CT0

ZCC
July 20, 2025

Lecture 0
On Rings and Modules
On Chain Complexes

Lecture 0

Abstract

Throughout these lectures, we discuss and construct explicit examples of cotorsion theory. The categories of modules and chain complexes over rings serve as our principal examples, wherein we demonstrate how set-theoretic techniques may be employed to construct cotorsion pairs. Throughout these lectures, we discuss and construct explicit examples of cotorsion theory. The categories of modules and chain complexes over rings serve as our principal examples, wherein we demonstrate how set-theoretic techniques may be employed to construct cotorsion pairs.

During the initial few lectures, we follow RHA-1 and RHA-2.

We also introduce certain aspects of the theory of chain complexes over rings, which might be atypical in most standard textbooks.

Notations

Throughout,

$R$: a unital (not necessarily commutative) ring,
$𝐌𝐨𝐝 _R \quad (≃ 𝐏𝐒𝐡_{𝐀𝐛}(R))$: the category of right $R$-modules,
$C(R)$: the category of chain complexes over $𝐌𝐨𝐝 _R$,
- our convension is $\cdots → X^k \xrightarrow {d^k} X^{k+1} → \cdots$,
- $Σ$ and $Σ ^{-1}$ denote the suspension and desuspension functors respectively.
$𝐏𝐫𝐨𝐣 (\mathcal{C})$: the class (or full subcategory) of projectives in $\mathcal{C}$,
$𝐈𝐧𝐣 (\mathcal{C})$: the class (or full subcategory) of injectives in $\mathcal{C}$.

Functors

We review some basic definitions and notions in homological algebra. We employ

$∐$ to denote coproducts,
$∏$ for products, and
$⨁$ specifically when $∐ \xrightarrow{∼} ∏$, For example, $⨁_{n ∈ ℤ} Σ^n M_n$ represents the sum of stalk complexes.

The symbols $\varinjlim$ and $\varprojlim$ are utilised for colimits (injective limits) and limits (projective limits), respectively; in particular, the superscript $fil$ in $\varinjlim^{fil}$ indicates the filtered colimits.

(Preserving, reflecting, and creating). We explain by functors between Abelian categories. Given $F : \mathcal{A} → \mathcal{B}$:

$F$ is said to preserve or commute with $\ker$, if any $0 → K \xrightarrow i X \xrightarrow f Y$ is exact in $\mathcal{A}$ implies $0 → F(K) \xrightarrow {F(i)} F(X) \xrightarrow {F(f)} F(Y)$ is exact in $\mathcal{B}$.
$F$ is said to reflect $\ker$, if any $0 → F(K) \xrightarrow {F(i)} F(X) \xrightarrow {F(f)} F(Y)$ is exact in $\mathcal{B}$ implies $0 → K \xrightarrow i X \xrightarrow f Y$ is exact in $\mathcal{A}$.
$F$ is said to create $\ker$, if any $0 → H \xrightarrow {φ } F(X) \xrightarrow {F(f)} F(Y)$ is exact in $\mathcal{B}$ implies the existence of the exact sequence $0 → K \xrightarrow i X \xrightarrow f Y$ in $\mathcal{A}$, such that $F(K) = H$ and $F(i) = φ$.

Demonstrate that

a fully faithful functor reflects limits;
the left/right adjoint functor preserves all limits/colimits, respectively;
for a Freydal luff sub-diagram $ι : Q' → Q$, the functor $ι ^∗ : 𝐅𝐮𝐧𝐜𝐭 (Q, \mathcal{C}) → 𝐅𝐮𝐧𝐜𝐭 (Q', \mathcal{C})$ creates all limits and colimits,
- corollary: the (co)limits in $C(R)$ are computed termwise!
the forgetful functor $𝐌𝐨𝐝 _R → 𝐒𝐞𝐭𝐬$ creates all limits and filtered colimits,
- corollary: the filtered colimits and limits in $𝐌𝐨𝐝 _R$ are computed elementwise!

(Exactness). Left/right exact functors are precisely those preserving finite limits/colimits, respectively. A functor is termed exact if it is both left and right exact.

(Grothendieck’s AB conditions). See stacks, or n-lab.

The conditions AB3, AB3$^∗$, AB4, AB4$^∗$, and AB5 are standard in both $𝐌𝐨𝐝 _R$ and $C(R)$. We assert the isomorphisms such as $∏ \operatorname{cok}(f_∙ ) \xrightarrow ∼ \operatorname{cok}(∏ f_∙ )$ without further elaboration throughout these lectures.
The author refined the statement of AB6 in n-lab during a rather uneventful afternoon. Nevertheless, AB6 is not considered significant within the context of homological algebra.

Here are some well-known properties of filtered colimits (over $𝐌𝐨𝐝_R$, $C(R)$, or a general AB5 category). We take $ι _0 : M_{i_0} → \varinjlim ^{fil}_{i ∈ I} M_i$ as the structure morphism for filtered colimit.

Any element $x : \{∗\} → \varinjlim ^{fil}_{i ∈ I} M_i$ factors through some $ι_i$. (As every point is small).
$\ker (ι _{i}) = ⋃ _{α_{j,i}^{(k)} : i → j} \ker α _{j,i}^{(k)}$ (such set-theoretic union for modules is also a module).
- In particular, a monic filtered system has monic structure map.
Let $\{f_i : M_i ↪ N\}_{i ∈ I}$ be a collection of monomorphisms which is compatible with the filtered diagram $I$. Then $\varinjlim ^{fil}_{i ∈ I} M_i ↪ N$ is also monic.
$∑_{λ ∈ Λ } (V ∩ U_λ) = V ∩ ∑_{λ ∈ Λ } (U_λ)$. This is another equivalent definition of AB5 condition.
(Assuming enough $𝐏𝐫𝐨𝐣$). Let $F$ be a left adjoint. Then filtered colimit preserves its left derivatives, i.e., $\varinjlim ^{fil}(\mathrm L^k F)(M_∙) ≃ (\mathrm L^k F)(\varinjlim ^{fil}M_∙)$ for $k ≥ 0$.
- General left adjoint functors include colimits (e.g. coproducts), free functors (e.g. $X ↦ R^{(X)}$, sheafification), realisations (e.g. geometric/categorical realisation of simplicial sets), $(? ⊗ -)$, inverse images of either sheaves or presheaves (e.g. stalks, constant sheaves, or the change of category in general).

On Rings and Modules

We recollect some basic properties of ring and modules.

An Interesting Example on Transfinite Filtration

This demonstrates the distinctions between finitely generated modules and those $M$ for which $∐ (M, -) ↪ (M, ∐ (-))$ constitutes an isomorphism. An equivalent characterisation is provided via the cofinality of transfinite filtration.

We begin with an exercise in ring theory.

$M$ is finitely generated if and only if,

for any (non-empty) set of proper submodules $\{N_i\}_{i ∈ I}$ such that either $N_i ⊆ N_j$ or $N_i ⊇ N_j$ for any $i, j ∈ I$, the union $⋃ N_∙$ is never equal to $M$ itself.

($→$). We show $a → b$ by showing that $a ∧ (¬ b) → \text{false}$. By $(¬ b)$, there exists an ordered system of proper submodules $N_∙$ whose union is $M$. Let $\{x_i\}_{i=1}^n$ be a generating set of $M$ (by $(a)$). Thus, for each $i$, there exists $x_i ∈ N_{k_i}$. The proper submodule $⋃ _{i=1}^n N_{k_i} = \text{Largest}\{N_{k_i}\}$ yields a contradiction.

($←$). We show $b → a$ with $b ∧ (¬ a) → \text{false}$. Consider the collection of proper submodules $N_∙ = \{N ∣ M / N\ \text{ is not f.g.}\}$. The poset of proper submodules $N_∙$ is both

non-empty by $(¬ a)$, and
contains no maximal elements by construction.

We examine the partially ordered set $P := \{\text{chains in } N_∙\}$ with the binary relation $⊆$. Observe that each chain in $P$ possesses an upper bound, obtained by taking the common refinement. Therefore, by Zorn’s lemma, $N_∙$ contains a maximal inclusion chain of proper submodules, denoted by $C_∙$.
$C_∙$ contains no maximal elements; thus, $⋃ C_∙ ∉ N_∙$. In this case $M / (⋃ C_∙)$ is f.g.. Consequently, there exists a finitely generated submodule $L$ such that $L + ⋃ C_∙ = M$; however, $⋃ (L + C_∙) ≠ M$ by $(b)$. This leads to a contradiction.

In the above theorem, $N_∙$ is taken over arbitrary totally-ordered set. For colimit $⋃$, it suffices to consider a cofinal subsystem indexed by an ordinal. Hence,

$M$ is f.g., iff any transfinite composition of proper submodules,

$$\begin{equation} 0 = M_0 ⊊ M_1 ⊊ \cdots ⊊ M_γ ⊊ \cdots, \end{equation}$$

satisfy $(⋃ M_∙) ⊊ M$.

$M$ is f.g. if and only if every proper submodule is contained in a maximal submodule.

(→). Suppose that $M$ is f.g.. For any $M_0 ⊆ M$, $M / M_0$ is f.g.. Now it suffices to show that $(M / M_0)$ possesses a maximal submodule. If not, then there exists some transfinite inclusion of proper submodules whose union is $(M / M_0)$. This is a contradiction.
(←). Conversely, denote the non-empty set $$\begin{equation} \mathcal{S} := \{N ∣ N ⊊ M , N \ \text{is not contained in a maximal submodule}\}. \end{equation}$$

Note that $\mathcal{S}$ is closed under upward inclusion. The partially ordered set $\{\text{chains in} \ \mathcal{S}\}$ is closed under transfinite refinement. Hence, there exists a maximal chain $C_∙ ⊆ \mathcal{S}$. Clearly, $⋃ C_∙ = M$, so $M$ is not f.g..

This criterion admits an inductive proof, which reduces to the case that every ring possesses a maximal ideal.

We slightly revise such characterisation to the following FAKE finitely generated cases.

(FAKE f.g.). A module $M$ is FAKE finitely generated if any $ω$-composition of proper submodules,

$$\begin{equation} 0 = M_0 ⊊ M_1 ⊊ \cdots ⊊ M_n ⊊ \cdots \quad (n ∈ ℕ ), \end{equation}$$

satisfy $(⋃_{n ∈ ℕ} M_n) ⊊ M$.

We show FAKE f.g. $M$’s are precisely those $∐ (M, -) \xrightarrow ∼ (M, ∐ (-))$ in the rest of this subsection. We call such $M$ a compact object in the underlying additive category, i.e., $(M, -)$ is coproduct preserving

compact objects in general additive categories corresponds to $∐$-preserving $(M, -)$;
compact objects in AB5 category (e.g. $𝐌𝐨𝐝_R$) corresponds to $\varinjlim^{fil}$-preserving $(M, -)$ (e.g., f.p. modules).

Show the subfunctor $∐ (M, -) ↪ (M, ∐ (-))$ for $𝐌𝐨𝐝_R$ or $C(R)$.

If $M$ is FAKE f.g., then $M$ is compact in the underlying additive category of $𝐌𝐨𝐝_R$.

We deduce the lemma to countable case, that is, $(M, -)$ preserves arbitrary coproducts if $(M, -)$ preserves countable coproducts.

Assume $∐_{i ∈ I} (M, N_i ) \overset {≠}↪ (M, ∐_I N_{i} )$, we can find $F : M → ∐ _{i ∈ I}N_i$ such that $N_i ∩ \mathrm{im}(F) ≠ 0$ for infinitely many $N_i$’s. Hence, there is a countable coproduct $N = (∐ _{n ∈ ℕ} N_n) ⊕ (∐ _{\text{the rest}}N_∙)$ where the image of $F$ has non-zero intersection with each summand. This provides an example that $(M, -)$ does not preserve countable coproducts.

We take arbitrary $f$ in

$$\begin{equation} ∐_{n ∈ ℕ} (M,N_n) ↪ (M, ∐_{n ∈ ℕ} N_n) ∋ f = (f_n)_{n ∈ ℕ}. \end{equation}$$

Then $⋂ _{m ≥ n} \ker f_m$ is an ascending chain of submodules of $M$, satisfying

$$\begin{equation} \limsup (\ker f_∙) = ⋃_{n ≥ 0} ⋂ _{m ≥ n} \ker f_m = M. \end{equation}$$

By FAKE f.g. condition, all but finite $⋂ _{m ≥ n} \ker f_m ⊆ ⋂ _{m ≥ n+1} \ker f_m$ takes $⊊$. Hence, there is $n_0$ such that $⋂ _{m ≥ n_0} \ker f_m = M$. In this case, $f$ is also characterised in $⨁ _{n ≤ n_0}(M, N_n)$.

Show the converse statement: when there exists $⋃ M_∙ = M$ for some inclusion sequence

$$\begin{equation} 0 = M_0 ⊊ M_1 ⊊ \cdots ⊊ M_n ⊊ \cdots \quad (n ∈ ℕ ). \end{equation}$$

Show that $∐ (M, -) ≇ (M, ∐ (-))$.

Hint: Let $π _n : M ↠ M/M_n$ be quotient maps. The morphism $(π _n) ∈ (M, ∏ \frac{M}{M_n})$ lies in $(M, ∐ \frac{M}{M_n})$, since each support set $S_x := \{n ∣ π _n(x) ≠ 0\}$ is finite. However, one can never find $?$ along

$$\begin{equation} ? ∈ ∐(M,\frac{M}{M_n}) ↪ (M, ∐ \frac{M}{M_n}) ↪ (M, ∏ \frac{M}{M_n}) ∋ (π _n). \end{equation}$$

Notice that $(π _n) ∈ (M, ∏ \frac{M}{M_n}) ≃ ∏ (M, \frac{M}{M_n})$ never belongs to $∐ (M, \frac{M}{M_n})$.

There exists of $ω$-filtered ordinal, e.g., the first uncountable cardinal and $|ℝ|$ (these two coincides under continuum hypothesis).

We learn from analysis that $\mathcal{P}(X) = \mathrm{Hom}_{𝐒𝐞𝐭𝐬}(X,𝔽 _2)$ is an $𝔽_2$-algebra ($∅ = 0$ and $∩ = ×$ for convention). Let $X$ be uncountable. Show that

The collection $X_ω := \{Y ∣ |Y| ≤ ω \}$ is an ideal of $\mathcal{P}(X)$;
$X_ω$ is not f.g. as a $\mathcal{P}(X)$-module.
$X_ω$ is uncountable, but any proper submodule of $X_ω$ is countable.

Hence, $X_ω$ is FAKE f.g. but not f.g.. This construction shows that for any cardinal $κ$, there exists examples such that the generating set of some FAKE f.g. modules must be $≥ κ$.

On FAKE f.g. and f.g.. Show that

f.g. are FAKE f.g..
f.g. coincides FAKE f.g. for projective modules;
FAKE f.g. are closed under quotients (and summands);
FAKE f.g. are closed under extensions (and finite sums);
For ses $K ↪ X ↠ Y$ where $K$ is small in Y (e.g., radical inclusion), then $X$ is FAKE f.g. if so is $Y$.

For left modules, f.g. coincides FAKE f.g. over left perfect rings.

We show any FAKE f.g. module $M$ is f.g..

(by 5). the projective cover $P(M)$ is again FAKE f.g..
(by 2). such $P(M)$ must be f.g..

By $P(M) ↠ M$, $M$ is also f.g..

Over a left/right perfect ring, there are no need to distinguish between

f.g. and FAKE f.g. modules,
projective modules and flat modules,
…

The REAL Compact Modules

Say $M$ is compact, provided $(M, - )$ preserves all filtered colimits.

Compact objects are closed under finite colimits. In particular, f.g. projective modules and f.p. modules are compact.

By AB5, filtered colimits commutes with finite limits. By pentagon procedure,

$$\begin{aligned} (\varinjlim {}^{<∞} M_i, \varinjlim{}^{fil}_jN_j) & ≃ \varprojlim {}^{<\infty}(M_i, \varinjlim{}^{fil}_j N_j) ≃ \varprojlim {}^{<\infty} \varinjlim{}^{fil}_j (M_i, N_j)\\[6pt] &≃ \varinjlim{}^{fil}_j \varprojlim {}^{<\infty} (M_i, N_j) ≃ \varinjlim{}^{fil}_j (\varinjlim {}^{<\infty} M_i, N_j). \end{aligned}$$

We see $\varinjlim {}^{<∞} M_i$ preserves $\varinjlim^{fil}$, thus is compact.

By definition, $f: \text{compact} → \text{induction of small}$ is described by $f : \text{compact} → \text{small}$ locally. We wish $\text{induction of small} = \text{all objects}$.

Finite sets are exactly compact objects in $𝐒𝐞𝐭𝐬$, conversely, any set is a filtered colimit of its finite subsets (local compactness). It is reasonable for check whether a module is a filtered limit of the compact objects.

(Local compactness for modules). Every module $M$ is a filtered colimit of compact modules.

It suffices to show $M$ is a filtered colimit of f.p. modules. We take the canonical surjection $R^{(M)} ↠ M$, and construct the filtered system as follows.

The system is indexed by a finite subset $I \subseteq M$ along with a finite subset of $\mathcal S \subseteq S_I\subseteq R^I$ which generates $\ker [R^I \hookrightarrow R^{(M)} \twoheadrightarrow M]$.
We denote $\Phi _{(I, \mathcal S)}$ as an element, which stands for a chain complex $⟨\mathcal S⟩ \hookrightarrow R^I \to M$.
We define the partial order $\Phi_{(I, \mathcal S)} \hookrightarrow \Phi_{(J, \mathcal T)}$.

The poset is clearly filtered. We show the induced $\alpha : \varinjlim_{\Phi_{(I,\mathcal S)}}R^I\to M$ is an isomorphism.

$\alpha$ is surjective, since any $x \in M$ is characterised by $R^{(x)} \to M$;
$\alpha$ is injective. If $[p] \in \Phi_{(I, \mathcal S)}$ and $[q] \in \Phi _{(J, \mathcal T)}$ has the same image $m\in M$ , then we take $\Phi _{(I \cup J \cup \{m\}, \mathcal S \cup \mathcal T \cup \{p-e_m, q-e_m\})}$ , an upper bound of $\Phi_{ (I, \mathcal S)}$ and $\Phi _{(J, \mathcal T)}$ where $[p]$ and $[q]$ has the same image.

We show that f.p. modules are precisely compact modules. Indeed we have the following equivalent characterisation.

In general, $M$ is finitely presented iff the following equivalent definitions holds.

There is an exact sequence $R^m \to R^n \to M \to 0$,
$M \otimes-$ commutes with $\prod$,
$M \otimes R^{\Lambda} \simeq M^\Lambda$ for arbitrary index set $\Lambda$.
$(M, -)$ preserves filtered colimits.

($1 \to 4$). We write $M = \mathrm{cok}(\varphi)$. Now

$$ \begin{align*} \varinjlim (\mathrm{cok}(\varphi), X_\bullet)&\simeq\varinjlim\ker(\varphi, X_\bullet)\quad (\text{definition of cok})\\ &\simeq\ker \varinjlim(\varphi, X_\bullet)\quad (\text{AB5 for Ab})\\ &\simeq\ker (\varphi, \varinjlim X_\bullet)\quad (\text{holds for free modules})\\ &\simeq (\mathrm{cok}\varphi, \varinjlim X_\bullet)\quad (\text{definition of $\mathrm{cok}$}). \end{align*} $$

($4 \to 1$). $M$ is a filtered colimit of f.p. modules, i.e., $M = \varinjlim M_\bullet$. The isomorphism $1_M \in (M,M) \simeq \varinjlim (M, M_\bullet)$ shows that $1_M$ factors through some f.p. modules. Hence $M$ is f.p..
($1 \to 2$). We write $M = \mathrm{cok}(\varphi)$. Now

$$ \begin{align*} \prod ((\mathrm{cok}\varphi) \otimes X_\bullet) &\simeq\prod \mathrm{cok}(\varphi \otimes X_\bullet)\quad (\text{$\otimes$ is left adj})\\ &\simeq\mathrm{cok}\prod (\varphi \otimes X_\bullet)\quad (\text{AB4* for Ab})\\ &\simeq\mathrm{cok} (\varphi \otimes \prod X_\bullet)\quad (\text{holds for free modules})\\ &\simeq (\mathrm{cok} \varphi) \otimes \prod X_\bullet\quad (\text{$\otimes$ is left adj}). \end{align*} $$

($2 \to 3$). Clear.
($3 \to 1$). We track $1_M \in M^M \simeq M \otimes R^M$, and see that $M$ is f.g.. We take ses $0 \to K \to R^n \to M \to 0$ and obtain the commutative diagram of exact sequences

Here $K^K$ is a syzygy of the upper sequence. Hence, $K \otimes R^K \to K^K$ is surjective. We see $K$ is f.g. and thus $M$ is f.p..

Note that

$$\begin{equation} \text{f.p.} = \text{compact (紧对象)} ⊆ \text{f.g.} ⊆ \text{FAKE f.g.} \quad \color{gray}{= \ compact \ (紧模)}. \end{equation}$$

$𝐏𝐫𝐨𝐣$, $𝐈𝐧𝐣$ and $𝐅𝐥𝐚𝐭$

We summarise the tests for f.p. modules, proj. modules, inj. modules and flat modules at the very beginning.

$M$ is f.p., iff the following equivalent definitions hold.

$M ≃ \operatorname{cok}φ$, where $φ$ is a morphism in $𝐩𝐫𝐨𝐣$.
$(M, -)$ preserves filtered colimits.
$M ⊗ -$ preserves arbitrary products.

$M$ is projective, iff the following equivalent definitions hold.

$(M, -)$ is exact, $\mathrm{Ext}^1(M, -)=0$, etc.
$M$ is a summand of some free module $R^{(λ)}$;
$M$ is the coproduct of summands of $R^{(ω)}$ (the countably generated free module);
There exists a subset $\{(e_i, f_i)\}_{i ∈ I} ⊆ P × (P, R)$ such that any $x ∈ R$ is a finite sum of some $∑_{i ∈ I_0} e_i ⋅ f_i (x)$.

$M$ is injective, iff the following equivalent definitions hold.

$(-, M)$ exact, $\mathrm{Ext}^1(-, M)=0$; $\mathrm{Ext}^1(\text{quotient ring},M)=0$, etc.
$(i, M)$ is a surjection for any ideal inclusion $i : I → R$.
$M$ is a direct summand for $∏ (_RR)^+$, where $(-)^+ = (-, ℚ / ℤ )_{𝐀𝐛}$.

$M$ is flat iff the following equivalent definitions hold.

$M ⊗ i$ is injective for arbitrary $i ∈ \mathcal{I}$, where $\mathcal{I}$ can be either
1. all inclusions of submodule;
2. all inclusions of an ideal;
3. all inclusions of two f.g. submodules;
4. all inclusions of a f.g. ideal;
  - we remark that $M ⊗ I ↪ M ⊗ R$ is injective iff it is isomorphic to its image, that is, $M ⊗ I ≃ MI$. As $M ⊗ I ↠ MI$, this is equivalent to $M ⊗ I → MI$ being injective.
$\mathrm{Tor}_{≥ 1}(M, -) = 0$, or $\mathrm{Tor}_1 (M, -) = 0$;
For any matrix equation $𝐑 ⋅ 𝐦 = 𝐎$, there is some $𝐑$-matrix $R$ such that $𝐑 ∈ \ker (⋅ 𝐀)$ and $𝐦 ∈ \operatorname{im}(𝐀 ⋅ )$.
Any morphism from f.p. module to $M$ factors through some f.g. projective module.
$(X,-)$ lifts all surjections towards $M$, iff $X$ is f.p..
$M$ is a filtered colimit of f.p. modules.

$M$ is f.g. projective iff the following equivalent definitions hold.

$M$ is a summand of some $R^n$;
$M$ is projective, and $(M, -)$ preserves coproducts;
$(M, -)$ preserves all colimits;
$M ⊗ -$ preserves all limits.

We commence with projective modules.

Projective modules (defined by summand of free modules) is a coproduct of countably generated projective modules.

See examples and propositions in this subsection.

$P$ is a projective (summand of free), iff there exists a subset $\{(e_i, f_i)\}_{i ∈ I} ⊆ P × (P, R)$ such that any $x ∈ R$ is a finite sum of some $∑_{i ∈ I_0} e_i ⋅ f_i (x)$. We call this projective basis.

When … any $x ∈ R$ is a UNIQUE finite sum of … holds, $P$ is a free module.

When there exists a projective basis, we see the retract

$$\begin{equation} P \xrightarrow {p \ ↦ \ (e_i ⋅ f_i (p))_{i ∈ I}} R^{(I)} \xrightarrow {(r^i)_{i ∈ I}\ ↦ \ ∑ e_i ⋅ r^i} P. \end{equation}$$

Conversely, any retract $P ↪ R^{(J)} ↠ P$ yields such dual basis.

Show the rest of the equivalent definitions of projective modules. The proof is found in most textbooks.

Now we show equivalent definitions of injectives.

(Baer’s criterion). $M_R$ is injective iff any morphism from the ideal $I_R$ to $M_R$ extends along $I_R ↪ R_R$.

We show $(←)$ only. It suffices to check the lifting property for $L_R ↪ N_R$. Set

$$\begin{equation} \mathcal{S} := \{K ∣ L ⊆ K ⊆ N, \ (K, M) ↠ (L,M)\}. \end{equation}$$

$\mathcal{S}$ is non-empty ($L ∈ \mathcal{S}$), and closed under the union of any ascending chain. Hence $\mathcal{S}$ has a maximal element $Q$ by Zorn’s lemma.
We show $Q = N$. Assume there is some $n ∈ (Q ∖ N)$, then $I := \{a ∣ na ∈ Q\}$ is right ideal. In this case, any $φ : Q → M$ extends to

We see that $(n ⋅ R) + Q ∈ \mathcal{S}$ is greater thatn $Q$, a contradiction!

Injective modules $I_R$ are summands of the product $∏ (R_R)^+$.

Consider any surjection from projective module $p : (_RR)^{(I_R)^+} ↠ (I_R)^+$. We show in the subsection of characteristic modules that

$(\text{flat})^+$ is are injective modules, and
$\text{(surjection)}^+$ are injections, and
the evaluation $(-) → (-)^{++},\quad ? ↦ [f ↦ f(?)]$ is an injective natural transformation.

Hence $I_R ↪ I_R^{++} ↪ ((_RR)^{(I_R)^+})^+$. In particular, $I_R$ is a summand of $∏ _{(I_R)^+}({}_RR)^+$.

Show the rest of the equivalent definitions of injective modules. The proof is found in most textbooks.

Now it comes to the tricky stuff called flat modules. We show the following equivalent definitions (Baer’s criterion).

(Baer’s criterion). $M$ is flat iff $M ⊗ i$ is injective for arbitrary $i ∈ \mathcal{I}$, where $\mathcal{I}$ can be either

all inclusions of submodule;
all inclusions of two f.g. submodules;
all inclusions of an ideal;
all inclusions of a f.g. ideal.

We remark that $M ⊗ I ↪ M ⊗ R$ is injective iff it is isomorphic to its image, that is, $M ⊗ I ≃ MI$. As $M ⊗ I ↠ MI$, this is equivalent to $M ⊗ I → MI$ being injective.

(1 → 3). is clear. We show (1 ↔ 2), (3 ↔ 4), and (3 → 2) in steps.
(1 ↔ 2, and 3 ↔ 4). We show $2 → 1$ only. To check whether $∑ x_i ⊗ m_i$ and $∑ x_i' ⊗ m_i'$ has the same image under $i ⊗ 1_M$, it suffices to check over the f.g. submodules $⟨\{x_i\} ∪ \{x_i'\}⟩ → ⟨\{i(x_i)\} ∪ \{i(x_i')\}⟩$.

In short, to verify $1$, it suffices to verify a collection of $2$.

(3 → 2). We show $i ⊗ 1_M$ is injective for any inclusion of f.g. submodules $X ⊆ Y$. WLOG we assume $Y = X + ⟨x_0⟩$, hence $Y/X ≃ A/I$, and there is a lifting ($A$ is projective)

We obtain the commutative diagram by pulling back two epimorphisms:

Note that the middle row splits as $A ↠ A/I$ factors through $Y ↠ A/I$. Applying $M ⊗ -$ we obtain

Note that $H^{k-1}(\operatorname{cok}(g^∙)) ≃ H^{k+1}(\ker (g^∙))$ when $f^∙$ is a morphism between exact complexes, $M ⊗ i$ is an injection.

We show computational properties of flat modules.

$M$ is flat, iff for any matrix equation $𝐑 ⋅ 𝐦 = 𝐎$, there is some $R$-matrix $𝐀$ such that $𝐑 ∈ \ker (⋅ 𝐀)$ and $𝐦 ∈ \operatorname{im}(𝐀 ⋅ )$.

We use left modules here for convention of linear algebra.

When $M$ is flat, the ses $0 → \ker (𝐑 ⋅ ) → R^m \xrightarrow {𝐑 ⋅ } R^n → 0$ is $(M ⊗ -)$-exact:

$$\begin{equation} 0 → \ker (𝐑 ⋅ ) ⊗ M → M ^m \xrightarrow {𝐑 ⋅ } M^n → 0. \end{equation}$$

By $\ker = \operatorname{im}$, $𝐦 ∈ \ker (𝐑 ⋅)$ iff it is of the form $𝐀 ⋅ 𝐦'$ for some $𝐑 ⋅ 𝐀 = 𝐎$.
Conversely, if the matrix equation $𝐑 ⋅ 𝐦 = 𝐎$ splits by some $𝐀$, we shall show the flatness of $M$. By Baer’s criterion, it suffices to show $I ⊗ M ↪ IM$ for f.g. ideals. Suppose there exists $∑ a_i ⊗ m_i$ with zero image. Then we have $𝐚^T ⋅ 𝐦 = 0$. Now there is $𝐀$ s.t. $𝐚^T ⋅ 𝐀 = \mathbf 0 ^T$ and $𝐦 = 𝐀 ⋅ 𝐦 '$ for some $𝐦 '$. Hence,

$$\begin{equation} ∑ a_i ⊗ m_i = ∑ a_i ⊗ A_{i,j} m'_j = ∑ (a_i A_{i,j}) ⊗ m'_j = 0. \end{equation}$$

We are not satisfied with the above lemma, which is not homological enough.

$M$ is flat, iff any morphism from f.p. module to $M$ factors through some f.g. projective module.

We show ($→$). When $M$ is flat, we take arbitrary f.p. module $R^m \xrightarrow {𝐁 ⋅ } R^n ↠ X$. A morphism $φ : X → M$ is characteristed by

a set $\{[e_i] ↦ m_i\}_{i=1}^n$, where $[e_i]$ is the image of $e_i ∈ R^n$ in $X$.

By $0 → (X, M) \xrightarrow{⊆} M^n → M^m$, the morphism $φ : X → M$ is characterised by the column vector $𝐦 := (m_i)_{i=1}^n$, such that $(𝐁 ^T)_{m × n} ⋅ 𝐦 = \mathbf 0$. By flatness, there is $(𝐀^T)_{n × l}$ and $(𝐦') ∈ M^l$ such that $(𝐁 ^T)_{m × n} ⋅ (𝐀 ^T)_{n × l} = 𝐎$ and $(𝐀 ^T)_{n × l} ⋅ 𝐦 ' = 𝐦$. Now $𝐀 ⋅ : R^n → R^l$ is our desired factorisation.
The converse is nothing but a backwise verification; it is clear if one shows that flat modules are filtered colimits of f.g. projective modules.

In fact, we deduce an exact sequence with syzygy

$M$ is flat, iff exactly for f.p. modules $X$, $(X,-)$ lifts all surjections towards $M$.

Let $p : F ↠ M$ be a free precover. $X$ lifts all surjections towards $M$ iff $(X,p)$ is surjective. The rest is clear.

Demonstrate that a morphism $M → F$ from a f.p. module $M$ to a flat module $F$ factors through some f.g. projective module $P(M)$. Now, express $F$ as a filtered colimit of f.p. modules, i.e., $\varinjlim^{fil} M_∙ ≃ F$. Our construction shows $P(M_∙)$ is a subsystem of $M_∙$. Demonstrate that $\varinjlim ^{fil} P(M_∙) ≃ F$ via cofinality.

Flat modules are filtered colimits of f.g. free modules.

We show filtered colimits of f.g. projective modules are flat ones. Clearly, $(\varinjlim^{fil}P_∙) ⊗ -$ is an exact functor. The converse is shown above.

The following theorem is a characterization of f.g. projective modules, easy but functorial.

$M$ is f.g. projective iff the following equivalent definitions holds.

$M$ is a summand of some $R^n$;
$M$ is projective, and $(M, -)$ preserves coproducts;
$(M, -)$ preserves all colimits;
$M ⊗ -$ preserves all limits.

Clearly 1. implies the others.
(2 → 1). This is clear by criterion of FAKE f.g. modules; we provide a straightforward and down-to-earth proof instead. There exists some $N$ such that $R^{(λ)} = M ⊕ N$ is free. In particular, $M$ is f.g. iff $λ$ can be finite. Since taking summand preserves the additive natural isomorphism, it suffices to show

$(R^{(λ)}, -)$ is not $∐$-preserving when $λ$ is infinite.

We consider $(R^{(λ)}, R)^{(λ)} ↪ (R^{(λ)}, R^{(λ)}) ∋ 1_{R^{(λ)}}$, or equivalently, $∐_λ ∏_λ R ↪ ∏_λ ∐_λ R ∋ 1_R$ by tracking the image of the free basis.

$∐_λ ∏_λ R ⊆ R^{λ × λ}$ consists of $λ × λ$ matrices with finite many columns;
$∏_λ ∐_λ R ⊆ R^{λ × λ}$ consists of $λ × λ$ matrices with all finite length columns.

The diagonal matrix $1_{∏ R} ∈ R^{λ × λ}$ lies in $∏_λ ∐_λ R$, but not in $∐_λ ∏_λ R$,
(3 → 1). $(M, -)$ is an exact functor as it preserves finite limits. Hence $M$ is projective. By 2., $M$ has to be f.g. projective.
(4 → 1). $(M ⊗ -)$ is an exact functor as it preserves finite limits. Hence $M$ is flat. $M$ is also f.p. as $M ⊗ -$ preserves products. By forthcoming techniques of character modules, f.p. flat modules are f.g. projective.

In particular, f.g. projective modules are precisely those $M$ such that $M ⊗ -$ and $(M, -)$ preserves all limits and colimits.

Coherency

Inducting from f.g. and f.p., it is natural to come up with the ideal of projection length. We show the following atomatic propositions over the ses

$$\begin{equation} 0 → K → X → Y → 0. \end{equation}$$

When $K$ and $Y$ are both f.g. (resp. f.p.), then so is $X$.

By horseshoe lemma.

When $X$ is f.g., then $Y$ is f.g..

Trivial.

When $X$ is f.p. and $K$ is f.g., then $Y$ is f.p..

We prove without taking elements. Taking projective resolutions of $K$ and $X$ yields the chain map $ϕ : P_∙ (K) → P_∙ (X)$. We shall construct $P_∙(Y)$ by mapping cone.

Now $Y ≃ \operatorname{cok}(ϕ ) ≃ \operatorname{cok}[H_0(P_∙ (K)) \xrightarrow{H_0(P_∙ (ϕ ))} H_0(P_∙ (X))]$. This is exactly $\operatorname{cok}(ϕ _0, d)$ by long exact sequence.

When $X$ is f.g., $Y$ is f.p., then $K$ is f.g.. Hint: by mapping coCone.

These shows that

if $X$ f.g., then $Y$ f.g.;
if $K$ f.g., $Y$ f.g., then $X$ f.g.;
if $X$ f.p., $Y$ f.g., then $K$ f.g.;
if $K$ f.g., $X$ f.p., then $Y$ f.p.;
if $K$ f.p., $Y$ f.p., then $X$ f.p.;
induction …

Show that the full subcategory $\{M ∈ 𝐌𝐨𝐝_R ∣ M \ \text{f.g.}\}$ is Abelian, iff $R$ is Noetherian.

In particular, it is a Serre subcategory which is closed under subobjects and quotient objects.

In particular, $𝐦𝐨𝐝_R$ is not closed under kernel in general. To fix this, we may require f.p. is automatically $3$-term f.p. (and thus $∞$-f.p.). This comes with coherency.

$R$ is (right) coherent provided (right) flat $R$-modules are closed under arbitrary products.

TFAE:

$\mathrm{flat}(\mathrm{Mod}_R)$ is closed under $\prod$,
$\prod _{\lambda \in \Lambda}R_R=:R^\Lambda$ is flat,
every f.g. submodule (left ideal) of ${}_RR$ is f.p.,
the category consists of f.p. modules $\mathrm{mod}_R$ is Abelian,
f.p. modules are $\infty$-presented.

We show ($1 \to 2 \to 3 \to 1$) with a set-theoritic analysis; and $(3\to 4 \to 5 \to 4)$ with a homological analysis.

($1 \to 2$) is clear.

($2 \to 3$). For any f.g. left ideal $I$ and index set $\Lambda$, we show $I^\Lambda\simeq R ^\Lambda\otimes I$ (recall the equivalent definitions of f.p. modules). The flatness shows injection, and the f.g. shows surjection.

($3 \to 1$). We show f.g. $\iota: _RI\hookrightarrow R$ is preserved by $R^\Lambda$ via Baer’s criterion. Since f.p. = f.g. (by assumption), $\iota \otimes -$ commutes with $\prod$.

($3\to 4$). Note that f.p. modules are closed under extensions and cokernels. For ses $0 \to K \to X \to Y \to 0$ with $X$ and $Y$ f.p., we have $K$ f.g. in general. By horseshoe lemma, we obtain ses $0 \to \Omega_1 (K) \to \Omega_1 (X) \to \Omega_1 (Y) \to 0$ for $\Omega _1(X)$ and $\Omega_1 (Y)$ f.g.; by 3., $\Omega _1 (Y)$ and $\Omega _1(X)$ are f.p., hence $\Omega _1(K)$ is f.g.. This shows $K$ f.p..

($4 \to 5$). By definition, we can choose the first syzygy of any f.p. module as some f.p. modules.

($5 \to 3$). When $I \hookrightarrow R$ is f.g. ideal, $R/I$ is f.p., thus is $\infty$-presentable by assumption. Now $I$ is also $\infty$-presentable.

We gives an example of coherent rings which is not Noetherian.

The ring $R = ℤ [X_1, X_2, \ldots]$ is coherent but not Noetherian. The trivial representation $ℤ = R / (x_1, x_2, \ldots)$ is constructed by quotienting a $∞$-generated ideal. Hence $ℤ$ is f.g. rather than f.p.. To construct a finite presentation of a certian f.g. ideal, we adopt the same construction for that over $ℤ [x_1, \ldots , x_N]$ ($N ≫ 0$).

This example shows that the property of coherency preserved by some filtered colimits. Coherency is related to cofinality, rather than its actual size.

Show that the category of f.g. free modules $𝐟𝐫𝐞𝐞 _R$ has weak kernel, iff $R$ is coherent.

$A \xrightarrow i B$ is a weak kernel of $B \xrightarrow p C$, provided $\ker \mathrm{Hom}(-, p) = \operatorname{im} \mathrm{Hom}(-, i)$.

A right coherent ring shows

the interchanging of $\prod$ and flat objects;
the $\infty$-presentability of f.p. objects.

Left hereditary rings are right coherent (the analysis).

Character Module

The character module rises in the standard injective resolution $ℤ ↪ ℚ ↠ ℚ / ℤ$. This yields a functor functor $(-)^+ := \mathrm{Ext}^1_ℤ (-, ℤ) := (-, \mathbb Q / \mathbb Z)_{\mathbb Z}$, aka Pontryagin dual. Note that this functor is

faithful (as $\mathbb Q / \mathbb Z$ is a cogenerator) and
exact (as $\mathbb Q / \mathbb Z$ is injective).

Consequently, it preserves and reflects zero objects and exactness. In particular, $((-)^+)^{\mathrm{op}}$ preserves and reflects monomorphisms and epimorphisms.

We shall see this functor suprisingly transfers and reflects from injective modules to flat modules; the dual procedure applies for Noetherian rings.

We demonstrate that f.p. flat modules are projective via $(-)^+$. Observe that

the natural transformation $? ⊗ (-)^+ → (?, -)^+$ is isomorphic when $?$ is f.g. free. Note that $? ⊗ (M)^+ → (?, M)^+$ are right exact functors, the isomorphism holds when $?$ is f.p..

If $F$ is flat, then $F ⊗ (-)^+ ≃ (F ⊗ -)^+ ≃ (F,-)^+$ is exact. Since $(-)^+$ reflects exactness, it follows that $F$ is projective.

There are examples that f.g. flat modules are not projectve. An example is given in stacks project. A forthcoming example is the unique principal ideal of the vRN:

$$\begin{equation} I = \{φ ∈ \mathrm{End}_𝔽 (𝔽^{(ω)}) ∣ \dim _𝔽 \operatorname{im}(φ ) < ∞\} ↪ \mathrm{End}_𝔽 (𝔽^{(ω)}). \end{equation}$$

See discussion of von Neumann Ring.

Demonstrate that $F$ is flat if and only if $F^+$ is injective. In general:

$f.d. Y = i.d. Y^+$. The flat (respectively, injective) dimension is considered over $𝐌𝐨𝐝_R$ (respectively, ${}_R𝐌𝐨𝐝$).

Let $P(X)$ be a projective resolution of $X$. Then $(Y \otimes P^\bullet (X))^+ \simeq (P^\bullet (X), Y^+)$. Taking $H^\bullet$ groups yields $(\mathrm{Tor}_k(Y,X))^+ \simeq \mathrm{Ext}^k (X,Y^+)$. Note that $\mathrm{Tor}_d(Y,-)^+$ vanishes if and only if $\mathrm{Ext}^k(-,Y^+)$ does.

One may enquire whether $i.d. F \overset ? = f.d. F^+$.

Show the isomorphism $\mathrm{Tor}_k (M, N^+) ≃ (\mathrm{Ext}^k (M,N))^+$ for all $N$ and f.p. $M$.

Hint: recall that f.p. modules admit resolutions (finite or infinite) consisting of f.g. projective modules.
The difficulty arises with $(\mathrm{Ext}^k (\varinjlim ^{fil }M_∙ ,N))^+ ≇ \varinjlim^{fil}(\mathrm{Ext}^k (M_∙,N))^+$ in proving $i.d. F = f.d. F^+$. In fact, coherency is insufficient to establish such an isomorphism.

When $R$ is right coherent, we have $i.d. M_R ≥ f.d. {}_R(M^+)$.

We show that $M^+$ is flat when $M$ is injective.

Recall that $M^+$ is flat if and only if $M^+ ⊗ i$ is injective for any $i : I ↪ R$, the inclusion of f.g. ideal. By coherency, $I$ is f.p., hence $M^+ ⊗ I ≃ (I, M)^+$. Thus, $M^+ ⊗ i$ is injective if and only if $(i,M)^+$ is injective, which holds if and only if $(i, M)$ is surjective.

Suppose $θ$ is an injective resolution of $M$ of length $d$. Then $θ ^+$ is a flat resolution of $M^+$ of length $d$. This shows $f.d. M^+ ≤ i.d. M$.

We observe that Baer’s criterion for injective modules is verified over all ideals; the criterion for flat modules is verified over all f.g. ideals. Hence,

$f.d. {}_R(M^+) = i.d. (M_R)$ holds for right Noetherian rings.

Set $R = ℝ [ℝ]$, the polynomial ring over indeterminates $\{x_t\}_{t ∈ ℝ}$. Show that $ℝ [ℕ]$ lifts f.g. ideals; whereas it is not an injective module.

In general, one can show that $f.d. M ≤ f.d. M^{++}$. Hence, $$\begin{equation} f.d. {}_RM = i.d. (M^+)_R ≥ f.d. {}_R(M^{++}) ≥ f.d. {}_R M. \end{equation}$$ over right coherent rings.

Show that

evaluation $e_M: M → M^{++}, m ↦ [f ↦ f(m)]$ is injective.
Moreover, $M^+ \xrightarrow {e_{M^+}} M^{+++} \xrightarrow {(e_{M})^+}M^+$ is identical. Hence $M^+$ is a summand of $M^{+++}$.
In this case, $e_M ⊗ 1_X$ is injective for arbitrary $X$. Hint: equiv. to verify $(e_M ⊗ 1_X)^+ ≅ (1_X, (e_M)^+)$ is surjection, which is split epi!

$f.d. M ≤ f.d. M^{++}$.

WLOG assume $f.d. M^{++} = k$. For any flat resolution of $N$, we take homological truncation

$$\begin{equation} T_k^∙ := (0 → Ω _{k+1}(N) \overset ι → F_{k} (N) → \cdots → F_0(N) → N → 0). \end{equation}$$

Since $\mathrm{Tor}_{k+1} (M^{++}, N) = H_{k+1} (M ^{++} ⊗ T_k ^∙) = 0$, $M^{++} ⊗ ι$ is injective. The diagram

shows that $M ⊗ ι$ is again injective. Equivalently, $\mathrm{Tor}_{k+1}(M, N) = 0$. This shows that if $\mathrm{Tor}_∙ (M^{++}, -)$ vanishes at $(k, N)$, then so does $\mathrm{Tor}_{∙}(M, -)$. Hence, $f.d. M ≤ f.d. M^{++}$.

On Chain Complexes

Functors for Chain Complexes

(Chain complexes). We unfold $(X, d)$ into

Here $Z → X → ΣB$, $B → X → C$, $B → Z → H$ and $H → C → Σ B$ are ses (short exact sequences). The $ZXHC$ squares are both pushouts and pullbacks.

The functors $Z$, $B$, $C$ and $H$ preserve $∐$, $∏$, and $\varinjlim^{fil}$.

All $Z$, $B$, $C$ and $H$ are defined by kernels and cokernels, which are preserved by exact functors.

The adjunctions confer truncations and $C$, $Z$ with favourable properties. We readily verify the necessary aspects, thereby eschewing the deeper structures (e.g., Dold-Kan correspondence).

We set

$(-)^k : C(R) → 𝐌𝐨𝐝 _R, \quad X ↦ X^k$,
$\overline {(?)}: 𝐌𝐨𝐝 _R → C(R),\quad M ↦ [\cdots → 0 → \overset{\deg = 0} M = \overset{\deg = 1}M → 0 → \cdots]$;
$\underline {(?)} : 𝐌𝐨𝐝 _R → C(R), \quad M ↦ [\cdots → 0 → \overset{\deg = 0} M → 0 → \cdots]$.

Show the adjoint triple $(-)^1 ⊣ \overline {(-)} ⊣ (-)^0$, and extend it to an adjoint $∞$-ple.

Show that $C^0 ⊣ (-) ⊣ Z^0$. We shall derive these functors later.

Resolving Chain Complexes

We analysis the projectives and injectives.

$C(R)$ has enough projectives, consisting of null-homotopic complexes with projective components.

$⊇$ is clear. Conversely, we take projective object $P ∈ C(R)$, and find that

$P$ has projective components, since it lifts epimorphisms of stalk complexes;
$P$ is null-homotopic, since $0 → P → \mathrm{Cone}({1_P}) → Σ P → 0$ splits, yielding $1_P$ is null-homotopic.

$C(R)$ has enough injectives, consisting of null-homotopic complexes with injective components.

Demonstrate that acyclic projective complexes are not necessarily projective. Hint: Let $R := k[x] / (x^2)$, and consider $X := [\cdots \to R \xrightarrow{\,\cdot x\,} R \xrightarrow{\,\cdot x\,} R \to \cdots]$, which is acyclic and projective; however, $X$ is not null-homotopic, as $X \otimes (x)$ is not acyclic.

The adjunction $∞$-ple extends to $\mathrm{Ext}$-groups. For instance, $\mathrm{Ext}^k_R((-)^1, ?) ≃ \mathrm{Ext}^k_{C(R)}(-, \overline ?)$, and $\mathrm{Ext}^k_R(-, (?)^0) ≃ \mathrm{Ext}^k_{C(R)}(\overline -, ?)$.

All these functors preserves projective and injective resolutions.

The resolutions in related to $\underline {(-)}$ functors are much more complicated, since $\underline {(-)}$ does not preserve projectives and injectives in general. We begin with an easy example.

$Z^0(-) ≃ (\underline R, -)$, and $\mathrm R^k (Z^0) = H^k$ for $k ≥ 1$.

We choose the projective resolution of $\underline R$ as

$$\begin{equation} P^∙ (\underline R) := \cdots → Σ ^k \overline R → \cdots → Σ \overline R → \overline R ↠ \underline R → 0, \end{equation}$$

where $Ω ^k (\underline R) = Σ ^k (\underline R)$. We see that

$$\begin{equation} (P^∙(\underline R), X) = \cdots → 0 → X^0 \xrightarrow {d^0} X^1 \xrightarrow {d^1} X^2 → \cdots, \end{equation}$$

Taking $H^∙$ yields the desired results.

The result generalises to Abelian category with enough projectives.

Assume $\mathcal{A}$ has enough projectives. $\mathrm R^k (Z^0) = H^k$ for $k ≥ 1$.

We take Eilenberg-Cartan resolution of $X$, i.e., firstly take injective resolutions for $H$-groups and $B$-groups, and then construct the resolutions for $Z$, $C$ and $X$ in steps with horseshoe lemma. The resulting complex looks like

We denote this by $X ↪ \widetilde{I^0} → \widetilde{I^1} → \cdots$, and remend it into an injective resolutions by

$I^0$ is injective with $I^0(B^k ⊕ B^{k+1}) ⊕ I^0 (H^k ⊕ H^{k+1})$ on the $k$-th component,
$I^1$ is injective with $I^1(B^k ⊕ B^{k+1}) ⊕ I^1 (H^k ⊕ H^{k+1}) ⊕ I^0 (H^{k+1} ⊕ H^{k+2})$ on the $k$-th component,
$I^n$ is injective with $I^n(B^k ⊕ B^{k+1}) ⊕ ⨁_{0 ≤ s ≤ n} I^{n-s} (H^{k+s} ⊕ H^{k+s+1})$ on the $k$-th component,and so on.

Applying $Z^k(-)$ at $I^0 → I^1 → \cdots$, we obtain

The rows are injective resolutions of $Z^k$, $H^{k+1}$, etc. The cohomomology groups are computed as $Z^k$, $H^{k+1}$, etc.

It is noteworthy that the preceding computation admits an $\mathcal{X}$-relative version, provided that there exist sufficiently many monomorphisms from $\mathcal{A}$ to $\mathcal{X}$. Observe that a relative version of the horseshoe lemma is available, and consequently, a relative version of Eilenberg-Cartan resolutions is obtained.

Verify that the resolutions remain diagonal during the above procedure.

The adjoint $∞$-ple $\cdots ⊣ (-)^1 ⊣ \overline {(-)} ⊣ (-)^0⊣ \cdots$ extends to $\cdots ⊣ L ⊣ U ⊣ R ⊣ \cdots$, wherein the forgetful functor $U : C(R) \to 𝐆𝐫 (R)$ maps a chain complex to its underlying graded module. This construction demonstrates that $K(R)$ is a Frobenius exact category, with a significantly simpler verification. Refer to the generalisation concerning $N$-chain complexes.

Closed Monoidal Structure

We show that over the commutative ring $R$, $C(R)$ is symmetric closed monoidal.

(Tensor product). We define $X ^∙ ⊗ Y ^ ∙$ as an object in $C(R)$ by

$∐_{p+q = n} X^p ⊗ Y^q$ as the $n$-th component,
the differential $d_{X ⊗ Y} = (1_X ⊗ d_Y + d_X ⊗ 1_Y)$ satisfies Koszul sign rule ($(f ⊗ g) (x ⊗ y) = (-1)^{\deg g ⋅ \deg x}(fx ⊗ gy)$), i.e., $d_{X ⊗ Y} (x ⊗ y) = d_X (x) ⊗ y + (-1)^{\deg x} x ⊗ d_Y (y)$ for homogeneous elements $x$ and $y$.

Show the natural isomorphism $α _{X,Y,Z} : X ⊗ (Y ⊗ Z) \xrightarrow ∼ (X ⊗ Y) ⊗ Z$ as well as the pentagon identity. Hint: Koszul’s sign rule is compatible with associativity.

Show that $\underline R$ is unital element, i.e., there are natural isomorphisms $X ⊗ \underline R \xrightarrow ∼ X \xleftarrow ∼ \underline R ⊗ X$ satisfying the triangle identities.

The morphism $B _{X,Y} : X ⊗ Y \xrightarrow ∼ Y ⊗ X$ characterised by

$$\begin{equation} X^p ⊗ Y^q → (-1) ^{pq} ⋅ Y^q ⊗ X^p,\quad x ⊗ y ↦ (-1)^{pq} y ⊗ x \end{equation}$$

is a natural isomorphism satisfying the hexagon identity.

We take $τ : a ⊗ b ↦ b ⊗ a$ for modules, and wish there exist a some $ε : ℤ × ℤ → \{± 1\}$ along with the commutative diagram

We see from the above diagram that $ε (p,q) = (-1)^{pq}$ is a solution:

We finally show hexagon identity:

Observe that $(-1)^{pq} ⋅ (-1)^{qr} ⋅ (-1)^{rp} = 1$ as we keeping track of the signs at $X^p ⊗ Y^q ⊗ Z^r$.

We finally show the natural right adjoint of $X ⊗ -$ by construction.

($\mathcal{HOM}(-, ?)$). We define $\mathcal{HOM}(X, Y)$ as an object in $C(R)$ by

$(\mathcal{HOM}(X,Y))^p$ consists of morphisms in $(X, Y)_{\text{Graded-Module}}$ of degree $p$, i.e., homogeneous morphisms in $(X, Σ ^p Y)_{\text{Graded-Module}}$;
the differential for a homogeneous morphism $f$ is $[d,f] := d_Y ∘ f - (-1)^{\deg f} f∘ d_X$.

Clearly,

$Z^0(\mathcal{HOM}(X,Y)) = (X,Y)_{C(R)}$;
$B^0(\mathcal{HOM}(X,Y)) ⊆ (X,Y)_{C(R)}$ consists of null-homotopic morphisms;
$H^0(\mathcal{HOM}(X,Y)) = (X,Y)_{K(R)}$.

The adjunction $(X ⊗ -) ⊣ \mathcal{HOM}(X, -)$ is natural in $X$.

We show $\mathcal{HOM}(X ⊗ Y, Z) ≃ \mathcal{HOM}(X, \mathcal{HOM}(Y,Z))$ is natural in $X$, $Y$ and $Z$. Taking $Z^0$ yields the adjunction. Termwisely, we take $\deg f = k$ and find that

$$\begin{aligned} 𝐋𝐇𝐒 ∋ f & = \{∐_p X^p ⊗ Y^{n-p} → Z^{n+k}\}_{n ∈ ℤ}\\[6pt] (f_{n,p})& ≅ \{X^p ⊗ Y^{n-p} → Z^{n+k}\}_{p, n ∈ ℤ}\\[6pt] (f'_{p,n})& ≅ \{X^p → (Y^{n-p} , Z^{n+k})\}_{p, n ∈ ℤ}\\[6pt] & ≅ \{X^p → ∏_n (Y^{n-p} , Z^{n+k})\}_{p ∈ ℤ}\quad = f' ∈ 𝐑𝐇𝐒. \end{aligned}$$

Here $f_{n,p} ↔ f'_{p,n}$ are of degree $k$. We finally verify differentials via the identification

$$\begin{equation} f' ∈ (X^{-p}, (Y^{-q}, Z^r)) ≃ (X^{-p} ⊗ Y^{-q}, Z^r) ∋ f. \end{equation}$$

The adjunction shows $f'(-) ↔ [? ↦ f(-, ?)]$ ($\deg ? = {-q}$). Hence,

$$\begin{aligned} & \ \ \quad [d_{𝐋𝐇𝐒}, f'(-)]\\[6pt] & = d_{\mathcal{HOM}(Y,Z)} ∘ f'(-) - (-1)^{p+q+r} f'(-) ∘ d_X\\[6pt] & ↔ [? ↦ [d_{\mathcal{HOM}(Y,Z)},f(-, ?)]] - (-1)^{p+q+r} [? ↦ f(-, ?) ∘ d_X]\\[6pt] & = d_Z ∘ f - (-1)^{q+r} f ∘ d_Y - (-1)^{p+q+r} f ∘ d_X\\[6pt] & = d_Z ∘ f - (-1)^{p+q+r} (f ∘ d_Y + (-1)^p f ∘ d_X)\\[6pt] & = d_Z ∘ f - (-1)^{p+q+r} (f ∘ d_{X ⊗ Y})\\[6pt] & = [d_{𝐑𝐇𝐒}, f] \end{aligned}$$

Show by Yoneda lemma that, one has natural isomorphisms

$$\begin{equation} \mathcal{HOM}(\varinjlim X_∙, Y) ≃ \varprojlim \mathcal{HOM}( X_∙, Y); \quad \mathcal{HOM}(X, \varprojlim Y_∙) ≃ \varprojlim \mathcal{HOM}(X, Y_∙). \end{equation}$$

Show that $(X, ∏ Y_∙)_{\text{null-ho}} ≃ ∏ (X,Y_∙)_{\text{null-ho}}$, and $(∐ X_∙, Y)_{\text{null-ho}} ≃ ∏ (X_∙, Y)_{\text{null-ho}}$.

Show that $\mathcal{HOM}(X, \mathrm{Cone}(f)) ≃ \mathrm{Cone}(\mathcal{HOM}(X, f))$, and $\mathcal{HOM}(Σ ^{-1}\mathrm{Cone}(f), A) ≃ \mathrm{Cone}(\mathcal{HOM}(f, A))$

There is also an homotopical $\mathrm{Ext}$ functor.

($\mathrm{Ext}_{dw}^∙$). We define the bifunctors

$$\begin{equation} \mathrm{Ext}_{dw}^∙ : C(R)^{\mathrm{op}} × C(R) → 𝐌𝐨𝐝 _R,\quad \end{equation}$$

sending the pair $(X, Y)$ to $H^k (\mathcal {HOM}(X,Y)) = (X,Σ ^k Y)_K$.

Here $(-)_{dw}$ means degreewise split.

Show that $\mathrm{Ext}^k_{dw}$ is a subfunctor of usual $\mathrm{Ext}^k$. Hint: it suffices to show for $k = 1$, by dimension shift. The distinguished triangle of cone embedding

$$\begin{equation} Σ ^{-1}X \xrightarrow f Y ↪ \mathrm{Cone}(f) ↠ X \end{equation}$$

gives the correspondence (by TR1) $[f] ↔ [\mathrm{Cone}(f)]$. Here

$[f]$ runs through $H^0(\mathcal{HOM}(Σ^{-1}X, Y)) = H^1(\mathcal{HOM}(X, Y))$;
$\mathrm{Cone}(f)$ identifies termwise split ses in $\mathrm{Ext}^1(X,Y)$.

More on $\mathcal{HOM}$

Thanks to closed monoidal structure, we obtain the following canonical morphisms without verification.

(Composition). There is a morphism extended from composition of $\mathrm{Hom}_{C(R)}(-, ?)$:

$$\begin{equation} φ : \mathcal{HOM}(Z,Y) ⊗ \mathcal{HOM}(Y,X) → \mathcal{HOM}(Z,X). \end{equation}$$

In view of closed monoidal structure, the morphism comes from

In element, let $g ∈ \mathcal{HOM}(Y,Z)^q$ and $f ∈ \mathcal{HOM}(X,Y)^p$, then $(gf) ∈ \mathcal{HOM}(X,Z)^{p+q}$. One can verify the compatibility of differential:

$$\begin{aligned} d_{𝐋𝐇𝐒 }(g ⊗ f) & = [d,g] ⊗ f + (-1)^{|g|} g ⊗ [d,f]\\ & = (dg) ⊗ f - (-1)^{|g|} (gd)⊗ f + (-1)^{|g|} g⊗ (df) - (-1)^{|f|+|g|} g⊗ (fd) \\ & \overset{φ } → (dg) f - (-1)^{|g|} (gd)f + (-1)^{|g|} g (df) - (-1)^{|f|+|g|} g (fd) \\ & = d (gf) - (-1)^{|gf|} (gf) d\qquad = d_{𝐑𝐇𝐒 }(gf). \end{aligned}$$

Show the (canonical) natural transformation

$$\begin{aligned} A ⊗ \mathcal{HOM}(B,C) &→ \mathcal{HOM}(B, A ⊗ C), \\[6pt] ∐\limits_{p+t=n} ∏\limits_{r-q=t} A^p ⊗ (B^q , C^r) &↦ ∏\limits_{s-q = n} ∐\limits_{p+r=s} (B^q , A^p ⊗ C^r). \end{aligned}$$

It is an isomorphism, iff finiteness condition is satisfied, while either

$B^∙$ are f.g. projective modules, or
$B^∙$’s are f.p., and $A^∙$ are flat

holds for each $(p,q,r)$. In particular, the isomorphism holds when $B$ is bounded with f.g. projective terms.

Show the (canonical) natural transformation

$$\begin{aligned} A ⊗ \mathcal{HOM}(B,C) &→ \mathcal{HOM}(\mathcal{HOM}(A, B), C), \\[6pt] ∐\limits_{p+t=n} ∏\limits_{r-q=t} A^p ⊗ (B^q , C^r) &↦ ∏\limits_{r-s = n} ∏\limits_{q-p = s} ((A^p, B^q) , C^r). \end{aligned}$$

It is an isomorphism, iff finiteness condition is satisfied, while either

$B^∙$ are f.g. projective modules, or
$B^∙$’s are f.p., and $C^∙$ are injective

holds for each $(p,q,r)$. In particular, the isomorphism holds when $B$ is bounded with f.g. projective terms.

There are some simplified forms.

The evaluation takes the form

$$\begin{equation} \mathcal{HOM}(A,B) ⊗ A → B,\quad f ⊗ a ↦ f(a). \end{equation}$$

It is epic when any $b ∈ B^k$ lies in the image of some $∐ A^∙ → B^k$. In short, $B^k ∈ 𝐠𝐞𝐧 (∐ A^∙)$.

The coevaluation takes the form

$$\begin{equation} A → \mathcal{HOM}(\mathcal{HOM}(A,B), B),\quad a ↦ [f ↦ f(a)] \end{equation}$$

It is monic when any $a ∈ A^n$ is detected by some $A → ∏ B^∙$. In short, $A^n ∈ 𝐜𝐨𝐠𝐞𝐧 (∏ B^∙)$.

Let $φ : A → B$ be a chain map. Then there is a induced chain map

$$\begin{equation} φ ∈ \mathcal{HOM}(A, B) → \mathcal{HOM} (\mathcal{HOM}(X,A),\mathcal{HOM}(X,B)) ∋ φ _∗. \end{equation}$$

Show that the morphism is compatible with the differential, i.e.,

$$\begin{equation} d_{\mathcal{HOM}(\mathcal{HOM}(X,A),\mathcal{HOM}(X,B))}(φ _∗) = (d_{\mathcal{HOM}(A,B)}(φ ))_∗ . \end{equation}$$

Consider

$$\begin{aligned} \{d_{𝐋𝐇𝐒}(φ _∗ )\}(α) & = [d_{𝐋𝐇𝐒}, φ _∗ ](α)\\ &= d_{\mathcal{HOM}(X,B)} ∘ φ _∗ (α) - (-1)^{|φ _∗|} φ _∗ ∘ d_{\mathcal{HOM}(X,A)} (α)\\ &= d_{\mathcal{HOM}(X,B)} (φα) - (-1)^{|φ|} φ (d_{\mathcal{HOM}(X,A)} (α))\\ &= dφα - (-1)^{|φ α|}φ α d - (-1)^{|φ|} φ (dα - (-1)^{|α |}α d)\\ &= dφα - (-1)^{|φ|} φ dα\quad = \{d_{𝐑𝐇𝐒} (φ)\}_∗ (α) . \end{aligned}$$

Show that $d(ψ ^∗) = (d (ψ ))^∗$.

Thanks to functoriality, we claim for any chain complex $K$, and chain map/null homotopy/homotopy equivalence $f : X → Y$,

$f_∗ : \mathcal{HOM}(K, X) → \mathcal{HOM}(K, Y)$ is again a chain map/null homotopy/homotopy equivalence;
$f^∗: \mathcal{HOM}(Y, K) → \mathcal{HOM}(X, K)$ is again a chain map/null homotopy/homotopy equivalence;
$f ⊗ 1 : Y ⊗ K → X ⊗ K$ is again a chain map/null homotopy/homotopy equivalence.

A Discussion on the Homotopy Category $K(R)$

We present several formal approaches to computing the homotopy category $K(R)$. Here, $R$ may be replaced by a general Abelian category.

The following categories are isomorphic; that is, they possess the same class of objects and naturally isomorphic $\mathrm{Hom}$-sets.

The $H^0(\mathcal{HOM})$-construction. (Observe that the Hom complex is defined for any Abelian category.)
The localisation $C(R) [S^{-1}]$, where $S$ comprises homotopy equivalences, i.e., morphisms $f$ for which there exists a morphism $g$ such that $f ∘ g$ and $g ∘ f$ are homotopic to the identity morphisms.
The additive quotient $C(R) / \mathcal{N}$, where $\mathcal{N}$ consists of split acyclic complexes (that is, contractible complexes).
The homotopy category $C(R) / ∼$, where $∼$ is an additive categorical equivalence relation on $\mathrm{Hom}$-sets, with $f ∼ 0$ if and only if $f$ is null-homotopic.

We first demonstrate that $C(R) / ∼$ is well-defined. Note that $s : f ∼ g$ if and only if $s ∘ h : f ∘ h ∼ g ∘ h$, and the dual statement holds as well. We denote the category by $K_i$ $(i = 1,2,3,4)$ according to the constructions above.

$(K_1 = K_4)$. This is evident.
$(K_2 = K_4)$. The quotient $C \to K_4$ maps $S$ to isomorphisms, yielding $K_2 \to K_4$. To establish the converse functor, observe that the collection of homotopies $\{(s,f,g) ∣ s : f ∼ g : X \to Y\}$ is represented by $(\mathrm{Cyl}(1_X), Y)$ via $Σ X ⊕ X ⊕ X \xrightarrow {(s,f,g)} Y$. In this context, we have
- $f = [X \xrightarrow {e_2} \mathrm{Cyl}(1_X) \xrightarrow {(s,f,g)} Y]$;
- $g = [X \xrightarrow {e_3} \mathrm{Cyl}(1_X) \xrightarrow {(s,f,g)} Y]$.
Observe that $e_2 = e_2$ in $K_2$. Thus, whenever $f = g$ in $K_4$, it follows that $f = g$ in $K_2$.
$(K_3 = K_4)$. Since $K_4$ is additive, the zero morphisms are precisely those which factor through zero objects. This yields $K_3 \to K_4$. Conversely, when $f ∼ g : X \to Y$ in $K_4$, we have $f ∼ g$ in $K_3$ as it factors through the zero object $\mathrm{Cyl}(1_X)$. Therefore, $K_4 = K_3$.

It is unnecessary to demonstrate the additivity of the localisation $C → K_2$; this is not obvious if one tries to show this directly. Here is a note on additive localisation.

The exact category $(C, \mathcal{E})$, where $\mathcal{E}$ consists of ses split termwise, is Frobenius; that is, the contractible complexes are precisely the injectives and projectives. Consequently, $K / 𝐏𝐫𝐨𝐣$ is triangulated, where the desuspension is the syzygy.

A Module-like Analysis of $C(R)$

We view a chain complex $X ∈ C(R)$ as graded set for convention. In particular $X = ⨆_{n ∈ ℤ } X^n$ as underlying sets.

Show the free functor $𝐅𝐫𝐞𝐞 : 𝐒𝐞𝐭𝐬 → 𝐌𝐨𝐝 _R,\quad X ↦ R^{(X)}$ is left adjoint to the forgotful functor $U : 𝐌𝐨𝐝 _R → 𝐒𝐞𝐭𝐬$ sending every module to its under lying sets.

(Free object). The free object generated by a graded subset $\{S^n\}_{n ∈ ℤ }$ is defined as the chain complex

$$\begin{equation} \cdots \underset{\text{degree} \ n}{\underbrace{R^{(S^n)} ⊕ R^{(S^{n-1})}}} \xrightarrow{\binom {0 \ \ 0}{1 \ \ 0}} \underset{\text{degree} \ {n+1}}{\underbrace{R^{(S^{n+1})} ⊕ R^{(S^{n})}}} → \cdots. \end{equation}$$

One has the the free forgotful adjunction from graded sets to chain complexes.

We define the adjunction as a composition

$$\begin{equation} 𝐆𝐫𝐒𝐞𝐭𝐬 ⇆ 𝐆𝐫𝐌𝐨𝐝_R ⇆ C(R). \end{equation}$$

The first adjunction is nothing but a free-forgotful adjunction by degrees. The second adjunction is

$$\begin{equation} (M^\bullet , C^\bullet)_{\mathbf{GrSets}} \simeq((M^\bullet \oplus \Sigma^{-1}M, \binom {0 \ 0}{1 \ 0}), (C^\bullet, d^n))_{C(R)},\quad (f^n) \mapsto (f^n, d_C^{n-1} \circ f^n). \end{equation}$$

The verification:

The argument on projective objects is similar.

Free objects are projective, and projective objects are precisely direct summand of free objects.

Note that the forgotful functor $U: C(R) → 𝐒𝐞𝐭𝐬$ preserves epimorphisms. The following lifting problems are equivalent

Clearly every graded set is projective, then the lifting problem has solutions.
For any projective object $P$, the counit $𝐅𝐫𝐞𝐞 (U(P)) ↠ P$ is an epimorphism which splits. Hence, projective objects are summand of free objects.

In fact, it is shown in previous that projective (injective) objects are precisely the split acyclic complexes which is termwise projective (injective).

Show that the morphism category $𝐌𝐨𝐝 _R^ →$ is also a module category. Hint:

$$\begin{equation} [→, 𝐌𝐨𝐝 _R] ≃ [ → , [R^{\mathrm{op}}, 𝐀𝐛 ]] ≃ [R^{\mathrm{op}} \underset ? ⊗ → , 𝐀𝐛] ≃ [\binom{R \ R}{0 \ R}^{\mathrm{op}}, 𝐀𝐛]. \end{equation}$$

The finitely presented objects in $C(R)$ is precisely $C(𝐦𝐨𝐝 _R)$.

For $M ∈ C(R)$ compact, we see that every $M^i$ has to be f.p.. Notice that

$$\begin{aligned} (M^{k+1}, \varinjlim {}^{fil} X_∙)_R & ≃ (M, Σ ^k( \overline {\varinjlim {}^{fil} X_∙}))_R ≃ (M, \varinjlim {}^{fil} Σ ^k \overline {X_∙})_R\\ & ≃ \varinjlim {}^{fil} (M, Σ ^k \overline {X_∙})_R ≃ \varinjlim {}^{fil} (M^{k+1}, X_∙)_R. \end{aligned}$$

A morphism $f : X → \varinjlim ^{fil} C_∙$ identifies a collection of morphisms in $𝐌𝐨𝐝_R ^→$, which is again a module category. Since gluing procedure preserves and reflect isomorphisms, $f$ identifies a morphism in $\varinjlim ^{fil}(X, C_∙)$ via structure map.

Hence, a compact object need not to be f.g..

If we choose $\mathcal{HOM}$ instead, then compact objects are exactly bounded complexes with f.p. modules.

In general, $C(R)$ can be regarded as a full subcategory of $𝐌𝐨𝐝 _{(R^ℤ) [d] / I}$, where $I$ is generated by

$$\begin{equation} \{d^2\} ∪ \{[d,r]\}_{r ∈ R} ∪ \{d ∘ e_k - e_{k+1} ∘ d ∣ e_l ^2 = e_l, \ \text{and} \ (R^ℤ)e_l = R^{(l\text{-th})}\}_{k ∈ ℤ}. \end{equation}$$

A chain complex $M^∙$ is regarded as a graded module $⨁_{k ∈ ℤ }M^k$ (with relations).

This definition is rarely used in practice; in this case, a functorial indentification $𝐌𝐨𝐝 _A ≃ 𝐅𝐮𝐧𝐜𝐭 (A^{\mathrm{op}}, 𝐀𝐛)$ is more convenient.
This is not our/topological definition, at least $M^i ∩ M^j = ∅$ for our convention.

We are interested in DG-projective/DG-injective complexes in practice.

Double Complexes

For a general theory of complex resolution, we shall introduce $C(C(R))$, which is equivalent to the category of double complexes.

(Double complex, or bicomplex). A double complex is a collection of objects $\{C^{p,q}\}_{p,q ∈ ℤ}$ with two differentials $d_h$ (horizontal) and $d_v$ (vertical), such that

$$\begin{equation} d_h ∘ d_h = 0, \quad d_v ∘ d_v = 0, \quad\text{and} \ \ d_h ∘ d_v + d_v ∘ d_h = 0. \end{equation}$$

We write $DC(\mathcal{A})$ as the category of double complexes over $\mathcal{A}$.

Show that if $\mathcal{A}$ is AB3/AB4/AB5/AB3*/…/, has enough projectives/injectives, well powered, etc., then so is $DC(\mathcal{A})$.

(Total complexes). For either $☆ ∈ \{∐ , ∏\}$, we define the functor

$$\begin{equation} \mathrm{Tot}^☆ : DC(\mathcal{A}) → C(\mathcal{A}), \quad (C^{p,q}, d_h , d_v) ↦ (\underset{p+q = n}{☆}C^{p,q}, (d_v + d_h)). \end{equation}$$

Since ses for in either $DC(\mathcal{A})$ or $C(\mathcal{A})$ are verified termwise, $\mathrm{Tot}^∏$ (resp., $\mathrm{Tot}^∐$) is an exact functor when $\mathcal{A}$ is AB4* (resp., AB3*).

(Support). The support of $X^{∙ , ∙}$ consists of the pairs $(p,q) ∈ ℤ ^2$ such that $X^{p,q} ≠ 0$. In particular,

say $X^{∙ , ∙ }$ is finite, provided $\mathrm{Supp}(X)$ is finite;
say $X^{∙ , ∙ }$ is locally finite, provided $\{(p,q) ∈ \mathrm{Supp}(X) ∣ p + q = m\}$ is finite for every $m ∈ ℤ$;
- In this case, $\mathrm{Tot}^∐ (X) = \mathrm{Tot}^∏ (X)$. We write $\mathrm{Tot}(X)$ for simplicity.
say $X^{∙ , ∙ }$ is a $k$-th quadrant double complex if $\mathrm{Supp}(X)$ lies in the $k$-th quadrant under finite steps of translations;
we also use the expression including upper half plane double complex, $(m × n)$ double complexes, etc.

Here are some criterion on exactness.

Suppose $X$ is locally finite with exact rows (or columns). Then both $\mathrm{Tot}^∏(X)$ and $\mathrm{Tot}^∐(X)$ are exact.
Suppose $X$ lies in upper (or left) half plane with exact columns. Then $\mathrm{Tot}^∏(X)$ is exact.
Suppose $X$ lies in upper (or left) half plane with exact rows. Then $\mathrm{Tot}^∐(X)$ is exact.

It is always quicker to write a proof rather than to read a proof.

To remember this, just note that for complexes over upper half plane, $∐$ and column-exactness are not compatible; otherwise DG-projective complexes are meaningless!

We show some non-examples in accordance with the above proposition.

For $X ≠ 0$, consider the following double complexes.

We see $\mathrm{Tot}^∏ (\text{Left})$ and $\mathrm{Tot}^∐ (\text{Right})$ are not acyclic; while $\mathrm{Tot}^∏ (\text{Right})$ and $\mathrm{Tot}^∐ (\text{Left})$ are acyclic.

Some books refer $⊗$ and $\mathrm{Hom}$ as bifunctors of the type $C × C → DC$. Recall our convention is $C × C → DC \xrightarrow {\mathrm{Tot}^{☆}} C$, i.e.,

$$\begin{equation} (X ⊗ Y)^{∙ , ∙ } ↦ \underset{:= X ⊗ Y}{\underbrace{\mathrm{Tot}^{∐ }(X ⊗ Y)}},\quad (\mathrm{Hom}(X,Y))^{∙ , ∙ } ↦ \underset{:= \mathcal{HOM}(X,Y)}{\underbrace{\mathrm{Tot}^{∏ }(\mathrm{Hom}(X,Y))}}. \end{equation}$$

Bi-resolution shows the balanced property of $\mathrm{Ext}^k$, that is, there is a natural isomorphism

$$\begin{equation} \mathrm{Ext}^k(-, Y) (X) ≃ \mathrm{Ext}^k(X, -) (Y). \end{equation}$$

We take projective resolution $X$ and injective resolution $Y$. Consider the black double complex which is locally finite:

By criterion of exactness, we see $\binom{ {\color{red}\text{红}}\ {\color{purple}\text{紫}} }{\text{黑} \ {\color{blue}\text{蓝}}}$, $\binom{ {\color{red}\text{红}}}{\text{黑}}$, and $({\text{黑}\ {\color{blue}\text{蓝}}})$ are exact. By ses

$$\begin{equation} 0 → \text{黑} → ({\text{黑}\ {\color{blue}\text{蓝}}}) → {\color{blue}\text{蓝}} → 0, \end{equation}$$

Hence, $Σ\mathrm{Tot}(\text{黑}) ⇠ {\color{blue}\text{蓝} }$ and $Σ\mathrm{Tot}(\text{黑}^⊺) ⇠ {\color{red}\text{红} }$ are quasi-isomorphisms. We obtain

$$\begin{aligned} \mathrm{Ext}^k (-,Y)(X) & := H^k ({\color{red}\text{红}}) ≃ H^k (Σ \mathrm{Tot}(\text{黑}))\\[6pt] & ≃ H^k ({\color{blue}\text{蓝}}) =: \mathrm{Ext}^k (X,-)(Y). \end{aligned}$$

The balance comes from $H^∙ (\mathrm{Tot}^{☆}(X)) = H^∙ (\mathrm{Tot}^{☆}(X)^⊺ \ )$.

(Balance) Suppose $T : ⨉_{i=1}^n \mathcal{A}_i → \mathcal{B}$ is a (covariant) multi-functor between Abelian categories. Say it is balances, provided

For any injective $I_k ∈ 𝐈𝐧𝐣 (\mathcal{A}_k)$, the functor $$\begin{equation} T(X_1, \ldots, (-)_j, \ldots , X_{k-1}, I_k, X_{k+1}, \cdots, X_n) : \mathcal{A}_j → \mathcal{B} \end{equation}$$ is exact for arbitrary $\{X_k\}_{k ∉ \{i,j\}}$.

For balanced functors, the derivatives are free of the choice of the variables.

$((-)^{\mathrm{op}} ⊗(?)^{\mathrm{op}})^{\mathrm{op}}$ and $\mathrm{Hom}((-)^{\mathrm{op}},?)$ are balanced. The definition is introduced by Cartan-Eilenberg, and also by Weibel. They know no example of a balanced functor in three variables…