by $\mathcal{J}$. The result is that we may assume that $\Ob(\mathcal{I}) = \{x_1, \ldots, x_n\} \amalg \{y_1, \ldots, y_m\}$ with $n, m \geq 1$ such that all nonidentity morphisms in $\mathcal{I}$ are morphisms $f : x_i \to y_j$ for some $i$ and $j$. \medskip\noindent Suppose that $n > 1$. Since $\mathcal{I}$ is connected there exist indices $i_1, i_2$ and $j_0$ and morphisms $a : x_{i_1} \to y_{j_0}$ and $b : x_{i_2} \to y_{j_0}$. Consider the category $$ \mathcal{I}' = \{x\} \amalg \{x_1, \ldots, \hat x_{i_1}, \ldots, \hat x_{i_2}, \ldots x_n\} \amalg \{y_1, \ldots, y_m\} $$ with $$ \Mor_{\mathcal{I}'}(x, y_j) = \Mor_\mathcal{I}(x_{i_1}, y_j) \amalg \Mor_\mathcal{I}(x_{i_2}, y_j) $$ and all other morphism sets the same as in $\mathcal{I}$. For any functor $M : \mathcal{I} \to \mathcal{C}$ we can construct a functor $M' : \mathcal{I}' \to \mathcal{C}$ by setting $$ M'(x) = M(x_{i_1}) \times_{M(a), M(y_{j_0}), M(b)} M(x_{i_2}) $$ and for a morphism $f' : x \to y_j$ corresponding to, say, $f : x_{i_1} \to y_j$ we set $M'(f) = M(f) \circ \text{pr}_1$. Then the functor $M$ has a limit if and only if the functor $M'$ has a limit (proof omitted). Hence by induction we reduce to the case $n = 1$. \medskip\noindent If $n = 1$, then the limit of any $M : \mathcal{I} \to \mathcal{C}$ is the successive equalizer of pairs of maps $x_1 \to y_j$ hence exists by assumption.
Since products of pairs, fibre products, and equalizers are limits with nonempty index categories we see that (1) implies both (2) and (3). Assume (2). Then finite nonempty products and equalizers exist. Hence by Lemma \ref{lemma-limits-products-equalizers} we see that finite nonempty limits exist, i.e., (1) holds. Assume (3). If $a, b : A \to B$ are morphisms of $\mathcal{C}$, then the equalizer of $a, b$ is $$ (A \times_{a, B, b} A)\times_{(pr_1, pr_2), A \times A, \Delta} A. $$ Thus (3) implies (2), and the lemma is proved.
Since products of pairs, fibre products, equalizers, and final objects are limits over finite index categories we see that (1) implies both (2) and (3). By Lemma \ref{lemma-limits-products-equalizers} above we see that (2) implies (1). Assume (3). Note that the product $A \times B$ is the fibre product over the final object. If $a, b : A \to B$ are morphisms of $\mathcal{C}$, then the equalizer of $a, b$ is $$ (A \times_{a, B, b} A)\times_{(pr_1, pr_2), A \times A, \Delta} A. $$ Thus (3) implies (2) and the lemma is proved.