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 \input{preamble} % OK, start here. % \begin{document} \title{Derived Categories} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent We first discuss triangulated categories and localization in triangulated categories. Next, we prove that the homotopy category of complexes in an additive category is a triangulated category. Once this is done we define the derived category of an abelian category as the localization of the of homotopy category with respect to quasi-isomorphisms. A good reference is Verdier's thesis \cite{Verdier}. \section{Triangulated categories} \label{section-triangulated-categories} \noindent Triangulated categories are a convenient tool to describe the type of structure inherent in the derived category of an abelian category. Some references are \cite{Verdier}, \cite{KS}, and \cite{Neeman}. \section{The definition of a triangulated category} \label{section-triangulated-definitions} \noindent In this section we collect most of the definitions concerning triangulated and pre-triangulated categories. \begin{definition} \label{definition-triangle} Let $\mathcal{D}$ be an additive category. Let $[n] : \mathcal{D} \to \mathcal{D}$, $E \mapsto E[n]$ be a collection of additive functors indexed by $n \in \mathbf{Z}$ such that $[n] \circ [m] = [n + m]$ and $[0] = \text{id}$ (equality as functors). In this situation we define a {\it triangle} to be a sextuple $(X, Y, Z, f, g, h)$ where $X, Y, Z \in \Ob(\mathcal{D})$ and $f : X \to Y$, $g : Y \to Z$ and $h : Z \to X[1]$ are morphisms of $\mathcal{D}$. A {\it morphism of triangles} $(X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')$ is given by morphisms $a : X \to X'$, $b : Y \to Y'$ and $c : Z \to Z'$ of $\mathcal{D}$ such that $b \circ f = f' \circ a$, $c \circ g = g' \circ b$ and $a[1] \circ h = h' \circ c$. \end{definition} \noindent A morphism of triangles is visualized by the following commutative diagram $$\xymatrix{ X \ar[r] \ar[d]^a & Y \ar[r] \ar[d]^b & Z \ar[r] \ar[d]^c & X[1] \ar[d]^{a[1]} \\ X' \ar[r] & Y' \ar[r] & Z' \ar[r] & X'[1] }$$ Here is the definition of a triangulated category as given in Verdier's thesis. \begin{definition} \label{definition-triangulated-category} A {\it triangulated category} consists of a triple $(\mathcal{D}, \{[n]\}_{n\in \mathbf{Z}}, \mathcal{T})$ where \begin{enumerate} \item $\mathcal{D}$ is an additive category, \item $[n] : \mathcal{D} \to \mathcal{D}$, $E \mapsto E[n]$ is a collection of additive functors indexed by $n \in \mathbf{Z}$ such that $[n] \circ [m] = [n + m]$ and $[0] = \text{id}$ (equality as functors), and \item $\mathcal{T}$ is a set of triangles called the {\it distinguished triangles} \end{enumerate} subject to the following conditions \begin{enumerate} \item[TR1] Any triangle isomorphic to a distinguished triangle is a distinguished triangle. Any triangle of the form $(X, X, 0, \text{id}, 0, 0)$ is distinguished. For any morphism $f : X \to Y$ of $\mathcal{D}$ there exists a distinguished triangle of the form $(X, Y, Z, f, g, h)$. \item[TR2] The triangle $(X, Y, Z, f, g, h)$ is distinguished if and only if the triangle $(Y, Z, X[1], g, h, -f[1])$ is. \item[TR3] Given a solid diagram $$\xymatrix{ X \ar[r]^f \ar[d]^a & Y \ar[r]^g \ar[d]^b & Z \ar[r]^h \ar@{-->}[d] & X[1] \ar[d]^{a[1]} \\ X' \ar[r]^{f'} & Y' \ar[r]^{g'} & Z' \ar[r]^{h'} & X'[1] }$$ whose rows are distinguished triangles and which satisfies $b \circ f = f' \circ a$, there exists a morphism $c : Z \to Z'$ such that $(a, b, c)$ is a morphism of triangles. \item[TR4] Given objects $X$, $Y$, $Z$ of $\mathcal{D}$, and morphisms $f : X \to Y$, $g : Y \to Z$, and distinguished triangles $(X, Y, Q_1, f, p_1, d_1)$, $(X, Z, Q_2, g \circ f, p_2, d_2)$, and $(Y, Z, Q_3, g, p_3, d_3)$, there exist morphisms $a : Q_1 \to Q_2$ and $b : Q_2 \to Q_3$ such that \begin{enumerate} \item $(Q_1, Q_2, Q_3, a, b, p_1[1] \circ d_3)$ is a distinguished triangle, \item the triple $(\text{id}_X, g, a)$ is a morphism of triangles $(X, Y, Q_1, f, p_1, d_1) \to (X, Z, Q_2, g \circ f, p_2, d_2)$, and \item the triple $(f, \text{id}_Z, b)$ is a morphism of triangles $(X, Z, Q_2, g \circ f, p_2, d_2) \to (Y, Z, Q_3, g, p_3, d_3)$. \end{enumerate} \end{enumerate} We will call $(\mathcal{D}, [\ ], \mathcal{T})$ a {\it pre-triangulated category} if TR1, TR2 and TR3 hold.\footnote{We use $[\ ]$ as an abbreviation for the family $\{[n]\}_{n\in \mathbf{Z}}$.} \end{definition} \noindent The explanation of TR4 is that if you think of $Q_1$ as $Y/X$, $Q_2$ as $Z/X$ and $Q_3$ as $Z/Y$, then TR4(a) expresses the isomorphism $(Z/X)/(Y/X) \cong Z/Y$ and TR4(b) and TR4(c) express that we can compare the triangles $X \to Y \to Q_1 \to X[1]$ etc with morphisms of triangles. For a more precise reformulation of this idea see the proof of Lemma \ref{lemma-two-split-injections}. \medskip\noindent The sign in TR2 means that if $(X, Y, Z, f, g, h)$ is a distinguished triangle then in the long sequence \begin{equation} \label{equation-rotate} \ldots \to Z[-1] \xrightarrow{-h[-1]} X \xrightarrow{f} Y \xrightarrow{g} Z \xrightarrow{h} X[1] \xrightarrow{-f[1]} Y[1] \xrightarrow{-g[1]} Z[1] \to \ldots \end{equation} each four term sequence gives a distinguished triangle. \medskip\noindent As usual we abuse notation and we simply speak of a (pre-)triangulated category $\mathcal{D}$ without explicitly introducing notation for the additional data. The notion of a pre-triangulated category is useful in finding statements equivalent to TR4. \medskip\noindent We have the following definition of a triangulated functor. \begin{definition} \label{definition-exact-functor-triangulated-categories} Let $\mathcal{D}$, $\mathcal{D}'$ be pre-triangulated categories. An {\it exact functor}, or a {\it triangulated functor} from $\mathcal{D}$ to $\mathcal{D}'$ is a functor $F : \mathcal{D} \to \mathcal{D}'$ together with given functorial isomorphisms $\xi_X : F(X[1]) \to F(X)[1]$ such that for every distinguished triangle $(X, Y, Z, f, g, h)$ of $\mathcal{D}$ the triangle $(F(X), F(Y), F(Z), F(f), F(g), \xi_X \circ F(h))$ is a distinguished triangle of $\mathcal{D}'$. \end{definition} \noindent An exact functor is additive, see Lemma \ref{lemma-exact-functor-additive}. When we say two triangulated categories are equivalent we mean that they are equivalent in the $2$-category of triangulated categories. A $2$-morphism $a : (F, \xi) \to (F', \xi')$ in this $2$-category is simply a transformation of functors $a : F \to F'$ which is compatible with $\xi$ and $\xi'$, i.e., $$\xymatrix{ F \circ [1] \ar[r]_\xi \ar[d]_{a \star 1} & [1] \circ F \ar[d]^{1 \star a} \\ F' \circ [1] \ar[r]^{\xi'} & [1] \circ F' }$$ commutes. \begin{definition} \label{definition-triangulated-subcategory} Let $(\mathcal{D}, [\ ], \mathcal{T})$ be a pre-triangulated category. A {\it pre-triangulated subcategory}\footnote{This definition may be nonstandard. If $\mathcal{D}'$ is a full subcategory then $\mathcal{T}'$ is the intersection of the set of triangles in $\mathcal{D}'$ with $\mathcal{T}$, see Lemma \ref{lemma-triangulated-subcategory}. In this case we drop $\mathcal{T}'$ from the notation.} is a pair $(\mathcal{D}', \mathcal{T}')$ such that \begin{enumerate} \item $\mathcal{D}'$ is an additive subcategory of $\mathcal{D}$ which is preserved under $[1]$ and $[-1]$, \item $\mathcal{T}' \subset \mathcal{T}$ is a subset such that for every $(X, Y, Z, f, g, h) \in \mathcal{T}'$ we have $X, Y, Z \in \Ob(\mathcal{D}')$ and $f, g, h \in \text{Arrows}(\mathcal{D}')$, and \item $(\mathcal{D}', [\ ], \mathcal{T}')$ is a pre-triangulated category. \end{enumerate} If $\mathcal{D}$ is a triangulated category, then we say $(\mathcal{D}', \mathcal{T}')$ is a {\it triangulated subcategory} if it is a pre-triangulated subcategory and $(\mathcal{D}', [\ ], \mathcal{T}')$ is a triangulated category. \end{definition} \noindent In this situation the inclusion functor $\mathcal{D}' \to \mathcal{D}$ is an exact functor with $\xi_X : X[1] \to X[1]$ given by the identity on $X[1]$. \medskip\noindent We will see in Lemma \ref{lemma-composition-zero} that for a distinguished triangle $(X, Y, Z, f, g, h)$ in a pre-triangulated category the composition $g \circ f : X \to Z$ is zero. Thus the sequence (\ref{equation-rotate}) is a complex. A homological functor is one that turns this complex into a long exact sequence. \begin{definition} \label{definition-homological} Let $\mathcal{D}$ be a pre-triangulated category. Let $\mathcal{A}$ be an abelian category. An additive functor $H : \mathcal{D} \to \mathcal{A}$ is called {\it homological} if for every distinguished triangle $(X, Y, Z, f, g, h)$ the sequence $$H(X) \to H(Y) \to H(Z)$$ is exact in the abelian category $\mathcal{A}$. An additive functor $H : \mathcal{D}^{opp} \to \mathcal{A}$ is called {\it cohomological} if the corresponding functor $\mathcal{D} \to \mathcal{A}^{opp}$ is homological. \end{definition} \noindent If $H : \mathcal{D} \to \mathcal{A}$ is a homological functor we often write $H^n(X) = H(X[n])$ so that $H(X) = H^0(X)$. Our discussion of TR2 above implies that a distinguished triangle $(X, Y, Z, f, g, h)$ determines a long exact sequence \begin{equation} \label{equation-long-exact-cohomology-sequence} \xymatrix@C=3pc{ H^{-1}(Z) \ar[r]^{H(h[-1])} & H^0(X) \ar[r]^{H(f)} & H^0(Y) \ar[r]^{H(g)} & H^0(Z) \ar[r]^{H(h)} & H^1(X) } \end{equation} This will be called the {\it long exact sequence} associated to the distinguished triangle and the homological functor. As indicated we will not use any signs for the morphisms in the long exact sequence. This has the side effect that maps in the long exact sequence associated to the rotation (TR2) of a distinguished triangle differ from the maps in the sequence above by some signs. \begin{definition} \label{definition-delta-functor} Let $\mathcal{A}$ be an abelian category. Let $\mathcal{D}$ be a triangulated category. A {\it $\delta$-functor from $\mathcal{A}$ to $\mathcal{D}$} is given by a functor $G : \mathcal{A} \to \mathcal{D}$ and a rule which assigns to every short exact sequence $$0 \to A \xrightarrow{a} B \xrightarrow{b} C \to 0$$ a morphism $\delta = \delta_{A \to B \to C} : G(C) \to G(A)[1]$ such that \begin{enumerate} \item the triangle $(G(A), G(B), G(C), G(a), G(b), \delta_{A \to B \to C})$ is a distinguished triangle of $\mathcal{D}$ for any short exact sequence as above, and \item for every morphism $(A \to B \to C) \to (A' \to B' \to C')$ of short exact sequences the diagram $$\xymatrix{ G(C) \ar[d] \ar[rr]_{\delta_{A \to B \to C}} & & G(A)[1] \ar[d] \\ G(C') \ar[rr]^{\delta_{A' \to B' \to C'}} & & G(A')[1] }$$ is commutative. \end{enumerate} In this situation we call $(G(A), G(B), G(C), G(a), G(b), \delta_{A \to B \to C})$ the {\it image of the short exact sequence under the given $\delta$-functor}. \end{definition} \noindent Note how a $\delta$-functor comes equipped with additional structure. Strictly speaking it does not make sense to say that a given functor $\mathcal{A} \to \mathcal{D}$ is a $\delta$-functor, but we will often do so anyway. \section{Elementary results on triangulated categories} \label{section-elementary-results} \noindent Most of the results in this section are proved for pre-triangulated categories and a fortiori hold in any triangulated category. \begin{lemma} \label{lemma-composition-zero} Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. Then $g \circ f = 0$, $h \circ g = 0$ and $f[1] \circ h = 0$. \end{lemma} \begin{proof} By TR1 we know $(X, X, 0, 1, 0, 0)$ is a distinguished triangle. Apply TR3 to $$\xymatrix{ X \ar[r] \ar[d]^1 & X \ar[r] \ar[d]^f & 0 \ar[r] \ar@{-->}[d] & X[1] \ar[d]^{1[1]} \\ X \ar[r]^f & Y \ar[r]^g & Z \ar[r]^h & X[1] }$$ Of course the dotted arrow is the zero map. Hence the commutativity of the diagram implies that $g \circ f = 0$. For the other cases rotate the triangle, i.e., apply TR2. \end{proof} \begin{lemma} \label{lemma-representable-homological} Let $\mathcal{D}$ be a pre-triangulated category. For any object $W$ of $\mathcal{D}$ the functor $\Hom_\mathcal{D}(W, -)$ is homological, and the functor $\Hom_\mathcal{D}(-, W)$ is cohomological. \end{lemma} \begin{proof} Consider a distinguished triangle $(X, Y, Z, f, g, h)$. We have already seen that $g \circ f = 0$, see Lemma \ref{lemma-composition-zero}. Suppose $a : W \to Y$ is a morphism such that $g \circ a = 0$. Then we get a commutative diagram $$\xymatrix{ W \ar[r]_1 \ar@{..>}[d]^b & W \ar[r] \ar[d]^a & 0 \ar[r] \ar[d]^0 & W[1] \ar@{..>}[d]^{b[1]} \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] }$$ Both rows are distinguished triangles (use TR1 for the top row). Hence we can fill the dotted arrow $b$ (first rotate using TR2, then apply TR3, and then rotate back). This proves the lemma. \end{proof} \begin{lemma} \label{lemma-third-isomorphism-triangle} Let $\mathcal{D}$ be a pre-triangulated category. Let $$(a, b, c) : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')$$ be a morphism of distinguished triangles. If two among $a, b, c$ are isomorphisms so is the third. \end{lemma} \begin{proof} Assume that $a$ and $c$ are isomorphisms. For any object $W$ of $\mathcal{D}$ write $H_W( - ) = \Hom_\mathcal{D}(W, -)$. Then we get a commutative diagram of abelian groups $$\xymatrix{ H_W(Z[-1]) \ar[r] \ar[d] & H_W(X) \ar[r] \ar[d] & H_W(Y) \ar[r] \ar[d] & H_W(Z) \ar[r] \ar[d] & H_W(X[1]) \ar[d] \\ H_W(Z'[-1]) \ar[r] & H_W(X') \ar[r] & H_W(Y') \ar[r] & H_W(Z') \ar[r] & H_W(X'[1]) }$$ By assumption the right two and left two vertical arrows are bijective. As $H_W$ is homological by Lemma \ref{lemma-representable-homological} and the five lemma (Homology, Lemma \ref{homology-lemma-five-lemma}) it follows that the middle vertical arrow is an isomorphism. Hence by Yoneda's lemma, see Categories, Lemma \ref{categories-lemma-yoneda} we see that $b$ is an isomorphism. This implies the other cases by rotating (using TR2). \end{proof} \begin{remark} \label{remark-special-triangles} Let $\mathcal{D}$ be an additive category with translation functors $[n]$ as in Definition \ref{definition-triangle}. Let us call a triangle $(X, Y, Z, f, g, h)$ {\it special}\footnote{This is nonstandard notation.} if for every object $W$ of $\mathcal{D}$ the long sequence of abelian groups $$\ldots \to \Hom_\mathcal{D}(W, X) \to \Hom_\mathcal{D}(W, Y) \to \Hom_\mathcal{D}(W, Z) \to \Hom_\mathcal{D}(W, X[1]) \to \ldots$$ is exact. The proof of Lemma \ref{lemma-third-isomorphism-triangle} shows that if $$(a, b, c) : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')$$ is a morphism of special triangles and if two among $a, b, c$ are isomorphisms so is the third. There is a dual statement for {\it co-special} triangles, i.e., triangles which turn into long exact sequences on applying the functor $\Hom_\mathcal{D}(-, W)$. Thus distinguished triangles are special and co-special, but in general there are many more (co-)special triangles, then there are distinguished triangles. \end{remark} \begin{lemma} \label{lemma-third-map-square-zero} Let $\mathcal{D}$ be a pre-triangulated category. Let $$(0, b, 0), (0, b', 0) : (X, Y, Z, f, g, h) \to (X, Y, Z, f, g, h)$$ be endomorphisms of a distinguished triangle. Then $bb' = 0$. \end{lemma} \begin{proof} Picture $$\xymatrix{ X \ar[r] \ar[d]^0 & Y \ar[r] \ar[d]^{b, b'} \ar@{..>}[ld]^\alpha & Z \ar[r] \ar[d]^0 \ar@{..>}[ld]^\beta & X[1] \ar[d]^0 \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] }$$ Applying Lemma \ref{lemma-representable-homological} we find dotted arrows $\alpha$ and $\beta$ such that $b' = f \circ \alpha$ and $b = \beta \circ g$. Then $bb' = \beta \circ g \circ f \circ \alpha = 0$ as $g \circ f = 0$ by Lemma \ref{lemma-composition-zero}. \end{proof} \begin{lemma} \label{lemma-third-map-idempotent} Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. If $$\xymatrix{ Z \ar[r]_h \ar[d]_c & X[1] \ar[d]^{a[1]} \\ Z \ar[r]^h & X[1] }$$ is commutative and $a^2 = a$, $c^2 = c$, then there exists a morphism $b : Y \to Y$ with $b^2 = b$ such that $(a, b, c)$ is an endomorphism of the triangle $(X, Y, Z, f, g, h)$. \end{lemma} \begin{proof} By TR3 there exists a morphism $b'$ such that $(a, b', c)$ is an endomorphism of $(X, Y, Z, f, g, h)$. Then $(0, (b')^2 - b', 0)$ is also an endomorphism. By Lemma \ref{lemma-third-map-square-zero} we see that $(b')^2 - b'$ has square zero. Set $b = b' - (2b' - 1)((b')^2 - b') = 3(b')^2 - 2(b')^3$. A computation shows that $(a, b, c)$ is an endomorphism and that $b^2 - b = (4(b')^2 - 4b' - 3)((b')^2 - b')^2 = 0$. \end{proof} \begin{lemma} \label{lemma-cone-triangle-unique-isomorphism} Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. There exists a distinguished triangle $(X, Y, Z, f, g, h)$ which is unique up to (nonunique) isomorphism of triangles. More precisely, given a second such distinguished triangle $(X, Y, Z', f, g', h')$ there exists an isomorphism $$(1, 1, c) : (X, Y, Z, f, g, h) \longrightarrow (X, Y, Z', f, g', h')$$ \end{lemma} \begin{proof} Existence by TR1. Uniqueness up to isomorphism by TR3 and Lemma \ref{lemma-third-isomorphism-triangle}. \end{proof} \begin{lemma} \label{lemma-third-object-zero} Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. The following are equivalent \begin{enumerate} \item $f$ is an isomorphism, \item $(X, Y, 0, f, 0, 0)$ is a distinguished triangle, and \item for any distinguished triangle $(X, Y, Z, f, g, h)$ we have $Z = 0$. \end{enumerate} \end{lemma} \begin{proof} By TR1 the triangle $(X, X, 0, 1, 0, 0)$ is distinguished. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. By TR3 there is a map of distinguished triangles $(1, f, 0) : (X, X, 0) \to (X, Y, Z)$. If $f$ is an isomorphism, then $(1, f, 0)$ is an isomorphism of triangles by Lemma \ref{lemma-third-isomorphism-triangle} and $Z = 0$. Conversely, if $Z = 0$, then $(1, f, 0)$ is an isomorphism of triangles as well, hence $f$ is an isomorphism. \end{proof} \begin{lemma} \label{lemma-direct-sum-triangles} Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ be triangles. The following are equivalent \begin{enumerate} \item $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')$ is a distinguished triangle, \item both $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ are distinguished triangles. \end{enumerate} \end{lemma} \begin{proof} Assume (2). By TR1 we may choose a distinguished triangle $(X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$. By TR3 we can find morphisms of distinguished triangles $(X, Y, Z, f, g, h) \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$ and $(X', Y', Z', f', g', h') \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$. Taking the direct sum of these morphisms we obtain a morphism of triangles $$\xymatrix{ (X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \ar[d]^{(1, 1, c)} \\ (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h''). }$$ In the terminology of Remark \ref{remark-special-triangles} this is a map of special triangles (because a direct sum of special triangles is special) and we conclude that $c$ is an isomorphism. Thus (1) holds. \medskip\noindent Assume (1). We will show that $(X, Y, Z, f, g, h)$ is a distinguished triangle. First observe that $(X, Y, Z, f, g, h)$ is a special triangle (terminology from Remark \ref{remark-special-triangles}) as a direct summand of the distinguished hence special triangle $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')$. Using TR1 let $(X, Y, Q, f, g'', h'')$ be a distinguished triangle. By TR3 there exists a morphism of distinguished triangles $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \to (X, Y, Q, f, g'', h'')$. Composing this with the inclusion map we get a morphism of triangles $$(1, 1, c) : (X, Y, Z, f, g, h) \longrightarrow (X, Y, Q, f, g'', h'')$$ By Remark \ref{remark-special-triangles} we find that $c$ is an isomorphism and we conclude that (2) holds. \end{proof} \begin{lemma} \label{lemma-split} Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. \begin{enumerate} \item If $h = 0$, then there exists a right inverse $s : Z \to Y$ to $g$. \item For any right inverse $s : Z \to Y$ of $g$ the map $f \oplus s : X \oplus Z \to Y$ is an isomorphism. \item For any objects $X', Z'$ of $\mathcal{D}$ the triangle $(X', X' \oplus Z', Z', (1, 0), (0, 1), 0)$ is distinguished. \end{enumerate} \end{lemma} \begin{proof} To see (1) use that $\Hom_\mathcal{D}(Z, Y) \to \Hom_\mathcal{D}(Z, Z) \to \Hom_\mathcal{D}(Z, X[1])$ is exact by Lemma \ref{lemma-representable-homological}. By the same token, if $s$ is as in (2), then $h = 0$ and the sequence $$0 \to \Hom_\mathcal{D}(W, X) \to \Hom_\mathcal{D}(W, Y) \to \Hom_\mathcal{D}(W, Z) \to 0$$ is split exact (split by $s : Z \to Y$). Hence by Yoneda's lemma we see that $X \oplus Z \to Y$ is an isomorphism. The last assertion follows from TR1 and Lemma \ref{lemma-direct-sum-triangles}. \end{proof} \begin{lemma} \label{lemma-when-split} Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. The following are equivalent \begin{enumerate} \item $f$ has a kernel, \item $f$ has a cokernel, \item $f$ is isomorphic to a map $K \oplus Z \to Z \oplus Q$ induced by $\text{id}_Z$. \end{enumerate} \end{lemma} \begin{proof} Any morphism isomorphic to a map of the form $X' \oplus Z \to Z \oplus Y'$ has both a kernel and a cokernel. Hence (3) $\Rightarrow$ (1), (2). Next we prove (1) $\Rightarrow$ (3). Suppose first that $f : X \to Y$ is a monomorphism, i.e., its kernel is zero. By TR1 there exists a distinguished triangle $(X, Y, Z, f, g, h)$. By Lemma \ref{lemma-composition-zero} the composition $f \circ h[-1] = 0$. As $f$ is a monomorphism we see that $h[-1] = 0$ and hence $h = 0$. Then Lemma \ref{lemma-split} implies that $Y = X \oplus Z$, i.e., we see that (3) holds. Next, assume $f$ has a kernel $K$. As $K \to X$ is a monomorphism we conclude $X = K \oplus X'$ and $f|_{X'} : X' \to Y$ is a monomorphism. Hence $Y = X' \oplus Y'$ and we win. The implication (2) $\Rightarrow$ (3) is dual to this. \end{proof} \begin{lemma} \label{lemma-products-sums-shifts-triangles} Let $\mathcal{D}$ be a pre-triangulated category. Let $I$ be a set. \begin{enumerate} \item Let $X_i$, $i \in I$ be a family of objects of $\mathcal{D}$. \begin{enumerate} \item If $\prod X_i$ exists, then $(\prod X_i)[1] = \prod X_i[1]$. \item If $\bigoplus X_i$ exists, then $(\bigoplus X_i)[1] = \bigoplus X_i[1]$. \end{enumerate} \item Let $X_i \to Y_i \to Z_i \to X_i[1]$ be a family of distinguished triangles of $\mathcal{D}$. \begin{enumerate} \item If $\prod X_i$, $\prod Y_i$, $\prod Z_i$ exist, then $\prod X_i \to \prod Y_i \to \prod Z_i \to \prod X_i[1]$ is a distinguished triangle. \item If $\bigoplus X_i$, $\bigoplus Y_i$, $\bigoplus Z_i$ exist, then $\bigoplus X_i \to \bigoplus Y_i \to \bigoplus Z_i \to \bigoplus X_i[1]$ is a distinguished triangle. \end{enumerate} \end{enumerate} \end{lemma} \begin{proof} Part (1) is true because $[1]$ is an autoequivalence of $\mathcal{D}$ and because direct sums and products are defined in terms of the category structure. Let us prove (2)(a). Choose a distinguished triangle $\prod X_i \to \prod Y_i \to Z \to \prod X_i[1]$. For each $j$ we can use TR3 to choose a morphism $p_j : Z \to Z_j$ fitting into a morphism of distinguished triangles with the projection maps $\prod X_i \to X_j$ and $\prod Y_i \to Y_j$. Using the definition of products we obtain a map $\prod p_i : Z \to \prod Z_i$ fitting into a morphism of triangles from the distinguished triangle to the triangle made out of the products. Observe that the product'' triangle $\prod X_i \to \prod Y_i \to \prod Z_i \to \prod X_i[1]$ is special in the terminology of Remark \ref{remark-special-triangles} because products of exact sequences of abelian groups are exact. Hence Remark \ref{remark-special-triangles} shows that the morphism of triangles is an isomorphism and we conclude by TR1. The proof of (2)(b) is dual. \end{proof} \begin{lemma} \label{lemma-projectors-have-images-triangulated} Let $\mathcal{D}$ be a pre-triangulated category. If $\mathcal{D}$ has countable products, then $\mathcal{D}$ is Karoubian. If $\mathcal{D}$ has countable coproducts, then $\mathcal{D}$ is Karoubian. \end{lemma} \begin{proof} Assume $\mathcal{D}$ has countable products. By Homology, Lemma \ref{homology-lemma-projectors-have-images} it suffices to check that morphisms which have a right inverse have kernels. Any morphism which has a right inverse is an epimorphism, hence has a kernel by Lemma \ref{lemma-when-split}. The second statement is dual to the first. \end{proof} \noindent The following lemma makes it slightly easier to prove that a pre-triangulated category is triangulated. \begin{lemma} \label{lemma-easier-axiom-four} Let $\mathcal{D}$ be a pre-triangulated category. In order to prove TR4 it suffices to show that given any pair of composable morphisms $f : X \to Y$ and $g : Y \to Z$ there exist \begin{enumerate} \item isomorphisms $i : X' \to X$, $j : Y' \to Y$ and $k : Z' \to Z$, and then setting $f' = j^{-1}fi : X' \to Y'$ and $g' = k^{-1}gj : Y' \to Z'$ there exist \item distinguished triangles $(X', Y', Q_1, f', p_1, d_1)$, $(X', Z', Q_2, g' \circ f', p_2, d_2)$ and $(Y', Z', Q_3, g', p_3, d_3)$, such that the assertion of TR4 holds. \end{enumerate} \end{lemma} \begin{proof} The replacement of $X, Y, Z$ by $X', Y', Z'$ is harmless by our definition of distinguished triangles and their isomorphisms. The lemma follows from the fact that the distinguished triangles $(X', Y', Q_1, f', p_1, d_1)$, $(X', Z', Q_2, g' \circ f', p_2, d_2)$ and $(Y', Z', Q_3, g', p_3, d_3)$ are unique up to isomorphism by Lemma \ref{lemma-cone-triangle-unique-isomorphism}. \end{proof} \begin{lemma} \label{lemma-triangulated-subcategory} Let $\mathcal{D}$ be a pre-triangulated category. Assume that $\mathcal{D}'$ is an additive full subcategory of $\mathcal{D}$. The following are equivalent \begin{enumerate} \item there exists a set of triangles $\mathcal{T}'$ such that $(\mathcal{D}', \mathcal{T}')$ is a pre-triangulated subcategory of $\mathcal{D}$, \item $\mathcal{D}'$ is preserved under $[1], [-1]$ and given any morphism $f : X \to Y$ in $\mathcal{D}'$ there exists a distinguished triangle $(X, Y, Z, f, g, h)$ in $\mathcal{D}$ such that $Z$ is isomorphic to an object of $\mathcal{D}'$. \end{enumerate} In this case $\mathcal{T}'$ as in (1) is the set of distinguished triangles $(X, Y, Z, f, g, h)$ of $\mathcal{D}$ such that $X, Y, Z \in \Ob(\mathcal{D}')$. Finally, if $\mathcal{D}$ is a triangulated category, then (1) and (2) are also equivalent to \begin{enumerate} \item[(3)] $\mathcal{D}'$ is a triangulated subcategory. \end{enumerate} \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-exact-functor-additive} An exact functor of pre-triangulated categories is additive. \end{lemma} \begin{proof} Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of pre-triangulated categories. Since $(0, 0, 0, 1_0, 1_0, 0)$ is a distinguished triangle of $\mathcal{D}$ the triangle $$(F(0), F(0), F(0), 1_{F(0)}, 1_{F(0)}, F(0))$$ is distinguished in $\mathcal{D}'$. This implies that $1_{F(0)} \circ 1_{F(0)}$ is zero, see Lemma \ref{lemma-composition-zero}. Hence $F(0)$ is the zero object of $\mathcal{D}'$. This also implies that $F$ applied to any zero morphism is zero (since a morphism in an additive category is zero if and only if it factors through the zero object). Next, using that $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is a distinguished triangle, we see that $(F(X), F(X \oplus Y), F(Y), F(1, 0), F(0, 1), 0)$ is one too. This implies that the map $F(1, 0) \oplus F(0, 1) : F(X) \oplus F(Y) \to F(X \oplus Y)$ is an isomorphism, see Lemma \ref{lemma-split}. We omit the rest of the argument. \end{proof} \begin{lemma} \label{lemma-exact-equivalence} Let $F : \mathcal{D} \to \mathcal{D}'$ be a fully faithful exact functor of pre-triangulated categories. Then a triangle $(X, Y, Z, f, g, h)$ of $\mathcal{D}$ is distinguished if and only if $(F(X), F(Y), F(Z), F(f), F(g), F(h))$ is distinguished in $\mathcal{D}'$. \end{lemma} \begin{proof} The only if'' part is clear. Assume $(F(X), F(Y), F(Z))$ is distinguished in $\mathcal{D}'$. Pick a distinguished triangle $(X, Y, Z', f, g', h')$ in $\mathcal{D}$. By Lemma \ref{lemma-cone-triangle-unique-isomorphism} there exists an isomorphism of triangles $$(1, 1, c') : (F(X), F(Y), F(Z)) \longrightarrow (F(X), F(Y), F(Z')).$$ Since $F$ is fully faithful, there exists a morphism $c : Z \to Z'$ such that $F(c) = c'$. Then $(1, 1, c)$ is an isomorphism between $(X, Y, Z)$ and $(X, Y, Z')$. Hence $(X, Y, Z)$ is distinguished by TR1. \end{proof} \begin{lemma} \label{lemma-composition-exact} Let $\mathcal{D}, \mathcal{D}', \mathcal{D}''$ be pre-triangulated categories. Let $F : \mathcal{D} \to \mathcal{D}'$ and $F' : \mathcal{D}' \to \mathcal{D}''$ be exact functors. Then $F' \circ F$ is an exact functor. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-exact-compose-homological-functor} Let $\mathcal{D}$ be a pre-triangulated category. Let $\mathcal{A}$ be an abelian category. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor. \begin{enumerate} \item Let $\mathcal{D}'$ be a pre-triangulated category. Let $F : \mathcal{D}' \to \mathcal{D}$ be an exact functor. Then the composition $G \circ F$ is a homological functor as well. \item Let $\mathcal{A}'$ be an abelian category. Let $G : \mathcal{A} \to \mathcal{A}'$ be an exact functor. Then $G \circ H$ is a homological functor as well. \end{enumerate} \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-exact-compose-delta-functor} Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A}$ be an abelian category. Let $G : \mathcal{A} \to \mathcal{D}$ be a $\delta$-functor. \begin{enumerate} \item Let $\mathcal{D}'$ be a triangulated category. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor. Then the composition $F \circ G$ is a $\delta$-functor as well. \item Let $\mathcal{A}'$ be an abelian category. Let $H : \mathcal{A}' \to \mathcal{A}$ be an exact functor. Then $G \circ H$ is a $\delta$-functor as well. \end{enumerate} \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-compose-delta-functor-homological} Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A}$ be an abelian category. Let $G : \mathcal{A} \to \mathcal{D}$ be a $\delta$-functor. Let $H : \mathcal{D} \to \mathcal{B}$ be a homological functor. Assume that $H^{-1}(G(A)) = 0$ for all $A$ in $\mathcal{A}$. Then the collection $$\{H^n \circ G, H^n(\delta_{A \to B \to C})\}_{n \geq 0}$$ is a $\delta$-functor from $\mathcal{A} \to \mathcal{B}$, see Homology, Definition \ref{homology-definition-cohomological-delta-functor}. \end{lemma} \begin{proof} The notation signifies the following. If $0 \to A \xrightarrow{a} B \xrightarrow{b} C \to 0$ is a short exact sequence in $\mathcal{A}$, then $$\delta = \delta_{A \to B \to C} : G(C) \to G(A)[1]$$ is a morphism in $\mathcal{D}$ such that $(G(A), G(B), G(C), a, b, \delta)$ is a distinguished triangle, see Definition \ref{definition-delta-functor}. Then $H^n(\delta) : H^n(G(C)) \to H^n(G(A)[1]) = H^{n + 1}(G(A))$ is clearly functorial in the short exact sequence. Finally, the long exact cohomology sequence (\ref{equation-long-exact-cohomology-sequence}) combined with the vanishing of $H^{-1}(G(C))$ gives a long exact sequence $$0 \to H^0(G(A)) \to H^0(G(B)) \to H^0(G(C)) \xrightarrow{H^0(\delta)} H^1(G(A)) \to \ldots$$ in $\mathcal{B}$ as desired. \end{proof} \noindent The proof of the following result uses TR4. \begin{proposition} \label{proposition-9} Let $\mathcal{D}$ be a triangulated category. Any commutative diagram $$\xymatrix{ X \ar[r] \ar[d] & Y \ar[d] \\ X' \ar[r] & Y' }$$ can be extended to a diagram $$\xymatrix{ X \ar[r] \ar[d] & Y \ar[r] \ar[d] & Z \ar[r] \ar[d] & X[1] \ar[d] \\ X' \ar[r] \ar[d] & Y' \ar[r] \ar[d] & Z' \ar[r] \ar[d] & X'[1] \ar[d] \\ X'' \ar[r] \ar[d] & Y'' \ar[r] \ar[d] & Z'' \ar[r] \ar[d] & X''[1] \ar[d] \\ X[1] \ar[r] & Y[1] \ar[r] & Z[1] \ar[r] & X[2] }$$ where all the squares are commutative, except for the lower right square which is anticommutative. Moreover, each of the rows and columns are distinguished triangles. Finally, the morphisms on the bottom row (resp.\ right column) are obtained from the morphisms of the top row (resp.\ left column) by applying $[1]$. \end{proposition} \begin{proof} During this proof we avoid writing the arrows in order to make the proof legible. Choose distinguished triangles $(X, Y, Z)$, $(X', Y', Z')$, $(X, X', X'')$, $(Y, Y', Y'')$, and $(X, Y', A)$. Note that the morphism $X \to Y'$ is both equal to the composition $X \to Y \to Y'$ and equal to the composition $X \to X' \to Y'$. Hence, we can find morphisms \begin{enumerate} \item $a : Z \to A$ and $b : A \to Y''$, and \item $a' : X'' \to A$ and $b' : A \to Z'$ \end{enumerate} as in TR4. Denote $c : Y'' \to Z[1]$ the composition $Y'' \to Y[1] \to Z[1]$ and denote $c' : Z' \to X''[1]$ the composition $Z' \to X'[1] \to X''[1]$. The conclusion of our application TR4 are that \begin{enumerate} \item $(Z, A, Y'', a, b, c)$, $(X'', A, Z', a', b', c')$ are distinguished triangles, \item $(X, Y, Z) \to (X, Y', A)$, $(X, Y', A) \to (Y, Y', Y'')$, $(X, X', X'') \to (X, Y', A)$, $(X, Y', A) \to (X', Y', Z')$ are morphisms of triangles. \end{enumerate} First using that $(X, X', X'') \to (X, Y', A)$ and $(X, Y', A) \to (Y, Y', Y'')$. are morphisms of triangles we see the first of the diagrams $$\vcenter{ \xymatrix{ X' \ar[r] \ar[d] & Y' \ar[d] \\ X'' \ar[r]^{b \circ a'} \ar[d] & Y'' \ar[d] \\ X[1] \ar[r] & Y[1] } } \quad\text{and}\quad \vcenter{ \xymatrix{ Y \ar[r] \ar[d] & Z \ar[d]^{b' \circ a} \ar[r] & X[1] \ar[d] \\ Y' \ar[r] & Z' \ar[r] & X'[1] } }$$ is commutative. The second is commutative too using that $(X, Y, Z) \to (X, Y', A)$ and $(X, Y', A) \to (X', Y', Z')$ are morphisms of triangles. At this point we choose a distinguished triangle $(X'', Y'' , Z'')$ starting with the map $b \circ a' : X'' \to Y''$. \medskip\noindent Next we apply TR4 one more time to the morphisms $X'' \to A \to Y''$ and the triangles $(X'', A, Z', a', b', c')$, $(X'', Y'', Z'')$, and $(A, Y'', Z[1], b, c , -a[1])$ to get morphisms $a'' : Z' \to Z''$ and $b'' : Z'' \to Z[1]$. Then $(Z', Z'', Z[1], a'', b'', - b'[1] \circ a[1])$ is a distinguished triangle, hence also $(Z, Z', Z'', -b' \circ a, a'', -b'')$ and hence also $(Z, Z', Z'', b' \circ a, a'', b'')$. Moreover, $(X'', A, Z') \to (X'', Y'', Z'')$ and $(X'', Y'', Z'') \to (A, Y'', Z[1], b, c , -a[1])$ are morphisms of triangles. At this point we have defined all the distinguished triangles and all the morphisms, and all that's left is to verify some commutativity relations. \medskip\noindent To see that the middle square in the diagram commutes, note that the arrow $Y' \to Z'$ factors as $Y' \to A \to Z'$ because $(X, Y', A) \to (X', Y', Z')$ is a morphism of triangles. Similarly, the morphism $Y' \to Y''$ factors as $Y' \to A \to Y''$ because $(X, Y', A) \to (Y, Y', Y'')$ is a morphism of triangles. Hence the middle square commutes because the square with sides $(A, Z', Z'', Y'')$ commutes as $(X'', A, Z') \to (X'', Y'', Z'')$ is a morphism of triangles (by TR4). The square with sides $(Y'', Z'', Y[1], Z[1])$ commutes because $(X'', Y'', Z'') \to (A, Y'', Z[1], b, c , -a[1])$ is a morphism of triangles and $c : Y'' \to Z[1]$ is the composition $Y'' \to Y[1] \to Z[1]$. The square with sides $(Z', X'[1], X''[1], Z'')$ is commutative because $(X'', A, Z') \to (X'', Y'', Z'')$ is a morphism of triangles and $c' : Z' \to X''[1]$ is the composition $Z' \to X'[1] \to X''[1]$. Finally, we have to show that the square with sides $(Z'', X''[1], Z[1], X[2])$ anticommutes. This holds because $(X'', Y'', Z'') \to (A, Y'', Z[1], b, c , -a[1])$ is a morphism of triangles and we're done. \end{proof} \section{Localization of triangulated categories} \label{section-localization} \noindent In order to construct the derived category starting from the homotopy category of complexes, we will use a localization process. \begin{definition} \label{definition-localization} Let $\mathcal{D}$ be a pre-triangulated category. We say a multiplicative system $S$ is {\it compatible with the triangulated structure} if the following two conditions hold: \begin{enumerate} \item[MS5] For $s \in S$ we have $s[n] \in S$ for all $n \in \mathbf{Z}$. \item[MS6] Given a solid commutative square $$\xymatrix{ X \ar[r] \ar[d]^s & Y \ar[r] \ar[d]^{s'} & Z \ar[r] \ar@{-->}[d] & X[1] \ar[d]^{s[1]} \\ X' \ar[r] & Y' \ar[r] & Z' \ar[r] & X'[1] }$$ whose rows are distinguished triangles with $s, s' \in S$ there exists a morphism $s'' : Z \to Z'$ in $S$ such that $(s, s', s'')$ is a morphism of triangles. \end{enumerate} \end{definition} \noindent It turns out that these axioms are not independent of the axioms defining multiplicative systems. \begin{lemma} \label{lemma-localization-conditions} Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a set of morphisms of $\mathcal{D}$ and assume that axioms MS1, MS5, MS6 hold (see Categories, Definition \ref{categories-definition-multiplicative-system} and Definition \ref{definition-localization}). Then MS2 holds. \end{lemma} \begin{proof} Suppose that $f : X \to Y$ is a morphism of $\mathcal{D}$ and $t : X \to X'$ an element of $S$. Choose a distinguished triangle $(X, Y, Z, f, g, h)$. Next, choose a distinguished triangle $(X', Y', Z, f', g', t[1] \circ h)$ (here we use TR1 and TR2). By MS5, MS6 (and TR2 to rotate) we can find the dotted arrow in the commutative diagram $$\xymatrix{ X \ar[r] \ar[d]^t & Y \ar[r] \ar@{..>}[d]^{s'} & Z \ar[r] \ar[d]^1 & X[1] \ar[d]^{t[1]} \\ X' \ar[r] & Y' \ar[r] & Z \ar[r] & X'[1] }$$ with moreover $s' \in S$. This proves LMS2. The proof of RMS2 is dual. \end{proof} \begin{lemma} \label{lemma-triangle-functor-localize} Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of pre-triangulated categories. Let $$S = \{f \in \text{Arrows}(\mathcal{D}) \mid F(f)\text{ is an isomorphism}\}$$ Then $S$ is a saturated (see Categories, Definition \ref{categories-definition-saturated-multiplicative-system}) multiplicative system compatible with the triangulated structure on $\mathcal{D}$. \end{lemma} \begin{proof} We have to prove axioms MS1 -- MS6, see Categories, Definitions \ref{categories-definition-multiplicative-system} and \ref{categories-definition-saturated-multiplicative-system} and Definition \ref{definition-localization}. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and Lemma \ref{lemma-third-isomorphism-triangle}. By Lemma \ref{lemma-localization-conditions} we conclude that MS2 holds. To finish the proof we have to show that MS3 holds. To do this let $f, g : X \to Y$ be morphisms of $\mathcal{D}$, and let $t : Z \to X$ be an element of $S$ such that $f \circ t = g \circ t$. As $\mathcal{D}$ is additive this simply means that $a \circ t = 0$ with $a = f - g$. Choose a distinguished triangle $(Z, X, Q, t, d, h)$ using TR1. Since $a \circ t = 0$ we see by Lemma \ref{lemma-representable-homological} there exists a morphism $i : Q \to Y$ such that $i \circ d = a$. Finally, using TR1 again we can choose a triangle $(Q, Y, W, i, j, k)$. Here is a picture $$\xymatrix{ Z \ar[r]_t & X \ar[r]_d \ar[d]^1 & Q \ar[r] \ar[d]^i & Z[1] \\ & X \ar[r]_a & Y \ar[d]^j \\ & & W }$$ OK, and now we apply the functor $F$ to this diagram. Since $t \in S$ we see that $F(Q) = 0$, see Lemma \ref{lemma-third-object-zero}. Hence $F(j)$ is an isomorphism by the same lemma, i.e., $j \in S$. Finally, $j \circ a = j \circ i \circ d = 0$ as $j \circ i = 0$. Thus $j \circ f = j \circ g$ and we see that LMS3 holds. The proof of RMS3 is dual. \end{proof} \begin{lemma} \label{lemma-homological-functor-localize} Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor between a pre-triangulated category and an abelian category. Let $$S = \{f \in \text{Arrows}(\mathcal{D}) \mid H^i(f)\text{ is an isomorphism for all }i \in \mathbf{Z}\}$$ Then $S$ is a saturated (see Categories, Definition \ref{categories-definition-saturated-multiplicative-system}) multiplicative system compatible with the triangulated structure on $\mathcal{D}$. \end{lemma} \begin{proof} We have to prove axioms MS1 -- MS6, see Categories, Definitions \ref{categories-definition-multiplicative-system} and \ref{categories-definition-saturated-multiplicative-system} and Definition \ref{definition-localization}. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and the long exact cohomology sequence (\ref{equation-long-exact-cohomology-sequence}). By Lemma \ref{lemma-localization-conditions} we conclude that MS2 holds. To finish the proof we have to show that MS3 holds. To do this let $f, g : X \to Y$ be morphisms of $\mathcal{D}$, and let $t : Z \to X$ be an element of $S$ such that $f \circ t = g \circ t$. As $\mathcal{D}$ is additive this simply means that $a \circ t = 0$ with $a = f - g$. Choose a distinguished triangle $(Z, X, Q, t, g, h)$ using TR1 and TR2. Since $a \circ t = 0$ we see by Lemma \ref{lemma-representable-homological} there exists a morphism $i : Q \to Y$ such that $i \circ g = a$. Finally, using TR1 again we can choose a triangle $(Q, Y, W, i, j, k)$. Here is a picture $$\xymatrix{ Z \ar[r]_t & X \ar[r]_g \ar[d]^1 & Q \ar[r] \ar[d]^i & Z[1] \\ & X \ar[r]_a & Y \ar[d]^j \\ & & W }$$ OK, and now we apply the functors $H^i$ to this diagram. Since $t \in S$ we see that $H^i(Q) = 0$ by the long exact cohomology sequence (\ref{equation-long-exact-cohomology-sequence}). Hence $H^i(j)$ is an isomorphism for all $i$ by the same argument, i.e., $j \in S$. Finally, $j \circ a = j \circ i \circ g = 0$ as $j \circ i = 0$. Thus $j \circ f = j \circ g$ and we see that LMS3 holds. The proof of RMS3 is dual. \end{proof} \begin{proposition} \label{proposition-construct-localization} Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a multiplicative system compatible with the triangulated structure. Then there exists a unique structure of a pre-triangulated category on $S^{-1}\mathcal{D}$ such that the localization functor $Q : \mathcal{D} \to S^{-1}\mathcal{D}$ is exact. Moreover, if $\mathcal{D}$ is a triangulated category, so is $S^{-1}\mathcal{D}$. \end{proposition} \begin{proof} We have seen that $S^{-1}\mathcal{D}$ is an additive category and that the localization functor $Q$ is additive in Homology, Lemma \ref{homology-lemma-localization-additive}, It is clear that we may define $Q(X)[n] = Q(X[n])$ since $\mathcal{S}$ is preserved under the shift functors $[n]$ by MS5. Finally, we say a triangle of $S^{-1}\mathcal{D}$ is distinguished if it is isomorphic to the image of a distinguished triangle under the localization functor $Q$. \medskip\noindent Proof of TR1. The only thing to prove here is that if $a : Q(X) \to Q(Y)$ is a morphism of $S^{-1}\mathcal{D}$, then $a$ fits into a distinguish triangle. Write $a = Q(s)^{-1} \circ Q(f)$ for some $s : Y \to Y'$ in $S$ and $f : X \to Y'$. Choose a distinguished triangle $(X, Y', Z, f, g, h)$ in $\mathcal{D}$. Then we see that $(Q(X), Q(Y), Q(Z), a, Q(g) \circ Q(s), Q(h))$ is a distinguished triangle of $S^{-1}\mathcal{D}$. \medskip\noindent Proof of TR2. This is immediate from the definitions. \medskip\noindent Proof of TR3. Note that the existence of the dotted arrow which is required to exist may be proven after replacing the two triangles by isomorphic triangles. Hence we may assume given distinguished triangles $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ of $\mathcal{D}$ and a commutative diagram $$\xymatrix{ Q(X) \ar[r]_{Q(f)} \ar[d]_a & Q(Y) \ar[d]^b \\ Q(X') \ar[r]^{Q(f')} & Q(Y') }$$ in $S^{-1}\mathcal{D}$. Now we apply Categories, Lemma \ref{categories-lemma-left-localization-diagram} to find a morphism $f'' : X'' \to Y''$ in $\mathcal{D}$ and a commutative diagram $$\xymatrix{ X \ar[d]_f \ar[r]_k & X'' \ar[d]^{f''} & X' \ar[d]^{f'} \ar[l]^s \\ Y \ar[r]^l & Y'' & Y' \ar[l]_t }$$ in $\mathcal{D}$ with $s, t \in S$ and $a = s^{-1}k$, $b = t^{-1}l$. At this point we can use TR3 for $\mathcal{D}$ and MS6 to find a commutative diagram $$\xymatrix{ X \ar[r] \ar[d]^k & Y \ar[r] \ar[d]^l & Z \ar[r] \ar[d]^m & X[1] \ar[d]^{g[1]} \\ X'' \ar[r] & Y'' \ar[r] & Z'' \ar[r] & X''[1] \\ X' \ar[r] \ar[u]_s & Y' \ar[r] \ar[u]_t & Z' \ar[r] \ar[u]_r & X'[1] \ar[u]_{s[1]} }$$ with $r \in S$. It follows that setting $c = Q(r)^{-1}Q(m)$ we obtain the desired morphism of triangles $$\xymatrix{ (Q(X), Q(Y), Q(Z), Q(f), Q(g), Q(h)) \ar[d]^{(a, b, c)} \\ (Q(X'), Q(Y'), Q(Z'), Q(f'), Q(g'), Q(h')) }$$ \medskip\noindent This proves the first statement of the lemma. If $\mathcal{D}$ is also a triangulated category, then we still have to prove TR4 in order to show that $S^{-1}\mathcal{D}$ is triangulated as well. To do this we reduce by Lemma \ref{lemma-easier-axiom-four} to the following statement: Given composable morphisms $a : Q(X) \to Q(Y)$ and $b : Q(Y) \to Q(Z)$ we have to produce an octahedron after possibly replacing $Q(X), Q(Y), Q(Z)$ by isomorphic objects. To do this we may first replace $Y$ by an object such that $a = Q(f)$ for some morphism $f : X \to Y$ in $\mathcal{D}$. (More precisely, write $a = s^{-1}f$ with $s : Y \to Y'$ in $S$ and $f : X \to Y'$. Then replace $Y$ by $Y'$.) After this we similarly replace $Z$ by an object such that $b = Q(g)$ for some morphism $g : Y \to Z$. Now we can find distinguished triangles $(X, Y, Q_1, f, p_1, d_1)$, $(X, Z, Q_2, g \circ f, p_2, d_2)$, and $(Y, Z, Q_3, g, p_3, d_3)$ in $\mathcal{D}$ (by TR1), and morphisms $a : Q_1 \to Q_2$ and $b : Q_2 \to Q_3$ as in TR4. Then it is immediately verified that applying the functor $Q$ to all these data gives a corresponding structure in $S^{-1}\mathcal{D}$ \end{proof} \noindent The universal property of the localization of a triangulated category is as follows (we formulate this for pre-triangulated categories, hence it holds a fortiori for triangulated categories). \begin{lemma} \label{lemma-universal-property-localization} Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a multiplicative system compatible with the triangulated category. Let $Q : \mathcal{D} \to S^{-1}\mathcal{D}$ be the localization functor, see Proposition \ref{proposition-construct-localization}. \begin{enumerate} \item If $H : \mathcal{D} \to \mathcal{A}$ is a homological functor into an abelian category $\mathcal{A}$ such that $H(s)$ is an isomorphism for all $s \in S$, then the unique factorization $H' : S^{-1}\mathcal{D} \to \mathcal{A}$ such that $H = H' \circ Q$ (see Categories, Lemma \ref{categories-lemma-properties-left-localization}) is a homological functor too. \item If $F : \mathcal{D} \to \mathcal{D}'$ is an exact functor into a pre-triangulated category $\mathcal{D}'$ such that $F(s)$ is an isomorphism for all $s \in S$, then the unique factorization $F' : S^{-1}\mathcal{D} \to \mathcal{D}'$ such that $F = F' \circ Q$ (see Categories, Lemma \ref{categories-lemma-properties-left-localization}) is an exact functor too. \end{enumerate} \end{lemma} \begin{proof} This lemma proves itself. Details omitted. \end{proof} \noindent The following lemma describes the kernel (see Definition \ref{definition-kernel-category}) of the localization functor. \begin{lemma} \label{lemma-kernel-localization} Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a multiplicative system compatible with the triangulated structure. Let $Z$ be an object of $\mathcal{D}$. The following are equivalent \begin{enumerate} \item $Q(Z) = 0$ in $S^{-1}\mathcal{D}$, \item there exists $Z' \in \Ob(\mathcal{D})$ such that $0 : Z \to Z'$ is an element of $S$, \item there exists $Z' \in \Ob(\mathcal{D})$ such that $0 : Z' \to Z$ is an element of $S$, and \item there exists an object $Z'$ and a distinguished triangle $(X, Y, Z \oplus Z', f, g, h)$ such that $f \in S$. \end{enumerate} If $S$ is saturated, then these are also equivalent to \begin{enumerate} \item[(4)] the morphism $0 \to Z$ is an element of $S$, \item[(5)] the morphism $Z \to 0$ is an element of $S$, \item[(6)] there exists a distinguished triangle $(X, Y, Z, f, g, h)$ such that $f \in S$. \end{enumerate} \end{lemma} \begin{proof} The equivalence of (1), (2), and (3) is Homology, Lemma \ref{homology-lemma-kernel-localization}. If (2) holds, then $(Z'[-1], Z'[-1] \oplus Z, Z, (1, 0), (0, 1), 0)$ is a distinguished triangle (see Lemma \ref{lemma-split}) with $0 \in S$''. By rotating we conclude that (4) holds. If $(X, Y, Z \oplus Z', f, g, h)$ is a distinguished triangle with $f \in S$ then $Q(f)$ is an isomorphism hence $Q(Z \oplus Z') = 0$ hence $Q(Z) = 0$. Thus (1) -- (4) are all equivalent. \medskip\noindent Next, assume that $S$ is saturated. Note that each of (4), (5), (6) implies one of the equivalent conditions (1) -- (4). Suppose that $Q(Z) = 0$. Then $0 \to Z$ is a morphism of $\mathcal{D}$ which becomes an isomorphism in $S^{-1}\mathcal{D}$. According to Categories, Lemma \ref{categories-lemma-what-gets-inverted} the fact that $S$ is saturated implies that $0 \to Z$ is in $S$. Hence (1) $\Rightarrow$ (4). Dually (1) $\Rightarrow$ (5). Finally, if $0 \to Z$ is in $S$, then the triangle $(0, Z, Z, 0, \text{id}_Z, 0)$ is distinguished by TR1 and TR2 and is a triangle as in (4). \end{proof} \begin{lemma} \label{lemma-limit-triangles} Let $\mathcal{D}$ be a triangulated category. Let $S$ be a saturated multiplicative system in $\mathcal{D}$ that is compatible with the triangulated structure. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle in $\mathcal{D}$. Consider the category of morphisms of triangles $$\mathcal{I} = \{(s, s', s'') : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h') \mid s, s', s'' \in S\}$$ Then $\mathcal{I}$ is a filtered category and the functors $\mathcal{I} \to X/S$, $\mathcal{I} \to Y/S$, and $\mathcal{I} \to Z/S$ are cofinal. \end{lemma} \begin{proof} We strongly suggest the reader skip the proof of this lemma and instead work it out on a napkin. \medskip\noindent The first remark is that using rotation of distinguished triangles (TR2) gives an equivalence of categories between $\mathcal{I}$ and the corresponding category for the distinguished triangle $(Y, Z, X[1], g, h, -f[1])$. Using this we see for example that if we prove the functor $\mathcal{I} \to X/S$ is cofinal, then the same thing is true for the functors $\mathcal{I} \to Y/S$ and $\mathcal{I} \to Z/S$. \medskip\noindent Note that if $s : X \to X'$ is a morphism of $S$, then using MS2 we can find $s' : Y \to Y'$ and $f' : X' \to Y'$ such that $f' \circ s = s' \circ f$, whereupon we can use MS6 to complete this into an object of $\mathcal{I}$. Hence the functor $\mathcal{I} \to X/S$ is surjective on objects. Using rotation as above this implies the same thing is true for the functors $\mathcal{I} \to Y/S$ and $\mathcal{I} \to Z/S$. \medskip\noindent Suppose given objects $s_1 : X \to X_1$ and $s_2 : X \to X_2$ in $X/S$ and a morphism $a : X_1 \to X_2$ in $X/S$. Since $S$ is saturated, we see that $a \in S$, see Categories, Lemma \ref{categories-lemma-what-gets-inverted}. By the argument of the previous paragraph we can complete $s_1 : X \to X_1$ to an object $(s_1, s'_1, s''_1) : (X, Y, Z, f, g, h) \to (X_1, Y_1, Z_1, f_1, g_1, h_1)$ in $\mathcal{I}$. Then we can repeat and find $(a, b, c) : (X_1, Y_1, Z_1, f_1, g_1, h_1) \to (X_2, Y_2, Z_2, f_2, g_2, h_2)$ with $a, b, c \in S$ completing the given $a : X_1 \to X_2$. But then $(a, b, c)$ is a morphism in $\mathcal{I}$. In this way we conclude that the functor $\mathcal{I} \to X/S$ is also surjective on arrows. Using rotation as above, this implies the same thing is true for the functors $\mathcal{I} \to Y/S$ and $\mathcal{I} \to Z/S$. \medskip\noindent The category $\mathcal{I}$ is nonempty as the identity provides an object. This proves the condition (1) of the definition of a filtered category, see Categories, Definition \ref{categories-definition-directed}. \medskip\noindent We check condition (2) of Categories, Definition \ref{categories-definition-directed} for the category $\mathcal{I}$. Suppose given objects $(s_1, s'_1, s''_1) : (X, Y, Z, f, g, h) \to (X_1, Y_1, Z_1, f_1, g_1, h_1)$ and $(s_2, s'_2, s''_2) : (X, Y, Z, f, g, h) \to (X_2, Y_2, Z_2, f_2, g_2, h_2)$ in $\mathcal{I}$. We want to find an object of $\mathcal{I}$ which is the target of an arrow from both $(X_1, Y_1, Z_1, f_1, g_1, h_1)$ and $(X_2, Y_2, Z_2, f_2, g_2, h_2)$. By Categories, Remark \ref{categories-remark-left-localization-morphisms-colimit} the categories $X/S$, $Y/S$, $Z/S$ are filtered. Thus we can find $X \to X_3$ in $X/S$ and morphisms $s : X_2 \to X_3$ and $a : X_1 \to X_3$. By the above we can find a morphism $(s, s', s'') : (X_2, Y_2, Z_2, f_2, g_2, h_2) \to (X_3, Y_3, Z_3, f_3, g_3, h_3)$ with $s', s'' \in S$. After replacing $(X_2, Y_2, Z_2)$ by $(X_3, Y_3, Z_3)$ we may assume that there exists a morphism $a : X_1 \to X_2$ in $X/S$. Repeating the argument for $Y$ and $Z$ (by rotating as above) we may assume there is a morphism $a : X_1 \to X_2$ in $X/S$, $b : Y_1 \to Y_2$ in $Y/S$, and $c : Z_1 \to Z_2$ in $Z/S$. However, these morphisms do not necessarily give rise to a morphism of distinguished triangles. On the other hand, the necessary diagrams do commute in $S^{-1}\mathcal{D}$. Hence we see (for example) that there exists a morphism $s'_2 : Y_2 \to Y_3$ in $S$ such that $s'_2 \circ f_2 \circ a = s'_2 \circ b \circ f_1$. Another replacement of $(X_2, Y_2, Z_2)$ as above then gets us to the situation where $f_2 \circ a = b \circ f_1$. Rotating and applying the same argument two more times we see that we may assume $(a, b, c)$ is a morphism of triangles. This proves condition (2). \medskip\noindent Next we check condition (3) of Categories, Definition \ref{categories-definition-directed}. Suppose $(s_1, s_1', s_1'') : (X, Y, Z) \to (X_1, Y_1, Z_1)$ and $(s_2, s_2', s_2'') : (X, Y, Z) \to (X_2, Y_2, Z_2)$ are objects of $\mathcal{I}$, and suppose $(a, b, c), (a', b', c')$ are two morphisms between them. Since $a \circ s_1 = a' \circ s_1$ there exists a morphism $s_3 : X_2 \to X_3$ such that $s_3 \circ a = s_3 \circ a'$. Using the surjectivity statement we can complete this to a morphism of triangles $(s_3, s_3', s_3'') : (X_2, Y_2, Z_2) \to (X_3, Y_3, Z_3)$ with $s_3, s_3', s_3'' \in S$. Thus $(s_3 \circ s_2, s_3' \circ s_2', s_3'' \circ s_2'') : (X, Y, Z) \to (X_3, Y_3, Z_3)$ is also an object of $\mathcal{I}$ and after composing the maps $(a, b, c), (a', b', c')$ with $(s_3, s_3', s_3'')$ we obtain $a = a'$. By rotating we may do the same to get $b = b'$ and $c = c'$. \medskip\noindent Finally, we check that $\mathcal{I} \to X/S$ is cofinal, see Categories, Definition \ref{categories-definition-cofinal}. The first condition is true as the functor is surjective. Suppose that we have an object $s : X \to X'$ in $X/S$ and two objects $(s_1, s'_1, s''_1) : (X, Y, Z, f, g, h) \to (X_1, Y_1, Z_1, f_1, g_1, h_1)$ and $(s_2, s'_2, s''_2) : (X, Y, Z, f, g, h) \to (X_2, Y_2, Z_2, f_2, g_2, h_2)$ in $\mathcal{I}$ as well as morphisms $t_1 : X' \to X_1$ and $t_2 : X' \to X_2$ in $X/S$. By property (2) of $\mathcal{I}$ proved above we can find morphisms $(s_3, s'_3, s''_3) : (X_1, Y_1, Z_1, f_1, g_1, h_1) \to (X_3, Y_3, Z_3, f_3, g_3, h_3)$ and $(s_4, s'_4, s''_4) : (X_2, Y_2, Z_2, f_2, g_2, h_2) \to (X_3, Y_3, Z_3, f_3, g_3, h_3)$ in $\mathcal{I}$. We would be done if the compositions $X' \to X_1 \to X_3$ and $X' \to X_1 \to X_3$ where equal (see displayed equation in Categories, Definition \ref{categories-definition-cofinal}). If not, then, because $X/S$ is filtered, we can choose a morphism $X_3 \to X_4$ in $S$ such that the compositions $X' \to X_1 \to X_3 \to X_4$ and $X' \to X_1 \to X_3 \to X_4$ are equal. Then we finally complete $X_3 \to X_4$ to a morphism $(X_3, Y_3, Z_3) \to (X_4, Y_4, Z_4)$ in $\mathcal{I}$ and compose with that morphism to see that the result is true. \end{proof} \section{Quotients of triangulated categories} \label{section-quotients} \noindent Given a triangulated category and a triangulated subcategory we can construct another triangulated category by taking the quotient''. The construction uses a localization. This is similar to the quotient of an abelian category by a Serre subcategory, see Homology, Section \ref{homology-section-serre-subcategories}. Before we do the actual construction we briefly discuss kernels of exact functors. \begin{definition} \label{definition-saturated} Let $\mathcal{D}$ be a pre-triangulated category. We say a full pre-triangulated subcategory $\mathcal{D}'$ of $\mathcal{D}$ is {\it saturated} if whenever $X \oplus Y$ is isomorphic to an object of $\mathcal{D}'$ then both $X$ and $Y$ are isomorphic to objects of $\mathcal{D}'$. \end{definition} \noindent A saturated triangulated subcategory is sometimes called a {\it thick triangulated subcategory}. In some references, this is only used for strictly full triangulated subcategories (and sometimes the definition is written such that it implies strictness). There is another notion, that of an {\it \'epaisse triangulated subcategory}. The definition is that given a commutative diagram $$\xymatrix{ & S \ar[rd] \\ X \ar[ru] \ar[rr] & & Y \ar[r] & T \ar[r] & X[1] }$$ where the second line is a distinguished triangle and $S$ and $T$ isomorphic to objects of $\mathcal{D}'$, then also $X$ and $Y$ are isomorphic to objects of $\mathcal{D}'$. It turns out that this is equivalent to being saturated (this is elementary and can be found in \cite{Rickard-derived}) and the notion of a saturated category is easier to work with. \begin{lemma} \label{lemma-triangle-functor-kernel} Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of pre-triangulated categories. Let $\mathcal{D}''$ be the full subcategory of $\mathcal{D}$ with objects $$\Ob(\mathcal{D}'') = \{X \in \Ob(\mathcal{D}) \mid F(X) = 0\}$$ Then $\mathcal{D}''$ is a strictly full saturated pre-triangulated subcategory of $\mathcal{D}$. If $\mathcal{D}$ is a triangulated category, then $\mathcal{D}''$ is a triangulated subcategory. \end{lemma} \begin{proof} It is clear that $\mathcal{D}''$ is preserved under $[1]$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $F(X) = F(Y) = 0$, then also $F(Z) = 0$ as $(F(X), F(Y), F(Z), F(f), F(g), F(h))$ is distinguished. Hence we may apply Lemma \ref{lemma-triangulated-subcategory} to see that $\mathcal{D}''$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The final assertion of being saturated follows from $F(X) \oplus F(Y) = 0 \Rightarrow F(X) = F(Y) = 0$. \end{proof} \begin{lemma} \label{lemma-homological-functor-kernel} Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor of a pre-triangulated category into an abelian category. Let $\mathcal{D}'$ be the full subcategory of $\mathcal{D}$ with objects $$\Ob(\mathcal{D}') = \{X \in \Ob(\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \in \mathbf{Z}\}$$ Then $\mathcal{D}'$ is a strictly full saturated pre-triangulated subcategory of $\mathcal{D}$. If $\mathcal{D}$ is a triangulated category, then $\mathcal{D}'$ is a triangulated subcategory. \end{lemma} \begin{proof} It is clear that $\mathcal{D}'$ is preserved under $[1]$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $H(X[n]) = H(Y[n]) = 0$ for all $n$, then also $H(Z[n]) = 0$ for all $n$ by the long exact sequence (\ref{equation-long-exact-cohomology-sequence}). Hence we may apply Lemma \ref{lemma-triangulated-subcategory} to see that $\mathcal{D}'$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The assertion of being saturated follows from \begin{align*} H((X \oplus Y)[n]) = 0 & \Rightarrow H(X[n] \oplus Y[n]) = 0 \\ & \Rightarrow H(X[n]) \oplus H(Y[n]) = 0 \\ & \Rightarrow H(X[n]) = H(Y[n]) = 0 \end{align*} for all $n \in \mathbf{Z}$. \end{proof} \begin{lemma} \label{lemma-homological-functor-bounded} Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor of a pre-triangulated category into an abelian category. Let $\mathcal{D}_H^{+}, \mathcal{D}_H^{-}, \mathcal{D}_H^b$ be the full subcategory of $\mathcal{D}$ with objects $$\begin{matrix} \Ob(\mathcal{D}_H^{+}) = \{X \in \Ob(\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \ll 0\} \\ \Ob(\mathcal{D}_H^{-}) = \{X \in \Ob(\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \gg 0\} \\ \Ob(\mathcal{D}_H^b) = \{X \in \Ob(\mathcal{D}) \mid H(X[n]) = 0\text{ for all }|n| \gg 0\} \end{matrix}$$ Each of these is a strictly full saturated pre-triangulated subcategory of $\mathcal{D}$. If $\mathcal{D}$ is a triangulated category, then each is a triangulated subcategory. \end{lemma} \begin{proof} Let us prove this for $\mathcal{D}_H^{+}$. It is clear that it is preserved under $[1]$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $H(X[n]) = H(Y[n]) = 0$ for all $n \ll 0$, then also $H(Z[n]) = 0$ for all $n \ll 0$ by the long exact sequence (\ref{equation-long-exact-cohomology-sequence}). Hence we may apply Lemma \ref{lemma-triangulated-subcategory} to see that $\mathcal{D}_H^{+}$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The assertion of being saturated follows from \begin{align*} H((X \oplus Y)[n]) = 0 & \Rightarrow H(X[n] \oplus Y[n]) = 0 \\ & \Rightarrow H(X[n]) \oplus H(Y[n]) = 0 \\ & \Rightarrow H(X[n]) = H(Y[n]) = 0 \end{align*} for all $n \in \mathbf{Z}$. \end{proof} \begin{definition} \label{definition-kernel-category} Let $\mathcal{D}$ be a (pre-)triangulated category. \begin{enumerate} \item Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor. The {\it kernel of $F$} is the strictly full saturated (pre-)triangulated subcategory described in Lemma \ref{lemma-triangle-functor-kernel}. \item Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor. The {\it kernel of $H$} is the strictly full saturated (pre-)triangulated subcategory described in Lemma \ref{lemma-homological-functor-kernel}. \end{enumerate} These are sometimes denoted $\Ker(F)$ or $\Ker(H)$. \end{definition} \noindent The proof of the following lemma uses TR4. \begin{lemma} \label{lemma-construct-multiplicative-system} Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{D}' \subset \mathcal{D}$ be a full triangulated subcategory. Set \begin{equation} \label{equation-multiplicative-system} S = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle }\\ (X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with } Z\text{ isomorphic to an object of }\mathcal{D}' \end{matrix} \right\} \end{equation} Then $S$ is a multiplicative system compatible with the triangulated structure on $\mathcal{D}$. In this situation the following are equivalent \begin{enumerate} \item $S$ is a saturated multiplicative system, \item $\mathcal{D}'$ is a saturated triangulated subcategory. \end{enumerate} \end{lemma} \begin{proof} To prove the first assertion we have to prove that MS1, MS2, MS3 and MS5, MS6 hold. \medskip\noindent Proof of MS1. It is clear that identities are in $S$ because $(X, X, 0, 1, 0, 0)$ is distinguished for every object $X$ of $\mathcal{D}$ and because $0$ is an object of $\mathcal{D}'$. Let $f : X \to Y$ and $g : Y \to Z$ be composable morphisms contained in $S$. Choose distinguished triangles $(X, Y, Q_1, f, p_1, d_1)$, $(X, Z, Q_2, g \circ f, p_2, d_2)$, and $(Y, Z, Q_3, g, p_3, d_3)$. By assumption we know that $Q_1$ and $Q_3$ are isomorphic to objects of $\mathcal{D}'$. By TR4 we know there exists a distinguished triangle $(Q_1, Q_2, Q_3, a, b, c)$. Since $\mathcal{D}'$ is a triangulated subcategory we conclude that $Q_2$ is isomorphic to an object of $\mathcal{D}'$. Hence $g \circ f \in S$. \medskip\noindent Proof of MS3. Let $a : X \to Y$ be a morphism and let $t : Z \to X$ be an element of $S$ such that $a \circ t = 0$. To prove LMS3 it suffices to find an $s \in S$ such that $s \circ a = 0$, compare with the proof of Lemma \ref{lemma-triangle-functor-localize}. Choose a distinguished triangle $(Z, X, Q, t, g, h)$ using TR1 and TR2. Since $a \circ t = 0$ we see by Lemma \ref{lemma-representable-homological} there exists a morphism $i : Q \to Y$ such that $i \circ g = a$. Finally, using TR1 again we can choose a triangle $(Q, Y, W, i, s, k)$. Here is a picture $$\xymatrix{ Z \ar[r]_t & X \ar[r]_g \ar[d]^1 & Q \ar[r] \ar[d]^i & Z[1] \\ & X \ar[r]_a & Y \ar[d]^s \\ & & W }$$ Since $t \in S$ we see that $Q$ is isomorphic to an object of $\mathcal{D}'$. Hence $s \in S$. Finally, $s \circ a = s \circ i \circ g = 0$ as $s \circ i = 0$ by Lemma \ref{lemma-composition-zero}. We conclude that LMS3 holds. The proof of RMS3 is dual. \medskip\noindent Proof of MS5. Follows as distinguished triangles and $\mathcal{D}'$ are stable under translations \medskip\noindent Proof of MS6. Suppose given a commutative diagram $$\xymatrix{ X \ar[r] \ar[d]^s & Y \ar[d]^{s'} \\ X' \ar[r] & Y' }$$ with $s, s' \in S$. By Proposition \ref{proposition-9} we can extend this to a nine square diagram. As $s, s'$ are elements of $S$ we see that $X'', Y''$ are isomorphic to objects of $\mathcal{D}'$. Since $\mathcal{D}'$ is a full triangulated subcategory we see that $Z''$ is also isomorphic to an object of $\mathcal{D}'$. Whence the morphism $Z \to Z'$ is an element of $S$. This proves MS6. \medskip\noindent MS2 is a formal consequence of MS1, MS5, and MS6, see Lemma \ref{lemma-localization-conditions}. This finishes the proof of the first assertion of the lemma. \medskip\noindent Let's assume that $S$ is saturated. (In the following we will use rotation of distinguished triangles without further mention.) Let $X \oplus Y$ be an object isomorphic to an object of $\mathcal{D}'$. Consider the morphism $f : 0 \to X$. The composition $0 \to X \to X \oplus Y$ is an element of $S$ as $(0, X \oplus Y, X \oplus Y, 0, 1, 0)$ is a distinguished triangle. The composition $Y[-1] \to 0 \to X$ is an element of $S$ as $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is a distinguished triangle, see Lemma \ref{lemma-split}. Hence $0 \to X$ is an element of $S$ (as $S$ is saturated). Thus $X$ is isomorphic to an object of $\mathcal{D}'$ as desired. \medskip\noindent Finally, assume $\mathcal{D}'$ is a saturated triangulated subcategory. Let $$W \xrightarrow{h} X \xrightarrow{g} Y \xrightarrow{f} Z$$ be composable morphisms of $\mathcal{D}$ such that $fg, gh \in S$. We will build up a picture of objects as in the diagram below. $$\xymatrix{ & & Q_{12} \ar[rd] & & Q_{23} \ar[rd] \\ & Q_1 \ar[ld]_{\! + \! 1} \ar[ru] & & Q_2 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \ar[ru] & & Q_3 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \\ W \ar[rr] & & X \ar[lu] \ar[rr] & & Y \ar[lu] \ar[rr] & & Z \ar[lu] }$$ First choose distinguished triangles $(W, X, Q_1)$, $(X, Y, Q_2)$, $(Y, Z, Q_3)$ $(W, Y, Q_{12})$, and $(X, Z, Q_{23})$. Denote $s : Q_2 \to Q_1[1]$ the composition $Q_2 \to X[1] \to Q_1[1]$. Denote $t : Q_3 \to Q_2[1]$ the composition $Q_3 \to Y[1] \to Q_2[1]$. By TR4 applied to the composition $W \to X \to Y$ and the composition $X \to Y \to Z$ there exist distinguished triangles $(Q_1, Q_{12}, Q_2)$ and $(Q_2, Q_{23}, Q_3)$ which use the morphisms $s$ and $t$. The objects $Q_{12}$ and $Q_{23}$ are isomorphic to objects of $\mathcal{D}'$ as $W \to Y$ and $X \to Z$ are assumed in $S$. Hence also $s[1]t$ is an element of $S$ as $S$ is closed under compositions and shifts. Note that $s[1]t = 0$ as $Y[1] \to Q_2[1] \to X[2]$ is zero, see Lemma \ref{lemma-composition-zero}. Hence $Q_3[1] \oplus Q_1[2]$ is isomorphic to an object of $\mathcal{D}'$, see Lemma \ref{lemma-split}. By assumption on $\mathcal{D}'$ we conclude that $Q_3$ and $Q_1$ are isomorphic to objects of $\mathcal{D}'$. Looking at the distinguished triangle $(Q_1, Q_{12}, Q_2)$ we conclude that $Q_2$ is also isomorphic to an object of $\mathcal{D}'$. Looking at the distinguished triangle $(X, Y, Q_2)$ we finally conclude that $g \in S$. (It is also follows that $h, f \in S$, but we don't need this.) \end{proof} \begin{definition} \label{definition-quotient-category} Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ be a full triangulated subcategory. We define the {\it quotient category $\mathcal{D}/\mathcal{B}$} by the formula $\mathcal{D}/\mathcal{B} = S^{-1}\mathcal{D}$, where $S$ is the multiplicative system of $\mathcal{D}$ associated to $\mathcal{B}$ via Lemma \ref{lemma-construct-multiplicative-system}. The localization functor $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ is called the {\it quotient functor} in this case. \end{definition} \noindent Note that the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ is an exact functor of triangulated categories, see Proposition \ref{proposition-construct-localization}. The universal property of this construction is the following. \begin{lemma} \label{lemma-universal-property-quotient} \begin{slogan} The universal property of the Verdier quotient. \end{slogan} Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ be a full triangulated subcategory of $\mathcal{D}$. Let $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ be the quotient functor. \begin{enumerate} \item If $H : \mathcal{D} \to \mathcal{A}$ is a homological functor into an abelian category $\mathcal{A}$ such that $\mathcal{B} \subset \Ker(H)$ then there exists a unique factorization $H' : \mathcal{D}/\mathcal{B} \to \mathcal{A}$ such that $H = H' \circ Q$ and $H'$ is a homological functor too. \item If $F : \mathcal{D} \to \mathcal{D}'$ is an exact functor into a pre-triangulated category $\mathcal{D}'$ such that $\mathcal{B} \subset \Ker(F)$ then there exists a unique factorization $F' : \mathcal{D}/\mathcal{B} \to \mathcal{D}'$ such that $F = F' \circ Q$ and $F'$ is an exact functor too. \end{enumerate} \end{lemma} \begin{proof} This lemma follows from Lemma \ref{lemma-universal-property-localization}. Namely, if $f : X \to Y$ is a morphism of $\mathcal{D}$ such that for some distinguished triangle $(X, Y, Z, f, g, h)$ the object $Z$ is isomorphic to an object of $\mathcal{B}$, then $H(f)$, resp.\ $F(f)$ is an isomorphism under the assumptions of (1), resp.\ (2). Details omitted. \end{proof} \noindent The kernel of the quotient functor can be described as follows. \begin{lemma} \label{lemma-kernel-quotient} Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ be a full triangulated subcategory. The kernel of the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ is the strictly full subcategory of $\mathcal{D}$ whose objects are $$\Ob(\Ker(Q)) = \left\{ \begin{matrix} Z \in \Ob(\mathcal{D}) \text{ such that there exists a }Z' \in \Ob(\mathcal{D}) \\ \text{ such that }Z \oplus Z'\text{ is isomorphic to an object of }\mathcal{B} \end{matrix} \right\}$$ In other words it is the smallest strictly full saturated triangulated subcategory of $\mathcal{D}$ containing $\mathcal{B}$. \end{lemma} \begin{proof} First note that the kernel is automatically a strictly full triangulated subcategory containing summands of any of its objects, see Lemma \ref{lemma-triangle-functor-kernel}. The description of its objects follows from the definitions and Lemma \ref{lemma-kernel-localization} part (4). \end{proof} \noindent Let $\mathcal{D}$ be a triangulated category. At this point we have constructions which induce order preserving maps between \begin{enumerate} \item the partially ordered set of multiplicative systems $S$ in $\mathcal{D}$ compatible with the triangulated structure, and \item the partially ordered set of full triangulated subcategories $\mathcal{B} \subset \mathcal{D}$. \end{enumerate} Namely, the constructions are given by $S \mapsto \mathcal{B}(S) = \Ker(Q : \mathcal{D} \to S^{-1}\mathcal{D})$ and $\mathcal{B} \mapsto S(\mathcal{B})$ where $S(\mathcal{B})$ is the multiplicative set of (\ref{equation-multiplicative-system}), i.e., $$S(\mathcal{B}) = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle }\\ (X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with } Z\text{ isomorphic to an object of }\mathcal{B} \end{matrix} \right\}$$ Note that it is not the case that these operations are mutually inverse. \begin{lemma} \label{lemma-operations} Let $\mathcal{D}$ be a triangulated category. The operations described above have the following properties \begin{enumerate} \item $S(\mathcal{B}(S))$ is the saturation'' of $S$, i.e., it is the smallest saturated multiplicative system in $\mathcal{D}$ containing $S$, and \item $\mathcal{B}(S(\mathcal{B}))$ is the saturation'' of $\mathcal{B}$, i.e., it is the smallest strictly full saturated triangulated subcategory of $\mathcal{D}$ containing $\mathcal{B}$. \end{enumerate} In particular, the constructions define mutually inverse maps between the (partially ordered) set of saturated multiplicative systems in $\mathcal{D}$ compatible with the triangulated structure on $\mathcal{D}$ and the (partially ordered) set of strictly full saturated triangulated subcategories of $\mathcal{D}$. \end{lemma} \begin{proof} First, let's start with a full triangulated subcategory $\mathcal{B}$. Then $\mathcal{B}(S(\mathcal{B})) = \Ker(Q : \mathcal{D} \to \mathcal{D}/\mathcal{B})$ and hence (2) is the content of Lemma \ref{lemma-kernel-quotient}. \medskip\noindent Next, suppose that $S$ is multiplicative system in $\mathcal{D}$ compatible with the triangulation on $\mathcal{D}$. Then $\mathcal{B}(S) = \Ker(Q : \mathcal{D} \to S^{-1}\mathcal{D})$. Hence (using Lemma \ref{lemma-third-object-zero} in the localized category) \begin{align*} S(\mathcal{B}(S)) & = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished}\\ \text{triangle }(X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with }Q(Z) = 0 \end{matrix} \right\} \\ & = \{f \in \text{Arrows}(\mathcal{D}) \mid Q(f)\text{ is an isomorphism}\} \\ & = \hat S = S' \end{align*} in the notation of Categories, Lemma \ref{categories-lemma-what-gets-inverted}. The final statement of that lemma finishes the proof. \end{proof} \begin{lemma} \label{lemma-acyclic-general} Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor from a triangulated category $\mathcal{D}$ to an abelian category $\mathcal{A}$, see Definition \ref{definition-homological}. The subcategory $\Ker(H)$ of $\mathcal{D}$ is a strictly full saturated triangulated subcategory of $\mathcal{D}$ whose corresponding saturated multiplicative system (see Lemma \ref{lemma-operations}) is the set $$S = \{f \in \text{Arrows}(\mathcal{D}) \mid H^i(f)\text{ is an isomorphism for all }i \in \mathbf{Z}\}.$$ The functor $H$ factors through the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\Ker(H)$. \end{lemma} \begin{proof} The category $\Ker(H)$ is a strictly full saturated triangulated subcategory of $\mathcal{D}$ by Lemma \ref{lemma-homological-functor-kernel}. The set $S$ is a saturated multiplicative system compatible with the triangulated structure by Lemma \ref{lemma-homological-functor-localize}. Recall that the multiplicative system corresponding to $\Ker(H)$ is the set $$\left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle }\\ (X, Y, Z, f, g, h)\text{ with } H^i(Z) = 0 \text{ for all }i \end{matrix} \right\}$$ By the long exact cohomology sequence, see (\ref{equation-long-exact-cohomology-sequence}), it is clear that $f$ is an element of this set if and only if $f$ is an element of $S$. Finally, the factorization of $H$ through $Q$ is a consequence of Lemma \ref{lemma-universal-property-quotient}. \end{proof} \noindent It is clear that in the lemma above the factorization of $H$ through $\mathcal{D}/\Ker(H)$ is the universal factorization. Namely, if $F : \mathcal{D} \to \mathcal{D}'$ is an exact functor of triangulated categories and if there exists a homological functor $H' : \mathcal{D}' \to \mathcal{A}$ such that $H \cong H' \circ F$, then $F$ factors through the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\Ker(H)$. \section{Adjoints for exact functors} \label{section-adjoints} \noindent Results on adjoint functors between triangulated categories. \begin{lemma} \label{lemma-adjoint-is-exact} Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor between triangulated categories. If $F$ has a right adjoint, then it is an exact functor. \end{lemma} \begin{proof} Let $G$ be a right adjoint. Let $X$ be an object of $\mathcal{D}$ and $A$ an object of $\mathcal{D}'$. Since $F$ is an exact functor we see that \begin{align*} \Mor_\mathcal{D}(X, G(A[1]) & = \Mor_{\mathcal{D}'}(F(X), A[1]) \\ & = \Mor_{\mathcal{D}'}(F(X)[-1], A) \\ & = \Mor_{\mathcal{D}'}(F(X[-1]), A) \\ & = \Mor_\mathcal{D}(X[-1], G(A)) \\ & = \Mor_\mathcal{D}(X, G(A)[1]) \end{align*} By Yoneda's lemma (Categories, Lemma \ref{categories-lemma-yoneda}) we obtain a canonical isomorphism $G(A)[1] = G(A[1])$. Let $A \to B \to C \to A[1]$ be a distinguished triangle in $\mathcal{D}'$. Choose a distinguished triangle $$G(A) \to G(B) \to X \to G(A)[1]$$ in $\mathcal{D}$. Then $F(G(A)) \to F(G(B)) \to F(X) \to F(G(A))[1]$ is a distinguished triangle in $\mathcal{D}'$. By TR3 we can choose a morphism of distinguished triangles $$\xymatrix{ F(G(A)) \ar[r] \ar[d] & F(G(B)) \ar[r] \ar[d] & F(X) \ar[r] \ar[d] & F(G(A))[1] \ar[d] \\ A \ar[r] & B \ar[r] & C \ar[r] & A[1] }$$ Since $G$ is the adjoint the new morphism determines a morphism $X \to G(C)$ such that the diagram $$\xymatrix{ G(A) \ar[r] \ar[d] & G(B) \ar[r] \ar[d] & X \ar[r] \ar[d] & G(A)[1] \ar[d] \\ G(A) \ar[r] & G(B) \ar[r] & G(C) \ar[r] & G(A)[1] }$$ commutes. Applying the homological functor $\Hom_{\mathcal{D}'}(W, -)$ for an object $W$ of $\mathcal{D}'$ we deduce from the $5$ lemma that $$\Hom_{\mathcal{D}'}(W, X) \to \Hom_{\mathcal{D}'}(W, G(C))$$ is a bijection and using the Yoneda lemma once more we conclude that $X \to G(C)$ is an isomorphism. Hence we conclude that $G(A) \to G(B) \to G(C) \to G(A)[1]$ is a distinguished triangle which is what we wanted to show. \end{proof} \begin{lemma} \label{lemma-fully-faithful-adjoint-kernel-zero} Let $\mathcal{D}$, $\mathcal{D}'$ be triangulated categories. Let $F : \mathcal{D} \to \mathcal{D}'$ and $G : \mathcal{D}' \to \mathcal{D}$ be functors. Assume that \begin{enumerate} \item $F$ and $G$ are exact functors, \item $F$ is fully faithful, \item $G$ is a right adjoint to $F$, and \item the kernel of $G$ is zero. \end{enumerate} Then $F$ is an equivalence of categories. \end{lemma} \begin{proof} Since $F$ is fully faithful the adjunction map $\text{id} \to G \circ F$ is an isomorphism (Categories, Lemma \ref{categories-lemma-adjoint-fully-faithful}). Let $X$ be an object of $\mathcal{D}'$. Choose a distinguished triangle $$F(G(X)) \to X \to Y \to F(G(X))[1]$$ in $\mathcal{D}'$. Applying $G$ and using that $G(F(G(X))) = G(X)$ we find a distinguished triangle $$G(X) \to G(X) \to G(Y) \to G(X)[1]$$ Hence $G(Y) = 0$. Thus $Y = 0$. Thus $F(G(X)) \to X$ is an isomorphism. \end{proof} \section{The homotopy category} \label{section-homotopy} \noindent Let $\mathcal{A}$ be an additive category. The homotopy category $K(\mathcal{A})$ of $\mathcal{A}$ is the category of complexes of $\mathcal{A}$ with morphisms given by morphisms of complexes up to homotopy. Here is the formal definition. \begin{definition} \label{definition-complexes-notation} Let $\mathcal{A}$ be an additive category. \begin{enumerate} \item We set $\text{Comp}(\mathcal{A}) = \text{CoCh}(\mathcal{A})$ be the {\it category of (cochain) complexes}. \item A complex $K^\bullet$ is said to be {\it bounded below} if $K^n = 0$ for all $n \ll 0$. \item A complex $K^\bullet$ is said to be {\it bounded above} if $K^n = 0$ for all $n \gg 0$. \item A complex $K^\bullet$ is said to be {\it bounded} if $K^n = 0$ for all $|n| \gg 0$. \item We let $\text{Comp}^{+}(\mathcal{A})$, $\text{Comp}^{-}(\mathcal{A})$, resp.\ $\text{Comp}^b(\mathcal{A})$ be the full subcategory of $\text{Comp}(\mathcal{A})$ whose objects are the complexes which are bounded below, bounded above, resp.\ bounded. \item We let $K(\mathcal{A})$ be the category with the same objects as $\text{Comp}(\mathcal{A})$ but as morphisms homotopy classes of maps of complexes (see Homology, Lemma \ref{homology-lemma-compose-homotopy-cochain}). \item We let $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, resp.\ $K^b(\mathcal{A})$ be the full subcategory of $K(\mathcal{A})$ whose objects are bounded below, bounded above, resp.\ bounded complexes of $\mathcal{A}$. \end{enumerate} \end{definition} \noindent It will turn out that the categories $K(\mathcal{A})$, $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^b(\mathcal{A})$ are triangulated categories. To prove this we first develop some machinery related to cones and split exact sequences. \section{Cones and termwise split sequences} \label{section-cones} \noindent 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 \ref{homology-definition-shift-cochain}, give rise to functors $[n] : K(\mathcal{A}) \to K(\mathcal{A})$ such that $[n] \circ [m] = [n + m]$ and $[0] = \text{id}$. \begin{definition} \label{definition-cone} Let $\mathcal{A}$ be an additive category. Let $f : K^\bullet \to L^\bullet$ be a morphism of complexes of $\mathcal{A}$. The {\it 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 $L^n \to C(f)^n \to K^{n + 1}$. \end{definition} \noindent 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. \begin{lemma} \label{lemma-functorial-cone} 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 $K(\mathcal{A})$. \end{lemma} \begin{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 $p_2 \circ c = a \circ p_1$. \end{proof} \noindent Note that the morphism $c : C(f_1)^\bullet \to C(f_2)^\bullet$ constructed in the proof of Lemma \ref{lemma-functorial-cone} in general depends on the chosen homotopy $h$ between $f_2 \circ a$ and $b \circ f_1$. \begin{lemma} \label{lemma-map-from-cone} 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 $g$ factors through a morphism $C(f)^\bullet \to M^\bullet$ of $K(\mathcal{A})$. \end{lemma} \begin{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 \ref{lemma-functorial-cone} is the desired map. \end{proof} \noindent Note that the morphism $C(f)^\bullet \to M^\bullet$ constructed in the proof of Lemma \ref{lemma-map-from-cone} in general depends on the chosen homotopy. \begin{definition} \label{definition-termwise-split-map} Let $\mathcal{A}$ be an additive category. A {\it 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 {\it 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. \end{definition} \begin{lemma} \label{lemma-make-commute-map} 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. \end{lemma} \begin{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. \end{proof} \noindent 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. \begin{lemma} \label{lemma-make-injective} 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 \begin{enumerate} \item $\tilde \alpha$ is a termwise split injection (see Definition \ref{definition-termwise-split-map}), \item 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 $\text{id}_{\tilde L^\bullet}$. \end{enumerate} Moreover, if both $K^\bullet$ and $L^\bullet$ are in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^b(\mathcal{A})$, then so is $\tilde L^\bullet$. \end{lemma} \begin{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. \end{proof} \begin{remark} \label{remark-compute-modules} 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. \end{remark} \begin{lemma} \label{lemma-make-surjective} 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 \begin{enumerate} \item $\tilde \alpha$ is a termwise split surjection (see Definition \ref{definition-termwise-split-map}), \item 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 $\text{id}_{\tilde K^\bullet}$. \end{enumerate} Moreover, if both $K^\bullet$ and $L^\bullet$ are in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^b(\mathcal{A})$, then so is $\tilde K^\bullet$. \end{lemma} \begin{proof} Dual to Lemma \ref{lemma-make-injective}. 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. \end{proof} \begin{definition} \label{definition-split-ses} Let $\mathcal{A}$ be an additive category. A {\it termwise split 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 \ref{homology-lemma-ses-termwise-split-cochain} 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 {\it triangle associated to the termwise split sequence of complexes}. \end{definition} \begin{lemma} \label{lemma-triangle-independent-splittings} 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 \ref{definition-split-ses}. 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 $K(\mathcal{A})$. \end{lemma} \begin{proof} The statement simply means that $\delta$ and $\delta'$ are homotopic maps of complexes. This is Homology, Lemma \ref{homology-lemma-ses-termwise-split-homotopy-cochain}. \end{proof} \begin{remark} \label{remark-make-commute} 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 {\bf 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 (\ref{examples-equation-commutes-up-to-homotopy}). 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. \end{remark} \begin{lemma} \label{lemma-nilpotent} 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 $K(\mathcal{A})$. \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 $\Im(b^n) \subset \Im(A_2^n \to B_2^n)$ and $\Ker((b')^n) \supset \Im(A_2^n \to B_2^n)$. Then $b \circ b' = 0$ as a map of complexes. \end{proof} \begin{lemma} \label{lemma-third-isomorphism} 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 $c$. \end{lemma} \begin{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 \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 $K(\mathcal{A})$ 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 \ref{lemma-nilpotent}. Hence $c$ is an isomorphism in $K(\mathcal{A})$ with inverse $2 - c$. \end{proof} \noindent 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! \begin{lemma} \label{lemma-the-same-up-to-isomorphisms} Let $\mathcal{A}$ be an additive category. \begin{enumerate} \item 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 $K(\mathcal{A})$. \item 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 $K^\bullet \to \tilde L^\bullet \to M^\bullet$. \end{enumerate} \end{lemma} \begin{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 \ref{homology-lemma-ses-termwise-split-cochain}. The composition $C^\bullet \to C(f)^\bullet \to C^\bullet$ is the identity. The composition $C(f)^\bullet \to C^\bullet \to C(f)^\bullet$ 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 \ref{definition-cone}) 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. \medskip\noindent Proof of (2). We let $\tilde f : K^\bullet \to \tilde L^\bullet$, $s : L^\bullet \to \tilde L^\bullet$ and $\pi : L^\bullet \to L^\bullet$ be as in Lemma \ref{lemma-make-injective}. By Lemmas \ref{lemma-functorial-cone} and \ref{lemma-third-isomorphism} 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). \end{proof} \begin{lemma} \label{lemma-sequence-maps-split} 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 $B_i^\bullet$. \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 $B_n^\bullet$. Apply Lemma \ref{lemma-make-injective} 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. \end{proof} \begin{lemma} \label{lemma-rotate-triangle} 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 $(C^\bullet[-1], A^\bullet, C(\delta[-1])^\bullet, \delta[-1], i, p)$. \end{lemma} \begin{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. \end{proof} \begin{lemma} \label{lemma-rotate-cone} 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 $f$. \end{lemma} \begin{proof} Immediate from the definitions. \end{proof} \section{Distinguished triangles in the homotopy category} \label{section-homotopy-triangulated} \noindent Since we want our boundary maps in long exact sequences of cohomology to be given by the maps in the snake lemma without signs we define distinguished triangles in the homotopy category as follows. \begin{definition} \label{definition-distinguished-triangle} Let $\mathcal{A}$ be an additive category. A triangle $(X, Y, Z, f, g, h)$ of $K(\mathcal{A})$ is called a {\it distinguished triangle of $K(\mathcal{A})$} if it is isomorphic to the triangle associated to a termwise split exact sequence of complexes, see Definition \ref{definition-split-ses}. Same definition for $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^b(\mathcal{A})$. \end{definition} \noindent Note that according to Lemma \ref{lemma-the-same-up-to-isomorphisms} a triangle of the form $(K^\bullet, L^\bullet, C(f)^\bullet, f, i, -p)$ is a distinguished triangle. This does indeed lead to a triangulated category, see Proposition \ref{proposition-homotopy-category-triangulated}. Before we can prove the proposition we need one more lemma in order to be able to prove TR4. \begin{lemma} \label{lemma-two-split-injections} Let $\mathcal{A}$ be an additive category. Suppose that $\alpha : A^\bullet \to B^\bullet$ and $\beta : B^\bullet \to C^\bullet$ are split injections of complexes. Then there exist distinguished triangles $(A^\bullet, B^\bullet, Q_1^\bullet, \alpha, p_1, d_1)$, $(A^\bullet, C^\bullet, Q_2^\bullet, \beta \circ \alpha, p_2, d_2)$ and $(B^\bullet, C^\bullet, Q_3^\bullet, \beta, p_3, d_3)$ for which TR4 holds. \end{lemma} \begin{proof} Say $\pi_1^n : B^n \to A^n$, and $\pi_3^n : C^n \to B^n$ are the splittings. Then also $A^\bullet \to C^\bullet$ is a split injection with splittings $\pi_2^n = \pi_1^n \circ \pi_3^n$. Let us write $Q_1^\bullet$, $Q_2^\bullet$ and $Q_3^\bullet$ for the quotient'' complexes. In other words, $Q_1^n = \Ker(\pi_1^n)$, $Q_3^n = \Ker(\pi_3^n)$ and $Q_2^n = \Ker(\pi_2^n)$. Note that the kernels exist. Then $B^n = A^n \oplus Q_1^n$ and $C_n = B^n \oplus Q_3^n$, where we think of $A^n$ as a subobject of $B^n$ and so on. This implies $C^n = A^n \oplus Q_1^n \oplus Q_3^n$. Note that $\pi_2^n = \pi_1^n \circ \pi_3^n$ is zero on both $Q_1^n$ and $Q_3^n$. Hence $Q_2^n = Q_1^n \oplus Q_3^n$. Consider the commutative diagram $$\begin{matrix} 0 & \to & A^\bullet & \to & B^\bullet & \to & Q_1^\bullet & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & \\ 0 & \to & A^\bullet & \to & C^\bullet & \to & Q_2^\bullet & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & \\ 0 & \to & B^\bullet & \to & C^\bullet & \to & Q_3^\bullet & \to & 0 \end{matrix}$$ The rows of this diagram are termwise split 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 $$(A^\bullet \to B^\bullet \to Q_1^\bullet \to A^\bullet[1]) \longrightarrow (A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet[1])$$ and $$(A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet[1]) \longrightarrow (B^\bullet \to C^\bullet \to Q_3^\bullet \to B^\bullet[1]).$$ Note that the splittings $Q_3^n \to C^n$ of the bottom split sequence in the diagram provides a splitting for the split sequence $0 \to Q_1^\bullet \to Q_2^\bullet \to Q_3^\bullet \to 0$ upon composing with $C^n \to Q_2^n$. It follows easily from this that the morphism $\delta : Q_3^\bullet \to Q_1^\bullet[1]$ in the corresponding distinguished triangle $$(Q_1^\bullet \to Q_2^\bullet \to Q_3^\bullet \to Q_1^\bullet[1])$$ is equal to the composition $Q_3^\bullet \to B^\bullet[1] \to Q_1^\bullet[1]$. Hence we get a structure as in the conclusion of axiom TR4. \end{proof} \begin{proposition} \label{proposition-homotopy-category-triangulated} Let $\mathcal{A}$ be an additive category. The category $K(\mathcal{A})$ of complexes up to homotopy with its natural translation functors and distinguished triangles as defined above is a triangulated category. \end{proposition} \begin{proof} Proof of TR1. By definition every triangle isomorphic to a distinguished one is distinguished. Also, any triangle $(A^\bullet, A^\bullet, 0, 1, 0, 0)$ is distinguished since $0 \to A^\bullet \to A^\bullet \to 0 \to 0$ is a termwise split sequence of complexes. Finally, given any morphism of complexes $f : K^\bullet \to L^\bullet$ 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 a termwise split sequence of complexes $A^\bullet \to B^\bullet \to C^\bullet$ such that the associated triangle $(A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta)$ 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 $(C^\bullet[-1], A^\bullet, B^\bullet, -\delta[-1], \alpha, \beta)$. It follows from Lemma \ref{lemma-rotate-triangle} that the triangle $(C^\bullet[-1], A^\bullet, B^\bullet, \delta[-1], \alpha, \beta)$ is isomorphic to $(C^\bullet[-1], A^\bullet, C(\delta[-1])^\bullet, \delta[-1], i, p)$. Precomposing the previous isomorphism of triangles with $-1$ on $Y$ it follows that $(X, Y, Z, f, g, h)$ is isomorphic to $(C^\bullet[-1], A^\bullet, C(\delta[-1])^\bullet, \delta[-1], i, -p)$. 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 $(K^\bullet, L^\bullet, C(f), f, i, -p)$ for some morphism of complexes $f$. Then the rotated triangle $(Y, Z, X[1], g, h, -f[1])$ is isomorphic to $(L^\bullet, C(f), K^\bullet[1], i, -p, -f[1])$ which is isomorphic to the triangle $(L^\bullet, C(f), K^\bullet[1], i, p, f[1])$. 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 $f, f', a, b$. \medskip\noindent Proof of TR4. At this point we know that $K(\mathcal{A})$ is a pre-triangulated category. Hence we can use Lemma \ref{lemma-easier-axiom-four}. Let $A^\bullet \to B^\bullet$ and $B^\bullet \to C^\bullet$ be composable morphisms of $K(\mathcal{A})$. By Lemma \ref{lemma-sequence-maps-split} we may assume that $A^\bullet \to B^\bullet$ and $B^\bullet \to C^\bullet$ are split injective morphisms. In this case the result follows from Lemma \ref{lemma-two-split-injections}. \end{proof} \begin{remark} \label{remark-boundedness-conditions-triangulated} Let $\mathcal{A}$ be an additive category. Exactly the same proof as the proof of Proposition \ref{proposition-homotopy-category-triangulated} shows that the categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^b(\mathcal{A})$ are triangulated categories. Namely, the cone of a morphism between bounded (above, below) is bounded (above, below). But we prove below that these are triangulated subcategories of $K(\mathcal{A})$ which gives another proof. \end{remark} \begin{lemma} \label{lemma-bounded-triangulated-subcategories} Let $\mathcal{A}$ be an additive category. The categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^b(\mathcal{A})$ are full triangulated subcategories of $K(\mathcal{A})$. \end{lemma} \begin{proof} Each of the categories mentioned is a full additive subcategory. We use the criterion of Lemma \ref{lemma-triangulated-subcategory} to show that they are triangulated subcategories. It is clear that each of the categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^b(\mathcal{A})$ is preserved under the shift functors $[1], [-1]$. Finally, suppose that $f : A^\bullet \to B^\bullet$ is a morphism in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^b(\mathcal{A})$. Then $(A^\bullet, B^\bullet, C(f)^\bullet, f, i, -p)$ is a distinguished triangle of $K(\mathcal{A})$ with $C(f)^\bullet \in K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^b(\mathcal{A})$ as is clear from the construction of the cone. Thus the lemma is proved. (Alternatively, $K^\bullet \to L^\bullet$ is isomorphic to an termwise split injection of complexes in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^b(\mathcal{A})$, see Lemma \ref{lemma-make-injective} and then one can directly take the associated distinguished triangle.) \end{proof} \begin{lemma} \label{lemma-additive-exact-homotopy-category} Let $\mathcal{A}$, $\mathcal{B}$ be additive categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor. The induced functors $$\begin{matrix} F : K(\mathcal{A}) \longrightarrow K(\mathcal{B}) \\ F : K^{+}(\mathcal{A}) \longrightarrow K^{+}(\mathcal{B}) \\ F : K^{-}(\mathcal{A}) \longrightarrow K^{-}(\mathcal{B}) \\ F : K^b(\mathcal{A}) \longrightarrow K^b(\mathcal{B}) \end{matrix}$$ are exact functors of triangulated categories. \end{lemma} \begin{proof} Suppose $A^\bullet \to B^\bullet \to C^\bullet$ is a termwise split sequence of complexes of $\mathcal{A}$ with splittings $(s^n, \pi^n)$ and associated morphism $\delta : C^\bullet \to A^\bullet[1]$, see Definition \ref{definition-split-ses}. Then $F(A^\bullet) \to F(B^\bullet) \to F(C^\bullet)$ is a termwise split sequence of complexes with splittings $(F(s^n), F(\pi^n))$ and associated morphism $F(\delta) : F(C^\bullet) \to F(A^\bullet)[1]$. Thus $F$ transforms distinguished triangles into distinguished triangles. \end{proof} \section{Derived categories} \label{section-derived-categories} \noindent In this section we construct the derived category of an abelian category $\mathcal{A}$ by inverting the quasi-isomorphisms in $K(\mathcal{A})$. Before we do this recall that the functors $H^i : \text{Comp}(\mathcal{A}) \to \mathcal{A}$ factor through $K(\mathcal{A})$, see Homology, Lemma \ref{homology-lemma-map-cohomology-homotopy-cochain}. Moreover, in Homology, Definition \ref{homology-definition-cohomology-shift} we have defined identifications $H^i(K^\bullet[n]) = H^{i + n}(K^\bullet)$. At this point it makes sense to redefine $$H^i(K^\bullet) = H^0(K^\bullet[i])$$ in order to avoid confusion and possible sign errors. \begin{lemma} \label{lemma-cohomology-homological} Let $\mathcal{A}$ be an abelian category. The functor $$H^0 : K(\mathcal{A}) \longrightarrow \mathcal{A}$$ is homological. \end{lemma} \begin{proof} Because $H^0$ is a functor, and by our definition of distinguished triangles it suffices to prove that given a termwise split short exact sequence of complexes $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ the sequence $H^0(A^\bullet) \to H^0(B^\bullet) \to H^0(C^\bullet)$ is exact. This follows from Homology, Lemma \ref{homology-lemma-long-exact-sequence-cochain}. \end{proof} \noindent In particular, this lemma implies that a distinguished triangle $(X, Y, Z, f, g, h)$ in $K(\mathcal{A})$ gives rise to a long exact cohomology sequence \begin{equation} \label{equation-long-exact-cohomology-sequence-D} \xymatrix{ \ldots \ar[r] & H^i(X) \ar[r]^{H^i(f)} & H^i(Y) \ar[r]^{H^i(g)} & H^i(Z) \ar[r]^{H^i(h)} & H^{i + 1}(X) \ar[r] & \ldots } \end{equation} see (\ref{equation-long-exact-cohomology-sequence}). Moreover, there is a compatibility with the long exact sequence of cohomology associated to a short exact sequence of complexes (insert future reference here). For example, if $(A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta)$ is the distinguished triangle associated to a termwise split exact sequence of complexes (see Definition \ref{definition-split-ses}), then the cohomology sequence above agrees with the one defined using the snake lemma, see Homology, Lemma \ref{homology-lemma-long-exact-sequence-cochain} and for agreement of sequences, see Homology, Lemma \ref{homology-lemma-ses-termwise-split-long-cochain}. \medskip\noindent Recall that a complex $K^\bullet$ is {\it acyclic} if $H^i(K^\bullet) = 0$ for all $i \in \mathbf{Z}$. Moreover, recall that a morphism of complexes $f : K^\bullet \to L^\bullet$ is a {\it quasi-isomorphism} if and only if $H^i(f)$ is an isomorphism for all $i$. See Homology, Definition \ref{homology-definition-quasi-isomorphism-cochain}. \begin{lemma} \label{lemma-acyclic} Let $\mathcal{A}$ be an abelian category. The full subcategory $\text{Ac}(\mathcal{A})$ of $K(\mathcal{A})$ consisting of acyclic complexes is a strictly full saturated triangulated subcategory of $K(\mathcal{A})$. The corresponding saturated multiplicative system (see Lemma \ref{lemma-operations}) of $K(\mathcal{A})$ is the set $\text{Qis}(\mathcal{A})$ of quasi-isomorphisms. In particular, the kernel of the localization functor $Q : K(\mathcal{A}) \to \text{Qis}(\mathcal{A})^{-1}K(\mathcal{A})$ is $\text{Ac}(\mathcal{A})$ and the functor $H^0$ factors through $Q$. \end{lemma} \begin{proof} We know that $H^0$ is a homological functor by Lemma \ref{lemma-cohomology-homological}. Thus this lemma is a special case of Lemma \ref{lemma-acyclic-general}. \end{proof} \begin{definition} \label{definition-unbounded-derived-category} Let $\mathcal{A}$ be an abelian category. Let $\text{Ac}(\mathcal{A})$ and $\text{Qis}(\mathcal{A})$ be as in Lemma \ref{lemma-acyclic}. The {\it derived category of $\mathcal{A}$} is the triangulated category $$D(\mathcal{A}) = K(\mathcal{A})/\text{Ac}(\mathcal{A}) = \text{Qis}(\mathcal{A})^{-1} K(\mathcal{A}).$$ We denote $H^0 : D(\mathcal{A}) \to \mathcal{A}$ the unique functor whose composition with the quotient functor gives back the functor $H^0$ defined above. Using Lemma \ref{lemma-homological-functor-bounded} we introduce the strictly full saturated triangulated subcategories $D^{+}(\mathcal{A}), D^{-}(\mathcal{A}), D^b(\mathcal{A})$ whose sets of objects are $$\begin{matrix} \Ob(D^{+}(\mathcal{A})) = \{X \in \Ob(D(\mathcal{A})) \mid H^n(X) = 0\text{ for all }n \ll 0\} \\ \Ob(D^{-}(\mathcal{A})) = \{X \in \Ob(D(\mathcal{A})) \mid H^n(X) = 0\text{ for all }n \gg 0\} \\ \Ob(D^b(\mathcal{A})) = \{X \in \Ob(D(\mathcal{A})) \mid H^n(X) = 0\text{ for all }|n| \gg 0\} \end{matrix}$$ The category $D^b(\mathcal{A})$ is called the {\it bounded derived category} of $\mathcal{A}$.