The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 014D. Beware of the difference between the letter 'O' and the digit '0'.