The Stacks project

Proof. Choose a diagram

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

2. $1 = b s$ and hence $b \text{d}(s) = 0$,

3. $1 = a \pi + s b$ and hence $a \text{d}(\pi ) + \text{d}(s) b = 0$,

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

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

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

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

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

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. $\square$

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

(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

(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 $y_ i \to c(f_ i) \to x_ i[1]$. $\square$

Proof. Consider the admissible short exact sequence given by axiom (C).

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

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

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

as desired. The proof for (2) is omitted. $\square$

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

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

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

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

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

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

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. $\square$

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

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

since $g\tilde{b} = 0$ and $\tilde{b}'f = 0$ as consequences of the two commuting squares. $\square$

Proof. Suppose $\delta $ is defined by the splitting

and $\delta '$ is defined by the splitting with $\pi ',s'$ in place of $\pi ,s$. Then

since $bs' = bs = 1_ z$ and $\pi s = 0$. Similarly,

Since $\delta = \pi d(s)$ and $\delta ' = \pi 'd(s')$ as constructed in Lemma 22.27.1, we may compute

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. $\square$

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

Proof. We have a diagram of the form

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

where we have used equation (6) of Lemma 22.27.1 for the first equality and the preceeding remark for the second. Similarly, we obtain $d(\tilde{p}) = i\delta [-1]$. Hence

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:

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. $\square$

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

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 $K(\mathcal{A})$:

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. $\square$

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

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)

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

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

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,

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

Using these identities, we compute

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

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

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). $\square$

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

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:

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. $\square$

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,

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

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:

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

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

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

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

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

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. $\square$

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 $F(x) \to F(y) \to F(z)$. $\square$