hcju/stacksqa huggingface dataset:
To prove the existence of the derived functor $- \otimes_R^\mathbf{L} N^\bullet$ we use the general theory developed in Derived Categories, Section \ref{derived-section-derived-functors}. Set $\mathcal{D} = K(\text{Mod}_R)$ and $\mathcal{D}' = D(A)$. Let us write $F : \mathcal{D} \to \mathcal{D}'$ the exact functor of triangulated categories defined by the rule $F(M^\bullet) = \text{Tot}(M^\bullet \otimes_R N^\bullet)$. To prove the stated properties of $F$ use Lemmas \ref{lemma-derived-tor-homotopy} and \ref{lemma-derived-tor-exact}. We let $S$ be the set of quasi-isomorphisms in $\mathcal{D} = K(\text{Mod}_R)$. This gives a situation as in Derived Categories, Situation \ref{derived-situation-derived-functor} so that Derived Categories, Definition \ref{derived-definition-right-derived-functor-defined} applies. We claim that $LF$ is everywhere defined. This follows from Derived Categories, Lemma \ref{derived-lemma-find-existence-computes} with $\mathcal{P} \subset \Ob(\mathcal{D})$ the collection of K-flat complexes: (1) follows from Lemma \ref{lemma-K-flat-resolution} and (2) follows from Lemma \ref{lemma-derived-tor-quasi-isomorphism-other-side}. Thus we obtain a derived functor $$ LF : D(R) = S^{-1}\mathcal{D} \longrightarrow \mathcal{D}' = D(A) $$ see Derived Categories, Equation (\ref{derived-equation-everywhere}). Finally, Derived Categories, Lemma \ref{derived-lemma-find-existence-computes} guarantees that $LF(K^\bullet) = F(K^\bullet) = \text{Tot}(K^\bullet \otimes_R N^\bullet)$ when $K^\bullet$ is K-flat, i.e., $LF$ is indeed computed in the way described above. Moreover, by Lemma \ref{lemma-base-change-K-flat} the complex $K^\bullet \otimes_R A$ is a K-flat complex of $A$-modules. Hence $$ (K^\bullet \otimes_R^\mathbf{L} A) \otimes_A^\mathbf{L} N^\bullet = \text{Tot}((K^\bullet \otimes_R A) \otimes_A N^\bullet) = \text{Tot}(K^\bullet \otimes_A N^\bullet) = K^\bullet \otimes_A^\mathbf{L} N^\bullet $$ which proves the final statement of the lemma.
Choose a K-flat complex $P^\bullet$ of $B$-modules and a quasi-isomorphism $P^\bullet \to N^\bullet$ (Lemma \ref{lemma-K-flat-resolution}). Let $M^\bullet$ be a K-flat complex of $A$-modules representing an arbitrary object of $D(A)$. Then we see that $$ (M^\bullet \otimes_A^\mathbf{L} P^\bullet) \otimes_B^\mathbf{L} K^\bullet \longrightarrow (M^\bullet \otimes_A^\mathbf{L} N^\bullet) \otimes_B^\mathbf{L} K^\bullet $$ is an isomorphism by Lemma \ref{lemma-functoriality-derived-base-change} applied to the material inside the brackets. By Lemmas \ref{lemma-base-change-K-flat} and \ref{lemma-tensor-product-K-flat} the complex $$ \text{Tot}(M^\bullet \otimes_A P^\bullet) = \text{Tot}((M^\bullet \otimes_R A) \otimes_A P^\bullet $$ is K-flat as a complex of $B$-modules and it represents the derived tensor product in $D(B)$ by construction. Hence we see that $(M^\bullet \otimes_A^\mathbf{L} P^\bullet) \otimes_B^\mathbf{L} K^\bullet$ is represented by the complex $$ \text{Tot}(\text{Tot}(M^\bullet \otimes_A P^\bullet)\otimes_B K^\bullet) = \text{Tot}(M^\bullet \otimes_A \text{Tot}(P^\bullet \otimes_B K^\bullet)) $$ of $C$-modules. Equality by Homology, Remark \ref{homology-remark-triple-complex}. Going back the way we came we see that this is equal to $$ M^\bullet \otimes_A^\mathbf{L} (P^\bullet \otimes_B^\mathbf{L} K^\bullet) \longleftarrow M^\bullet \otimes_A^\mathbf{L} (N^\bullet \otimes_B^\mathbf{L} K^\bullet) $$ The arrow is an isomorphism by definition of the functor $-\otimes_B^\mathbf{L} K^\bullet$. All of these constructions are functorial in the complex $M^\bullet$ and hence we obtain our isomorphism of functors. \medskip\noindent By the above we have the first equality in $$ (M \otimes_A^\mathbf{L} N) \otimes_B^\mathbf{L} K = M \otimes_A^\mathbf{L} (N \otimes_B^\mathbf{L} K) = (M \otimes_A^\mathbf{L} C) \otimes_C^\mathbf{L} (N \otimes_B^\mathbf{L} K) $$ The second equality follows from the final statement of Lemma \ref{lemma-derived-base-change}. The same thing allows us to write $N \otimes_B^\mathbf{L} K = (N \otimes_B^\mathbf{L} C) \otimes_C^\mathbf{L} K$ and substituting we get \begin{align*} (M \otimes_A^\mathbf{L} N) \otimes_B^\mathbf{L} K & = (M \otimes_A^\mathbf{L} C) \otimes_C^\mathbf{L} ((N \otimes_B^\mathbf{L} C) \otimes_C^\mathbf{L} K) \\ & = (M \otimes_A^\mathbf{L} C) \otimes_C^\mathbf{L} (K \otimes_C^\mathbf{L} (N \otimes_B^\mathbf{L} C)) \\ & = ((M \otimes_A^\mathbf{L} C) \otimes_C^\mathbf{L} K) \otimes_C^\mathbf{L} (N \otimes_B^\mathbf{L} C)) \\ & = (M \otimes_C^\mathbf{L} K) \otimes_C^\mathbf{L} (N \otimes_B^\mathbf{L} C) \end{align*} by Lemmas \ref{lemma-flip-douoble-tensor-product} and \ref{lemma-triple-tensor-product} as well as the previously mentioned lemma.
By Algebra, Section \ref{algebra-section-functoriality-tor} there are canonical maps $$ \text{Tor}_i^R(A, B) \longrightarrow \text{Tor}_i^{R'}(A \otimes_R R', B \otimes_R R') \longrightarrow \text{Tor}_i^{R'}(A', B') $$ These induce a map from left to right in the formula of the lemma. \medskip\noindent Take a free resolution $F_\bullet \to A$ of $A$ as an $R$-module. Then we see that $F_\bullet \otimes_R R'$ is a resolution of $A \otimes_R R'$. Hence $\text{Tor}_i^{R'}(A \otimes_R R', B \otimes_R R')$ is computed by $F_\bullet \otimes_R B \otimes_R R'$. By our assumption that $R'$ is flat over $R$, this computes $\text{Tor}_i^R(A, B) \otimes_R R'$. Thus $\text{Tor}_i^{R'}(A \otimes_R R', B \otimes_R R') = \text{Tor}_i^R(A, B) \otimes_R R'$ (uses only flatness of $R'$ over $R$). \medskip\noindent By Lazard's theorem (Algebra, Theorem \ref{algebra-theorem-lazard}) we can write $A'$, resp.\ $B'$ as a filtered colimit of finite free $A \otimes_R R'$, resp.\ $B \otimes_R R'$-modules. Say $A' = \colim M_i$ and $B' = \colim N_j$. The result above gives $$ \text{Tor}_i^{R'}(M_i, N_j) = \text{Tor}_i^R(A, B) \otimes_{A \otimes_R B} (M_i \otimes_{R'} N_j) $$ as one can see by writing everything out in terms of bases. Taking the colimit we get the result of the lemma.
Let $\mathfrak s$ be a prime of $A \otimes_R B$ as in (3). The equality $$ \text{Tor}_i^R(A, B)_\mathfrak s = \text{Tor}_i^{R_\mathfrak r}(A_\mathfrak p, B_\mathfrak q)_\mathfrak s $$ where $\mathfrak p = A \cap \mathfrak s$, $\mathfrak q = B \cap \mathfrak s$ and $\mathfrak r = R \cap \mathfrak s$ follows from Lemma \ref{lemma-tor-independent-flat}. Hence (2) implies (3). Since we can test the vanishing of modules by localizing at primes (Algebra, Lemma \ref{algebra-lemma-characterize-zero-local}) we conclude that (3) implies (1). For (1) $\Rightarrow$ (2) we use that $$ \text{Tor}_i^{R_\mathfrak r}(A_\mathfrak p, B_\mathfrak q) = \text{Tor}_i^R(A, B) \otimes_{(A \otimes_R B)} (A_\mathfrak p \otimes_{R_{\mathfrak r}} B_\mathfrak q) $$ again by Lemma \ref{lemma-tor-independent-flat}.
Proof of (1). Choose $\alpha : P^\bullet \to K^\bullet$ with $P^\bullet$ a bounded complex of finite free modules such that $H^i(\alpha)$ is an isomorphism for $i > m + 1$ and surjective for $i = m + 1$. We may replace $P^\bullet$ by $\sigma_{\geq m + 1}P^\bullet$ and hence we may assume that $P^i = 0$ for $i < m + 1$. Choose $\beta : E^\bullet \to L^\bullet$ with $E^\bullet$ a bounded complex of finite free modules such that $H^i(\beta)$ is an isomorphism for $i > m$ and surjective for $i = m$. By Derived Categories, Lemma \ref{derived-lemma-lift-map} we can find a map $\alpha : P^\bullet \to E^\bullet$ such that the diagram $$ \xymatrix{ K^\bullet \ar[r] & L^\bullet \\ P^\bullet \ar[u] \ar[r]^\alpha & E^\bullet \ar[u] } $$ is commutative in $D(R)$. The cone $C(\alpha)^\bullet$ is a bounded complex of finite free $R$-modules, and the commutativity of the diagram implies that there exists a morphism of distinguished triangles $$ (P^\bullet, E^\bullet, C(\alpha)^\bullet) \longrightarrow (K^\bullet, L^\bullet, M^\bullet). $$ It follows from the induced map on long exact cohomology sequences and Homology, Lemmas \ref{homology-lemma-four-lemma} and \ref{homology-lemma-five-lemma} that $C(\alpha)^\bullet \to M^\bullet$ induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. Hence $M^\bullet$ is $m$-pseudo-coherent. \medskip\noindent Assertions (2) and (3) follow from (1) by rotating the distinguished triangle.
If $M$ is of finite type (resp.\ of finite presentation), then $M$ is $0$-pseudo-coherent (resp.\ $(-1)$-pseudo-coherent) as follows from the discussion preceding Definition \ref{definition-pseudo-coherent}. Conversely, if $M$ is $0$-pseudo-coherent, then $M = H^0(M[0])$ is of finite type by Lemma \ref{lemma-finite-cohomology}. If $M$ is $(-1)$-pseudo-coherent, then it is $0$-pseudo-coherent hence of finite type. Choose a surjection $R^{\oplus a} \to M$ and denote $K = \Ker(R^{\oplus a} \to M)$. By Lemma \ref{lemma-cone-pseudo-coherent} we see that $K$ is $0$-pseudo-coherent, hence of finite type, whence $M$ is of finite presentation. \medskip\noindent To prove the third and fourth statement use induction and an argument similar to the above (details omitted).
It suffices to prove (3). Set $M = H^{m + 1}(K^\bullet)$. Note that $\tau_{\geq m + 1}K^\bullet$ is quasi-isomorphic to $M[- m - 1]$. By Lemma \ref{lemma-n-pseudo-module} we see that $M[- m - 1]$ is $m$-pseudo-coherent. Since we have the distinguished triangle $$ (\tau_{\leq m}K^\bullet, K^\bullet, \tau_{\geq m + 1}K^\bullet) $$ (Derived Categories, Remark \ref{derived-remark-truncation-distinguished-triangle}) by Lemma \ref{lemma-cone-pseudo-coherent} it suffices to prove that $\tau_{\leq m}K^\bullet$ is pseudo-coherent. By assumption $H^m(\tau_{\leq m}K^\bullet)$ is a finite type $R$-module. Hence we can find a finite free $R$-module $E$ and a map $E \to \Ker(d_K^m)$ such that the composition $E \to \Ker(d_K^m) \to H^m(\tau_{\leq m}K^\bullet)$ is surjective. Then $E[-m] \to \tau_{\leq m}K^\bullet$ witnesses the fact that $\tau_{\leq m}K^\bullet$ is $m$-pseudo-coherent.
In this proof we drop the superscript ${}^\bullet$. Assume that $K \oplus L$ is $m$-pseudo-coherent. It is clear that $K, L \in D^{-}(R)$. Note that there is a distinguished triangle $$ (K \oplus L, K \oplus L, L \oplus L[1]) = (K, K, 0) \oplus (L, L, L \oplus L[1]) $$ see Derived Categories, Lemma \ref{derived-lemma-direct-sum-triangles}. By Lemma \ref{lemma-cone-pseudo-coherent} we see that $L \oplus L[1]$ is $m$-pseudo-coherent. Hence also $L[1] \oplus L[2]$ is $m$-pseudo-coherent. By induction $L[n] \oplus L[n + 1]$ is $m$-pseudo-coherent. By Lemma \ref{lemma-recognize-pseudo-coherent} we see that $L[n]$ is $m$-pseudo-coherent for large $n$. Hence working backwards, using the distinguished triangles $$ (L[n], L[n] \oplus L[n - 1], L[n - 1]) $$ we conclude that $L[n], L[n - 1], \ldots, L$ are $m$-pseudo-coherent as desired. The pseudo-coherent case follows from this and Lemma \ref{lemma-pseudo-coherent}.
We may replace $K^\bullet$ by $\sigma_{\geq m - 1}K^\bullet$ (for example) and hence assume that $K^\bullet$ is bounded. Then the complex $K^\bullet$ is $m$-pseudo-coherent as each $K^i[-i]$ is $m$-pseudo-coherent by induction on the length of the complex: use Lemma \ref{lemma-cone-pseudo-coherent} and the stupid truncations. For the final statement, it suffices to prove that $K^\bullet$ is $m$-pseudo-coherent for all $m \in \mathbf{Z}$, see Lemma \ref{lemma-pseudo-coherent}. This follows from the first part.
Assume $K^\bullet$ is an object of $D^{-}(R)$ such that each $H^i(K^\bullet)$ is $(m - i)$-pseudo-coherent. Let $n$ be the largest integer such that $H^n(K^\bullet)$ is nonzero. We will prove the lemma by induction on $n$. If $n < m$, then $K^\bullet$ is $m$-pseudo-coherent by Lemma \ref{lemma-recognize-pseudo-coherent}. If $n \geq m$, then we have the distinguished triangle $$ (\tau_{\leq n - 1}K^\bullet, K^\bullet, H^n(K^\bullet)[-n]) $$ (Derived Categories, Remark \ref{derived-remark-truncation-distinguished-triangle}) Since $H^n(K^\bullet)[-n]$ is $m$-pseudo-coherent by assumption, we can use Lemma \ref{lemma-cone-pseudo-coherent} to see that it suffices to prove that $\tau_{\leq n - 1}K^\bullet$ is $m$-pseudo-coherent. By induction on $n$ we win. (The pseudo-coherent case follows from this and Lemma \ref{lemma-pseudo-coherent}.)
Assume (1). Choose a bounded complex of finite free $B$-modules $E^\bullet$ and a map $\alpha : E^\bullet \to K^\bullet$ which is an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. Consider the distinguished triangle $(E^\bullet, K^\bullet, C(\alpha)^\bullet)$. By Lemma \ref{lemma-recognize-pseudo-coherent} $C(\alpha)^\bullet$ is $m$-pseudo-coherent as a complex of $A$-modules. Hence it suffices to prove that $E^\bullet$ is pseudo-coherent as a complex of $A$-modules, which follows from Lemma \ref{lemma-complex-pseudo-coherent-modules}. The pseudo-coherent case of (1) $\Rightarrow$ (2) follows from this and Lemma \ref{lemma-pseudo-coherent}. \medskip\noindent Assume (2). Let $n$ be the largest integer such that $H^n(K^\bullet) \not = 0$. We will prove that $K^\bullet$ is $m$-pseudo-coherent as a complex of $B$-modules by induction on $n - m$. The case $n < m$ follows from Lemma \ref{lemma-recognize-pseudo-coherent}. Choose a bounded complex of finite free $A$-modules $E^\bullet$ and a map $\alpha : E^\bullet \to K^\bullet$ which is an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. Consider the induced map of complexes $$ \alpha \otimes 1 : E^\bullet \otimes_A B \to K^\bullet. $$ Note that $C(\alpha \otimes 1)^\bullet$ is acyclic in degrees $\geq n$ as $H^n(E) \to H^n(E^\bullet \otimes_A B) \to H^n(K^\bullet)$ is surjective by construction and since $H^i(E^\bullet \otimes_A B) = 0$ for $i > n$ by the spectral sequence of Example \ref{example-tor}. On the other hand, $C(\alpha \otimes 1)^\bullet$ is $m$-pseudo-coherent as a complex of $A$-modules because both $K^\bullet$ and $E^\bullet \otimes_A B$ (see Lemma \ref{lemma-complex-pseudo-coherent-modules}) are so, see Lemma \ref{lemma-cone-pseudo-coherent}. Hence by induction we see that $C(\alpha \otimes 1)^\bullet$ is $m$-pseudo-coherent as a complex of $B$-modules. Finally another application of Lemma \ref{lemma-cone-pseudo-coherent} shows that $K^\bullet$ is $m$-pseudo-coherent as a complex of $B$-modules (as clearly $E^\bullet \otimes_A B$ is pseudo-coherent as a complex of $B$-modules). The pseudo-coherent case of (2) $\Rightarrow$ (1) follows from this and Lemma \ref{lemma-pseudo-coherent}.
First we note that the statement of the lemma makes sense as $K^\bullet$ is bounded above and hence $K^\bullet \otimes_A^{\mathbf{L}} B$ is defined by Equation (\ref{equation-derived-tensor-algebra}). Having said this, choose a bounded complex $E^\bullet$ of finite free $A$-modules and $\alpha : E^\bullet \to K^\bullet$ with $H^i(\alpha)$ an isomorphism for $i > m$ and surjective for $i = m$. Then the cone $C(\alpha)^\bullet$ is acyclic in degrees $\geq m$. Since $-\otimes_A^{\mathbf{L}} B$ is an exact functor we get a distinguished triangle $$ (E^\bullet \otimes_A^{\mathbf{L}} B, K^\bullet \otimes_A^{\mathbf{L}} B, C(\alpha)^\bullet \otimes_A^{\mathbf{L}} B) $$ of complexes of $B$-modules. By the dual to Derived Categories, Lemma \ref{derived-lemma-negative-vanishing} we see that $H^i(C(\alpha)^\bullet \otimes_A^{\mathbf{L}} B) = 0$ for $i \geq m$. Since $E^\bullet$ is a complex of projective $R$-modules we see that $E^\bullet \otimes_A^{\mathbf{L}} B = E^\bullet \otimes_A B$ and hence $$ E^\bullet \otimes_A B \longrightarrow K^\bullet \otimes_A^{\mathbf{L}} B $$ is a morphism of complexes of $B$-modules that witnesses the fact that $K^\bullet \otimes_A^{\mathbf{L}} B$ is $m$-pseudo-coherent. The case of pseudo-coherent complexes follows from the case of $m$-pseudo-coherent complexes via Lemma \ref{lemma-pseudo-coherent}.
We will use without further mention that $- \otimes_R R_{f_i}$ is an exact functor and that therefore $$ H^i(K^\bullet)_{f_i} = H^i(K^\bullet) \otimes_R R_{f_i} = H^i(K^\bullet \otimes_R R_{f_i}). $$ Assume $K^\bullet \otimes_R R_{f_i}$ is $m$-pseudo-coherent for $i = 1, \ldots, r$. Let $n \in \mathbf{Z}$ be the largest integer such that $H^n(K^\bullet \otimes_R R_{f_i})$ is nonzero for some $i$. This implies in particular that $H^i(K^\bullet) = 0$ for $i > n$ (and that $H^n(K^\bullet) \not = 0$) see Algebra, Lemma \ref{algebra-lemma-cover}. We will prove the lemma by induction on $n - m$. If $n < m$, then the lemma is true by Lemma \ref{lemma-recognize-pseudo-coherent}. If $n \geq m$, then $H^n(K^\bullet)_{f_i}$ is a finite $R_{f_i}$-module for each $i$, see Lemma \ref{lemma-finite-cohomology}. Hence $H^n(K^\bullet)$ is a finite $R$-module, see Algebra, Lemma \ref{algebra-lemma-cover}. Choose a finite free $R$-module $E$ and a surjection $E \to H^n(K^\bullet)$. As $E$ is projective we can lift this to a map of complexes $\alpha : E[-n] \to K^\bullet$. Then the cone $C(\alpha)^\bullet$ has vanishing cohomology in degrees $\geq n$. On the other hand, the complexes $C(\alpha)^\bullet \otimes_R R_{f_i}$ are $m$-pseudo-coherent for each $i$, see Lemma \ref{lemma-cone-pseudo-coherent}. Hence by induction we see that $C(\alpha)^\bullet$ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma \ref{lemma-cone-pseudo-coherent} once more we conclude.
We will use without further mention that $- \otimes_R R'$ is an exact functor and that therefore $$ H^i(K^\bullet) \otimes_R R' = H^i(K^\bullet \otimes_R R'). $$ Assume $K^\bullet \otimes_R R'$ is $m$-pseudo-coherent. Let $n \in \mathbf{Z}$ be the largest integer such that $H^n(K^\bullet)$ is nonzero; then $n$ is also the largest integer such that $H^n(K^\bullet \otimes_R R')$ is nonzero. We will prove the lemma by induction on $n - m$. If $n < m$, then the lemma is true by Lemma \ref{lemma-recognize-pseudo-coherent}. If $n \geq m$, then $H^n(K^\bullet) \otimes_R R'$ is a finite $R'$-module, see Lemma \ref{lemma-finite-cohomology}. Hence $H^n(K^\bullet)$ is a finite $R$-module, see Algebra, Lemma \ref{algebra-lemma-descend-properties-modules}. Choose a finite free $R$-module $E$ and a surjection $E \to H^n(K^\bullet)$. As $E$ is projective we can lift this to a map of complexes $\alpha : E[-n] \to K^\bullet$. Then the cone $C(\alpha)^\bullet$ has vanishing cohomology in degrees $\geq n$. On the other hand, the complex $C(\alpha)^\bullet \otimes_R R'$ is $m$-pseudo-coherent, see Lemma \ref{lemma-cone-pseudo-coherent}. Hence by induction we see that $C(\alpha)^\bullet$ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma \ref{lemma-cone-pseudo-coherent} once more we conclude.
Proof of (1). We may assume there exist bounded complexes $K^\bullet$ and $L^\bullet$ of finite free $R$-modules and maps $\alpha : K^\bullet \to K$ and $\beta : L^\bullet \to L$ with $H^i(\alpha)$ and isomorphism for $i > n$ and surjective for $i = n$ and with $H^i(\beta)$ and isomorphism for $i > m$ and surjective for $i = m$. Then the map $$ \alpha \otimes^\mathbf{L} \beta : \text{Tot}(K^\bullet \otimes_R L^\bullet) \to K \otimes_R^\mathbf{L} L $$ induces isomorphisms on cohomology in degree $i$ for $i > t$ and a surjection for $i = t$. This follows from the spectral sequence of tors (details omitted). Part (2) follows from part (1) and Lemma \ref{lemma-pseudo-coherent}.
In Algebra, Lemma \ref{algebra-lemma-resolution-by-finite-free} we have seen that any finite $R$-module is pseudo-coherent. On the other hand, a pseudo-coherent module is finite, see Lemma \ref{lemma-n-pseudo-module}. Hence (3) holds. Suppose that $K^\bullet$ is an $m$-pseudo-coherent complex. Then there exists a bounded complex of finite free $R$-modules $E^\bullet$ such that $H^i(K^\bullet)$ is isomorphic to $H^i(E^\bullet)$ for $i > m$ and such that $H^m(K^\bullet)$ is a quotient of $H^m(E^\bullet)$. Thus it is clear that each $H^i(K^\bullet)$, $i \geq m$ is a finite module. The converse implication in (1) follows from Lemma \ref{lemma-cohomology-pseudo-coherent} and part (3). Part (2) follows from (1) and Lemma \ref{lemma-pseudo-coherent}.
Recall that an $R$-module $M$ is coherent if and only if it is of finite presentation (Algebra, Lemma \ref{algebra-lemma-coherent-ring}). This explains the equivalence of (2) and (3). If so and if we choose an exact sequence $0 \to N \to R^{\oplus m} \to M \to 0$, then $N$ is coherent by Algebra, Lemma \ref{algebra-lemma-coherent}. Thus in this case, repeating this procedure with $N$ we find a resolution $$ \ldots \to R^{\oplus n} \to R^{\oplus m} \to M \to 0 $$ by finite free $R$-modules. In other words, $M$ is pseudo-coherent. The equivalence of (1) and (2) follows from this and Lemmas \ref{lemma-cohomology-pseudo-coherent} and \ref{lemma-n-pseudo-module}. The final assertion follows from the equivalence of (1) and (2) combined with Lemma \ref{lemma-pseudo-coherent}.
By definition we can find a distinguished triangle $$ E \to K \to L \to E[1] $$ in $D(R)$ such that $E$ is represented by a bounded complex of finite free $R$-modules and such that $H^i(L) = 0$ for $i \geq m$. Then $\Ext^n_R(L, N) = 0$ for any $R$-module $N$ and $n \leq -m$, see Derived Categories, Lemma \ref{derived-lemma-negative-exts}. By the long exact sequence of $\Ext$ associated to the distinguished triangle we see that $\Ext^n_R(K, N) \to \Ext^n_R(E, N)$ is an isomorphism for $n < -m$ and injective for $n = -m$. Thus it suffices to prove that $M \mapsto \Ext_R^n(E, M)$ commutes with filtered colimits when $E$ can be represented by a bounded complex of finite free $R$-modules $E^\bullet$. The modules $\Ext^n_R(E, M)$ are computed by the complex $\Hom_R(E^\bullet, M)$, see Derived Categories, Lemma \ref{derived-lemma-morphisms-from-projective-complex}. The functor $M \mapsto \Hom_R(E^p, M)$ commutes with filtered colimits as $E^p$ is finite free. Thus $\Hom_R(E^\bullet, M) = \colim \Hom_R(E^\bullet, M_i)$ as complexes. Since filtered colimits are exact (Algebra, Lemma \ref{algebra-lemma-directed-colimit-exact}) we conclude.
One implication was shown in Lemma \ref{lemma-pseudo-coherence-colimit-ext}. Assume for any filtered colimit $M = \colim M_i$ of $R$-modules we have $\colim \Ext^n_R(K, M_i) = \Ext^n_R(K, M)$ for $n < -m$ and $\colim \Ext^{-m}_R(K, M_i) \to \Ext^{-m}_R(K, M)$ is injective. We will show $K$ is $m$-pseudo-coherent. \medskip\noindent Let $t$ be the maximal integer such that $H^t(K)$ is nonzero. We will use induction on $t$. If $t < m$, then $K$ is $m$-pseudo-coherent by Lemma \ref{lemma-recognize-pseudo-coherent}. If $t \geq m$, then since $\Hom_R(H^t(K), M) = \Ext^{-t}_R(K, M)$ we conclude that $\colim \Hom_R(H^t(K), M_i) \to \Hom_R(H^t(K), M)$ is injective for any filtered colimit $M = \colim M_i$. This implies that $H^t(K)$ is a finite $R$-module by Algebra, Lemma \ref{algebra-lemma-characterize-finite-module-hom}. Choose a finite free $R$-module $F$ and a surjection $F \to H^t(K)$. We can lift this to a morphism $F[-t] \to K$ in $D(R)$ and choose a distinguished triangle $$ F[-t] \to K \to L \to F[-t + 1] $$ in $D(R)$. Then $H^i(L) = 0$ for $i \geq t$. Moreover, the long exact sequence of $\Ext$ associated to this distinguished triangle shows that $L$ inherts the assumption we made on $K$ by a small argument we omit. By induction on $t$ we conclude that $L$ is $m$-pseudo-coherent. Hence $K$ is $m$-pseudo-coherent by Lemma \ref{lemma-cone-pseudo-coherent}.