
## 13.9 Cones and termwise split sequences

Let $\mathcal{A}$ be an additive category, and let $K(\mathcal{A})$ denote the category of complexes of $\mathcal{A}$ with morphisms given by morphisms of complexes up to homotopy. Note that the shift functors $[n]$ on complexes, see Homology, Definition 12.13.7, give rise to functors $[n] : K(\mathcal{A}) \to K(\mathcal{A})$ such that $[n] \circ [m] = [n + m]$ and $[0] = \text{id}$.

Definition 13.9.1. Let $\mathcal{A}$ be an additive category. Let $f : K^\bullet \to L^\bullet$ be a morphism of complexes of $\mathcal{A}$. The cone of $f$ is the complex $C(f)^\bullet$ given by $C(f)^ n = L^ n \oplus K^{n + 1}$ and differential

$d_{C(f)}^ n = \left( \begin{matrix} d^ n_ L & f^{n + 1} \\ 0 & -d_ K^{n + 1} \end{matrix} \right)$

It comes equipped with canonical morphisms of complexes $i : L^\bullet \to C(f)^\bullet$ and $p : C(f)^\bullet \to K^\bullet [1]$ induced by the obvious maps $L^ n \to C(f)^ n \to K^{n + 1}$.

In other words $(K, L, C(f), f, i, p)$ forms a triangle:

$K^\bullet \to L^\bullet \to C(f)^\bullet \to K^\bullet [1]$

The formation of this triangle is functorial in the following sense.

Lemma 13.9.2. Suppose that

$\xymatrix{ K_1^\bullet \ar[r]_{f_1} \ar[d]_ a & L_1^\bullet \ar[d]^ b \\ K_2^\bullet \ar[r]^{f_2} & L_2^\bullet }$

is a diagram of morphisms of complexes which is commutative up to homotopy. Then there exists a morphism $c : C(f_1)^\bullet \to C(f_2)^\bullet$ which gives rise to a morphism of triangles $(a, b, c) : (K_1^\bullet , L_1^\bullet , C(f_1)^\bullet , f_1, i_1, p_1) \to (K_2^\bullet , L_2^\bullet , C(f_2)^\bullet , f_2, i_2, p_2)$ of $K(\mathcal{A})$.

Proof. Let $h^ n : K_1^ n \to L_2^{n - 1}$ be a family of morphisms such that $b \circ f_1 - f_2 \circ a= d \circ h + h \circ d$. Define $c^ n$ by the matrix

$c^ n = \left( \begin{matrix} b^ n & h^{n + 1} \\ 0 & a^{n + 1} \end{matrix} \right) : L_1^ n \oplus K_1^{n + 1} \to L_2^ n \oplus K_2^{n + 1}$

A matrix computation show that $c$ is a morphism of complexes. It is trivial that $c \circ i_1 = i_2 \circ b$, and it is trivial also to check that $p_2 \circ c = a \circ p_1$. $\square$

Note that the morphism $c : C(f_1)^\bullet \to C(f_2)^\bullet$ constructed in the proof of Lemma 13.9.2 in general depends on the chosen homotopy $h$ between $f_2 \circ a$ and $b \circ f_1$.

Lemma 13.9.3. Suppose that $f: K^\bullet \to L^\bullet$ and $g : L^\bullet \to M^\bullet$ are morphisms of complexes such that $g \circ f$ is homotopic to zero. Then

1. $g$ factors through a morphism $C(f)^\bullet \to M^\bullet$, and

2. $f$ factors through a morphism $K^\bullet \to C(g)^\bullet [-1]$.

Proof. The assumptions say that the diagram

$\xymatrix{ K^\bullet \ar[r]_ f \ar[d] & L^\bullet \ar[d]^ g \\ 0 \ar[r] & M^\bullet }$

commutes up to homotopy. Since the cone on $0 \to M^\bullet$ is $M^\bullet$ the map $C(f)^\bullet \to C(0 \to M^\bullet ) = M^\bullet$ of Lemma 13.9.2 is the map in (1). The cone on $K^\bullet \to 0$ is $K^\bullet [1]$ and applying Lemma 13.9.2 gives a map $K^\bullet [1] \to C(g)^\bullet$. Applying $[-1]$ we obtain the map in (2). $\square$

Note that the morphisms $C(f)^\bullet \to M^\bullet$ and $K^\bullet \to C(g)^\bullet [-1]$ constructed in the proof of Lemma 13.9.3 in general depend on the chosen homotopy.

Definition 13.9.4. Let $\mathcal{A}$ be an additive category. A termwise split injection $\alpha : A^\bullet \to B^\bullet$ is a morphism of complexes such that each $A^ n \to B^ n$ is isomorphic to the inclusion of a direct summand. A termwise split surjection $\beta : B^\bullet \to C^\bullet$ is a morphism of complexes such that each $B^ n \to C^ n$ is isomorphic to the projection onto a direct summand.

Lemma 13.9.5. Let $\mathcal{A}$ be an additive category. Let

$\xymatrix{ A^\bullet \ar[r]_ f \ar[d]_ a & B^\bullet \ar[d]^ b \\ C^\bullet \ar[r]^ g & D^\bullet }$

be a diagram of morphisms of complexes commuting up to homotopy. If $f$ is a split injection, then $b$ is homotopic to a morphism which makes the diagram commute. If $g$ is a split surjection, then $a$ is homotopic to a morphism which makes the diagram commute.

Proof. Let $h^ n : A^ n \to D^{n - 1}$ be a collection of morphisms such that $bf - ga = dh + hd$. Suppose that $\pi ^ n : B^ n \to A^ n$ are morphisms splitting the morphisms $f^ n$. Take $b' = b - dh\pi - h\pi d$. Suppose $s^ n : D^ n \to C^ n$ are morphisms splitting the morphisms $g^ n : C^ n \to D^ n$. Take $a' = a + dsh + shd$. Computations omitted. $\square$

The following lemma can be used to replace a morphism of complexes by a morphism where in each degree the map is the injection of a direct summand.

Lemma 13.9.6. Let $\mathcal{A}$ be an additive category. Let $\alpha : K^\bullet \to L^\bullet$ be a morphism of complexes of $\mathcal{A}$. There exists a factorization

$\xymatrix{ K^\bullet \ar[r]^{\tilde\alpha } \ar@/_1pc/[rr]_\alpha & \tilde L^\bullet \ar[r]^\pi & L^\bullet }$

such that

1. $\tilde\alpha$ is a termwise split injection (see Definition 13.9.4),

2. there is a map of complexes $s : L^\bullet \to \tilde L^\bullet$ such that $\pi \circ s = \text{id}_{L^\bullet }$ and such that $s \circ \pi$ is homotopic to $\text{id}_{\tilde L^\bullet }$.

Moreover, if both $K^\bullet$ and $L^\bullet$ are in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$, then so is $\tilde L^\bullet$.

Proof. We set

$\tilde L^ n = L^ n \oplus K^ n \oplus K^{n + 1}$

and we define

$d^ n_{\tilde L} = \left( \begin{matrix} d^ n_ L & 0 & 0 \\ 0 & d^ n_ K & \text{id}_{K^{n + 1}} \\ 0 & 0 & -d^{n + 1}_ K \end{matrix} \right)$

In other words, $\tilde L^\bullet = L^\bullet \oplus C(1_{K^\bullet })$. Moreover, we set

$\tilde\alpha = \left( \begin{matrix} \alpha \\ \text{id}_{K^ n} \\ 0 \end{matrix} \right)$

which is clearly a split injection. It is also clear that it defines a morphism of complexes. We define

$\pi = \left( \begin{matrix} \text{id}_{L^ n} & 0 & 0 \end{matrix} \right)$

so that clearly $\pi \circ \tilde\alpha = \alpha$. We set

$s = \left( \begin{matrix} \text{id}_{L^ n} \\ 0 \\ 0 \end{matrix} \right)$

so that $\pi \circ s = \text{id}_{L^\bullet }$. Finally, let $h^ n : \tilde L^ n \to \tilde L^{n - 1}$ be the map which maps the summand $K^ n$ of $\tilde L^ n$ via the identity morphism to the summand $K^ n$ of $\tilde L^{n - 1}$. Then it is a trivial matter (see computations in remark below) to prove that

$\text{id}_{\tilde L^\bullet } - s \circ \pi = d \circ h + h \circ d$

which finishes the proof of the lemma. $\square$

Remark 13.9.7. To see the last displayed equality in the proof above we can argue with elements as follows. We have $s\pi (l, k, k^{+}) = (l, 0, 0)$. Hence the morphism of the left hand side maps $(l, k, k^{+})$ to $(0, k, k^{+})$. On the other hand $h(l, k, k^{+}) = (0, 0, k)$ and $d(l, k, k^{+}) = (dl, dk + k^{+}, -dk^{+})$. Hence $(dh + hd)(l, k, k^{+}) = d(0, 0, k) + h(dl, dk + k^{+}, -dk^{+}) = (0, k, -dk) + (0, 0, dk + k^{+}) = (0, k, k^{+})$ as desired.

Lemma 13.9.8. Let $\mathcal{A}$ be an additive category. Let $\alpha : K^\bullet \to L^\bullet$ be a morphism of complexes of $\mathcal{A}$. There exists a factorization

$\xymatrix{ K^\bullet \ar[r]^ i \ar@/_1pc/[rr]_\alpha & \tilde K^\bullet \ar[r]^{\tilde\alpha } & L^\bullet }$

such that

1. $\tilde\alpha$ is a termwise split surjection (see Definition 13.9.4),

2. there is a map of complexes $s : \tilde K^\bullet \to K^\bullet$ such that $s \circ i = \text{id}_{K^\bullet }$ and such that $i \circ s$ is homotopic to $\text{id}_{\tilde K^\bullet }$.

Moreover, if both $K^\bullet$ and $L^\bullet$ are in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$, then so is $\tilde K^\bullet$.

Proof. Dual to Lemma 13.9.6. Take

$\tilde K^ n = K^ n \oplus L^{n - 1} \oplus L^ n$

and we define

$d^ n_{\tilde K} = \left( \begin{matrix} d^ n_ K & 0 & 0 \\ 0 & - d^{n - 1}_ L & \text{id}_{L^ n} \\ 0 & 0 & d^ n_ L \end{matrix} \right)$

in other words $\tilde K^\bullet = K^\bullet \oplus C(1_{L^\bullet [-1]})$. Moreover, we set

$\tilde\alpha = \left( \begin{matrix} \alpha & 0 & \text{id}_{L^ n} \end{matrix} \right)$

which is clearly a split surjection. It is also clear that it defines a morphism of complexes. We define

$i = \left( \begin{matrix} \text{id}_{K^ n} \\ 0 \\ 0 \end{matrix} \right)$

so that clearly $\tilde\alpha \circ i = \alpha$. We set

$s = \left( \begin{matrix} \text{id}_{K^ n} & 0 & 0 \end{matrix} \right)$

so that $s \circ i = \text{id}_{K^\bullet }$. Finally, let $h^ n : \tilde K^ n \to \tilde K^{n - 1}$ be the map which maps the summand $L^{n - 1}$ of $\tilde K^ n$ via the identity morphism to the summand $L^{n - 1}$ of $\tilde K^{n - 1}$. Then it is a trivial matter to prove that

$\text{id}_{\tilde K^\bullet } - i \circ s = d \circ h + h \circ d$

which finishes the proof of the lemma. $\square$

Definition 13.9.9. Let $\mathcal{A}$ be an additive category. A termwise split exact sequence of complexes of $\mathcal{A}$ is a complex of complexes

$0 \to A^\bullet \xrightarrow {\alpha } B^\bullet \xrightarrow {\beta } C^\bullet \to 0$

together with given direct sum decompositions $B^ n = A^ n \oplus C^ n$ compatible with $\alpha ^ n$ and $\beta ^ n$. We often write $s^ n : C^ n \to B^ n$ and $\pi ^ n : B^ n \to A^ n$ for the maps induced by the direct sum decompositions. According to Homology, Lemma 12.13.10 we get an associated morphism of complexes

$\delta : C^\bullet \longrightarrow A^\bullet [1]$

which in degree $n$ is the map $\pi ^{n + 1} \circ d_ B^ n \circ s^ n$. In other words $(A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta )$ forms a triangle

$A^\bullet \to B^\bullet \to C^\bullet \to A^\bullet [1]$

This will be the triangle associated to the termwise split sequence of complexes.

Lemma 13.9.10. Let $\mathcal{A}$ be an additive category. Let $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ be termwise split exact sequences as in Definition 13.9.9. Let $(\pi ')^ n$, $(s')^ n$ be a second collection of splittings. Denote $\delta ' : C^\bullet \longrightarrow A^\bullet [1]$ the morphism associated to this second set of splittings. Then

$(1, 1, 1) : (A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta ) \longrightarrow (A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta ')$

is an isomorphism of triangles in $K(\mathcal{A})$.

Proof. The statement simply means that $\delta$ and $\delta '$ are homotopic maps of complexes. This is Homology, Lemma 12.13.12. $\square$

Remark 13.9.11. Let $\mathcal{A}$ be an additive category. Let $0 \to A_ i^\bullet \to B_ i^\bullet \to C_ i^\bullet \to 0$, $i = 1, 2$ be termwise split exact sequences. Suppose that $a : A_1^\bullet \to A_2^\bullet$, $b : B_1^\bullet \to B_2^\bullet$, and $c : C_1^\bullet \to C_2^\bullet$ are morphisms of complexes such that

$\xymatrix{ A_1^\bullet \ar[d]_ a \ar[r] & B_1^\bullet \ar[r] \ar[d]_ b & C_1^\bullet \ar[d]_ c \\ A_2^\bullet \ar[r] & B_2^\bullet \ar[r] & C_2^\bullet }$

commutes in $K(\mathcal{A})$. In general, there does not exist a morphism $b' : B_1^\bullet \to B_2^\bullet$ which is homotopic to $b$ such that the diagram above commutes in the category of complexes. Namely, consider Examples, Equation (102.56.0.1). If we could replace the middle map there by a homotopic one such that the diagram commutes, then we would have additivity of traces which we do not.

Lemma 13.9.12. Let $\mathcal{A}$ be an additive category. Let $0 \to A_ i^\bullet \to B_ i^\bullet \to C_ i^\bullet \to 0$, $i = 1, 2, 3$ be termwise split exact sequences of complexes. Let $b : B_1^\bullet \to B_2^\bullet$ and $b' : B_2^\bullet \to B_3^\bullet$ be morphisms of complexes such that

$\vcenter { \xymatrix{ A_1^\bullet \ar[d]_0 \ar[r] & B_1^\bullet \ar[r] \ar[d]_ b & C_1^\bullet \ar[d]_0 \\ A_2^\bullet \ar[r] & B_2^\bullet \ar[r] & C_2^\bullet } } \quad \text{and}\quad \vcenter { \xymatrix{ A_2^\bullet \ar[d]^0 \ar[r] & B_2^\bullet \ar[r] \ar[d]^{b'} & C_2^\bullet \ar[d]^0 \\ A_3^\bullet \ar[r] & B_3^\bullet \ar[r] & C_3^\bullet } }$

commute in $K(\mathcal{A})$. Then $b' \circ b = 0$ in $K(\mathcal{A})$.

Proof. By Lemma 13.9.5 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 $\mathop{\mathrm{Im}}(b^ n) \subset \mathop{\mathrm{Im}}(A_2^ n \to B_2^ n)$ and $\mathop{\mathrm{Ker}}((b')^ n) \supset \mathop{\mathrm{Im}}(A_2^ n \to B_2^ n)$. Then $b \circ b' = 0$ as a map of complexes. $\square$

Lemma 13.9.13. Let $\mathcal{A}$ be an additive category. Let $f_1 : K_1^\bullet \to L_1^\bullet$ and $f_2 : K_2^\bullet \to L_2^\bullet$ be morphisms of complexes. Let

$(a, b, c) : (K_1^\bullet , L_1^\bullet , C(f_1)^\bullet , f_1, i_1, p_1) \longrightarrow (K_2^\bullet , L_2^\bullet , C(f_2)^\bullet , f_2, i_2, p_2)$

be any morphism of triangles of $K(\mathcal{A})$. If $a$ and $b$ are homotopy equivalences then so is $c$.

Proof. Let $a^{-1} : K_2^\bullet \to K_1^\bullet$ be a morphism of complexes which is inverse to $a$ in $K(\mathcal{A})$. Let $b^{-1} : L_2^\bullet \to L_1^\bullet$ be a morphism of complexes which is inverse to $b$ in $K(\mathcal{A})$. Let $c' : C(f_2)^\bullet \to C(f_1)^\bullet$ be the morphism from Lemma 13.9.2 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 $K(\mathcal{A})$ then we win. Hence it suffices to prove the following: Given a morphism of triangles $(1, 1, c) : (K^\bullet , L^\bullet , C(f)^\bullet , f, i, p)$ in $K(\mathcal{A})$ the morphism $c$ is an isomorphism in $K(\mathcal{A})$. By assumption the two squares in the diagram

$\xymatrix{ L^\bullet \ar[r] \ar[d]_1 & C(f)^\bullet \ar[r] \ar[d]_ c & K^\bullet [1] \ar[d]_1 \\ L^\bullet \ar[r] & C(f)^\bullet \ar[r] & K^\bullet [1] }$

commute up to homotopy. By construction of $C(f)^\bullet$ the rows form termwise split sequences of complexes. Thus we see that $(c - 1)^2 = 0$ in $K(\mathcal{A})$ by Lemma 13.9.12. Hence $c$ is an isomorphism in $K(\mathcal{A})$ with inverse $2 - c$. $\square$

Hence if $a$ and $b$ are homotopy equivalences then the resulting morphism of triangles is an isomorphism of triangles in $K(\mathcal{A})$. It turns out that the collection of triangles of $K(\mathcal{A})$ given by cones and the collection of triangles of $K(\mathcal{A})$ given by termwise split sequences of complexes are the same up to isomorphisms, at least up to sign!

Lemma 13.9.14. Let $\mathcal{A}$ be an additive category.

1. Given a termwise split sequence of complexes $(\alpha : A^\bullet \to B^\bullet , \beta : B^\bullet \to C^\bullet , s^ n, \pi ^ n)$ there exists a homotopy equivalence $C(\alpha )^\bullet \to C^\bullet$ such that the diagram

$\xymatrix{ A^\bullet \ar[r] \ar[d] & B^\bullet \ar[d] \ar[r] & C(\alpha )^\bullet \ar[r]_{-p} \ar[d] & A^\bullet [1] \ar[d] \\ A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet \ar[r]^\delta & A^\bullet [1] }$

defines an isomorphism of triangles in $K(\mathcal{A})$.

2. Given a morphism of complexes $f : K^\bullet \to L^\bullet$ there exists an isomorphism of triangles

$\xymatrix{ K^\bullet \ar[r] \ar[d] & \tilde L^\bullet \ar[d] \ar[r] & M^\bullet \ar[r]_{\delta } \ar[d] & K^\bullet [1] \ar[d] \\ K^\bullet \ar[r] & L^\bullet \ar[r] & C(f)^\bullet \ar[r]^{-p} & K^\bullet [1] }$

where the upper triangle is the triangle associated to a termwise split exact sequence $K^\bullet \to \tilde L^\bullet \to M^\bullet$.

Proof. Proof of (1). We have $C(\alpha )^ n = B^ n \oplus A^{n + 1}$ and we simply define $C(\alpha )^ n \to C^ n$ via the projection onto $B^ n$ followed by $\beta ^ n$. This defines a morphism of complexes because the compositions $A^{n + 1} \to B^{n + 1} \to C^{n + 1}$ are zero. To get a homotopy inverse we take $C^\bullet \to C(\alpha )^\bullet$ given by $(s^ n , -\delta ^ n)$ in degree $n$. This is a morphism of complexes because the morphism $\delta ^ n$ can be characterized as the unique morphism $C^ n \to A^{n + 1}$ such that $d \circ s^ n - s^{n + 1} \circ d = \alpha \circ \delta ^ n$, see proof of Homology, Lemma 12.13.10. The composition $C^\bullet \to C(\alpha )^\bullet \to C^\bullet$ is the identity. The composition $C(\alpha )^\bullet \to C^\bullet \to C(\alpha )^\bullet$ is equal to the morphism

$\left( \begin{matrix} s^ n \circ \beta ^ n & 0 \\ -\delta ^ n \circ \beta ^ n & 0 \end{matrix} \right)$

To see that this is homotopic to the identity map use the homotopy $h^ n : C(\alpha )^ n \to C(\alpha )^{n - 1}$ given by the matrix

$\left( \begin{matrix} 0 & 0 \\ \pi ^ n & 0 \end{matrix} \right) : C(\alpha )^ n = B^ n \oplus A^{n + 1} \to B^{n - 1} \oplus A^ n = C(\alpha )^{n - 1}$

It is trivial to verify that

$\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) - \left( \begin{matrix} s^ n \\ -\delta ^ n \end{matrix} \right) \left( \begin{matrix} \beta ^ n & 0 \end{matrix} \right) = \left( \begin{matrix} d & \alpha ^ n \\ 0 & -d \end{matrix} \right) \left( \begin{matrix} 0 & 0 \\ \pi ^ n & 0 \end{matrix} \right) + \left( \begin{matrix} 0 & 0 \\ \pi ^{n + 1} & 0 \end{matrix} \right) \left( \begin{matrix} d & \alpha ^{n + 1} \\ 0 & -d \end{matrix} \right)$

To finish the proof of (1) we have to show that the morphisms $-p : C(\alpha )^\bullet \to A^\bullet [1]$ (see Definition 13.9.1) and $C(\alpha )^\bullet \to C^\bullet \to A^\bullet [1]$ agree up to homotopy. This is clear from the above. Namely, we can use the homotopy inverse $(s, -\delta ) : C^\bullet \to C(\alpha )^\bullet$ and check instead that the two maps $C^\bullet \to A^\bullet [1]$ agree. And note that $p \circ (s, -\delta ) = -\delta$ as desired.

Proof of (2). We let $\tilde f : K^\bullet \to \tilde L^\bullet$, $s : L^\bullet \to \tilde L^\bullet$ and $\pi : \tilde L^\bullet \to L^\bullet$ be as in Lemma 13.9.6. By Lemmas 13.9.2 and 13.9.13 the triangles $(K^\bullet , L^\bullet , C(f), i, p)$ and $(K^\bullet , \tilde L^\bullet , C(\tilde f), \tilde i, \tilde p)$ are isomorphic. Note that we can compose isomorphisms of triangles. Thus we may replace $L^\bullet$ by $\tilde L^\bullet$ and $f$ by $\tilde f$. In other words we may assume that $f$ is a termwise split injection. In this case the result follows from part (1). $\square$

Lemma 13.9.15. Let $\mathcal{A}$ be an additive category. Let $A_1^\bullet \to A_2^\bullet \to \ldots \to A_ n^\bullet$ be a sequence of composable morphisms of complexes. There exists a commutative diagram

$\xymatrix{ A_1^\bullet \ar[r] & A_2^\bullet \ar[r] & \ldots \ar[r] & A_ n^\bullet \\ B_1^\bullet \ar[r] \ar[u] & B_2^\bullet \ar[r] \ar[u] & \ldots \ar[r] & B_ n^\bullet \ar[u] }$

such that each morphism $B_ i^\bullet \to B_{i + 1}^\bullet$ is a split injection and each $B_ i^\bullet \to A_ i^\bullet$ is a homotopy equivalence. Moreover, if all $A_ i^\bullet$ are in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$, then so are the $B_ i^\bullet$.

Proof. The case $n = 1$ is without content. Lemma 13.9.6 is the case $n = 2$. Suppose we have constructed the diagram except for $B_ n^\bullet$. Apply Lemma 13.9.6 to the composition $B_{n - 1}^\bullet \to A_{n - 1}^\bullet \to A_ n^\bullet$. The result is a factorization $B_{n - 1}^\bullet \to B_ n^\bullet \to A_ n^\bullet$ as desired. $\square$

Lemma 13.9.16. Let $\mathcal{A}$ be an additive category. Let $(\alpha : A^\bullet \to B^\bullet , \beta : B^\bullet \to C^\bullet , s^ n, \pi ^ n)$ be a termwise split sequence of complexes. Let $(A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta )$ be the associated triangle. Then the triangle $(C^\bullet [-1], A^\bullet , B^\bullet , \delta [-1], \alpha , \beta )$ is isomorphic to the triangle $(C^\bullet [-1], A^\bullet , C(\delta [-1])^\bullet , \delta [-1], i, p)$.

Proof. We write $B^ n = A^ n \oplus C^ n$ and we identify $\alpha ^ n$ and $\beta ^ n$ with the natural inclusion and projection maps. By construction of $\delta$ we have

$d_ B^ n = \left( \begin{matrix} d_ A^ n & \delta ^ n \\ 0 & d_ C^ n \end{matrix} \right)$

On the other hand the cone of $\delta [-1] : C^\bullet [-1] \to A^\bullet$ is given as $C(\delta [-1])^ n = A^ n \oplus C^ n$ with differential identical with the matrix above! Whence the lemma. $\square$

Lemma 13.9.17. Let $\mathcal{A}$ be an additive category. Let $f : K^\bullet \to L^\bullet$ be a morphism of complexes. The triangle $(L^\bullet , C(f)^\bullet , K^\bullet [1], i, p, f[1])$ is the triangle associated to the termwise split sequence

$0 \to L^\bullet \to C(f)^\bullet \to K^\bullet [1] \to 0$

coming from the definition of the cone of $f$.

Proof. Immediate from the definitions. $\square$

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