Permalink
Find file
Fetching contributors…
Cannot retrieve contributors at this time
6283 lines (5690 sloc) 217 KB
\input{preamble}
% OK, start here.
%
\begin{document}
\title{Differential Graded Algebra}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
In this chapter we talk about differential graded algebras, modules,
categories, etc. A basic reference is \cite{Keller-Deriving}.
A survey paper is \cite{Keller-survey}.
\medskip\noindent
Since we do not worry about length of exposition in the Stacks project
we first develop the material in the setting of categories of differential
graded modules. After that we redo the constructions in the setting of
differential graded modules over differential graded categories.
\section{Conventions}
\label{section-conventions}
\noindent
In this chapter we hold on to the convention that {\it ring} means
commutative ring with $1$. If $R$ is a ring, then an {\it $R$-algebra $A$}
will be an $R$-module $A$ endowed with an $R$-bilinear map $A \times A \to A$
(multiplication) such that multiplication is associative and has a unit.
In other words, these are unital associative $R$-algebras
such that the structure map $R \to A$ maps into the center of $A$.
\section{Differential graded algebras}
\label{section-dga}
\noindent
Just the definitions.
\begin{definition}
\label{definition-dga}
Let $R$ be a commutative ring. A {\it differential graded algebra over $R$}
is either
\begin{enumerate}
\item a chain complex $A_\bullet$ of $R$-modules endowed with
$R$-bilinear maps $A_n \times A_m \to A_{n + m}$,
$(a, b) \mapsto ab$ such that
$$
\text{d}_{n + m}(ab) = \text{d}_n(a)b + (-1)^n a\text{d}_m(b)
$$
and such that $\bigoplus A_n$ becomes an associative and unital
$R$-algebra, or
\item a cochain complex $A^\bullet$ of $R$-modules endowed with
$R$-bilinear maps $A^n \times A^m \to A^{n + m}$, $(a, b) \mapsto ab$
such that
$$
\text{d}^{n + m}(ab) = \text{d}^n(a)b + (-1)^n a\text{d}^m(b)
$$
and such that $\bigoplus A^n$ becomes an associative and unital $R$-algebra.
\end{enumerate}
\end{definition}
\noindent
We often just write $A = \bigoplus A_n$ or $A = \bigoplus A^n$ and
think of this as an associative unital $R$-algebra endowed with a
$\mathbf{Z}$-grading and an $R$-linear operator $\text{d}$ whose square
is zero and which satisfies the Leibniz rule as explained above. In this case
we often say ``Let $(A, \text{d})$ be a differential graded algebra''.
\begin{definition}
\label{definition-homomorphism-dga}
A {\it homomorphism of differential graded algebras}
$f : (A, \text{d}) \to (B, \text{d})$ is an algebra map $f : A \to B$
compatible with the gradings and $\text{d}$.
\end{definition}
\begin{definition}
\label{definition-opposite-dga}
Let $R$ be a ring. Let $(A, \text{d})$ be a differential graded algebra
over $R$. The {\it opposite differential graded algebra} is the differential
graded algebra $(A^{opp}, \text{d})$ over $R$ where $A^{opp} = A$
as an $R$-module, $\text{d} = \text{d}$, and multiplication is
given by
$$
a \cdot_{opp} b = (-1)^{\deg(a)\deg(b)} b a
$$
for homogeneous elements $a, b \in A$.
\end{definition}
\noindent
This makes sense because
\begin{align*}
\text{d}(a \cdot_{opp} b)
& =
(-1)^{\deg(a)\deg(b)} \text{d}(b a) \\
& =
(-1)^{\deg(a)\deg(b)} \text{d}(b) a +
(-1)^{\deg(a)\deg(b) + \deg(b)}b\text{d}(a) \\
& =
(-1)^{\deg(a)}a \cdot_{opp} \text{d}(b) + \text{d}(a) \cdot_{opp} b
\end{align*}
as desired.
\begin{definition}
\label{definition-cdga}
A differential graded algebra $(A, \text{d})$ is {\it commutative} if
$ab = (-1)^{nm}ba$ for $a$ in degree $n$ and $b$ in degree $m$.
We say $A$ is {\it strictly commutative} if in addition $a^2 = 0$
for $\deg(a)$ odd.
\end{definition}
\noindent
The following definition makes sense in general but is perhaps
``correct'' only when tensoring commutative differential graded
algebras.
\begin{definition}
\label{definition-tensor-product}
Let $R$ be a ring.
Let $(A, \text{d})$, $(B, \text{d})$ be differential graded algebras over $R$.
The {\it tensor product differential graded algebra} of $A$ and $B$
is the algebra $A \otimes_R B$ with multiplication defined by
$$
(a \otimes b)(a' \otimes b') = (-1)^{\deg(a')\deg(b)} aa' \otimes bb'
$$
endowed with differential $\text{d}$ defined by the rule
$\text{d}(a \otimes b) = \text{d}(a) \otimes b + (-1)^m a \otimes \text{d}(b)$
where $m = \deg(a)$.
\end{definition}
\begin{lemma}
\label{lemma-total-complex-tensor-product}
Let $R$ be a ring.
Let $(A, \text{d})$, $(B, \text{d})$ be differential graded algebras over $R$.
Denote $A^\bullet$, $B^\bullet$ the underlying cochain complexes.
As cochain complexes of $R$-modules we have
$$
(A \otimes_R B)^\bullet = \text{Tot}(A^\bullet \otimes_R B^\bullet).
$$
\end{lemma}
\begin{proof}
Recall that the differential of the total complex is given by
$\text{d}_1^{p, q} + (-1)^p \text{d}_2^{p, q}$ on $A^p \otimes_R B^q$.
And this is exactly the same as the rule for the differential
on $A \otimes_R B$ in
Definition \ref{definition-tensor-product}.
\end{proof}
\section{Differential graded modules}
\label{section-modules}
\noindent
Just the definitions.
\begin{definition}
\label{definition-dgm}
Let $R$ be a ring.
Let $(A, \text{d})$ be a differential graded algebra over $R$.
A (right) {\it differential graded module} $M$ over $A$ is a right $A$-module
$M$ which has a grading $M = \bigoplus M^n$ and a differential $\text{d}$
such that $M^n A^m \subset M^{n + m}$, such that
$\text{d}(M^n) \subset M^{n + 1}$, and such that
$$
\text{d}(ma) = \text{d}(m)a + (-1)^n m\text{d}(a)
$$
for $a \in A$ and $m \in M^n$. A
{\it homomorphism of differential graded modules} $f : M \to N$
is an $A$-module map compatible with gradings and differentials.
The category of (right) differential graded $A$-modules is denoted
$\text{Mod}_{(A, \text{d})}$.
\end{definition}
\noindent
Note that we can think of $M$ as a cochain complex $M^\bullet$
of (right) $R$-modules. Namely, for $r \in R$ we have $\text{d}(r) = 0$
and $r$ maps to a degree $0$ element of $A$, hence
$\text{d}(mr) = \text{d}(m)r$.
\medskip\noindent
We can define {\it left differential graded $A$-modules} in exactly the same
manner. If $M$ is a left $A$-module, then we can think of $M$ as a
right $A^{opp}$-module with multiplication $\cdot_{opp}$ defined by
the rule
$$
m \cdot_{opp} a = (-1)^{\deg(a)\deg(m)} a m
$$
for $a$ and $m$ homogeneous. The category of left differential graded
$A$-modules is equivalent to the category of right differential
graded $A^{opp}$-modules. We prefer to work with right modules
(essentially because of what happens in Example \ref{example-dgm-dg-cat}), but
the reader is free to switch to left modules if (s)he so desires.
\begin{lemma}
\label{lemma-dgm-abelian}
Let $(A, d)$ be a differential graded algebra. The category
$\text{Mod}_{(A, \text{d})}$ is abelian and has arbitrary limits and colimits.
\end{lemma}
\begin{proof}
Kernels and cokernels commute with taking underlying $A$-modules.
Similarly for direct sums and colimits. In other words, these operations
in $\text{Mod}_{(A, \text{d})}$ commute with the forgetful functor to the
category of $A$-modules. This is not the case for products and limits.
Namely, if $N_i$, $i \in I$ is a family of
differential graded $A$-modules, then the product $\prod N_i$ in
$\text{Mod}_{(A, \text{d})}$ is given by setting $(\prod N_i)^n = \prod N_i^n$
and $\prod N_i = \bigoplus_n (\prod N_i)^n$. Thus we see that the product
does commute with the forgetful functor to the category of graded $A$-modules.
A category with products and equalizers has limits, see
Categories, Lemma \ref{categories-lemma-limits-products-equalizers}.
\end{proof}
\noindent
Thus, if $(A, \text{d})$ is a differential graded
algebra over $R$, then there is an exact functor
$$
\text{Mod}_{(A, \text{d})} \longrightarrow \text{Comp}(R)
$$
of abelian categories. For a differential graded module $M$ the
cohomology groups $H^n(M)$ are defined as the cohomology of the
corresponding complex of $R$-modules. Therefore, a short exact
sequence $0 \to K \to L \to M \to 0$ of differential graded modules
gives rise to a long exact sequence
\begin{equation}
\label{equation-les}
H^n(K) \to H^n(L) \to H^n(M) \to H^{n + 1}(K)
\end{equation}
of cohomology modules, see
Homology, Lemma \ref{homology-lemma-long-exact-sequence-cochain}.
\medskip\noindent
Moreover, from now on we borrow all the terminology used for
complexes of modules. For example, we say that a differential
graded $A$-module $M$ is {\it acyclic} if $H^k(M) = 0$ for
all $k \in \mathbf{Z}$. We say that a homomorphism $M \to N$
of differential graded $A$-modules is a {\it quasi-isomorphism}
if it induces isomorphisms $H^k(M) \to H^k(N)$ for all $k \in \mathbf{Z}$.
And so on and so forth.
\begin{definition}
\label{definition-shift}
Let $(A, \text{d})$ be a differential graded algebra.
Let $M$ be a differential graded module.
For any $k \in \mathbf{Z}$ we define the {\it $k$-shifted module}
$M[k]$ as follows
\begin{enumerate}
\item as $A$-module $M[k] = M$,
\item $M[k]^n = M^{n + k}$,
\item $\text{d}_{M[k]} = (-1)^k\text{d}_M$.
\end{enumerate}
For a morphism $f : M \to N$ of differential graded $A$-modules
we let $f[k] : M[k] \to N[k]$ be the map equal to $f$ on underlying
$A$-modules. This defines a functor
$[k] : \text{Mod}_{(A, \text{d})} \to \text{Mod}_{(A, \text{d})}$.
\end{definition}
\noindent
The remarks in Homology, Section \ref{homology-section-homotopy-shift} apply.
In particular, we will identify the cohomology groups of all shifts
$M[k]$ without the intervention of signs.
\medskip\noindent
At this point we have enough structure to talk about {\it triangles},
see Derived Categories, Definition \ref{derived-definition-triangle}.
In fact, our next goal is to develop enough theory to be able to
state and prove that the homotopy category of differential graded
modules is a triangulated category. First we define the homotopy category.
\section{The homotopy category}
\label{section-homotopy}
\noindent
Our homotopies take into account the $A$-module structure and the
grading, but not the differential (of course).
\begin{definition}
\label{definition-homotopy}
Let $(A, \text{d})$ be a differential graded algebra. Let
$f, g : M \to N$ be homomorphisms of differential graded $A$-modules.
A {\it homotopy between $f$ and $g$} is an $A$-module map $h : M \to N$
such that
\begin{enumerate}
\item $h(M^n) \subset N^{n - 1}$ for all $n$, and
\item $f(x) - g(x) = \text{d}_N(h(x)) + h(\text{d}_M(x))$ for
all $x \in M$.
\end{enumerate}
If a homotopy exists, then we say $f$ and $g$ are {\it homotopic}.
\end{definition}
\noindent
Thus $h$ is compatible with the $A$-module structure and the grading
but not with the differential. If $f = g$ and $h$ is a homotopy
as in the definition, then $h$ defines a morphism $h : M \to N[-1]$
in $\text{Mod}_{(A, \text{d})}$.
\begin{lemma}
\label{lemma-compose-homotopy}
Let $(A, \text{d})$ be a differential graded algebra.
Let $f, g : L \to M$ be homomorphisms of differential graded $A$-modules.
Suppose given further homomorphisms $a : K \to L$, and $c : M \to N$.
If $h : L \to M$ is an $A$-module map which defines a homotopy between
$f$ and $g$, then $c \circ h \circ a$ defines a homotopy between
$c \circ f \circ a$ and $c \circ g \circ a$.
\end{lemma}
\begin{proof}
Immediate from Homology, Lemma \ref{homology-lemma-compose-homotopy-cochain}.
\end{proof}
\noindent
This lemma allows us to define the homotopy category as follows.
\begin{definition}
\label{definition-complexes-notation}
Let $(A, \text{d})$ be a differential graded algebra.
The {\it homotopy category}, denoted $K(\text{Mod}_{(A, \text{d})})$, is
the category whose objects are the objects of
$\text{Mod}_{(A, \text{d})}$ and whose morphisms are homotopy classes
of homomorphisms of differential graded $A$-modules.
\end{definition}
\noindent
The notation $K(\text{Mod}_{(A, \text{d})})$ is not standard but at least is
consistent with the use of $K(-)$ in other places of the Stacks project.
\begin{lemma}
\label{lemma-homotopy-direct-sums}
Let $(A, \text{d})$ be a differential graded algebra.
The homotopy category $K(\text{Mod}_{(A, \text{d})})$
has direct sums and products.
\end{lemma}
\begin{proof}
Omitted. Hint: Just use the direct sums and products as in
Lemma \ref{lemma-dgm-abelian}. This works because we saw that
these functors commute with the forgetful functor to the category
of graded $A$-modules and because $\prod$ is an exact functor
on the category of families of abelian groups.
\end{proof}
\section{Cones}
\label{section-cones}
\noindent
We introduce cones for the category of differential graded modules.
\begin{definition}
\label{definition-cone}
Let $(A, \text{d})$ be a differential graded algebra.
Let $f : K \to L$ be a homomorphism of differential graded $A$-modules.
The {\it cone} of $f$ is the differential graded $A$-module
$C(f)$ given by $C(f) = L \oplus K$ with grading
$C(f)^n = L^n \oplus K^{n + 1}$ and
differential
$$
d_{C(f)} =
\left(
\begin{matrix}
\text{d}_L & f \\
0 & -\text{d}_K
\end{matrix}
\right)
$$
It comes equipped with canonical morphisms of complexes $i : L \to C(f)$
and $p : C(f) \to K[1]$ induced by the obvious maps $L \to C(f)$
and $C(f) \to K$.
\end{definition}
\noindent
The formation of the cone triangle is functorial in the following sense.
\begin{lemma}
\label{lemma-functorial-cone}
Let $(A, \text{d})$ be a differential graded algebra.
Suppose that
$$
\xymatrix{
K_1 \ar[r]_{f_1} \ar[d]_a & L_1 \ar[d]^b \\
K_2 \ar[r]^{f_2} & L_2
}
$$
is a diagram of homomorphisms of differential graded $A$-modules which is
commutative up to homotopy.
Then there exists a morphism $c : C(f_1) \to C(f_2)$ which gives rise to
a morphism of triangles
$$
(a, b, c) : (K_1, L_1, C(f_1), f_1, i_1, p_1) \to
(K_1, L_1, C(f_1), f_2, i_2, p_2)
$$
in $K(\text{Mod}_{(A, \text{d})})$.
\end{lemma}
\begin{proof}
Let $h : K_1 \to L_2$ be a homotopy between $f_2 \circ a$ and $b \circ f_1$.
Define $c$ by the matrix
$$
c =
\left(
\begin{matrix}
b & h \\
0 & a
\end{matrix}
\right) :
L_1 \oplus K_1 \to L_2 \oplus K_2
$$
A matrix computation show that $c$ is a morphism of differential
graded modules. It is trivial that $c \circ i_1 = i_2 \circ b$, and it is
trivial also to check that $p_2 \circ c = a \circ p_1$.
\end{proof}
\section{Admissible short exact sequences}
\label{section-admissible}
\noindent
An admissible short exact sequence is the analogue of termwise split exact
sequences in the setting of differential graded modules.
\begin{definition}
\label{definition-admissible-ses}
Let $(A, \text{d})$ be a differential graded algebra.
\begin{enumerate}
\item A homomorphism $K \to L$ of differential graded $A$-modules
is an {\it admissible monomorphism} if there exists a graded $A$-module
map $L \to K$ which is left inverse to $K \to L$.
\item A homomorphism $L \to M$ of differential graded $A$-modules
is an {\it admissible epimorphism} if there exists a graded $A$-module
map $M \to L$ which is right inverse to $L \to M$.
\item A short exact sequence $0 \to K \to L \to M \to 0$ of differential
graded $A$-modules is an {\it admissible short exact sequence}
if it is split as a sequence of graded $A$-modules.
\end{enumerate}
\end{definition}
\noindent
Thus the splittings are compatible with all the data except for
the differentials. Given an admissible short exact sequence we
obtain a triangle; this is the reason that we require our splittings
to be compatible with the $A$-module structure.
\begin{lemma}
\label{lemma-admissible-ses}
Let $(A, \text{d})$ be a differential graded algebra.
Let $0 \to K \to L \to M \to 0$ be an admissible short exact sequence
of differential graded $A$-modules. Let $s : M \to L$ and $\pi : L \to K$
be splittings such that $\Ker(\pi) = \Im(s)$.
Then we obtain a morphism
$$
\delta = \pi \circ \text{d}_L \circ s : M \to K[1]
$$
of $\text{Mod}_{(A, \text{d})}$ which induces the boundary maps
in the long exact sequence of cohomology (\ref{equation-les}).
\end{lemma}
\begin{proof}
The map $\pi \circ \text{d}_L \circ s$ is compatible with the $A$-module
structure and the gradings by construction. It is compatible with
differentials by Homology, Lemmas
\ref{homology-lemma-ses-termwise-split-cochain}.
Let $R$ be the ring that $A$ is a differential graded algebra over.
The equality of maps is a statement about $R$-modules. Hence this
follows from Homology, Lemmas
\ref{homology-lemma-ses-termwise-split-cochain} and
\ref{homology-lemma-ses-termwise-split-long-cochain}.
\end{proof}
\begin{lemma}
\label{lemma-make-commute-map}
Let $(A, \text{d})$ be a differential graded algebra. Let
$$
\xymatrix{
K \ar[r]_f \ar[d]_a & L \ar[d]^b \\
M \ar[r]^g & N
}
$$
be a diagram of homomorphisms of differential graded $A$-modules
commuting up to homotopy.
\begin{enumerate}
\item If $f$ is an admissible monomorphism, then $b$ is homotopic to a
homomorphism which makes the diagram commute.
\item If $g$ is an admissible epimorphism, then $a$ is homotopic to a
morphism which makes the diagram commute.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $h : K \to N$ be a homotopy between $bf$ and $ga$, i.e.,
$bf - ga = \text{d}h + h\text{d}$. Suppose that $\pi : L \to K$
is a graded $A$-module map left inverse to $f$. Take
$b' = b - \text{d}h\pi - h\pi \text{d}$.
Suppose $s : N \to M$ is a graded $A$-module map right inverse to $g$.
Take $a' = a + \text{d}sh + sh\text{d}$.
Computations omitted.
\end{proof}
\begin{lemma}
\label{lemma-make-injective}
Let $(A, \text{d})$ be a differential graded algebra.
Let $\alpha : K \to L$ be a homomorphism of differential graded
$A$-modules. There exists a factorization
$$
\xymatrix{
K \ar[r]^{\tilde \alpha} \ar@/_1pc/[rr]_\alpha &
\tilde L \ar[r]^\pi & L
}
$$
in $\text{Mod}_{(A, \text{d})}$ such that
\begin{enumerate}
\item $\tilde \alpha$ is an admissible monomorphism (see
Definition \ref{definition-admissible-ses}),
\item there is a morphism $s : L \to \tilde L$
such that $\pi \circ s = \text{id}_L$ and such that
$s \circ \pi$ is homotopic to $\text{id}_{\tilde L}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The proof is identical to the proof of
Derived Categories, Lemma \ref{derived-lemma-make-injective}.
Namely, we set $\tilde L = L \oplus C(1_K)$ and we use elementary
properties of the cone construction.
\end{proof}
\begin{lemma}
\label{lemma-sequence-maps-split}
Let $(A, \text{d})$ be a differential graded algebra.
Let $L_1 \to L_2 \to \ldots \to L_n$
be a sequence of composable homomorphisms of
differential graded $A$-modules.
There exists a commutative diagram
$$
\xymatrix{
L_1 \ar[r] &
L_2 \ar[r] &
\ldots \ar[r] &
L_n \\
M_1 \ar[r] \ar[u] &
M_2 \ar[r] \ar[u] &
\ldots \ar[r] &
M_n \ar[u]
}
$$
in $\text{Mod}_{(A, \text{d})}$ such that each $M_i \to M_{i + 1}$
is an admissible monomorphism and each $M_i \to L_i$
is a homotopy equivalence.
\end{lemma}
\begin{proof}
The case $n = 1$ is without content.
Lemma \ref{lemma-make-injective} is the case $n = 2$.
Suppose we have constructed the diagram
except for $M_n$. Apply Lemma \ref{lemma-make-injective} to
the composition $M_{n - 1} \to L_{n - 1} \to L_n$.
The result is a factorization $M_{n - 1} \to M_n \to L_n$
as desired.
\end{proof}
\begin{lemma}
\label{lemma-nilpotent}
Let $(A, \text{d})$ be a differential graded algebra.
Let $0 \to K_i \to L_i \to M_i \to 0$, $i = 1, 2, 3$
be admissible short exact sequence of differential graded $A$-modules.
Let $b : L_1 \to L_2$ and $b' : L_2 \to L_3$
be homomorphisms of differential graded modules such that
$$
\vcenter{
\xymatrix{
K_1 \ar[d]_0 \ar[r] &
L_1 \ar[r] \ar[d]_b &
M_1 \ar[d]_0 \\
K_2 \ar[r] & L_2 \ar[r] & M_2
}
}
\quad\text{and}\quad
\vcenter{
\xymatrix{
K_2 \ar[d]^0 \ar[r] &
L_2 \ar[r] \ar[d]^{b'} &
M_2 \ar[d]^0 \\
K_3 \ar[r] & L_3 \ar[r] & M_3
}
}
$$
commute up to homotopy. Then $b' \circ b$ is homotopic to $0$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-make-commute-map} we can replace $b$ and $b'$ by
homotopic maps such that the right square of the left diagram commutes
and the left square of the right diagram commutes. In other words, we have
$\Im(b) \subset \Im(K_2 \to L_2)$ and
$\Ker((b')^n) \supset \Im(K_2 \to L_2)$.
Then $b \circ b' = 0$ as a map of modules.
\end{proof}
\section{Distinguished triangles}
\label{section-distinguished}
\noindent
The following lemma produces our distinguished triangles.
\begin{lemma}
\label{lemma-triangle-independent-splittings}
Let $(A, \text{d})$ be a differential graded algebra. Let
$0 \to K \to L \to M \to 0$ be an admissible short exact sequence
of differential graded $A$-modules. The triangle
\begin{equation}
\label{equation-triangle-associated-to-admissible-ses}
K \to L \to M \xrightarrow{\delta} K[1]
\end{equation}
with $\delta$ as in Lemma \ref{lemma-admissible-ses} is, up to canonical
isomorphism in $K(\text{Mod}_{(A, \text{d})})$, independent of the choices
made in Lemma \ref{lemma-admissible-ses}.
\end{lemma}
\begin{proof}
Namely, let $(s', \pi')$ be a second choice of splittings as in
Lemma \ref{lemma-admissible-ses}. Then we claim that $\delta$ and $\delta'$
are homotopic. Namely, write $s' = s + \alpha \circ h$ and
$\pi' = \pi + g \circ \beta$ for some unique homomorphisms
of $A$-modules $h : M \to K$ and $g : M \to K$ of degree $-1$.
Then $g = -h$ and $g$ is a homotopy between $\delta$ and $\delta'$.
The computations are done in the proof of
Homology, Lemma \ref{homology-lemma-ses-termwise-split-homotopy-cochain}.
\end{proof}
\begin{definition}
\label{definition-distinguished-triangle}
Let $(A, \text{d})$ be a differential graded algebra.
\begin{enumerate}
\item If $0 \to K \to L \to M \to 0$ is an admissible short exact sequence
of differential graded $A$-modules, then the {\it triangle associated
to $0 \to K \to L \to M \to 0$} is the triangle
(\ref{equation-triangle-associated-to-admissible-ses})
of $K(\text{Mod}_{(A, \text{d})})$.
\item A triangle of $K(\text{Mod}_{(A, \text{d})})$ is called a
{\it distinguished triangle} if it is isomorphic to a triangle
associated to an admissible short exact sequence
of differential graded $A$-modules.
\end{enumerate}
\end{definition}
\section{Cones and distinguished triangles}
\label{section-cones-and-triangles}
\noindent
Let $(A, \text{d})$ be a differential graded algebra.
Let $f : K \to L$ be a homomorphism of differential graded $A$-modules.
Then $(K, L, C(f), f, i, p)$ forms a triangle:
$$
K \to L \to C(f) \to K[1]
$$
in $\text{Mod}_{(A, \text{d})}$ and hence in $K(\text{Mod}_{(A, \text{d})})$.
Cones are {\bf not} distinguished triangles in general, but the difference
is a sign or a rotation (your choice). Here are two precise statements.
\begin{lemma}
\label{lemma-rotate-cone}
Let $(A, \text{d})$ be a differential graded algebra.
Let $f : K \to L$ be a homomorphism of differential graded modules.
The triangle $(L, C(f), K[1], i, p, f[1])$ is
the triangle associated to the admissible short exact sequence
$$
0 \to L \to C(f) \to K[1] \to 0
$$
coming from the definition of the cone of $f$.
\end{lemma}
\begin{proof}
Immediate from the definitions.
\end{proof}
\begin{lemma}
\label{lemma-rotate-triangle}
Let $(A, \text{d})$ be a differential graded algebra.
Let $\alpha : K \to L$ and $\beta : L \to M$
define an admissible short exact sequence
$$
0 \to K \to L \to M \to 0
$$
of differential graded $A$-modules.
Let $(K, L, M, \alpha, \beta, \delta)$
be the associated triangle. Then the triangles
$$
(M[-1], K, L, \delta[-1], \alpha, \beta)
\quad\text{and}\quad
(M[-1], K, C(\delta[-1]), \delta[-1], i, p)
$$
are isomorphic.
\end{lemma}
\begin{proof}
Using a choice of splittings we write $L = K \oplus M$ and we identify
$\alpha$ and $\beta$ with the natural inclusion and projection maps.
By construction of $\delta$ we have
$$
d_B =
\left(
\begin{matrix}
d_K & \delta \\
0 & d_M
\end{matrix}
\right)
$$
On the other hand the cone of $\delta[-1] : M[-1] \to K$
is given as $C(\delta[-1]) = K \oplus M$ with differential identical
with the matrix above! Whence the lemma.
\end{proof}
\begin{lemma}
\label{lemma-third-isomorphism}
Let $(A, \text{d})$ be a differential graded algebra.
Let $f_1 : K_1 \to L_1$ and $f_2 : K_2 \to L_2$ be homomorphisms of
differential graded $A$-modules. Let
$$
(a, b, c) :
(K_1, L_1, C(f_1), f_1, i_1, p_1)
\longrightarrow
(K_1, L_1, C(f_1), f_2, i_2, p_2)
$$
be any morphism of triangles of $K(\text{Mod}_{(A, \text{d})})$.
If $a$ and $b$ are homotopy equivalences then so is $c$.
\end{lemma}
\begin{proof}
Let $a^{-1} : K_2 \to K_1$ be a homomorphism of differential graded $A$-modules
which is inverse to $a$ in $K(\text{Mod}_{(A, \text{d})})$.
Let $b^{-1} : L_2 \to L_1$ be a homomorphism of differential graded $A$-modules
which is inverse to $b$ in $K(\text{Mod}_{(A, \text{d})})$.
Let $c' : C(f_2) \to C(f_1)$ be the morphism from
Lemma \ref{lemma-functorial-cone} applied to
$f_1 \circ a^{-1} = b^{-1} \circ f_2$.
If we can show that $c \circ c'$ and $c' \circ c$ are isomorphisms in
$K(\text{Mod}_{(A, \text{d})})$
then we win. Hence it suffices to prove the following: Given
a morphism of triangles
$(1, 1, c) : (K, L, C(f), f, i, p)$
in $K(\text{Mod}_{(A, \text{d})})$ the morphism $c$ is an isomorphism
in $K(\text{Mod}_{(A, \text{d})})$.
By assumption the two squares in the diagram
$$
\xymatrix{
L \ar[r] \ar[d]_1 &
C(f) \ar[r] \ar[d]_c &
K[1] \ar[d]_1 \\
L \ar[r] &
C(f) \ar[r] &
K[1]
}
$$
commute up to homotopy. By construction of $C(f)$ the rows
form admissible short exact sequences. Thus we see that
$(c - 1)^2 = 0$ in $K(\text{Mod}_{(A, \text{d})})$ by
Lemma \ref{lemma-nilpotent}.
Hence $c$ is an isomorphism in $K(\text{Mod}_{(A, \text{d})})$
with inverse $2 - c$.
\end{proof}
\noindent
The following lemma shows that the collection of triangles of the homotopy
category given by cones and the distinguished triangles are the same
up to isomorphisms, at least up to sign!
\begin{lemma}
\label{lemma-the-same-up-to-isomorphisms}
Let $(A, \text{d})$ be a differential graded algebra.
\begin{enumerate}
\item Given an admissible short exact sequence
$0 \to K \xrightarrow{\alpha} L \to M \to 0$
of differential graded $A$-modules there exists a homotopy equivalence
$C(\alpha) \to M$ such that the diagram
$$
\xymatrix{
K \ar[r] \ar[d] & L \ar[d] \ar[r] &
C(\alpha) \ar[r]_{-p} \ar[d] & K[1] \ar[d] \\
K \ar[r]^\alpha & L \ar[r]^\beta &
M \ar[r]^\delta & K[1]
}
$$
defines an isomorphism of triangles in $K(\text{Mod}_{(A, \text{d})})$.
\item Given a morphism of complexes $f : K \to L$
there exists an isomorphism of triangles
$$
\xymatrix{
K \ar[r] \ar[d] & \tilde L \ar[d] \ar[r] &
M \ar[r]_{\delta} \ar[d] & K[1] \ar[d] \\
K \ar[r] & L \ar[r] &
C(f) \ar[r]^{-p} & K[1]
}
$$
where the upper triangle is the triangle associated to a
admissible short exact sequence $K \to \tilde L \to M$.
\end{enumerate}
\end{lemma}
\begin{proof}
Proof of (1). We have $C(\alpha) = L \oplus K$ and we simply define
$C(\alpha) \to M$ via the projection onto $L$ followed by $\beta$.
This defines a morphism of differential graded modules because the
compositions $K^{n + 1} \to L^{n + 1} \to M^{n + 1}$ are zero.
Choose splittings $s : M \to L$ and $\pi : L \to K$ with
$\Ker(\pi) = \Im(s)$ and set
$\delta = \pi \circ \text{d}_L \circ s$ as usual.
To get a homotopy inverse we take
$M \to C(\alpha)$ given by $(s , -\delta)$. This is compatible with
differentials because $\delta^n$ can be characterized as the
unique map $M^n \to K^{n + 1}$ such that
$\text{d} \circ s^n - s^{n + 1} \circ \text{d} = \alpha \circ \delta^n$,
see proof of
Homology, Lemma \ref{homology-lemma-ses-termwise-split-cochain}.
The composition $M \to C(f) \to M$ is the identity.
The composition $C(f) \to M \to C(f)$ is equal to the morphism
$$
\left(
\begin{matrix}
s \circ \beta & 0 \\
-\delta \circ \beta & 0
\end{matrix}
\right)
$$
To see that this is homotopic to the identity map
use the homotopy $h : C(\alpha) \to C(\alpha)$
given by the matrix
$$
\left(
\begin{matrix}
0 & 0 \\
\pi & 0
\end{matrix}
\right) :
C(\alpha) = L \oplus K
\to
L \oplus K = C(\alpha)
$$
It is trivial to verify that
$$
\left(
\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}
\right)
-
\left(
\begin{matrix}
s \\
-\delta
\end{matrix}
\right)
\left(
\begin{matrix}
\beta & 0
\end{matrix}
\right)
=
\left(
\begin{matrix}
\text{d} & \alpha \\
0 & -\text{d}
\end{matrix}
\right)
\left(
\begin{matrix}
0 & 0 \\
\pi & 0
\end{matrix}
\right)
+
\left(
\begin{matrix}
0 & 0 \\
\pi & 0
\end{matrix}
\right)
\left(
\begin{matrix}
\text{d} & \alpha \\
0 & -\text{d}
\end{matrix}
\right)
$$
To finish the proof of (1) we have to show that the morphisms
$-p : C(\alpha) \to K[1]$ (see
Definition \ref{definition-cone})
and $C(\alpha) \to M \to K[1]$ agree up
to homotopy. This is clear from the above. Namely, we can use the homotopy
inverse $(s, -\delta) : M \to C(\alpha)$
and check instead that the two maps
$M \to K[1]$ agree. And note that
$p \circ (s, -\delta) = -\delta$ as desired.
\medskip\noindent
Proof of (2). We let $\tilde f : K \to \tilde L$,
$s : L \to \tilde L$
and $\pi : L \to L$ be as in
Lemma \ref{lemma-make-injective}. By
Lemmas \ref{lemma-functorial-cone} and \ref{lemma-third-isomorphism}
the triangles $(K, L, C(f), i, p)$ and
$(K, \tilde L, C(\tilde f), \tilde i, \tilde p)$
are isomorphic. Note that we can compose isomorphisms of
triangles. Thus we may replace $L$ by
$\tilde L$ and $f$ by $\tilde f$. In other words
we may assume that $f$ is an admissible monomorphism.
In this case the result follows from part (1).
\end{proof}
\section{The homotopy category is triangulated}
\label{section-homotopy-triangulated}
\noindent
We first prove that it is pre-triangulated.
\begin{lemma}
\label{lemma-homotopy-category-pre-triangulated}
Let $(A, \text{d})$ be a differential graded algebra.
The homotopy category $K(\text{Mod}_{(A, \text{d})})$
with its natural translation functors and distinguished triangles
is a pre-triangulated category.
\end{lemma}
\begin{proof}
Proof of TR1. By definition every triangle isomorphic to a distinguished
one is distinguished. Also, any triangle $(K, K, 0, 1, 0, 0)$
is distinguished since $0 \to K \to K \to 0 \to 0$ is
an admissible short exact sequence. Finally, given any homomorphism
$f : K \to L$ of differential graded $A$-modules the triangle
$(K, L, C(f), f, i, -p)$ is distinguished by
Lemma \ref{lemma-the-same-up-to-isomorphisms}.
\medskip\noindent
Proof of TR2. Let $(X, Y, Z, f, g, h)$ be a triangle.
Assume $(Y, Z, X[1], g, h, -f[1])$ is distinguished.
Then there exists an admissible short exact sequence
$0 \to K \to L \to M \to 0$ such that the associated
triangle $(K, L, M, \alpha, \beta, \delta)$
is isomorphic to $(Y, Z, X[1], g, h, -f[1])$. Rotating back we see
that $(X, Y, Z, f, g, h)$ is isomorphic to
$(M[-1], K, L, -\delta[-1], \alpha, \beta)$.
It follows from Lemma \ref{lemma-rotate-triangle} that the triangle
$(M[-1], K, L, \delta[-1], \alpha, \beta)$
is isomorphic to
$(M[-1], K, C(\delta[-1]), \delta[-1], i, p)$.
Precomposing the previous isomorphism of triangles with $-1$ on $Y$
it follows that $(X, Y, Z, f, g, h)$ is isomorphic to
$(M[-1], K, C(\delta[-1]), \delta[-1], i, -p)$.
Hence it is distinguished by
Lemma \ref{lemma-the-same-up-to-isomorphisms}.
On the other hand, suppose that $(X, Y, Z, f, g, h)$ is distinguished.
By Lemma \ref{lemma-the-same-up-to-isomorphisms} this means that it is
isomorphic to a triangle of the form
$(K, L, C(f), f, i, -p)$ for some morphism $f$ of
$\text{Mod}_{(A, \text{d})}$. Then the rotated triangle
$(Y, Z, X[1], g, h, -f[1])$ is
isomorphic to $(L, C(f), K[1], i, -p, -f[1])$ which is
isomorphic to the triangle
$(L, C(f), K[1], i, p, f[1])$.
By Lemma \ref{lemma-rotate-cone} this triangle is distinguished.
Hence $(Y, Z, X[1], g, h, -f[1])$ is distinguished as desired.
\medskip\noindent
Proof of TR3. Let $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$
be distinguished triangles of $K(\mathcal{A})$ and let $a : X \to X'$
and $b : Y \to Y'$ be morphisms such that $f' \circ a = b \circ f$. By
Lemma \ref{lemma-the-same-up-to-isomorphisms} we may assume that
$(X, Y, Z, f, g, h) = (X, Y, C(f), f, i, -p)$ and
$(X', Y', Z', f', g', h') = (X', Y', C(f'), f', i', -p')$.
At this point we simply apply Lemma \ref{lemma-functorial-cone}
to the commutative diagram given by $f, f', a, b$.
\end{proof}
\noindent
Before we prove TR4 in general we prove it in a special case.
\begin{lemma}
\label{lemma-two-split-injections}
Let $(A, \text{d})$ be a differential graded algebra. Suppose that
$\alpha : K \to L$ and $\beta : L \to M$ are admissible monomorphisms
of differential graded $A$-modules. Then there exist distinguished triangles
$(K, L, Q_1, \alpha, p_1, d_1)$, $(K, M, Q_2, \beta \circ \alpha, p_2, d_2)$
and $(L, M, Q_3, \beta, p_3, d_3)$ for which TR4 holds.
\end{lemma}
\begin{proof}
Say $\pi_1 : L \to K$ and $\pi_3 : M \to L$ are homomorphisms
of graded $A$-modules which are left inverse to $\alpha$ and $\beta$.
Then also $K \to M$ is an admissible monomorphism with left
inverse $\pi_2 = \pi_1 \circ \pi_3$. Let us write $Q_1$, $Q_2$
and $Q_3$ for the cokernels of $K \to L$, $K \to M$, and $L \to M$.
Then we obtain identifications (as graded $A$-modules)
$Q_1 = \Ker(\pi_1)$, $Q_3 = \Ker(\pi_3)$ and
$Q_2 = \Ker(\pi_2)$. Then $L = K \oplus Q_1$ and
$M = L \oplus Q_3$ as graded $A$-modules. This implies
$M = K \oplus Q_1 \oplus Q_3$. Note that $\pi_2 = \pi_1 \circ \pi_3$
is zero on both $Q_1$ and $Q_3$. Hence $Q_2 = Q_1 \oplus Q_3$.
Consider the commutative diagram
$$
\begin{matrix}
0 & \to & K & \to & L & \to & Q_1 & \to & 0 \\
& & \downarrow & & \downarrow & & \downarrow & \\
0 & \to & K & \to & M & \to & Q_2 & \to & 0 \\
& & \downarrow & & \downarrow & & \downarrow & \\
0 & \to & L & \to & M & \to & Q_3 & \to & 0
\end{matrix}
$$
The rows of this diagram are admissible short exact sequences, and
hence determine distinguished triangles by definition. Moreover
downward arrows in the diagram above are compatible with the chosen
splittings and hence define morphisms of triangles
$$
(K \to L \to Q_1 \to K[1])
\longrightarrow
(K \to M \to Q_2 \to K[1])
$$
and
$$
(K \to M \to Q_2 \to K[1])
\longrightarrow
(L \to M \to Q_3 \to L[1]).
$$
Note that the splittings $Q_3 \to M$ of the bottom sequence in the
diagram provides a splitting for the split sequence
$0 \to Q_1 \to Q_2 \to Q_3 \to 0$ upon composing with $M \to Q_2$.
It follows easily from this that the morphism $\delta : Q_3 \to Q_1[1]$
in the corresponding distinguished triangle
$$
(Q_1 \to Q_2 \to Q_3 \to Q_1[1])
$$
is equal to the composition $Q_3 \to L[1] \to Q_1[1]$.
Hence we get a structure as in the conclusion of axiom TR4.
\end{proof}
\noindent
Here is the final result.
\begin{proposition}
\label{proposition-homotopy-category-triangulated}
Let $(A, \text{d})$ be a differential graded algebra. The homotopy category
$K(\text{Mod}_{(A, \text{d})})$ of differential graded $A$-modules with its
natural translation functors and distinguished triangles is a triangulated
category.
\end{proposition}
\begin{proof}
We know that $K(\text{Mod}_{(A, \text{d})})$ is a pre-triangulated category.
Hence it suffices to prove TR4 and to prove it we can use
Derived Categories, Lemma \ref{derived-lemma-easier-axiom-four}.
Let $K \to L$ and $L \to M$ be composable morphisms of
$K(\text{Mod}_{(A, \text{d})})$. By
Lemma \ref{lemma-sequence-maps-split} we may assume that
$K \to L$ and $L \to M$ are admissible monomorphisms.
In this case the result follows from
Lemma \ref{lemma-two-split-injections}.
\end{proof}
\section{Projective modules over algebras}
\label{section-projectives-over-algebras}
\noindent
In this section we discuss projective modules over algebras
and over graded algebras. Thus it is the analogue of
Algebra, Section \ref{algebra-section-projective}
in the setting of this chapter.
\medskip\noindent
{\bf Algebras and modules.} Let $R$ be a ring and let $A$ be an
$R$-algebra, see Section \ref{section-conventions} for our conventions.
It is clear that $A$ is a projective right $A$-module since
$\Hom_A(A, M) = M$ for any right $A$-module $M$ (and thus $\Hom_A(A, -)$
is exact). Conversely, let $P$ be a projective right $A$-module. Then
we can choose a surjection
$\bigoplus_{i \in I} A \to P$ by choosing a set $\{p_i\}_{i \in I}$
of generators of $P$ over $A$. Since $P$ is projective there is a
left inverse to the surjection, and we find that $P$ is isomorphic
to a direct summand of a free module, exactly as in the commutative case
(Algebra, Lemma \ref{algebra-lemma-characterize-projective}).
\medskip\noindent
{\bf Graded algebras and modules.}
Let $R$ be a ring. Let $A$ be a graded algebra over $R$.
Let $\text{Mod}_A$ denote the category of graded right $A$-modules.
For an integer $k$ let $A[k]$ denote the shift of $A$.
For an graded right $A$-module we have
$$
\Hom_{\text{Mod}_A}(A[k], M) = M^{-k}
$$
As the functor $M \mapsto M^{-k}$ is exact on $\text{Mod}_A$ we
conclude that $A[k]$ is a projective object of $\text{Mod}_A$.
Conversely, suppose that $P$ is a projective object of $\text{Mod}_A$.
By choosing a set of homogeneous generators of $P$ as an $A$-module,
we can find a surjection
$$
\bigoplus\nolimits_{i \in I} A[k_i] \longrightarrow P
$$
Thus we conclude that a projective object of $\text{Mod}_A$ is
a direct summand of a direct sum of the shifts $A[k]$.
\medskip\noindent
If $(A, \text{d})$ is a differential graded algebra and $P$ is
an object of $\text{Mod}_{(A, \text{d})}$ then we say
{\it $P$ is projective as a graded $A$-module} or sometimes
{\it $P$ is graded projective} to mean
that $P$ is a projective object of the abelian category $\text{Mod}_A$
of graded $A$-modules.
\begin{lemma}
\label{lemma-target-graded-projective}
Let $(A, \text{d})$ be a differential graded algebra.
Let $M \to P$ be a surjective homomorphism of differential graded
$A$-modules. If $P$ is projective as a graded $A$-module, then
$M \to P$ is an admissible epimorphism.
\end{lemma}
\begin{proof}
This is immediate from the definitions.
\end{proof}
\begin{lemma}
\label{lemma-hom-from-shift-free}
Let $(A, d)$ be a differential graded algebra. Then we have
$$
\Hom_{\text{Mod}_{(A, \text{d})}}(A[k], M) =
\Ker(\text{d} : M^{-k} \to M^{-k + 1})
$$
and
$$
\Hom_{K(\text{Mod}_{(A, \text{d})})}(A[k], M) = H^{-k}(M)
$$
for any differential graded $A$-module $M$.
\end{lemma}
\begin{proof}
This is clear from the discussion above.
\end{proof}
\section{Injective modules over algebras}
\label{section-modules-noncommutative}
\noindent
In this section we discuss injective modules over algebras
and over graded algebras. Thus it is the analogue of
More on Algebra, Section \ref{more-algebra-section-injectives-modules}
in the setting of this chapter.
\medskip\noindent
{\bf Algebras and modules.} Let $R$ be a ring and let $A$ be an
$R$-algebra, see Section \ref{section-conventions} for our conventions.
For a right $A$-module $M$ we set
$$
M^\vee = \Hom_\mathbf{Z}(M, \mathbf{Q}/\mathbf{Z})
$$
which we think of as a left $A$-module by the multiplication
$(a f)(x) = f(xa)$. Namely, $((ab)f)(x) = f(xab) = (bf)(xa) = (a(bf))(x)$.
Conversely, if $M$ is a left $A$-module, then $M^\vee$ is a right
$A$-module. Since $\mathbf{Q}/\mathbf{Z}$ is an injective abelian
group (More on Algebra, Lemma \ref{more-algebra-lemma-injective-abelian}), the
functor $M \mapsto M^\vee$ is exact
(More on Algebra, Lemma \ref{more-algebra-lemma-vee-exact}).
Moreover, the evaluation map $M \to (M^\vee)^\vee$ is
injective for all modules $M$
(More on Algebra, Lemma \ref{more-algebra-lemma-ev-injective}).
\medskip\noindent
We claim that $A^\vee$ is an injective right $A$-module. Namely, given
a right $A$-module $N$ we have
$$
\Hom_A(N, A^\vee) =
\Hom_A(N, \Hom_\mathbf{Z}(A, \mathbf{Q}/\mathbf{Z})) = N^\vee
$$
and we conclude because the functor $N \mapsto N^\vee$ is exact.
The second equality holds because
$$
\Hom_\mathbf{Z}(N, \Hom_\mathbf{Z}(A, \mathbf{Q}/\mathbf{Z})) =
\Hom_\mathbf{Z}(N \otimes_\mathbf{Z} A, \mathbf{Q}/\mathbf{Z})
$$
by Algebra, Lemma \ref{algebra-lemma-hom-from-tensor-product}.
Inside this module $A$-linearity exactly picks out the bilinear maps
$\varphi : N \times A \to \mathbf{Q}/\mathbf{Z}$ which
have the same value on $x \otimes a$ and $xa \otimes 1$, i.e.,
come from elements of $N^\vee$.
\medskip\noindent
Finally, for every right $A$-module $M$ we can choose a surjection
$\bigoplus_{i \in I} A \to M^\vee$ to get an injection
$M \to (M^\vee)^\vee \to \prod_{i \in I} A^\vee$.
\medskip\noindent
We conclude
\begin{enumerate}
\item the category of $A$-modules has enough injectives,
\item $A^\vee$ is an injective $A$-module, and
\item every $A$-module injects into a product of copies of $A^\vee$.
\end{enumerate}
\noindent
{\bf Graded algebras and modules.}
Let $R$ be a ring. Let $A$ be a graded algebra over $R$.
If $M$ is a graded $A$-module we set
$$
M^\vee =
\bigoplus\nolimits_{n \in \mathbf{Z}}
\Hom_\mathbf{Z}(M^{-n}, \mathbf{Q}/\mathbf{Z}) =
\bigoplus\nolimits_{n \in \mathbf{Z}} (M^{-n})^\vee
$$
as a graded $R$-module with the $A$-module structure defined as above
(for homogeneous elements). This again switches left and right modules.
On the category of graded $A$-modules the functor $M \mapsto M^\vee$
is exact (check on graded pieces). Moreover, the evaluation map
$M \to (M^\vee)^\vee$ is injective as before (because we can check
this on the graded pieces).
\medskip\noindent
We claim that $A^\vee$ is an injective object of the category
$\text{Mod}_A$ of graded right $A$-modules. Namely, given a graded
right $A$-module $N$ we have
$$
\Hom_{\text{Mod}_A}(N, A^\vee) =
\Hom_{\text{Mod}_A}(
N, \bigoplus \Hom_\mathbf{Z}(A^{-n}, \mathbf{Q}/\mathbf{Z})) = (N^0)^\vee
$$
and we conclude because the functor $N \mapsto (N^0)^\vee = (N^\vee)^0$
is exact. To see that the second equality holds we use the equalities
$$
\Hom_\mathbf{Z}(N^n, \Hom_\mathbf{Z}(A^{-n}, \mathbf{Q}/\mathbf{Z})) =
\Hom_\mathbf{Z}(N^n \otimes_\mathbf{Z} A^{-n}, \mathbf{Q}/\mathbf{Z})
$$
of Algebra, Lemma \ref{algebra-lemma-hom-from-tensor-product}.
Thus an element of $\Hom_{\text{Mod}_A}(N, A^\vee)$ corresponds
to a family of $\mathbf{Z}$-bilinear maps
$\psi_n : N^n \times A^{-n} \to \mathbf{Q}/\mathbf{Z}$ such that
$\psi_n(x, a) = \psi_0(xa, 1)$ for all $x \in N^n$ and $a \in A^{-n}$.
Moreover, $\psi_0(x, a) = \psi_0(xa, 1)$ for all $x \in N^0$, $a \in A^0$.
It follows that the maps $\psi_n$ are determined by $\psi_0$ and that
$\psi_0(x, a) = \varphi(xa)$ for a unique element $\varphi \in (N^0)^\vee$.
\medskip\noindent
Finally, for every graded right $A$-module $M$ we can choose a surjection
(of graded left $A$-modules)
$$
\bigoplus\nolimits_{i \in I} A[k_i] \to M^\vee
$$
where $A[k_i]$ denotes the shift of $A$ by $k_i \in \mathbf{Z}$.
(We do this by choosing homogeneous generators for $M^\vee$.)
In this way we get an injection
$$
M \to (M^\vee)^\vee \to \prod A[k_i]^\vee = \prod A^\vee[-k_i]
$$
Observe that the products in the formula above are products in the
category of graded modules (in other words, take products in each degree
and then take the direct sum of the pieces).
\medskip\noindent
We conclude that
\begin{enumerate}
\item the category of graded $A$-modules has enough injectives,
\item for every $k \in \mathbf{Z}$ the module $A^\vee[k]$ is injective, and
\item every $A$-module injects into a product in the category of graded
modules of copies of shifts $A^\vee[k]$.
\end{enumerate}
If $(A, \text{d})$ is a differential graded algebra and $I$ is
an object of $\text{Mod}_{(A, \text{d})}$ then we say
{\it $I$ is injective as a graded $A$-module} to mean
that $I$ is a injective object of the abelian category $\text{Mod}_A$
of graded $A$-modules.
\begin{lemma}
\label{lemma-source-graded-injective}
Let $(A, \text{d})$ be a differential graded algebra.
Let $I \to M$ be an injective homomorphism of differential graded
$A$-modules. If $I$ is an injective object of the category
of graded $A$-modules, then $I \to M$ is an admissible monomorphism.
\end{lemma}
\begin{proof}
This is immediate from the definitions.
\end{proof}
\noindent
Let $(A, \text{d})$ be a differential graded algebra. If $M$ is a
{\bf left} differential graded $A$-module, then we will endow $M^\vee$
(with its graded module structure as above) with a right differential
graded module structure by setting
$$
\text{d}_{M^\vee}(f) = - (-1)^n f \circ \text{d}_M^{-n - 1}
\quad\text{in }(M^\vee)^{n + 1}
$$
for $f \in (M^\vee)^n = \Hom_\mathbf{Z}(M^{-n}, \mathbf{Q}/\mathbf{Z})$
and $\text{d}_M^{-n - 1} : M^{-n - 1} \to M^{-n}$ the differential
of $M$\footnote{The sign rule is analogous to the one in
Example \ref{example-dgm-dg-cat}, although there we are working with
right modules and the same sign rule taken there does not work for
left modules. Sigh!}.
We will show by a computation that this works.
Namely, if $a \in A^m$, $x \in M^{-n - m - 1}$ and $f \in (M^\vee)^n$,
then we have
\begin{align*}
\text{d}_{M^\vee}(f a)(x) & =
- (-1)^{n + m} (f a)(\text{d}_M(x)) \\
& =
- (-1)^{n + m} f(a\text{d}_M(x)) \\
& =
-(-1)^n f(\text{d}_M(ax) - \text{d}(a)x) \\
& =
-(-1)^n[-(-1)^n \text{d}_{M^\vee}(f)(ax) - (f\text{d}(a))(x)] \\
& =
(\text{d}_{M^\vee}(f)a)(x) + (-1)^n (f\text{d}(a))(x)
\end{align*}
the third equality because
$\text{d}_M(ax) = \text{d}(a)x + (-1)^m a\text{d}_M(x)$.
In other words we have
$\text{d}_{M^\vee}(fa) = \text{d}_{M^\vee}(f)a + (-1)^n f \text{d}(a)$
as desired.
\medskip\noindent
If $M$ is a {\bf right} differential graded module, then the sign rule above
does not work. The problem seems to be that in defining the left $A$-module
structure on $M^\vee$ our conventions for graded modules above defines $af$
to be the element of $(M^\vee)^{n + m}$ such that $(af)(x) = f(xa)$ for
$f \in (M^\vee)^n$, $a \in A^m$ and $x \in M^{-n - m}$ which in some sense
is the ``wrong'' thing to do if $m$ is odd. Anyway, instead of changing
the sign rule for the module structure, we fix the problem by using
$$
\text{d}_{M^\vee}(f) = (-1)^n f \circ \text{d}_M^{-n - 1}
$$
when $M$ is a right differential graded $A$-module. The computation for
$a \in A^m$, $x \in M^{-n - m - 1}$ and $f \in (M^\vee)^n$ then becomes
\begin{align*}
\text{d}_{M^\vee}(a f)(x) & =
(-1)^{n + m} (f a)(\text{d}_M(x)) \\
& =
(-1)^{n + m} f(\text{d}_M(x)a) \\
& =
(-1)^{n + m} f(\text{d}_M(ax) - (-1)^{m + n + 1} x\text{d}(a)) \\
& =
(-1)^m \text{d}_{M^\vee}(f)(ax) + f(x\text{d}(a)) \\
& =
(-1)^m (a\text{d}_{M^\vee}(f))(x) + (\text{d}(a)f)(x)
\end{align*}
the third equality because
$\text{d}_M(xa) = \text{d}_M(x)a + (-1)^{n + m + 1} x\text{d}(a)$.
In other words, we have
$\text{d}_{M^\vee}(af) = \text{d}(a) f + (-1)^ma\text{d}_{M^\vee}(f)$
as desired.
\medskip\noindent
We leave it to the reader to show that with the conventions above
there is a natural evaluation map $M \to (M^\vee)^\vee$ in the
category of differential graded modules if $M$ is either a differential
graded left module or a differential graded right module. This works
because the sign choices above cancel out and the differentials of
$((M^\vee)^\vee$ are the natural maps
$((M^n)^\vee)^\vee \to ((M^{n + 1})^\vee)^\vee$.
\begin{lemma}
\label{lemma-map-into-dual}
Let $(A, \text{d})$ be a differential graded algebra. If
$M$ is a left differential graded $A$-module and $N$ is a
right differential graded $A$-module, then
$$
\Hom_{\text{Mod}_{(A, \text{d})}}(N, M^\vee)
$$
is isomorphic to the set of sequences $(\psi_n)$ of $\mathbf{Z}$-bilinear
pairings
$$
\psi_n : N^n \times M^{-n} \longrightarrow \mathbf{Q}/\mathbf{Z}
$$
such that $\psi_{n + m}(y, ax) = \psi_{n + m}(ya, x)$ for all
$y \in N^n$, $x \in M^{-m}$, and $a \in A^{m - n}$ and such that
$\psi_{n + 1}(\text{d}(y), x) + (-1)^n \psi_n(y, \text{d}(x)) = 0$
for all $y \in N^n$ and $x \in M^{-n - 1}$.
\end{lemma}
\begin{proof}
If $f \in \Hom_{\text{Mod}_{(A, \text{d})}}(N, M^\vee)$, then we
map this to the sequence of pairings defined by
$\psi_n(y, x) = f(y)(x)$. It is a computation (omitted) to see
that these pairings satisfy the conditions as in the lemma.
For the converse, use
Algebra, Lemma \ref{algebra-lemma-hom-from-tensor-product} to turn a sequence
of pairings into a map $f : N \to M^\vee$.
\end{proof}
\begin{lemma}
\label{lemma-hom-into-shift-dual-free}
Let $(A, \text{d})$ be a differential graded algebra. Then we have
$$
\Hom_{\text{Mod}_{(A, \text{d})}}(M, A^\vee[k]) =
\Ker(\text{d} : (M^\vee)^k \to (M^\vee)^{k + 1})
$$
and
$$
\Hom_{K(\text{Mod}_{(A, \text{d})})}(M, A^\vee[k]) = H^k(M^\vee)
$$
for any differential graded $A$-module $M$.
\end{lemma}
\begin{proof}
This is clear from the discussion above.
\end{proof}
\section{P-resolutions}
\label{section-P-resolutions}
\noindent
This section is the analogue of
Derived Categories, Section \ref{derived-section-unbounded}.
\medskip\noindent
Let $(A, \text{d})$ be a differential graded algebra.
Let $P$ be a differential graded $A$-module. We say $P$
{\it has property (P)} if it there exists a filtration
$$
0 = F_{-1}P \subset F_0P \subset F_1P \subset \ldots \subset P
$$
by differential graded submodules such that
\begin{enumerate}
\item $P = \bigcup F_pP$,
\item the inclusions $F_iP \to F_{i + 1}P$ are admissible
monomorphisms,
\item the quotients $F_{i + 1}P/F_iP$ are isomorphic as differential
graded $A$-modules to a direct sum of $A[k]$.
\end{enumerate}
In fact, condition (2) is a consequence of condition (3), see
Lemma \ref{lemma-target-graded-projective}. Moreover, the reader
can verify that as a graded $A$-module $P$ will be isomorphic to a
direct sum of shifts of $A$.
\begin{lemma}
\label{lemma-property-P-sequence}
Let $(A, \text{d})$ be a differential graded algebra.
Let $P$ be a differential graded $A$-module. If $F_\bullet$
is a filtration as in property (P), then we obtain an
admissible short exact sequence
$$
0 \to
\bigoplus\nolimits F_iP \to
\bigoplus\nolimits F_iP \to P \to 0
$$
of differential graded $A$-modules.
\end{lemma}
\begin{proof}
The second map is the direct sum of the inclusion maps.
The first map on the summand $F_iP$ of the source is the sum
of the identity $F_iP \to F_iP$ and the negative of the inclusion
map $F_iP \to F_{i + 1}P$. Choose homomorphisms $s_i : F_{i + 1}P \to F_iP$
of graded $A$-modules which are left inverse to the inclusion maps.
Composing gives maps $s_{j, i} : F_jP \to F_iP$ for all $j > i$.
Then a left inverse of the first arrow maps $x \in F_jP$ to
$(s_{j, 0}(x), s_{j, 1}(x), \ldots, s_{j, j - 1}(x), 0, \ldots)$
in $\bigoplus F_iP$.
\end{proof}
\noindent
The following lemma shows that differential graded modules with
property (P) are the dual notion to K-injective modules
(i.e., they are K-projective in some sense). See
Derived Categories, Definition \ref{derived-definition-K-injective}.
\begin{lemma}
\label{lemma-property-P-K-projective}
Let $(A, \text{d})$ be a differential graded algebra.
Let $P$ be a differential graded $A$-module with property (P).
Then
$$
\Hom_{K(\text{Mod}_{(A, \text{d})})}(P, N) = 0
$$
for all acyclic differential graded $A$-modules $N$.
\end{lemma}
\begin{proof}
We will use that $K(\text{Mod}_{(A, \text{d})})$ is a triangulated
category (Proposition \ref{proposition-homotopy-category-triangulated}).
Let $F_\bullet$ be a filtration on $P$ as in property (P).
The short exact sequence of Lemma \ref{lemma-property-P-sequence}
produces a distinguished triangle. Hence by
Derived Categories, Lemma \ref{derived-lemma-representable-homological}
it suffices to show that
$$
\Hom_{K(\text{Mod}_{(A, \text{d})})}(F_iP, N) = 0
$$
for all acyclic differential graded $A$-modules $N$ and all $i$.
Each of the differential graded modules $F_iP$ has a finite filtration
by admissible monomorphisms, whose graded pieces are direct sums
of shifts $A[k]$. Thus it suffices to prove that
$$
\Hom_{K(\text{Mod}_{(A, \text{d})})}(A[k], N) = 0
$$
for all acyclic differential graded $A$-modules $N$ and all $k$.
This follows from Lemma \ref{lemma-hom-from-shift-free}.
\end{proof}
\begin{lemma}
\label{lemma-good-quotient}
Let $(A, \text{d})$ be a differential graded algebra.
Let $M$ be a differential graded $A$-module. There exists a homomorphism
$P \to M$ of differential graded $A$-modules with the following
properties
\begin{enumerate}
\item $P \to M$ is surjective,
\item $\Ker(\text{d}_P) \to \Ker(\text{d}_M)$ is surjective, and
\item $P$ sits in an admissible short exact sequence
$0 \to P' \to P \to P'' \to 0$ where $P'$, $P''$ are direct sums
of shifts of $A$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $P_k$ be the free $A$-module with generators $x, y$ in degrees
$k$ and $k + 1$. Define the structure of a differential graded
$A$-module on $P_k$ by setting $\text{d}(x) = y$ and $\text{d}(y) = 0$.
For every element $m \in M^k$ there is a homomorphism
$P_k \to M$ sending $x$ to $m$ and $y$ to $\text{d}(m)$.
Thus we see that there is a surjection from a direct sum
of copies of $P_k$ to $M$. This clearly produces $P \to M$
having properties (1) and (3). To obtain property (2) note
that if $m \in \Ker(\text{d}_M)$ has degree $k$, then there is a map
$A[k] \to M$ mapping $1$ to $m$. Hence we can achieve (2) by adding
a direct sum of copies of shifts of $A$.
\end{proof}
\begin{lemma}
\label{lemma-resolve}
Let $(A, \text{d})$ be a differential graded algebra.
Let $M$ be a differential graded $A$-module. There exists a homomorphism
$P \to M$ of differential graded $A$-modules such that
\begin{enumerate}
\item $P \to M$ is a quasi-isomorphism, and
\item $P$ has property (P).
\end{enumerate}
\end{lemma}
\begin{proof}
Set $M = M_0$. We inductively choose short exact sequences
$$
0 \to M_{i + 1} \to P_i \to M_i \to 0
$$
where the maps $P_i \to M_i$ are chosen as in Lemma \ref{lemma-good-quotient}.
This gives a ``resolution''
$$
\ldots \to P_2 \xrightarrow{f_2} P_1 \xrightarrow{f_1} P_0 \to M \to 0
$$
Then we set
$$
P = \bigoplus\nolimits_{i \geq 0} P_i
$$
as an $A$-module with grading given by
$P^n = \bigoplus_{a + b = n} P_{-a}^b$ and
differential (as in the construction of the total complex associated
to a double complex) by
$$
\text{d}_P(x) = f_{-a}(x) + (-1)^a \text{d}_{P_{-a}}(x)
$$
for $x \in P_{-a}^b$. With these conventions $P$ is indeed a differential
graded $A$-module. Recalling that each $P_i$ has a two step filtration
$0 \to P_i' \to P_i \to P_i'' \to 0$ we set
$$
F_{2i}P = \bigoplus\nolimits_{i \geq j \geq 0} P_j
\subset
\bigoplus\nolimits_{i \geq 0} P_i = P
$$
and we add $P'_{i + 1}$ to $F_{2i}P$ to get $F_{2i + 1}$.
These are differential graded submodules and the successive quotients
are direct sums of shifts of $A$. By
Lemma \ref{lemma-target-graded-projective} we see that
the inclusions $F_iP \to F_{i + 1}P$ are admissible monomorphisms.
Finally, we have to show that the map $P \to M$ (given by the
augmentation $P_0 \to M$) is a quasi-isomorphism. This follows from
Homology, Lemma \ref{homology-lemma-good-resolution-gives-qis}.
\end{proof}
\section{I-resolutions}
\label{section-I-resolutions}
\noindent
This section is the dual of the section on P-resolutions.
\medskip\noindent
Let $(A, \text{d})$ be a differential graded algebra.
Let $I$ be a differential graded $A$-module. We say $I$
{\it has property (I)} if it there exists a filtration
$$
I = F_0I \supset F_1I \supset F_2I \supset \ldots \supset 0
$$
by differential graded submodules such that
\begin{enumerate}
\item $I = \lim I/F_pI$,
\item the maps $I/F_{i + 1}I \to I/F_iI$ are admissible epimorphisms,
\item the quotients $F_iI/F_{i + 1}I$ are isomorphic as differential
graded $A$-modules to products of $A^\vee[k]$.
\end{enumerate}
In fact, condition (2) is a consequence of condition (3), see
Lemma \ref{lemma-source-graded-injective}. The reader can verify that as
a graded module $I$ will be isomorphic to a product of $A^\vee[k]$.
\begin{lemma}
\label{lemma-property-I-sequence}
Let $(A, \text{d})$ be a differential graded algebra.
Let $I$ be a differential graded $A$-module. If $F_\bullet$
is a filtration as in property (I), then we obtain an
admissible short exact sequence
$$
0 \to I \to
\prod\nolimits I/F_iI \to
\prod\nolimits I/F_iI \to 0
$$
of differential graded $A$-modules.
\end{lemma}
\begin{proof}
Omitted. Hint: This is dual to Lemma \ref{lemma-property-P-sequence}.
\end{proof}
\noindent
The following lemma shows that differential graded modules with
property (I) are the analogue of K-injective modules. See
Derived Categories, Definition \ref{derived-definition-K-injective}.
\begin{lemma}
\label{lemma-property-I-K-injective}
Let $(A, \text{d})$ be a differential graded algebra.
Let $I$ be a differential graded $A$-module with property (I).
Then
$$
\Hom_{K(\text{Mod}_{(A, \text{d})})}(N, I) = 0
$$
for all acyclic differential graded $A$-modules $N$.
\end{lemma}
\begin{proof}
We will use that $K(\text{Mod}_{(A, \text{d})})$ is a triangulated
category (Proposition \ref{proposition-homotopy-category-triangulated}).
Let $F_\bullet$ be a filtration on $I$ as in property (I).
The short exact sequence of Lemma \ref{lemma-property-I-sequence}
produces a distinguished triangle. Hence by
Derived Categories, Lemma \ref{derived-lemma-representable-homological}
it suffices to show that
$$
\Hom_{K(\text{Mod}_{(A, \text{d})})}(N, I/F_iI) = 0
$$
for all acyclic differential graded $A$-modules $N$ and all $i$.
Each of the differential graded modules $I/F_iI$ has a finite filtration
by admissible monomorphisms, whose graded pieces are
products of $A^\vee[k]$. Thus it suffices to prove that
$$
\Hom_{K(\text{Mod}_{(A, \text{d})})}(N, A^\vee[k]) = 0
$$
for all acyclic differential graded $A$-modules $N$ and all $k$.
This follows from Lemma \ref{lemma-hom-into-shift-dual-free}
and the fact that $(-)^\vee$ is an exact functor.
\end{proof}
\begin{lemma}
\label{lemma-good-sub}
Let $(A, \text{d})$ be a differential graded algebra.
Let $M$ be a differential graded $A$-module. There exists a homomorphism
$M \to I$ of differential graded $A$-modules with the following
properties
\begin{enumerate}
\item $M \to I$ is injective,
\item $\Coker(\text{d}_M) \to \Coker(\text{d}_I)$ is injective,
and
\item $I$ sits in an admissible short exact sequence
$0 \to I' \to I \to I'' \to 0$ where $I'$, $I''$ are products
of shifts of $A^\vee$.
\end{enumerate}
\end{lemma}
\begin{proof}
For every $k \in \mathbf{Z}$ let $Q_k$ be the free left $A$-module with
generators $x, y$ in degrees $k$ and $k + 1$. Define the structure of a
left differential graded $A$-module on $Q_k$ by setting $\text{d}(x) = y$
and $\text{d}(y) = 0$. Let $I_k = Q_{-k}^\vee$ be the ``dual'' right
differential graded $A$-module, see
Section \ref{section-modules-noncommutative}.
The next paragraph shows that we can embed $M$ into a product
of copies of $I_k$ (for varying $k$). The dual statement (that any
differential graded module is a quotient of a direct sum of of $P_k$'s)
is easy to prove (see proof of Lemma \ref{lemma-good-quotient})
and using double duals there should be a noncomputational
way to deduce what we want. Thus we suggest skipping the next paragraph.
\medskip\noindent
Given a $\mathbf{Z}$-linear map $\lambda : M^k \to \mathbf{Q}/\mathbf{Z}$
we construct pairings
$$
\psi_n : M^n \times Q_k^{-n} \longrightarrow \mathbf{Q}/\mathbf{Z}
$$
by setting
$$
\psi_n(m, ax + by) = \lambda(ma + (-1)^{k + 1}\text{d}(mb))
$$
for $m \in M^n$, $a \in A^{-n - k}$, and $b \in A^{-n - k - 1}$. We compute
\begin{align*}
\psi_{n + 1}(\text{d}(m), ax + by)
& =
\lambda\left(\text{d}(m)a + (-1)^{k + 1}\text{d}(\text{d}(m)b)\right) \\
& =
\lambda\left(\text{d}(m)a + (-1)^{k + n}\text{d}(m)\text{d}(b)\right)
\end{align*}
and because
$\text{d}(ax + by) = \text{d}(a)x + (-1)^{-n - k}ay + \text{d}(b)y$ we have
\begin{align*}
\psi_n(m, \text{d}(ax + by))
& =
\lambda\left(
m\text{d}(a) + (-1)^{k + 1}\text{d}(m((-1)^{-n - k}a + \text{d}(b)))
\right) \\
& =
\lambda\left(
m\text{d}(a) + (-1)^{-n + 1}\text{d}(ma) + (-1)^{k + 1}\text{d}(m)\text{d}(b)))
\right)
\end{align*}
and we see that
$$
\psi_{n + 1}(\text{d}(m), ax + by) + (-1)^n\psi_n(m, \text{d}(ax + by)) = 0
$$
Thus these pairings define a homomorphism
$f_\lambda : M \to I_k$ by Lemma \ref{lemma-map-into-dual}
such that the composition
$$
M^k \xrightarrow{f^k_\lambda} I_k^k = (Q_k^k)^\vee
\xrightarrow{\text{evaluation at }x} \mathbf{Q}/\mathbf{Z}
$$
is the given map $\lambda$. It is clear that we can find an embedding
into a product of copies of $I_k$'s by using a map of the form
$\prod f_\lambda$ for a suitable choice of the maps $\lambda$.
\medskip\noindent
The result of the previous paragraph produces $M \to I$
having properties (1) and (3). To obtain property (2), suppose
$\overline{m} \in \Coker(\text{d}_M)$ is a nonzero element of
degree $k$. Pick a map $\lambda : M^k \to \mathbf{Q}/\mathbf{Z}$
which vanishes on $\Im(M^{k - 1} \to M^k)$ but not on $m$. By
Lemma \ref{lemma-hom-into-shift-dual-free}
this corresponds to a homomorphism $M \to A^\vee[k]$ of
differential graded $A$-modules which does not vanish on $m$.
Hence we can achieve (2) by adding
a product of copies of shifts of $A^\vee$.
\end{proof}
\begin{lemma}
\label{lemma-right-resolution}
Let $(A, \text{d})$ be a differential graded algebra.
Let $M$ be a differential graded $A$-module. There exists a homomorphism
$M \to I$ of differential graded $A$-modules such that
\begin{enumerate}
\item $M \to I$ is a quasi-isomorphism, and
\item $I$ has property (I).
\end{enumerate}
\end{lemma}
\begin{proof}
Set $M = M_0$. We inductively choose short exact sequences
$$
0 \to M_i \to I_i \to M_{i + 1} \to 0
$$
where the maps $M_i \to I_i$ are chosen as in Lemma \ref{lemma-good-sub}.
This gives a ``resolution''
$$
0 \to M \to I_0 \xrightarrow{f_0} I_1 \xrightarrow{f_1} I_1 \to \ldots
$$
Then we set
$$
I = \prod\nolimits_{i \geq 0} I_i
$$
where we take the product in the category of graded $A$-modules
and differential defined by
$$
\text{d}_I(x) = f_a(x) + (-1)^a \text{d}_{I_a}(x)
$$
for $x \in I_a^b$. With these conventions $I$ is indeed a differential
graded $A$-module. Recalling that each $I_i$ has a two step filtration
$0 \to I_i' \to I_i \to I_i'' \to 0$ we set
$$
F_{2i}P = \prod\nolimits_{j \geq i} I_j
\subset
\prod\nolimits_{i \geq 0} I_i = I
$$
and we add a factor $I'_{i + 1}$ to $F_{2i}I$ to get $F_{2i + 1}I$.
These are differential graded submodules and the successive quotients
are products of shifts of $A^\vee$. By
Lemma \ref{lemma-source-graded-injective} we see that
the inclusions $F_{i + 1}I \to F_iI$ are admissible monomorphisms.
Finally, we have to show that the map $M \to I$ (given by the
augmentation $M \to I_0$) is a quasi-isomorphism. This follows from
Homology, Lemma \ref{homology-lemma-good-right-resolution-gives-qis}.
\end{proof}
\section{The derived category}
\label{section-derived}
\noindent
Recall that the notions of acyclic differential graded modules
and quasi-isomorphism of differential graded modules make sense
(see Section \ref{section-modules}).
\begin{lemma}
\label{lemma-acyclic}
Let $(A, \text{d})$ be a differential graded algebra.
The full subcategory $\text{Ac}$ of $K(\text{Mod}_{(A, \text{d})})$
consisting of acyclic modules is a strictly full saturated triangulated
subcategory of $K(\text{Mod}_{(A, \text{d})})$.
The corresponding saturated multiplicative system
(see Derived Categories, Lemma \ref{derived-lemma-operations})
of $K(\text{Mod}_{(A, \text{d})})$ is the class $\text{Qis}$
of quasi-isomorphisms. In particular, the kernel of the localization
functor
$$
Q : K(\text{Mod}_{(A, \text{d})}) \to
\text{Qis}^{-1}K(\text{Mod}_{(A, \text{d})})
$$
is $\text{Ac}$. Moreover, the functor $H^0$ factors through $Q$.
\end{lemma}
\begin{proof}
We know that $H^0$ is a homological functor by the long exact
sequence of homology (\ref{equation-les}).
The kernel of $H^0$ is the subcategory of acyclic objects and
the arrows with induce isomorphisms on all $H^i$ are the
quasi-isomorphisms. Thus this lemma is a special case of
Derived Categories, Lemma \ref{derived-lemma-acyclic-general}.
\medskip\noindent
Set theoretical remark. The construction of the localization in
Derived Categories, Proposition
\ref{derived-proposition-construct-localization}
assumes the given triangulated category is ``small'', i.e., that the
underlying collection of objects forms a set. Let $V_\alpha$ be a
partial universe (as in Sets, Section \ref{sets-section-sets-hierarchy})
containing $(A, \text{d})$ and where the cofinality of $\alpha$
is bigger than $\aleph_0$
(see Sets, Proposition \ref{sets-proposition-exist-ordinals-large-cofinality}).
Then we can consider the category $\text{Mod}_{(A, \text{d}), \alpha}$
of differential graded $A$-modules contained in $V_\alpha$.
A straightforward check shows that all the constructions used in
the proof of Proposition \ref{proposition-homotopy-category-triangulated}
work inside of $\text{Mod}_{(A, \text{d}), \alpha}$
(because at worst we take finite direct sums of differential graded modules).
Thus we obtain a triangulated category
$\text{Qis}_\alpha^{-1}K(\text{Mod}_{(A, \text{d}), \alpha})$.
We will see below that if $\beta > \alpha$, then the transition functors
$$
\text{Qis}_\alpha^{-1}K(\text{Mod}_{(A, \text{d}), \alpha})
\longrightarrow
\text{Qis}_\beta^{-1}K(\text{Mod}_{(A, \text{d}), \beta})
$$
are fully faithful as the morphism sets in the quotient categories
are computed by maps in the homotopy categories from P-resolutions
(the construction of a P-resolution in the proof of Lemma \ref{lemma-resolve}
takes countable direct sums as well as direct sums indexed over subsets
of the given module). The reader should therefore think of the category
of the lemma as the union of these subcategories.
\end{proof}
\noindent
Taking into account the set theoretical remark at the end of the
proof of the preceding lemma we define the derived category as follows.
\begin{definition}
\label{definition-unbounded-derived-category}
Let $(A, \text{d})$ be a differential graded algebra.
Let $\text{Ac}$ and $\text{Qis}$ be as in Lemma \ref{lemma-acyclic}.
The {\it derived category of $(A, \text{d})$} is the triangulated
category
$$
D(A, \text{d}) =
K(\text{Mod}_{(A, \text{d})})/\text{Ac} =
\text{Qis}^{-1}K(\text{Mod}_{(A, \text{d})}).
$$
We denote $H^0 : D(A, \text{d}) \to \text{Mod}_R$ the unique functor
whose composition with the quotient functor gives back the functor
$H^0$ defined above.
\end{definition}
\noindent
Here is the promised lemma computing morphism sets in the
derived category.
\begin{lemma}
\label{lemma-hom-derived}
Let $(A, \text{d})$ be a differential graded algebra.
Let $M$ and $N$ be differential graded $A$-modules.
\begin{enumerate}
\item Let $P \to M$ be a P-resolution as in
Lemma \ref{lemma-resolve}. Then
$$
\Hom_{D(A, \text{d})}(M, N) =
\Hom_{K(\text{Mod}_{(A, \text{d})})}(P, N)
$$
\item Let $N \to I$ be an I-resolution as in
Lemma \ref{lemma-right-resolution}. Then
$$
\Hom_{D(A, \text{d})}(M, N) =
\Hom_{K(\text{Mod}_{(A, \text{d})})}(M, I)
$$
\end{enumerate}
\end{lemma}
\begin{proof}
Let $P \to M$ be as in (1). Since $P \to M$ is a quasi-isomorphism we see that
$$
\Hom_{D(A, \text{d})}(P, N) = \Hom_{D(A, \text{d})}(M, N)
$$
by definition of the derived category. A morphism
$f : P \to N$ in $D(A, \text{d})$ is equal to
$s^{-1}f'$ where $f' : P \to N'$ is a morphism and
$s : N \to N'$ is a quasi-isomorphism. Choose a distinguished triangle
$$
N \to N' \to Q \to N[1]
$$
As $s$ is a quasi-isomorphism, we see that $Q$ is acyclic. Thus
$\Hom_{K(\text{Mod}_{(A, \text{d})})}(P, Q[k]) = 0$ for all $k$ by
Lemma \ref{lemma-property-P-K-projective}. Since
$\Hom_{K(\text{Mod}_{(A, \text{d})})}(P, -)$
is cohomological, we conclude that we can lift $f' : P \to N'$
uniquely to a morphism $f : P \to N$. This finishes the proof.
\medskip\noindent
The proof of (2) is dual to that of (1) using
Lemma \ref{lemma-property-I-K-injective} in stead of
Lemma \ref{lemma-property-P-K-projective}.
\end{proof}
\begin{lemma}
\label{lemma-derived-products}
Let $(A, \text{d})$ be a differential graded algebra. Then
\begin{enumerate}
\item $D(A, \text{d})$ has both direct sums and products,
\item direct sums are obtained by taking direct sums of differential graded
modules,
\item products are obtained by taking products of differential
graded modules.
\end{enumerate}
\end{lemma}
\begin{proof}
We will use that $\text{Mod}_{(A, \text{d})}$ is an abelian category
with arbitrary direct sums and products, and that these give rise
to direct sums and products in $K(\text{Mod}_{(A, \text{d})})$.
See Lemmas \ref{lemma-dgm-abelian} and \ref{lemma-homotopy-direct-sums}.
\medskip\noindent
Let $M_j$ be a family of differential graded $A$-modules.
Consider the graded direct sum $M = \bigoplus M_j$ which is a
differential graded $A$-module with the obvious.
For a differential graded $A$-module $N$ choose a quasi-isomorphism
$N \to I$ where $I$ is a differential graded $A$-module with property (I).
See Lemma \ref{lemma-right-resolution}.
Using Lemma \ref{lemma-hom-derived} we have
\begin{align*}
\Hom_{D(A, \text{d})}(M, N)
& =
\Hom_{K(A, \text{d})}(M, I) \\
& =
\prod \Hom_{K(A, \text{d})}(M_j, I) \\
& =
\prod \Hom_{D(A, \text{d})}(M_j, N)
\end{align*}
whence the existence of direct sums in $D(A, \text{d})$ as given in
part (2) of the lemma.
\medskip\noindent
Let $M_j$ be a family of differential graded $A$-modules.
Consider the product $M = \prod M_j$ of differential graded $A$-modules.
For a differential graded $A$-module $N$ choose a quasi-isomorphism
$P \to N$ where $P$ is a differential graded $A$-module with property (P).
See Lemma \ref{lemma-resolve}.
Using Lemma \ref{lemma-hom-derived} we have
\begin{align*}
\Hom_{D(A, \text{d})}(N, M)
& =
\Hom_{K(A, \text{d})}(P, M) \\
& =
\prod \Hom_{K(A, \text{d})}(P, M_j) \\
& =
\prod \Hom_{D(A, \text{d})}(N, M_j)
\end{align*}
whence the existence of direct sums in $D(A, \text{d})$ as given in
part (3) of the lemma.
\end{proof}
\section{The canonical delta-functor}
\label{section-canonical-delta-functor}
\noindent
Let $(A, \text{d})$ be a differential graded algebra.
Consider the functor
$\text{Mod}(\mathcal{A}) \to K(\text{Mod}_{(A, \text{d})})$.
This functor is {\bf not} a $\delta$-functor in general.
However, it turns out that the functor
$\text{Mod}_{(A, \text{d})} \to D(A, \text{d})$ is a
$\delta$-functor. In order to see this we have to define
the morphisms $\delta$ associated to a short exact sequence
$$
0 \to K \xrightarrow{a} L \xrightarrow{b} M \to 0
$$
in the abelian category $\text{Mod}_{(A, \text{d})}$.
Consider the cone $C(a)$ of the morphism $a$. We have $C(a) = L \oplus K$
and we define $q : C(a) \to M$ via the projection to $L$ followed
by $b$. Hence a homomorphism of differential graded $A$-modules
$$
q : C(a) \longrightarrow M.
$$
It is clear that $q \circ i = b$ where $i$ is as in
Definition \ref{definition-cone}.
Note that, as $a$ is injective, the kernel of $q$ is identified with the
cone of $\text{id}_K$ which is acyclic. Hence we see that
$q$ is a quasi-isomorphism. According to
Lemma \ref{lemma-the-same-up-to-isomorphisms}
the triangle
$$
(K, L, C(a), a, i, -p)
$$
is a distinguished triangle in $K(\text{Mod}_{(A, \text{d})})$.
As the localization functor
$K(\text{Mod}_{(A, \text{d})}) \to D(A, \text{d})$ is
exact we see that $(K, L, C(a), a, i, -p)$ is a distinguished
triangle in $D(A, \text{d})$. Since $q$ is a quasi-isomorphism
we see that $q$ is an isomorphism in $D(A, \text{d})$.
Hence we deduce that
$$
(K, L, M, a, b, -p \circ q^{-1})
$$
is a distinguished triangle of $D(A, \text{d})$.
This suggests the following lemma.
\begin{lemma}
\label{lemma-derived-canonical-delta-functor}
Let $(A, \text{d})$ be a differential graded algebra. The functor
$\text{Mod}_{(A, \text{d})} \to D(A, \text{d})$
defined has the natural structure of a $\delta$-functor, with
$$
\delta_{K \to L \to M} = - p \circ q^{-1}
$$
with $p$ and $q$ as explained above.
\end{lemma}
\begin{proof}
We have already seen that this choice leads to a distinguished
triangle whenever given a short exact sequence of complexes.
We have to show functoriality of this construction, see
Derived Categories, Definition \ref{derived-definition-delta-functor}.
This follows from Lemma \ref{lemma-functorial-cone} with a bit of
work. Compare with
Derived Categories, Lemma \ref{derived-lemma-derived-canonical-delta-functor}.
\end{proof}
\begin{lemma}
\label{lemma-homotopy-colimit}
Let $(A, \text{d})$ be a differential graded algebra. Let
$M_n$ be a system of differential graded modules. Then the derived
colimit $\text{hocolim} M_n$ in $D(A, \text{d})$ is represented
by the differential graded module $\colim M_n$.
\end{lemma}
\begin{proof}
Set $M = \colim M_n$. We have an exact sequence of differential graded modules
$$
0 \to \bigoplus M_n \to \bigoplus M_n \to M \to 0
$$
by Derived Categories, Lemma \ref{derived-lemma-compute-colimit}
(applied the the underlying complexes of abelian groups).
The direct sums are direct sums in $D(\mathcal{A})$ by
Lemma \ref{lemma-derived-products}.
Thus the result follows from the definition
of derived colimits in Derived Categories,
Definition \ref{derived-definition-derived-colimit}
and the fact that a short exact sequence of complexes
gives a distinguished triangle
(Lemma \ref{lemma-derived-canonical-delta-functor}).
\end{proof}
\section{Linear categories}
\label{section-linear}
\noindent
Just the definitions.
\begin{definition}
\label{definition-linear-category}
Let $R$ be a ring. An {\it $R$-linear category $\mathcal{A}$} is a category
where every morphism set is given the structure of an $R$-module
and where for $x, y, z \in \Ob(\mathcal{A})$ composition law
$$
\Hom_\mathcal{A}(y, z) \times \Hom_\mathcal{A}(x, y)
\longrightarrow
\Hom_\mathcal{A}(x, z)
$$
is $R$-bilinear.
\end{definition}
\noindent
Thus composition determines an $R$-linear map
$$
\Hom_\mathcal{A}(y, z) \otimes_R \Hom_\mathcal{A}(x, y)
\longrightarrow
\Hom_\mathcal{A}(x, z)
$$
of $R$-modules. Note that we do not assume $R$-linear categories to be
additive.
\begin{definition}
\label{definition-functor-linear-categories}
Let $R$ be a ring. A {\it functor of $R$-linear categories}, or an
{\it $R$-linear} is a functor $F : \mathcal{A} \to \mathcal{B}$
where for all objects $x, y$ of $\mathcal{A}$ the map
$F : \Hom_\mathcal{A}(x, y) \to \Hom_\mathcal{A}(F(x), F(y))$
is a homomorphism of $R$-modules.
\end{definition}
\section{Graded categories}
\label{section-graded}
\noindent
Just some definitions.
\begin{definition}
\label{definition-graded-category}
Let $R$ be a ring. A {\it graded category $\mathcal{A}$
over $R$} is a category where every morphism set is given the structure
of a graded $R$-module and where for
$x, y, z \in \Ob(\mathcal{A})$ composition is $R$-bilinear and induces
a homomorphism
$$
\Hom_\mathcal{A}(y, z) \otimes_R \Hom_\mathcal{A}(x, y)
\longrightarrow
\Hom_\mathcal{A}(x, z)
$$
of graded $R$-modules (i.e., preserving degrees).
\end{definition}
\noindent
In this situation we denote $\Hom_\mathcal{A}^i(x, y)$ the degree $i$
part of the graded object $\Hom_\mathcal{A}(x, y)$, so that
$$
\Hom_\mathcal{A}(x, y) =
\bigoplus\nolimits_{i \in \mathbf{Z}} \Hom_\mathcal{A}^i(x, y)
$$
is the direct sum decomposition into graded parts.
\begin{definition}
\label{definition-functor-graded-categories}
Let $R$ be a ring. A {\it functor of graded categories over $R$}, or a
{\it graded functor}
is a functor $F : \mathcal{A} \to \mathcal{B}$ where for all objects
$x, y$ of $\mathcal{A}$ the map
$F : \Hom_\mathcal{A}(x, y) \to \Hom_\mathcal{A}(F(x), F(y))$
is a homomorphism of graded $R$-modules.
\end{definition}
\noindent
Given a graded category we are often interested in the
corresponding ``usual'' category of maps of degree $0$.
Here is a formal definition.
\begin{definition}
\label{definition-H0-of-graded-category}
Let $R$ be a ring. Let $\mathcal{A}$ be a differential graded category
over $R$. We let {\it $\mathcal{A}^0$} be the category with the
same objects as $\mathcal{A}$ and with
$$
\Hom_{\mathcal{A}^0}(x, y) = \Hom^0_\mathcal{A}(x, y)
$$
the degree $0$ graded piece of the graded module of morphisms of
$\mathcal{A}$.
\end{definition}
\begin{definition}
\label{definition-graded-direct-sum}
Let $R$ be a ring. Let $\mathcal{A}$ be a graded category over $R$.
A direct sum $(x, y, z, i, j, p, q)$ in $\mathcal{A}$ (notation as in
Homology, Remark \ref{homology-remark-direct-sum})
is a {\it graded direct sum} if $i, j, p, q$ are homogeneous
of degree $0$.
\end{definition}
\begin{example}[Graded category of graded objects]
\label{example-graded-category-graded-objects}
Let $\mathcal{B}$ be an additive category. Recall that we have defined
the category $\text{Gr}(\mathcal{B})$ of graded objects of $\mathcal{B}$ in
Homology, Definition \ref{homology-definition-graded}.
In this example, we will construct a graded category
$\text{Gr}^{gr}(\mathcal{B})$ over $R = \mathbf{Z}$
whose associated category $\text{Gr}^{gr}(\mathcal{B})^0$
recovers $\text{Gr}(\mathcal{B})$.
As objects of $\text{Comp}^{gr}(\mathcal{B})$
we take graded objects of $\mathcal{B}$. Then, given graded objects
$A = (A^i)$ and $B = (B^i)$ of $\mathcal{B}$ we set
$$
\Hom_{\text{Gr}^{gr}(\mathcal{B})}(A, B) =
\bigoplus\nolimits_{n \in \mathbf{Z}} \Hom^n(A, B)
$$
where the graded piece of degree $n$ is the abelian group of homogeneous
maps of degree $n$ from $A$ to $B$
defined by the rule
$$
\Hom^n(A, B) = \Hom_{\text{Gr}(\mathcal{A})}(A, B[n]) =
\Hom_{\text{Gr}(\mathcal{A})}(A[-n], B)
$$
see Homology, Equation (\ref{homology-equation-hom-into-shift}).
Explicitly we have
$$
\Hom^n(A, B) = \prod\nolimits_{p + q = n} \Hom_\mathcal{B}(A^{-q}, B^p)
$$
(observe reversal of indices and observe that we have a product here and
not a direct sum). In other words, a degree $n$ morphism $f$
from $A$ to $B$ can be seen as a system $f = (f_{p, q})$ where
$p, q \in \mathbf{Z}$, $p + q = n$ with
$f_{p, q} : A^{-q} \to B^p$ a morphism of $\mathcal{B}$.
Given graded objects $A$, $B$, $C$ of $\mathcal{B}$
composition of morphisms in $\text{Gr}^{gr}(\mathcal{B})$ is defined
via the maps
$$
\Hom^m(B, C) \times \Hom^n(A, B) \longrightarrow \Hom^{n + m}(A, C)
$$
by simple composition $(g, f) \mapsto g \circ f$ of homogeneous
maps of graded objects. In terms of components we have
$$
(g \circ f)_{p, r} = g_{p, q} \circ f_{-q, r}
$$
where $q$ is such that $p + q = m$ and $-q + r = n$.
\end{example}
\begin{example}[Graded category of graded modules]
\label{example-gm-gr-cat}
Let $A$ be a $\mathbf{Z}$-graded algebra over a ring $R$. We will construct
a graded category $\text{Mod}^{gr}_A$ over $R$ whose associated category
$(\text{Mod}^{gr}_A)^0$ is the category of graded $A$-modules. As objects
of $\text{Mod}^{gr}_A$ we take right graded $A$-modules (see
Section \ref{section-projectives-over-algebras}). Given graded
$A$-modules $L$ and $M$ we set
$$
\Hom_{\text{Mod}^{gr}_A}(L, M) =
\bigoplus\nolimits_{n \in \mathbf{Z}} \Hom^n(L, M)
$$
where $\Hom^n(L, M)$ is the set of right $A$-module maps $L \to M$ which
are homogeneous of degree $n$, i.e., $f(L^i) \subset M^{i + n}$ for
all $i \in \mathbf{Z}$. In terms of components, we have that
$$
\Hom^n(L, M) \subset \prod\nolimits_{p + q = n} \Hom_R(L^{-q}, M^p)
$$
(observe reversal of indices) is the subset consisting of those
$f = (f_{p, q})$ such that
$$
f_{p, q}(m a) = f_{p - i, q + i}(m)a
$$
for $a \in A^i$ and $m \in L^{-q - i}$. For graded $A$-modules
$K$, $L$, $M$ we define composition in $\text{Mod}^{gr}_A$ via
the maps
$$
\Hom^m(L, M) \times \Hom^n(K, L) \longrightarrow \Hom^{n + m}(K, M)
$$
by simple composition of right $A$-module maps: $(g, f) \mapsto g \circ f$.
\end{example}
\begin{remark}
\label{remark-graded-shift-functors}
Let $R$ be a ring. Let $\mathcal{D}$ be an $R$-linear category endowed with a
collection of $R$-linear functors $[n] : \mathcal{D} \to \mathcal{D}$,
$x \mapsto x[n]$ indexed by $n \in \mathbf{Z}$ such that
$[n] \circ [m] = [n + m]$ and $[0] = \text{id}_\mathcal{D}$ (equality as
functors). This allows us to construct a graded category $\mathcal{D}^{gr}$
over $R$ with the same objects of $\mathcal{D}$ setting
$$
\Hom_{\mathcal{D}^{gr}}(x, y) =
\bigoplus\nolimits_{n \in \mathbf{Z}} \Hom_\mathcal{D}(x, y[n])
$$
for $x, y$ in $\mathcal{D}$. Observe that $(\mathcal{D}^{gr})^0 = \mathcal{D}$
(see Definition \ref{definition-H0-of-graded-category}). Moreover, the graded
category $\mathcal{D}^{gr}$ inherits $R$-linear graded functors $[n]$
satisfying $[n] \circ [m] = [n + m]$ and $[0] = \text{id}_{\mathcal{D}^{gr}}$
with the property that
$$
\Hom_{\mathcal{D}^{gr}}(x, y[n]) = \Hom_{\mathcal{D}^{gr}}(x, y)[n]
$$
as graded $R$-modules compatible with composition of morphisms.
\medskip\noindent
Conversely, suppose given a graded category $\mathcal{A}$ over $R$ endowed
with a collection of $R$-linear graded functors $[n]$
satisfying $[n] \circ [m] = [n + m]$ and $[0] = \text{id}_\mathcal{A}$
which are moreover equipped with isomorphisms
$$
\Hom_\mathcal{A}(x, y[n]) = \Hom_\mathcal{A}(x, y)[n]
$$
as graded $R$-modules compatible with composition of morphisms. Then
the reader easily shows that $\mathcal{A} = (\mathcal{A}^0)^{gr}$.
\medskip\noindent
Here are two examples of the relationship
$\mathcal{D} \leftrightarrow \mathcal{A}$ we established above:
\begin{enumerate}
\item Let $\mathcal{B}$ be an additive category. If
$\mathcal{D} = \text{Gr}(\mathcal{B})$, then
$\mathcal{A} = \text{Gr}^{gr}(\mathcal{B})$ as in
Example \ref{example-graded-category-graded-objects}.
\item If $A$ is a graded ring and $\mathcal{D} = \text{Mod}_A$
is the category of graded right $A$-modules, then
$\mathcal{A} = \text{Mod}^{gr}_A$, see Example \ref{example-gm-gr-cat}.
\end{enumerate}
\end{remark}
\section{Differential graded categories}
\label{section-dga-categories}
\noindent
Note that if $R$ is a ring, then $R$ is a differential graded algebra
over itself (with $R = R^0$ of course). In this case a differential
graded $R$-module is the same thing as a complex of $R$-modules.
In particular, given two differential graded $R$-modules $M$ and $N$
we denote $M \otimes_R N$ the differential graded $R$-module
corresponding to the total complex associated to the double
complex obtained by the tensor product of the complexes of $R$-modules
associated to $M$ and $N$.
\begin{definition}
\label{definition-dga-category}
Let $R$ be a ring. A {\it differential graded category $\mathcal{A}$
over $R$} is a category where every morphism set is given the structure
of a differential graded $R$-module and where for
$x, y, z \in \Ob(\mathcal{A})$ composition is $R$-bilinear and induces
a homomorphism
$$
\Hom_\mathcal{A}(y, z) \otimes_R \Hom_\mathcal{A}(x, y)
\longrightarrow
\Hom_\mathcal{A}(x, z)
$$
of differential graded $R$-modules.
\end{definition}
\noindent
The final condition of the definition signifies the following:
if $f \in \Hom_\mathcal{A}^n(x, y)$ and
$g \in \Hom_\mathcal{A}^m(y, z)$ are homogeneous
of degrees $n$ and $m$, then
$$
\text{d}(g \circ f) = \text{d}(g) \circ f + (-1)^mg \circ \text{d}(f)
$$
in $\Hom_\mathcal{A}^{n + m + 1}(x, z)$. This follows from the sign
rule for the differential on the total complex of a double complex, see
Homology, Definition \ref{homology-definition-associated-simple-complex}.
\begin{definition}
\label{definition-functor-dga-categories}
Let $R$ be a ring. A {\it functor of differential graded categories over $R$}
is a functor $F : \mathcal{A} \to \mathcal{B}$ where for all objects
$x, y$ of $\mathcal{A}$ the map
$F : \Hom_\mathcal{A}(x, y) \to \Hom_\mathcal{A}(F(x), F(y))$
is a homomorphism of differential graded $R$-modules.
\end{definition}
\noindent
Given a differential graded category we are often interested in the
corresponding categories of complexes and homotopy category.
Here is a formal definition.
\begin{definition}
\label{definition-homotopy-category-of-dga-category}
Let $R$ be a ring. Let $\mathcal{A}$ be a differential graded category
over $R$. Then we let
\begin{enumerate}
\item the {\it category of complexes of $\mathcal{A}$}\footnote{This may
be nonstandard terminology.} be the category
$\text{Comp}(\mathcal{A})$ whose objects are the same as the objects
of $\mathcal{A}$ and with
$$
\Hom_{\text{Comp}(\mathcal{A})}(x, y) =
\Ker(d : \Hom^0_\mathcal{A}(x, y) \to \Hom^0_\mathcal{A}(x, y))
$$
\item the {\it homotopy category of $\mathcal{A}$} be the category
$K(\mathcal{A})$ whose objects are the same as the objects
of $\mathcal{A}$ and with
$$
\Hom_{K(\mathcal{A})}(x, y) = H^0(\Hom_\mathcal{A}(x, y))
$$
\end{enumerate}
\end{definition}
\noindent
Our use of the symbol $K(\mathcal{A})$ is nonstandard, but at least
is compatible with the use of $K(-)$ in other chapters of the Stacks project.
\begin{definition}
\label{definition-dg-direct-sum}
Let $R$ be a ring. Let $\mathcal{A}$ be a differential graded category over
$R$. A direct sum $(x, y, z, i, j, p, q)$ in $\mathcal{A}$ (notation as in
Homology, Remark \ref{homology-remark-direct-sum})
is a {\it differential graded direct sum} if $i, j, p, q$ are homogeneous
of degree $0$ and closed, i.e., $\text{d}(i) = 0$, etc.
\end{definition}
\begin{lemma}
\label{lemma-functorial}
Let $R$ be a ring. A functor
$F : \mathcal{A} \to \mathcal{B}$
of differential graded categories over $R$
induces functors $\text{Comp}(\mathcal{A}) \to \text{Comp}(\mathcal{B})$
and $K(\mathcal{A}) \to K(\mathcal{B})$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{example}[Differential graded category of complexes]
\label{example-category-complexes}
Let $\mathcal{B}$ be an additive category. We will construct
a differential graded category $\text{Comp}^{dg}(\mathcal{B})$
over $R = \mathbf{Z}$ whose associated category of complexes
is $\text{Comp}(\mathcal{B})$ and whose associated homotopy
category is $K(\mathcal{B})$. As objects of $\text{Comp}^{dg}(\mathcal{B})$
we take complexes of $\mathcal{B}$. Given complexes
$A^\bullet$ and $B^\bullet$ of $\mathcal{B}$, we sometimes also
denote $A^\bullet$ and $B^\bullet$ the corresponding graded objects
of $\mathcal{B}$ (i.e., forget about the differential).
Using this abuse of notation, we set
$$
\Hom_{\text{Comp}^{dg}(\mathcal{B})}(A^\bullet, B^\bullet) =
\Hom_{\text{Gr}^{gr}(\mathcal{B})}(A^\bullet, B^\bullet)
$$
as a graded $\mathbf{Z}$-module where the right hand side is defined
in Example \ref{example-graded-category-graded-objects}.
In other words, the $n$th graded piece is
the abelian group of homogeneous morphism of degree $n$ of graded objects
$$
\Hom^n(A^\bullet, B^\bullet) =
\Hom_{\text{Gr}(\mathcal{B})}(A^\bullet, B^\bullet[n]) =
\prod\nolimits_{p + q = n} \Hom_\mathcal{B}(A^{-q}, B^p)
$$
(observe reversal of indices and observe we have a direct product
and not a direct sum). For an element
$f \in \Hom^n(A^\bullet, B^\bullet)$ of degree $n$ we set
$$
\text{d}(f) = \text{d}_B \circ f - (-1)^n f \circ \text{d}_A
$$
To make sense of this we think of $\text{d}_B$ and $\text{d}_A$
as maps of graded objects of $\mathcal{B}$ homogeneous of degree $1$
and we use composition in the category $\text{Gr}^{gr}(\mathcal{B})$
on the right hand side. In terms of components, if $f = (f_{p, q})$ with
$f_{p, q} : A^{-q} \to B^p$ we have
\begin{equation}
\label{equation-differential-hom-complex}
\text{d}(f_{p, q}) =
\text{d}_B \circ f_{p, q} + (-1)^{p + q + 1} f_{p, q} \circ \text{d}_A
\end{equation}
Note that the first term of this expression is in
$\Hom_\mathcal{B}(A^{-q}, B^{p + 1})$ and the second term is in
$\Hom_\mathcal{B}(A^{-q - 1}, B^p)$.
In other words, given $p + q = n + 1$ we have
$$
\text{d}(f)_{p, q} =
\text{d}_B \circ f_{p - 1, q} - (-1)^n f_{p, q - 1} \circ \text{d}_A
$$
with obvious notation. The reader checks\footnote{What may be useful
here is to think of the double complex $H^{\bullet, \bullet}$ with terms
$H^{p, q} = \Hom_\mathcal{B}(A^{-q}, B^p)$ and differentials
$d_1$ of degree $(1, 0)$ given by $\text{d}_B$ and $d_2$ of degree
$(0, 1)$ given by the contragredient of $d_A$. Up to sign and up to
replacing the direct sum by a direct product, the differential graded
$\mathbf{Z}$-module
$\Hom_{\text{Comp}^{dg}(\mathcal{B})}(A^\bullet, B^\bullet)$
is the total complex associated to $H^{\bullet, \bullet}$, see
Homology, Definition \ref{homology-definition-associated-simple-complex}.
To get the sign correct, change $d_2^{p, q} : H^{p, q} \to H^{p, q + 1}$ by
$(-1)^{q + 1}$ (after this change we still have a double complex).} that
\begin{enumerate}
\item $\text{d}$ has square zero,
\item an element $f$ in $\Hom^n(A^\bullet, B^\bullet)$
has $\text{d}(f) = 0$ if and only if the morphism
$f : A^\bullet \to B^\bullet[n]$ of graded objects of $\mathcal{B}$
is actually a map of complexes,
\item in particular, the category of complexes of
$\text{Comp}^{dg}(\mathcal{B})$ is equal to $\text{Comp}(\mathcal{B})$,
\item the morphism of complexes defined by $f$ as in (2)
is homotopy equivalent to zero if and only if $f = \text{d}(g)$
for some $g \in \Hom^{n - 1}(A^\bullet, B^\bullet)$.
\item in particular, we obtain a canonical isomorphism
$$
\Hom_{K(\mathcal{B})}(A^\bullet, B^\bullet)
\longrightarrow
H^0(\Hom_{\text{Comp}^{dg}(\mathcal{B})}(A^\bullet, B^\bullet))
$$
and the homotopy category of $\text{Comp}^{dg}(\mathcal{B})$ is equal to
$K(\mathcal{B})$.
\end{enumerate}
Given complexes $A^\bullet$, $B^\bullet$, $C^\bullet$ we define
composition
$$
\Hom^m(B^\bullet, C^\bullet) \times \Hom^n(A^\bullet, B^\bullet)
\longrightarrow
\Hom^{n + m}(A^\bullet, C^\bullet)
$$
by composition $(g, f) \mapsto g \circ f$ in the graded category
$\text{Gr}^{gr}(\mathcal{B})$, see
Example \ref{example-graded-category-graded-objects}.
This defines a map of differential graded modules as in
Definition \ref{definition-dga-category}
because
\begin{align*}
\text{d}(g \circ f) & =
\text{d}_C \circ g \circ f - (-1)^{n + m} g \circ f \circ \text{d}_A \\
& =
\left(\text{d}_C \circ g - (-1)^m g \circ \text{d}_B\right) \circ f +
(-1)^m g \circ \left(\text{d}_B \circ f - (-1)^n f \circ \text{d}_A\right) \\
& =
\text{d}(g) \circ f + (-1)^m g \circ \text{d}(f)
\end{align*}
as desired.
\end{example}
\begin{lemma}
\label{lemma-additive-functor-induces-dga-functor}
Let $F : \mathcal{B} \to \mathcal{B}'$ be an additive functor between
additive categories. Then $F$ induces a functor of differential
graded categories
$$
F : \text{Comp}^{dg}(\mathcal{B}) \to \text{Comp}^{dg}(\mathcal{B}')
$$
of Example \ref{example-category-complexes}
inducing the usual functors on the category of complexes and the
homotopy categories.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{example}[Differential graded category of differential graded modules]
\label{example-dgm-dg-cat}
Let $(A, \text{d})$ be a differential graded algebra over a ring $R$. We will
construct a differential graded category $\text{Mod}^{dg}_{(A, \text{d})}$
over $R$ whose category of complexes is $\text{Mod}_{(A, \text{d})}$ and
whose homotopy category is $K(\text{Mod}_{(A, \text{d})})$.
As objects of $\text{Mod}^{dg}_{(A, \text{d})}$
we take the differential graded $A$-modules. Given differential
graded $A$-modules $L$ and $M$ we set
$$
\Hom_{\text{Mod}^{dg}_{(A, \text{d})}}(L, M) =
\Hom_{\text{Mod}^{gr}_A}(L, M) = \bigoplus \Hom^n(L, M)
$$
as a graded $R$-module where the right hand side is defined as in
Example \ref{example-gm-gr-cat}. In other words, the $n$th graded piece
$\Hom^n(L, M)$ is the $R$-module of right $A$-module maps homogeneous
of degree $n$. For an element $f \in \Hom^n(L, M)$ we set
$$
\text{d}(f) = \text{d}_M \circ f - (-1)^n f \circ \text{d}_L
$$
To make sense of this we think of $\text{d}_M$ and $\text{d}_L$
as graded $R$-module maps and we use composition of graded
$R$-module maps. It is clear that $\text{d}(f)$ is homogeneous of
degree $n + 1$ as a graded $R$-module map, and it is linear
because
\begin{align*}
\text{d}(f)(xa)
& =
\text{d}_M(f(x) a) - (-1)^n f (\text{d}_L(xa)) \\
& =
\text{d}_M(f(x)) a + (-1)^{\deg(x) + n} f(x) \text{d}(a)
- (-1)^n f(\text{d}_L(x)) a - (-1)^{n + \deg(x)} f(x) \text{d}(a) \\
& = \text{d}(f)(x) a
\end{align*}
as desired (observe that this calculation would not work without the
sign in the definition of our differential on $\Hom$). Similar formulae
to those of Example \ref{example-category-complexes} hold for the
differential of $f$ in terms of components.
The reader checks (in the same way as in
Example \ref{example-category-complexes}) that
\begin{enumerate}
\item $\text{d}$ has square zero,
\item an element $f$ in $\Hom^n(L, M)$ has $\text{d}(f) = 0$ if and only if
$f : L \to M[n]$ is a homomorphism of differential graded $A$-modules,
\item in particular, the category of complexes of
$\text{Mod}^{dg}_{(A, \text{d})}$ is $\text{Mod}_{(A, \text{d})}$,
\item the homomorphism defined by $f$ as in (2) is homotopy equivalent
to zero if and only if $f = \text{d}(g)$ for some
$g \in \Hom^{n - 1}(L, M)$.
\item in particular, we obtain a canonical isomorphism
$$
\Hom_{K(\text{Mod}_{(A, \text{d})})}(L, M)
\longrightarrow
H^0(\Hom_{\text{Mod}^{dg}_{(A, \text{d})}}(L, M))
$$
and the homotopy category of $\text{Mod}^{dg}_{(A, \text{d})}$ is
$K(\text{Mod}_{(A, \text{d})})$.
\end{enumerate}
Given differential graded $A$-modules $K$, $L$, $M$ we define
composition
$$
\Hom^m(L, M) \times \Hom^n(K, L) \longrightarrow \Hom^{n + m}(K, M)
$$
by composition of homogeneous right $A$-module maps $(g, f) \mapsto g \circ f$.
This defines a map of differential graded modules as in
Definition \ref{definition-dga-category}
because
\begin{align*}
\text{d}(g \circ f) & =
\text{d}_M \circ g \circ f - (-1)^{n + m} g \circ f \circ \text{d}_K \\
& =
\left(\text{d}_M \circ g - (-1)^m g \circ \text{d}_L\right) \circ f +
(-1)^m g \circ \left(\text{d}_L \circ f - (-1)^n f \circ \text{d}_K\right) \\
& =
\text{d}(g) \circ f + (-1)^m g \circ \text{d}(f)
\end{align*}
as desired.
\end{example}
\begin{lemma}
\label{lemma-homomorphism-induces-dga-functor}
Let $\varphi : (A, \text{d}) \to (E, \text{d})$ be a homomorphism of
differential graded algebras. Then $\varphi$ induces a functor of differential
graded categories
$$
F :
\text{Mod}^{dg}_{(E, \text{d})}
\longrightarrow
\text{Mod}^{dg}_{(A, \text{d})}
$$
of Example \ref{example-dgm-dg-cat} inducing obvious restriction functors
on the categories of differential graded modules and homotopy categories.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-construction}
Let $R$ be a ring. Let $\mathcal{A}$ be a differential graded category
over $R$. Let $x$ be an object of $\mathcal{A}$. Let
$$
(E, \text{d}) = \Hom_\mathcal{A}(x, x)
$$
be the differential graded $R$-algebra of endomorphisms of $x$.
We obtain a functor
$$
\mathcal{A} \longrightarrow \text{Mod}^{dg}_{(E, \text{d})},\quad
y \longmapsto \Hom_\mathcal{A}(x, y)
$$
of differential graded categories by letting $E$ act on
$\Hom_\mathcal{A}(x, y)$ via composition in $\mathcal{A}$.
This functor induces functors
$$
\text{Comp}(\mathcal{A}) \to \text{Mod}_{(A, \text{d})}
\quad\text{and}\quad
K(\mathcal{A}) \to K(\text{Mod}_{(A, \text{d})})
$$
by an application of Lemma \ref{lemma-functorial}.
\end{lemma}
\begin{proof}
This lemma proves itself.
\end{proof}
\section{Obtaining triangulated categories}
\label{section-review}
\noindent
In this section we discuss the most general setup to which the arguments
proving Derived Categories, Proposition
\ref{derived-proposition-homotopy-category-triangulated} and
Proposition \ref{proposition-homotopy-category-triangulated} apply.
\medskip\noindent
Let $R$ be a ring. Let $\mathcal{A}$ be a differential graded category
over $R$. To make our argument work, we impose some axioms on $\mathcal{A}$:
\begin{enumerate}
\item[(A)] $\mathcal{A}$ has a zero object and differential
graded direct sums of two objects
(as in Definition \ref{definition-dg-direct-sum}).
\item[(B)] there are functors $[n] : \mathcal{A} \longrightarrow \mathcal{A}$
of differential graded categories such that
$[0] = \text{id}_\mathcal{A}$ and $[n + m] = [n] \circ [m]$
and given isomorphisms
$$
\Hom_\mathcal{A}(x, y[n]) = \Hom_\mathcal{A}(x, y)[n]
$$
of differential graded $R$-modules compatible with composition.
\end{enumerate}
\noindent
Given our differential graded category $\mathcal{A}$ we say
\begin{enumerate}
\item a sequence $x \to y \to z$ of morphisms of $\text{Comp}(\mathcal{A})$
is an {\it admissible short exact sequence} if there exists
an isomorphism $y \cong x \oplus z$ in the underlying graded category
such that $x \to z$ and $y \to z$ are (co)projections.
\item a morphism $x\to y$ of $\text{Comp}(\mathcal{A})$ is an
{\it admissible monomorphism} if it extends to an
admissible short exact sequence $x\to y\to z$.
\item a morphism $y\to z$ of $\text{Comp}(\mathcal{A})$ is an
{\it admissible epimorphism} if it extends to an
admissible short exact sequence $x\to y\to z$.
\end{enumerate}
The next lemma tells us an admissible short exact sequence gives a
triangle, provided we have axioms (A) and (B).
\begin{lemma}
\label{lemma-get-triangle}
Let $\mathcal{A}$ be a differential graded category satisfying
axioms (A) and (B). Given an admissible short exact sequence
$x \to y \to z$ we obtain (see proof) a triangle
$$
x \to y \to z \to x[1]
$$
in $\text{Comp}(\mathcal{A})$ with the property that any two compositions
in $z[-1] \to x \to y \to z \to x[1]$ are zero in $K(\mathcal{A})$.
\end{lemma}
\begin{proof}
Choose a diagram
$$
\xymatrix{
x \ar[rr]_1 \ar[rd]_a & & x \\
& y \ar[ru]_\pi \ar[rd]^b & \\
z \ar[rr]^1 \ar[ru]^s & & z
}
$$
giving the isomorphism of graded objects $y \cong x \oplus z$ as in the
definition of an admissible short exact sequence. Here are some equations
that hold in this situation
\begin{enumerate}
\item $1 = \pi a$ and hence $\text{d}(\pi) a = 0$,
\item $1 = b s$ and hence $b \text{d}(s) = 0$,
\item $1 = a \pi + s b$ and hence $a \text{d}(\pi) + \text{d}(s) b = 0$,
\item $\pi s = 0$ and hence $\text{d}(\pi)s + \pi \text{d}(s) = 0$,
\item $\text{d}(s) = a \pi \text{d}(s)$ because
$\text{d}(s) = (a \pi + s b)\text{d}(s)$ and $b\text{d}(s) = 0$,
\item $\text{d}(\pi) = \text{d}(\pi) s b$ because
$\text{d}(\pi) = \text{d}(\pi)(a \pi + s b)$ and $\text{d}(\pi)a = 0$,
\item $\text{d}(\pi \text{d}(s)) = 0$ because if we postcompose it
with the monomorphism $a$ we get
$\text{d}(a\pi \text{d}(s)) = \text{d}(\text{d}(s)) = 0$, and
\item $\text{d}(\text{d}(\pi)s) = 0$ as by (4) it is the negative
of $\text{d}(\pi\text{d}(s))$ which is $0$ by (7).
\end{enumerate}
We've used repeatedly that $\text{d}(a) = 0$, $\text{d}(b) = 0$,
and that $\text{d}(1) = 0$. By (7) we see that
$$
\delta = \pi \text{d}(s) = - \text{d}(\pi) s : z \to x[1]
$$
is a morphism in $\text{Comp}(\mathcal{A})$. By (5) we see that
the composition $a \delta = a \pi \text{d}(s) = \text{d}(s)$
is homotopic to zero. By (6) we see that the composition
$\delta b = - \text{d}(\pi)sb = \text{d}(-\pi)$ is homotopic to zero.
\end{proof}
\noindent
Besides axioms (A) and (B) we need an axiom concerning the existence of
cones. We formalize everything as follows.
\begin{situation}
\label{situation-ABC}
Here $R$ is a ring and $\mathcal{A}$ is a differential graded category
over $R$ having axioms (A), (B), and
\begin{enumerate}
\item[(C)] given an arrow $f : x \to y$ of degree $0$ with
$\text{d}(f) = 0$ there exists an admissible short exact sequence
$y \to c(f) \to x[1]$ in $\text{Comp}(\mathcal{A})$ such that the map
$x[1] \to y[1]$ of Lemma \ref{lemma-get-triangle} is equal to $f[1]$.
\end{enumerate}
\end{situation}
\noindent
We will call $c(f)$ a {\it cone} of the morphism $f$.
If (A), (B), and (C) hold, then
cones are functorial in a weak sense.
\begin{lemma}
\label{lemma-cone}
\begin{slogan}
The homotopy category is a triangulated category.
This lemma proves a part of the axioms of a triangulated category.
\end{slogan}
In Situation \ref{situation-ABC} suppose that
$$
\xymatrix{
x_1 \ar[r]_{f_1} \ar[d]_a & y_1 \ar[d]^b \\
x_2 \ar[r]^{f_2} & y_2
}
$$
is a diagram of $\text{Comp}(\mathcal{A})$ commutative up to homotopy.
Then there exists a morphism $c : c(f_1) \to c(f_2)$ which gives rise to
a morphism of triangles
$$
(a, b, c) : (x_1, y_1, c(f_1)) \to (x_1, y_1, c(f_1))
$$
in $K(\mathcal{A})$.
\end{lemma}
\begin{proof}
The assumption means there exists a morphism $h : x_1 \to y_2$ of degree
$-1$ such that $\text{d}(h) = b f_1 - f_2 a$. Choose isomorphisms
$c(f_i) = y_i \oplus x_i[1]$ of graded objects compatible with the
morphisms $y_i \to c(f_i) \to x_i[1]$. Let's denote
$a_i : y_i \to c(f_i)$, $b_i : c(f_i) \to x_i[1]$, $s_i : x_i[1] \to c(f_i)$,
and $\pi_i : c(f_i) \to y_i$ the given morphisms. Recall that
$x_i[1] \to y_i[1]$ is given by $\pi_i \text{d}(s_i)$. By axiom (C)
this means that
$$
f_i = \pi_i \text{d}(s_i) = - \text{d}(\pi_i) s_i
$$
(we identify $\Hom(x_i, y_i)$ with $\Hom(x_i[1], y_i[1])$
using the shift functor $[1]$).
Set $c = a_2 b \pi_1 + s_2 a b_1 + a_2hb$. Then, using the
equalities found in the proof of Lemma \ref{lemma-get-triangle}
we obtain
\begin{align*}
\text{d}(c)
& =
a_2 b \text{d}(\pi_1) + \text{d}(s_2) a b_1 + a_2 \text{d}(h) b_1 \\
& =
- a_2 b f_1 b_1 + a_2 f_2 a b_1 + a_2 (b f_1 - f_2 a) b_1 \\
& = 0
\end{align*}
(where we have used in particular that
$\text{d}(\pi_1) = \text{d}(\pi_1) s_1 b_1 = f_1 b_1$ and
$\text{d}(s_2) = a_2 \pi_2 \text{d}(s_2) = a_2 f_2$).
Thus $c$ is a degree $0$ morphism $c : c(f_1) \to c(f_2)$ of $\mathcal{A}$
compatible with the given morphisms $y_i \to c(f_i) \to x_i[1]$.
\end{proof}
\noindent
In Situation \ref{situation-ABC} we say that a triangle
$(x, y, z, f, g, h)$ in $K(\mathcal{A})$ is a
{\it distinguished triangle} if there exists an admissible
short exact sequence $x' \to y' \to z'$ such that
$(x, y, z, f, g, h)$ is isomorphic as a triangle in $K(\mathcal{A})$
to the triangle $(x', y', z', x' \to y', y' \to z', \delta)$
constructed in Lemma \ref{lemma-get-triangle}. We will show below that
$$
\boxed{
K(\mathcal{A})\text{ is a triangulated category}
}
$$
This result, although not as general as one might think, applies to a
number of natural generalizations of the cases covered so far in the
Stacks project. Here are some examples:
\begin{enumerate}
\item Let $(X, \mathcal{O}_X)$ be a ringed space. Let $(A, d)$ be a
sheaf of differential graded $\mathcal{O}_X$-algebras. Let
$\mathcal{A}$ be the differential graded category of differential
graded $A$-modules. Then $K(\mathcal{A})$ is a triangulated category.
\item Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $(A, d)$ be a
sheaf of differential graded $\mathcal{O}$-algebras. Let
$\mathcal{A}$ be the differential graded category of differential
graded $A$-modules. Then $K(\mathcal{A})$ is a triangulated category.
\item Two examples with a different flavor may be found in Examples, Section
\ref{examples-section-nongraded-differential-graded}.
\end{enumerate}
\noindent
The following simple lemma is a key to the construction.
\begin{lemma}
\label{lemma-id-cone-null}
In Situation \ref{situation-ABC}
given any object $x$ of $\mathcal{A}$, and the cone $C(1_x)$ of the
identity morphism $1_x : x \to x$, the identity morphism on
$C(1_x)$ is homotopic to zero.
\end{lemma}
\begin{proof}
Consider the admissible short exact sequence given by axiom (C).
$$
\xymatrix{
x \ar@<0.5ex>[r]^a &
C(1_x) \ar@<0.5ex>[l]^{\pi} \ar@<0.5ex>[r]^b &
x[1]\ar@<0.5ex>[l]^s
}
$$
Then by Lemma \ref{lemma-get-triangle}, identifying hom-sets under
shifting, we have $1_x=\pi d(s)=-d(\pi)s$ where $s$ is regarded as
a morphism in $\Hom_{\mathcal{A}}^{-1}(x,C(1_x))$. Therefore
$a=a\pi d(s)=d(s)$ using formula (5) of Lemma \ref{lemma-get-triangle},
and $b=-d(\pi)sb=-d(\pi)$ by formula (6) of Lemma \ref{lemma-get-triangle}.
Hence
$$
1_{C(1_x)} = a\pi + sb = d(s)\pi - sd(\pi) = d(s\pi)
$$
since $s$ is of degree $-1$.
\end{proof}
\noindent
A more general version of the above lemma will appear in
Lemma \ref{lemma-cone-homotopy}. The following lemma is the
analogue of Lemma \ref{lemma-make-commute-map}.
\begin{lemma}
\label{lemma-homo-change}
In Situation \ref{situation-ABC} given a diagram
$$
\xymatrix{x\ar[r]^f\ar[d]_a & y\ar[d]^b\\
z\ar[r]^g & w}
$$
in $\text{Comp}(\mathcal{A})$ commuting up to homotopy. Then
\begin{enumerate}
\item If $f$ is an admissible monomorphism, then $b$ is homotopic
to a morphism $b'$ which makes the diagram commute.
\item If $g$ is an admissible epimorphism, then $a$ is homotopic
to a morphism $a'$ which makes the diagram commute.
\end{enumerate}
\end{lemma}
\begin{proof}
To prove (1), observe that the hypothesis implies that there is some
$h\in\Hom_{\mathcal{A}}(x,w)$ of degree $-1$ such that $bf-ga=d(h)$.
Since $f$ is an admissible monomorphism, there is a morphism
$\pi : y \to x$ in the category $\mathcal{A}$ of degree $0$.
Let $b' = b - d(h\pi)$. Then
\begin{align*}
b'f = bf - d(h\pi)f
= &
bf - d(h\pi f) \quad (\text{since }d(f) = 0) \\
= &
bf-d(h) \\
= &
ga
\end{align*}
as desired. The proof for (2) is omitted.
\end{proof}
\noindent
The following lemma is the analogue of Lemma \ref{lemma-make-injective}.
\begin{lemma}
\label{lemma-factor}
In Situation \ref{situation-ABC} let $\alpha : x \to y$
be a morphism in $\text{Comp}(\mathcal{A})$. Then there exists
a factorization in $\text{Comp}(\mathcal{A})$:
$$
\xymatrix{
x \ar[r]^{\tilde{\alpha}} &
\tilde{y} \ar@<0.5ex>[r]^{\pi} &
y\ar@<0.5ex>[l]^s
}
$$
such that
\begin{enumerate}
\item $\tilde{\alpha}$ is an admissible monomorphism, and
$\pi\tilde{\alpha}=\alpha$.
\item There exists a morphism
$s:y\to\tilde{y}$ in $\text{Comp}(\mathcal{A})$
such that $\pi s=1_y$ and $s\pi$ is homotopic to $1_{\tilde{y}}$.
\end{enumerate}
\end{lemma}
\begin{proof}
By axiom (B), we may let $\tilde{y}$ be the differential graded direct
sum of $y$ and $C(1_x)$, i.e., there exists a diagram
$$
\xymatrix@C=3pc{
y \ar@<0.5ex>[r]^s &
y\oplus C(1_x) \ar@<0.5ex>[l]^{\pi} \ar@<0.5ex>[r]^{p} &
C(1_x)\ar@<0.5ex>[l]^t
}
$$
where all morphisms are of degree zero, and in
$\text{Comp}(\mathcal{A})$. Let $\tilde{y} = y \oplus C(1_x)$.
Then $1_{\tilde{y}} = s\pi + tp$. Consider now the diagram
$$
\xymatrix{
x \ar[r]^{\tilde{\alpha}} &
\tilde{y} \ar@<0.5ex>[r]^{\pi} &
y\ar@<0.5ex>[l]^s
}
$$
where $\tilde{\alpha}$ is induced by the morphism $x\xrightarrow{\alpha}y$
and the natural morphism $x\to C(1_x)$ fitting in the admissible
short exact sequence
$$
\xymatrix{
x \ar@<0.5ex>[r] &
C(1_x) \ar@<0.5ex>[l] \ar@<0.5ex>[r] &
x[1]\ar@<0.5ex>[l]
}
$$
So the morphism $C(1_x)\to x$ of degree 0 in this diagram,
together with the zero morphism $y\to x$, induces a degree-0
morphism $\beta : \tilde{y} \to x$. Then $\tilde{\alpha}$ is an
admissible monomorphism since it fits into the admissible short
exact sequence
$$
\xymatrix{
x\ar[r]^{\tilde{\alpha}} &
\tilde{y} \ar[r] &
x[1]
}
$$
Furthermore, $\pi\tilde{\alpha} = \alpha$ by the construction of
$\tilde{\alpha}$, and $\pi s = 1_y$ by the first diagram. It
remains to show that $s\pi$ is homotopic to $1_{\tilde{y}}$.
Write $1_x$ as $d(h)$ for some degree $-1$ map. Then, our
last statement follows from
\begin{align*}
1_{\tilde{y}} - s\pi
= &
tp \\
= &
t(dh)p\quad\text{(by Lemma \ref{lemma-id-cone-null})} \\
= &
d(thp)
\end{align*}
since $dt = dp = 0$, and $t$ is of degree zero.
\end{proof}
\noindent
The following lemma is the analogue of Lemma \ref{lemma-sequence-maps-split}.
\begin{lemma}
\label{lemma-analogue-sequence-maps-split}
In Situation \ref{situation-ABC}
let $x_1 \to x_2 \to \ldots \to x_n$
be a sequence of composable morphisms in $\text{Comp}(\mathcal{A})$.
Then there exists a commutative diagram in $\text{Comp}(\mathcal{A})$:
$$
\xymatrix{x_1\ar[r] & x_2\ar[r] & \ldots\ar[r] & x_n\\
y_1\ar[r]\ar[u] & y_2\ar[r]\ar[u] & \ldots\ar[r] & y_n\ar[u]}
$$
such that each $y_i\to y_{i+1}$ is an admissible monomorphism
and each $y_i\to x_i$ is a homotopy equivalence.
\end{lemma}
\begin{proof}
The case for $n=1$ is trivial: one simply takes $y_1 = x_1$ and the
identity morphism on $x_1$ is in particular a homotopy equivalence.
The case $n = 2$ is given by Lemma \ref{lemma-factor}. Suppose we have
constructed the diagram up to $x_{n - 1}$. We apply
Lemma \ref{lemma-factor} to the composition
$y_{n - 1} \to x_{n-1} \to x_n$ to obtain $y_n$. Then
$y_{n - 1} \to y_n$ will be an admissible monomorphism, and
$y_n \to x_n$ a homotopy equivalence.
\end{proof}
\noindent
The following lemma is the analogue of Lemma \ref{lemma-nilpotent}.
\begin{lemma}
\label{lemma-triseq}
In Situation \ref{situation-ABC} let $x_i \to y_i \to z_i$
be morphisms in $\mathcal{A}$ ($i=1,2,3$) such that
$x_2 \to y_2\to z_2$ is an admissible short exact sequence.
Let $b : y_1 \to y_2$ and $b' : y_2\to y_3$ be morphisms
in $\text{Comp}(\mathcal{A})$ such that
$$
\vcenter{
\xymatrix{
x_1 \ar[d]_0 \ar[r] &
y_1 \ar[r] \ar[d]_b &
z_1 \ar[d]_0 \\
x_2 \ar[r] & y_2 \ar[r] & z_2
}
}
\quad\text{and}\quad
\vcenter{
\xymatrix{
x_2 \ar[d]^0 \ar[r] &
y_2 \ar[r] \ar[d]^{b'} &
z_2 \ar[d]^0 \\
x_3 \ar[r] & y_3 \ar[r] & z_3
}
}
$$
commute up to homotopy. Then $b'\circ b$ is homotopic to $0$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-homo-change}, we can replace $b$ and $b'$
by homotopic maps $\tilde{b}$ and $\tilde{b}'$, such that the right
square of the left diagram commutes and the left square of the right
diagram commutes. Say $b = \tilde{b} + d(h)$ and $b'=\tilde{b}'+d(h')$
for degree $-1$ morphisms $h$ and $h'$ in $\mathcal{A}$. Hence
$$
b'b = \tilde{b}'\tilde{b} + d(\tilde{b}'h + h'\tilde{b} + h'd(h))
$$
since $d(\tilde{b})=d(\tilde{b}')=0$, i.e. $b'b$ is homotopic to
$\tilde{b}'\tilde{b}$. We now want to show that $\tilde{b}'\tilde{b}=0$.
Because $x_2\xrightarrow{f} y_2\xrightarrow{g} z_2$ is an admissible
short exact sequence, there exist degree $0$ morphisms
$\pi : y_2 \to x_2$ and $s : z_2 \to y_2$ such that
$\text{id}_{y_2} = f\pi + sg$. Therefore
$$
\tilde{b}'\tilde{b} = \tilde{b}'(f\pi + sg)\tilde{b} = 0
$$
since $g\tilde{b} = 0$ and $\tilde{b}'f = 0$ as consequences
of the two commuting squares.
\end{proof}
\noindent
The following lemma is the analogue of
Lemma \ref{lemma-triangle-independent-splittings}.
\begin{lemma}
\label{lemma-analogue-triangle-independent-splittings}
In Situation \ref{situation-ABC}
let $0 \to x \to y \to z \to 0$ be an admissible short
exact sequence in $\text{Comp}(\mathcal{A})$. The triangle
$$
\xymatrix{x\ar[r] & y\ar[r] & z\ar[r]^{\delta} & x[1]}
$$
with $\delta : z \to x[1]$ as defined in Lemma \ref{lemma-get-triangle}
is up to canonical isomorphism in $K(\mathcal{A})$, independent of the
choices made in Lemma \ref{lemma-get-triangle}.
\end{lemma}
\begin{proof}
Suppose $\delta$ is defined by the splitting
$$
\xymatrix{
x \ar@<0.5ex>[r]^{a} &
y \ar@<0.5ex>[r]^b\ar@<0.5ex>[l]^{\pi} &
z \ar@<0.5ex>[l]^s
}
$$
and $\delta'$ is defined by the splitting with $\pi',s'$
in place of $\pi,s$. Then
$$
s'-s = (a\pi + sb)(s'-s) = a\pi s'
$$
since $bs' = bs = 1_z$ and $\pi s = 0$. Similarly,
$$
\pi' - \pi = (\pi' - \pi)(a\pi + sb) = \pi'sb
$$
Since $\delta = \pi d(s)$ and $\delta' = \pi'd(s')$
as constructed in Lemma \ref{lemma-get-triangle}, we may compute
$$
\delta' = \pi'd(s') = (\pi + \pi'sb)d(s + a\pi s') = \delta + d(\pi s')
$$
using $\pi a = 1_x$, $ba = 0$, and $\pi'sbd(s') = \pi'sba\pi d(s') = 0$
by formula (5) in Lemma \ref{lemma-get-triangle}.
\end{proof}
\noindent
The following lemma is the analogue of Lemma \ref{lemma-rotate-cone}.
\begin{lemma}
\label{lemma-restate-axiom-c}
In Situation \ref{situation-ABC}
let $f: x \to y$ be a morphism in $\text{Comp}(\mathcal{A})$.
The triangle $(y, c(f), x[1], i, p, f[1])$ is the triangle associated
to the admissible short exact sequence
$$
\xymatrix{y\ar[r] & c(f) \ar[r] & x[1]}
$$
where the cone $c(f)$ is defined as in Lemma \ref{lemma-get-triangle}.
\end{lemma}
\begin{proof}
This follows from axiom (C).
\end{proof}
\noindent
The following lemma is the analogue of Lemma \ref{lemma-rotate-triangle}.
\begin{lemma}
\label{lemma-cone-rotate-isom}
In Situation \ref{situation-ABC} let $\alpha : x \to y$ and $\beta : y \to z$
define an admissible short exact sequence
$$
\xymatrix{
x \ar[r] &
y\ar[r] &
z
}
$$
in $\text{Comp}(\mathcal{A})$. Let $(x, y, z, \alpha, \beta, \delta)$
be the associated triangle in $K(\mathcal{A})$. Then, the triangles
$$
(z[-1], x, y, \delta[-1], \alpha, \beta)
\quad\text{and}\quad
(z[-1], x, c(\delta[-1]), \delta[-1], i, p)
$$
are isomorphic.
\end{lemma}
\begin{proof}
We have a diagram of the form
$$
\xymatrix{
z[-1]\ar[r]^{\delta[-1]}\ar[d]^1 &
x\ar@<0.5ex>[r]^{\alpha}\ar[d]^1 &
y\ar@<0.5ex>[r]^{\beta}\ar@{.>}[d]\ar@<0.5ex>[l]^{\tilde{\alpha}} &
z\ar[d]^1\ar@<0.5ex>[l]^{\tilde\beta} \\
z[-1] \ar[r]^{\delta[-1]} &
x\ar@<0.5ex>[r]^i &
c(\delta[-1]) \ar@<0.5ex>[r]^p\ar@<0.5ex>[l]^{\tilde i} &
z\ar@<0.5ex>[l]^{\tilde p}
}
$$
with splittings to $\alpha, \beta, i$, and $p$ given by
$\tilde{\alpha}, \tilde{\beta}, \tilde{i},$ and $\tilde{p}$ respectively.
Define a morphism $y \to c(\delta[-1])$ by
$i\tilde{\alpha} + \tilde{p}\beta$ and a morphism
$c(\delta[-1]) \to y$ by $\alpha \tilde{i} + \tilde{\beta} p$.
Let us first check that these define morphisms in $\text{Comp}(\mathcal{A})$.
We remark that by identities from Lemma \ref{lemma-get-triangle},
we have the relation
$\delta[-1] = \tilde{\alpha}d(\tilde{\beta}) = -d(\tilde{\alpha})\tilde{\beta}$
and the relation $\delta[-1] = \tilde{i}d(\tilde{p})$. Then
\begin{align*}
d(\tilde{\alpha})
& =
d(\tilde{\alpha})\tilde{\beta}\beta \\
& =
-\delta[-1]\beta
\end{align*}
where we have used equation (6) of
Lemma \ref{lemma-get-triangle} for the first equality and
the preceeding remark for the second. Similarly, we obtain
$d(\tilde{p}) = i\delta[-1]$. Hence
\begin{align*}
d(i\tilde{\alpha} + \tilde{p}\beta)
& =
d(i)\tilde{\alpha} + id(\tilde{\alpha}) +
d(\tilde{p})\beta + \tilde{p}d(\beta) \\
& =
id(\tilde{\alpha}) + d(\tilde{p})\beta \\
& =
-i\delta[-1]\beta + i\delta[-1]\beta \\
& =
0
\end{align*}
so $i\tilde{\alpha} + \tilde{p}\beta$ is indeed a morphism of
$\text{Comp}(\mathcal{A})$. By a similar calculation,
$\alpha \tilde{i} + \tilde{\beta} p$ is also a morphism of
$\text{Comp}(\mathcal{A})$. It is immediate that these morphisms
fit in the commutative diagram. We compute:
\begin{align*}
(i\tilde{\alpha} + \tilde{p}\beta)(\alpha \tilde{i} + \tilde{\beta} p)
& =
i\tilde{\alpha}\alpha\tilde{i} + i\tilde{\alpha}\tilde{\beta}p
+ \tilde{p}\beta\alpha\tilde{i} + \tilde{p}\beta\tilde{\beta}p \\
& =
i\tilde{i} + \tilde{p}p \\
& =
1_{c(\delta[-1])}
\end{align*}
where we have freely used the identities of
Lemma \ref{lemma-get-triangle}. Similarly, we compute
$(\alpha \tilde{i} + \tilde{\beta} p)(i\tilde{\alpha} + \tilde{p}\beta) = 1_y$,
so we conclude $y \cong c(\delta[-1])$. Hence, the two triangles in question
are isomorphic.
\end{proof}
\noindent
The following lemma is the analogue of
Lemma \ref{lemma-third-isomorphism}.
\begin{lemma}
\label{lemma-analogue-third-isomorphism}
In Situation \ref{situation-ABC} let $f_1 : x_1 \to y_1$ and
$f_2 : x_2 \to y_2$ be morphisms in $\text{Comp}(\mathcal{A})$. Let
$$
(a,b,c): (x_1,y_1,c(f_1), f_1, i_1, p_1) \to (x_2,y_2, c(f_2), f_2, i_1, p_1)
$$
be any morphism of triangles in $K(\mathcal{A})$.
If $a$ and $b$ are homotopy equivalences, then so is $c$.
\end{lemma}
\begin{proof}
Since $a$ and $b$ are homotopy equivalences, they are invertible in
$K(\mathcal{A})$ so let $a^{-1}$ and $b^{-1}$ denote their inverses
in $K(\mathcal{A})$, giving us a commutative diagram
$$
\xymatrix{
x_2\ar[d]^{a^{-1}}\ar[r]^{f_2} &
y_2\ar[d]^{b^{-1}}\ar[r]^{i_2} &
c(f_2)\ar[d]^{c'} \\
x_1\ar[r]^{f_1} &
y_1 \ar[r]^{i_1} &
c(f_1)
}
$$
where the map $c'$ is defined via Lemma \ref{lemma-cone} applied to the left
commutative box of the above diagram. Since the diagram commutes
in $K(\mathcal{A})$, it suffices by Lemma \ref{lemma-triseq} to
prove the following: given a morphism of triangle
$(1,1,c): (x,y,c(f),f,i,p)\to (x,y,c(f),f,i,p)$
in $K(\mathcal{A})$, the map $c$ is an isomorphism in
$K(\mathcal{A})$. We have the commutative diagrams in $K(\mathcal{A})$:
$$
\vcenter{
\xymatrix{
y\ar[d]^{1}\ar[r] &
c(f)\ar[d]^{c}\ar[r] &
x[1]\ar[d]^{1} \\
y\ar[r] &
c(f) \ar[r] &
x[1]
}
}
\quad\Rightarrow\quad
\vcenter{
\xymatrix{
y\ar[d]^{0}\ar[r] &
c(f)\ar[d]^{c-1}\ar[r] &
x[1]\ar[d]^{0} \\
y\ar[r] &
c(f) \ar[r] &
x[1]
}
}
$$
Since the rows are admissible short exact sequences, we obtain
the identity $(c-1)^2 = 0$ by Lemma \ref{lemma-triseq}, from
which we conclude that $2-c$ is inverse to $c$ in $K(\mathcal{A})$
so that $c$ is an isomorphism.