## 22.27 Obtaining triangulated categories

In this section we discuss the most general setup to which the arguments proving Derived Categories, Proposition 13.10.3 and Proposition 22.10.3 apply.

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 $\mathcal{A}$:

$\mathcal{A}$ has a zero object and differential graded direct sums of two objects (as in Definition 22.26.4).

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

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(x, y[n]) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(x, y)[n] \]

of differential graded $R$-modules compatible with composition.

Given our differential graded category $\mathcal{A}$ we say

a sequence $x \to y \to z$ of morphisms of $\text{Comp}(\mathcal{A})$ is an *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.

a morphism $x\to y$ of $\text{Comp}(\mathcal{A})$ is an *admissible monomorphism* if it extends to an admissible short exact sequence $x\to y\to z$.

a morphism $y\to z$ of $\text{Comp}(\mathcal{A})$ is an *admissible epimorphism* if it extends to an admissible short exact sequence $x\to y\to z$.

The next lemma tells us an admissible short exact sequence gives a triangle, provided we have axioms (A) and (B).

Lemma 22.27.1. 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 $K(\mathcal{A})$.

**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

$1 = \pi a$ and hence $\text{d}(\pi ) a = 0$,

$1 = b s$ and hence $b \text{d}(s) = 0$,

$1 = a \pi + s b$ and hence $a \text{d}(\pi ) + \text{d}(s) b = 0$,

$\pi s = 0$ and hence $\text{d}(\pi )s + \pi \text{d}(s) = 0$,

$\text{d}(s) = a \pi \text{d}(s)$ because $\text{d}(s) = (a \pi + s b)\text{d}(s)$ and $b\text{d}(s) = 0$,

$\text{d}(\pi ) = \text{d}(\pi ) s b$ because $\text{d}(\pi ) = \text{d}(\pi )(a \pi + s b)$ and $\text{d}(\pi )a = 0$,

$\text{d}(\pi \text{d}(s)) = 0$ 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

$\text{d}(\text{d}(\pi )s) = 0$ as by (4) it is the negative of $\text{d}(\pi \text{d}(s))$ which is $0$ by (7).

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.
$\square$

Besides axioms (A) and (B) we need an axiom concerning the existence of cones. We formalize everything as follows.

Situation 22.27.2. Here $R$ is a ring and $\mathcal{A}$ is a differential graded category over $R$ having axioms (A), (B), and

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 22.27.1 is equal to $f[1]$.

We will call $c(f)$ a *cone* of the morphism $f$. If (A), (B), and (C) hold, then cones are functorial in a weak sense.

slogan
Lemma 22.27.3. In Situation 22.27.2 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 $K(\mathcal{A})$.

**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 $\mathop{\mathrm{Hom}}\nolimits (x_ i, y_ i)$ with $\mathop{\mathrm{Hom}}\nolimits (x_ i[1], y_ i[1])$ 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 22.27.1 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 $y_ i \to c(f_ i) \to x_ i[1]$.
$\square$

In Situation 22.27.2 we say that a triangle $(x, y, z, f, g, h)$ in $K(\mathcal{A})$ is a *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 22.27.1. 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:

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.

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. See Differential Graded Sheaves, Proposition 24.22.4.

Two examples with a different flavor may be found in Examples, Section 110.70.

The following simple lemma is a key to the construction.

Lemma 22.27.4. In Situation 22.27.2 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.

**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 22.27.1, identifying hom-sets under shifting, we have $1_ x=\pi d(s)=-d(\pi )s$ where $s$ is regarded as a morphism in $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{A}}^{-1}(x,C(1_ x))$. Therefore $a=a\pi d(s)=d(s)$ using formula (5) of Lemma 22.27.1, and $b=-d(\pi )sb=-d(\pi )$ by formula (6) of Lemma 22.27.1. Hence

\[ 1_{C(1_ x)} = a\pi + sb = d(s)\pi - sd(\pi ) = d(s\pi ) \]

since $s$ is of degree $-1$.
$\square$

A more general version of the above lemma will appear in Lemma 22.27.13. The following lemma is the analogue of Lemma 22.7.3.

Lemma 22.27.5. In Situation 22.27.2 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

If $f$ is an admissible monomorphism, then $b$ is homotopic to a morphism $b'$ which makes the diagram commute.

If $g$ is an admissible epimorphism, then $a$ is homotopic to a morphism $a'$ which makes the diagram commute.

**Proof.**
To prove (1), observe that the hypothesis implies that there is some $h\in \mathop{\mathrm{Hom}}\nolimits _{\mathcal{A}}(x,w)$ 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.
$\square$

The following lemma is the analogue of Lemma 22.7.4.

Lemma 22.27.6. In Situation 22.27.2 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

$\tilde{\alpha }$ is an admissible monomorphism, and $\pi \tilde{\alpha }=\alpha $.

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 $1_{\tilde{y}}$.

**Proof.**
By axiom (A), 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 09QK)} \\ = & d(thp) \end{align*}

since $dt = dp = 0$, and $t$ is of degree zero.
$\square$

The following lemma is the analogue of Lemma 22.7.5.

Lemma 22.27.7. In Situation 22.27.2 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.

**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 22.27.6. Suppose we have constructed the diagram up to $x_{n - 1}$. We apply Lemma 22.27.6 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.
$\square$

The following lemma is the analogue of Lemma 22.7.6.

Lemma 22.27.8. In Situation 22.27.2 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 $0$.

**Proof.**
By Lemma 22.27.5, 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.
$\square$

The following lemma is the analogue of Lemma 22.8.1.

Lemma 22.27.9. In Situation 22.27.2 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 22.27.1 is up to canonical isomorphism in $K(\mathcal{A})$, independent of the choices made in Lemma 22.27.1.

**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 22.27.1, 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 22.27.1.
$\square$

The following lemma is the analogue of Lemma 22.9.1.

Lemma 22.27.10. In Situation 22.27.2 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 22.27.1.

**Proof.**
This follows from axiom (C).
$\square$

The following lemma is the analogue of Lemma 22.9.2.

Lemma 22.27.11. In Situation 22.27.2 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.

**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 22.27.1, 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 22.27.1 for the first equality and the preceding 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 22.27.1. 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.
$\square$

The following lemma is the analogue of Lemma 22.9.3.

Lemma 22.27.12. In Situation 22.27.2 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 $c$.

**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 22.27.3 applied to the left commutative box of the above diagram. Since the diagram commutes in $K(\mathcal{A})$, it suffices by Lemma 22.27.8 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 22.27.8, from which we conclude that $2-c$ is inverse to $c$ in $K(\mathcal{A})$ so that $c$ is an isomorphism.
$\square$

The following lemma is the analogue of Lemma 22.9.4.

Lemma 22.27.13. In Situation 22.27.2.

Given an admissible short exact sequence $x\xrightarrow {\alpha } y\xrightarrow {\beta } z$. Then there exists a homotopy equivalence $e:C(\alpha )\to z$ such that the diagram

22.27.13.1
\begin{equation} \label{dga-equation-cone-isom-triangle} \vcenter { \xymatrix{ x\ar[r]^{\alpha }\ar[d] & y\ar[r]^{b}\ar[d] & C(\alpha )\ar[r]^{-c}\ar@{.>}[d]^{e} & x[1]\ar[d] \\ x\ar[r]^{\alpha } & y\ar[r]^{\beta } & z\ar[r]^{\delta } & x[1] } } \end{equation}

defines an isomorphism of triangles in $K(\mathcal{A})$. Here $y\xrightarrow {b}C(\alpha )\xrightarrow {c}x[1]$ is the admissible short exact sequence given as in axiom (C).

Given a morphism $\alpha : x \to y$ in $\text{Comp}(\mathcal{A})$, let $x \xrightarrow {\tilde{\alpha }} \tilde{y} \to y$ be the factorization given as in Lemma 22.27.6, where the admissible monomorphism $x \xrightarrow {\tilde{\alpha }} y$ extends to the admissible short exact sequence

\[ \xymatrix{ x \ar[r]^{\tilde{\alpha }} & \tilde{y} \ar[r] & z } \]

Then there exists an isomorphism of triangles

\[ \xymatrix{ x \ar[r]^{\tilde{\alpha }} \ar[d] & \tilde{y} \ar[r] \ar[d] & z \ar[r]^{\delta } \ar@{.>}[d]^{e} & x[1] \ar[d] \\ x \ar[r]^{\alpha } & y \ar[r] & C(\alpha ) \ar[r]^{-c} & x[1] } \]

where the upper triangle is the triangle associated to the sequence $x \xrightarrow {\tilde{\alpha }} \tilde{y} \to z$.

**Proof.**
For (1), we consider the more complete diagram, *without* the sign change on $c$:

\[ \xymatrix{ x\ar@<0.5ex>[r]^{\alpha } \ar[d] & y\ar@<0.5ex>[l]^{\pi } \ar@<0.5ex>[r]^{b}\ar[d] & C(\alpha )\ar@<0.5ex>[l]^{p} \ar@<0.5ex>[r]^{c}\ar@{.>}@<0.5ex>[d]^{e} & x[1]\ar@<0.5ex>[l]^{\sigma } \ar[d]\ar@<0.5ex>[r]^{\alpha } & y[1]\ar@<0.5ex>[l]^{\pi } \\ x\ar@<0.5ex>[r]^{\alpha } & y\ar@<0.5ex>[r]^{\beta } \ar@<0.5ex>[l]^{\pi } & z\ar[r]^{\delta }\ar@<0.5ex>[l]^{s} \ar@{.>}@<0.5ex>[u]^{f} & x[1] } \]

where the admissible short exact sequence $x\xrightarrow {\alpha } y\xrightarrow {\beta } z$ is given the splitting $\pi $, $s$, and the admissible short exact sequence $y\xrightarrow {b}C(\alpha )\xrightarrow {c}x[1]$ is given the splitting $p$, $\sigma $. Note that (identifying hom-sets under shifting)

\[ \alpha = pd(\sigma ) = -d(p)\sigma ,\quad \delta = \pi d(s) = -d(\pi )s \]

by the construction in Lemma 22.27.1.

We define $e=\beta p$ and $f=bs-\sigma \delta $. We first check that they are morphisms in $\text{Comp}(\mathcal{A})$. To show that $d(e)=\beta d(p)$ vanishes, it suffices to show that $\beta d(p)b$ and $\beta d(p)\sigma $ both vanish, whereas

\[ \beta d(p)b = \beta d(pb) = \beta d(1_ y) = 0,\quad \beta d(p)\sigma = -\beta \alpha = 0 \]

Similarly, to check that $d(f)=bd(s)-d(\sigma )\delta $ vanishes, it suffices to check the post-compositions by $p$ and $c$ both vanish, whereas

\begin{align*} pbd(s) - pd(\sigma )\delta = & d(s)-\alpha \delta = d(s)-\alpha \pi d(s) = 0 \\ cbd(s)-cd(\sigma )\delta = & -cd(\sigma )\delta = -d(c\sigma )\delta = 0 \end{align*}

The commutativity of left two squares of the diagram 22.27.13.1 follows directly from definition. Before we prove the commutativity of the right square (up to homotopy), we first check that $e$ is a homotopy equivalence. Clearly,

\[ ef=\beta p (bs-\sigma \delta )=\beta s=1_ z \]

To check that $fe$ is homotopic to $1_{C(\alpha )}$, we first observe

\[ b\alpha = bpd(\alpha ) = d(\sigma ),\quad \alpha c = -d(p)\sigma c = -d(p),\quad d(\pi )p = d(\pi )s\beta p = -\delta \beta p \]

Using these identities, we compute

\begin{align*} 1_{C(\alpha )} = & bp + \sigma c \quad (\text{from }y \xrightarrow {b} C(\alpha ) \xrightarrow {c} x[1]) \\ = & b(\alpha \pi + s\beta )p + \sigma (\pi \alpha )c \quad (\text{from }x \xrightarrow {\alpha } y \xrightarrow {\beta } z) \\ = & d(\sigma )\pi p + bs\beta p - \sigma \pi d(p) \quad (\text{by the first two identities above}) \\ = & d(\sigma )\pi p + bs\beta p - \sigma \delta \beta p + \sigma \delta \beta p - \sigma \pi d(p) \\ = & (bs - \sigma \delta )\beta p + d(\sigma )\pi p - \sigma d(\pi )p - \sigma \pi d(p)\quad (\text{by the third identity above}) \\ = & fe + d(\sigma \pi p) \end{align*}

since $\sigma \in \mathop{\mathrm{Hom}}\nolimits ^{-1}(x, C(\alpha ))$ (cf. proof of Lemma 22.27.4). Hence $e$ and $f$ are homotopy inverses. Finally, to check that the right square of diagram 22.27.13.1 commutes up to homotopy, it suffices to check that $-cf=\delta $. This follows from

\[ -cf = -c(bs-\sigma \delta ) = c\sigma \delta = \delta \]

since $cb=0$.

For (2), consider the factorization $x\xrightarrow {\tilde{\alpha }}\tilde{y}\to y$ given as in Lemma 22.27.6, so the second morphism is a homotopy equivalence. By Lemmas 22.27.3 and 22.27.12, there exists an isomorphism of triangles between

\[ x \xrightarrow {\alpha } y \to C(\alpha ) \to x[1] \quad \text{and}\quad x \xrightarrow {\tilde{\alpha }} \tilde{y} \to C(\tilde{\alpha }) \to x[1] \]

Since we can compose isomorphisms of triangles, by replacing $\alpha $ by $\tilde{\alpha }$, $y$ by $\tilde{y}$, and $C(\alpha )$ by $C(\tilde{\alpha })$, we may assume $\alpha $ is an admissible monomorphism. In this case, the result follows from (1).
$\square$

The following lemma is the analogue of Lemma 22.10.1.

Lemma 22.27.14. In Situation 22.27.2 the homotopy category $K(\mathcal{A})$ with its natural translation functors and distinguished triangles is a pre-triangulated category.

**Proof.**
We will verify each of TR1, TR2, and TR3.

Proof of TR1. By definition every triangle isomorphic to a distinguished one is distinguished. Since

\[ \xymatrix{x\ar[r]^{1_ x} & x\ar[r] & 0} \]

is an admissible short exact sequence, $(x, x, 0, 1_ x, 0, 0)$ is a distinguished triangle. Moreover, given a morphism $\alpha : x \to y$ in $\text{Comp}(\mathcal{A})$, the triangle given by $(x, y, c(\alpha ), \alpha , i, -p)$ is distinguished by Lemma 22.27.13.

Proof of TR2. Let $(x,y,z,\alpha ,\beta ,\gamma )$ be a triangle and suppose $(y,z,x[1],\beta ,\gamma ,-\alpha [1])$ is distinguished. Then there exists an admissible short exact sequence $0 \to x' \to y' \to z' \to 0$ such that the associated triangle $(x',y',z',\alpha ',\beta ',\gamma ')$ is isomorphic to $(y,z,x[1],\beta ,\gamma ,-\alpha [1])$. After rotating, we conclude that $(x,y,z,\alpha ,\beta ,\gamma )$ is isomorphic to $(z'[-1],x',y', \gamma '[-1], \alpha ',\beta ')$. By Lemma 22.27.11, we deduce that $(z'[-1],x',y', \gamma '[-1], \alpha ',\beta ')$ is isomorphic to $(z'[-1],x',c(\gamma '[-1]), \gamma '[-1], i, p)$. Composing the two isomorphisms with sign changes as indicated in the following diagram:

\[ \xymatrix@C=3pc{ x\ar[r]^{\alpha }\ar[d] & y\ar[r]^{\beta }\ar[d] & z\ar[r]^{\gamma }\ar[d] & x[1]\ar[d] \\ z'[-1]\ar[r]^{-\gamma '[-1]}\ar[d]_{-1_{z'[-1]}} & x \ar[r]^{\alpha '}\ar@{=}[d] & y' \ar[r]^{\beta '} \ar[d] & z'\ar[d]^{-1_{z'}} \\ z'[-1]\ar[r]^{\gamma '[-1]} & x \ar[r]^{\alpha '} & c(\gamma '[-1]) \ar[r]^{-p} & z' } \]

We conclude that $(x,y,z,\alpha ,\beta ,\gamma )$ is distinguished by Lemma 22.27.13 (2). Conversely, suppose that $(x,y,z,\alpha ,\beta ,\gamma )$ is distinguished, so that by Lemma 22.27.13 (1), it is isomorphic to a triangle of the form $(x',y', c(\alpha '), \alpha ', i, -p)$ for some morphism $\alpha ': x' \to y'$ in $\text{Comp}(\mathcal{A})$. The rotated triangle $(y,z,x[1],\beta ,\gamma , -\alpha [1])$ is isomorphic to the triangle $(y',c(\alpha '), x'[1], i, -p, -\alpha [1])$ which is isomorphic to $(y',c(\alpha '), x'[1], i, p, \alpha [1])$. By Lemma 22.27.10, this triangle is distinguished, from which it follows that $(y,z,x[1], \beta ,\gamma , -\alpha [1])$ is distinguished.

Proof of TR3: Suppose $(x,y,z, \alpha ,\beta ,\gamma )$ and $(x',y',z',\alpha ',\beta ',\gamma ')$ are distinguished triangles of $\text{Comp}(\mathcal{A})$ and let $f: x \to x'$ and $g: y \to y'$ be morphisms such that $\alpha ' \circ f = g \circ \alpha $. By Lemma 22.27.13, we may assume that $(x,y,z,\alpha ,\beta ,\gamma )= (x,y,c(\alpha ),\alpha , i, -p)$ and $(x',y',z', \alpha ',\beta ',\gamma ')= (x',y',c(\alpha '), \alpha ',i',-p')$. Now apply Lemma 22.27.3 and we are done.
$\square$

The following lemma is the analogue of Lemma 22.10.2.

Lemma 22.27.15. In Situation 22.27.2 given admissible monomorphisms $x \xrightarrow {\alpha } y$, $y \xrightarrow {\beta } z$ in $\mathcal{A}$, there exist distinguished triangles $(x,y,q_1,\alpha ,p_1,\delta _1)$, $(x,z,q_2,\beta \alpha ,p_2,\delta _2)$ and $(y,z,q_3,\beta ,p_3,\delta _3)$ for which TR4 holds.

**Proof.**
Given admissible monomorphisms $x\xrightarrow {\alpha } y$ and $y\xrightarrow {\beta }z$, we can find distinguished triangles, via their extensions to admissible short exact sequences,

\[ \xymatrix{ x\ar@<0.5ex>[r]^{\alpha } & y\ar@<0.5ex>[l]^{\pi _1} \ar@<0.5ex>[r]^{p_1} & q_1 \ar[r]^{\delta _1} \ar@<0.5ex>[l]^{s_1} & x[1] } \]

\[ \xymatrix{ x\ar@<0.5ex>[r]^{\beta \alpha } & z\ar@<0.5ex>[l]^{\pi _1\pi _3} \ar@<0.5ex>[r]^{p_2} & q_2 \ar[r]^{\delta _2} \ar@<0.5ex>[l]^{s_2} & x[1] } \]

\[ \xymatrix{ y\ar@<0.5ex>[r]^{\beta } & z\ar@<0.5ex>[l]^{\pi _3} \ar@<0.5ex>[r]^{p_3} & q_3 \ar[r]^{\delta _3} \ar@<0.5ex>[l]^{s_3} & x[1] } \]

In these diagrams, the maps $\delta _ i$ are defined as $\delta _ i = \pi _ i d(s_ i)$ analogous to the maps defined in Lemma 22.27.1. They fit in the following solid commutative diagram

\[ \xymatrix@C=5pc@R=3pc{ x\ar@<0.5ex>[r]^{\alpha } \ar@<0.5ex>[dr]^{\beta \alpha } & y\ar@<0.5ex>[d]^{\beta } \ar@<0.5ex>[l]^{\pi _1} \ar@<0.5ex>[r]^{p_1} & q_1 \ar[r]^{\delta _1} \ar@<0.5ex>[l]^{s_1} \ar@{.>}[dd]^{p_2\beta s_1} & x[1] \\ & z \ar@<0.5ex>[u]^{\pi _3}\ar@<0.5ex>[d]^{p_3} \ar@<0.5ex>[dr]^{p_2} \ar@<0.5ex>[ul]^{\pi _1\pi _3} & & \\ & q_3\ar@<0.5ex>[u]^{s_3} \ar[d]^{\delta _3} & q_2 \ar@{.>}[l]^{p_3s_2} \ar@<0.5ex>[ul]^{s_2} \ar[dr]^{\delta _2} \\ & y[1] & & x[1]} \]

where we have defined the dashed arrows as indicated. Clearly, their composition $p_3s_2p_2\beta s_1 = 0$ since $s_2p_2 = 0$. We claim that they both are morphisms of $\text{Comp}(\mathcal{A})$. We can check this using equations in Lemma 22.27.1:

\[ d(p_2\beta s_1) = p_2\beta d(s_1) = p_2\beta \alpha \pi _1 d(s_1) = 0 \]

since $p_2\beta \alpha = 0$, and

\[ d(p_3s_2) = p_3d(s_2) = p_3\beta \alpha \pi _1\pi _3 d(s_2) = 0 \]

since $p_3\beta = 0$. To check that $q_1\to q_2\to q_3$ is an admissible short exact sequence, it remains to show that in the underlying graded category, $q_2 = q_1\oplus q_3$ with the above two morphisms as coprojection and projection. To do this, observe that in the underlying graded category $\mathcal{C}$, there hold

\[ y = x\oplus q_1,\quad z = y\oplus q_3 = x\oplus q_1\oplus q_3 \]

where $\pi _1\pi _3$ gives the projection morphism onto the first factor: $x\oplus q_1\oplus q_3\to z$. By axiom (A) on $\mathcal{A}$, $\mathcal{C}$ is an additive category, hence we may apply Homology, Lemma 12.3.10 and conclude that

\[ \mathop{\mathrm{Ker}}(\pi _1\pi _3) = q_1\oplus q_3 \]

in $\mathcal{C}$. Another application of Homology, Lemma 12.3.10 to $z = x\oplus q_2$ gives $\mathop{\mathrm{Ker}}(\pi _1\pi _3) = q_2$. Hence $q_2\cong q_1\oplus q_3$ in $\mathcal{C}$. It is clear that the dashed morphisms defined above give coprojection and projection.

Finally, we have to check that the morphism $\delta : q_3 \to q_1[1]$ induced by the admissible short exact sequence $q_1\to q_2\to q_3$ agrees with $p_1\delta _3$. By the construction in Lemma 22.27.1, the morphism $\delta $ is given by

\begin{align*} p_1\pi _3s_2d(p_2s_3) = & p_1\pi _3s_2p_2d(s_3) \\ = & p_1\pi _3(1-\beta \alpha \pi _1\pi _3)d(s_3) \\ = & p_1\pi _3d(s_3)\quad (\text{since }\pi _3\beta = 0) \\ = & p_1\delta _3 \end{align*}

as desired. The proof is complete.
$\square$

Putting everything together we finally obtain the analogue of Proposition 22.10.3.

Proposition 22.27.16. In Situation 22.27.2 the homotopy category $K(\mathcal{A})$ with its natural translation functors and distinguished triangles is a triangulated category.

**Proof.**
By Lemma 22.27.14 we know that $K(\mathcal{A})$ is pre-triangulated. Combining Lemmas 22.27.7 and 22.27.15 with Derived Categories, Lemma 13.4.15, we conclude that $K(\mathcal{A})$ is a triangulated category.
$\square$

Lemma 22.27.17. Let $R$ be a ring. Let $F : \mathcal{A} \to \mathcal{B}$ be a functor between differential graded categories over $R$ satisfying axioms (A), (B), and (C) such that $F(x[1]) = F(x)[1]$. Then $F$ induces an exact functor $K(\mathcal{A}) \to K(\mathcal{B})$ of triangulated categories.

**Proof.**
Namely, if $x \to y \to z$ is an admissible short exact sequence in $\text{Comp}(\mathcal{A})$, then $F(x) \to F(y) \to F(z)$ is an admissible short exact sequence in $\text{Comp}(\mathcal{B})$. Moreover, the “boundary” morphism $\delta = \pi \text{d}(s) : z \to x[1]$ constructed in Lemma 22.27.1 produces the morphism $F(\delta ) : F(z) \to F(x[1]) = F(x)[1]$ which is equal to the boundary map $F(\pi ) \text{d}(F(s))$ for the admissible short exact sequence $F(x) \to F(y) \to F(z)$.
$\square$

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