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aisejohan committed Dec 4, 2013
1 parent 5c7c237 commit 49beb56bd6f5f2473b2bd942b2a2753a04fff271
Showing with 39 additions and 39 deletions.
  1. +23 −23 derived.tex
  2. +16 −16 homology.tex
@@ -263,7 +263,7 @@ \section{The definition of a triangulated category}
\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 says that a distinguished triangle
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}
@@ -527,9 +527,14 @@ \section{Elementary results on triangulated categories}
\end{lemma}

\begin{proof}
Immediate from
Lemma \ref{lemma-cone-triangle-unique-isomorphism}
and TR1.
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}
@@ -565,14 +570,9 @@ \section{Elementary results on triangulated categories}
(X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'').
}
$$
Let $W$ be any object in $\mathcal{D}$ and apply the functor
$H_W = \Hom_\mathcal{D}(W, -)$ to this diagram.
By
Lemma \ref{lemma-representable-homological}
(applied three times)
we deduce that $H_W(c) : H_W(Z \oplus Z') \to H_W(Q)$
is an isomorphism. Hence $c$ is an isomorphism and we conclude that
(1) holds.
By Lemma \ref{lemma-third-isomorphism-triangle}
we find that $c$ is an isomorphism and we conclude
that (1) holds.

\medskip\noindent
Assume (1). We will show that $(X, Y, Z, f, g, h)$ is a distinguished
@@ -594,18 +594,18 @@ \section{Elementary results on triangulated categories}
\longrightarrow
(X, Y, Q, f, g'', h'')
$$
Applying $H_W$ and using the above we once again see that
$H_W(c) : H_W(Z) \to H_W(Q)$ is an isomorphism and we conclude that
$c$ is an isomorphism. Hence we win.
By Lemma \ref{lemma-third-isomorphism-triangle}
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 left inverse $s : Z \to Y$ to $g$.
\item For any left inverse $s : Z \to Y$ of $g$ the map
\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.
@@ -648,10 +648,10 @@ \section{Elementary results on triangulated categories}
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 $(X, Y, Z, f, g, h)$
and by
Lemma \ref{lemma-representable-homological}
we see that $h = 0$. Then
By TR1 there exists a distinguished triangle $(X, Y, Z, f, g, h)$.
By Lemma \ref{lemma-composition-zero} the composition
$h[-1] \circ f = 0$. 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
@@ -836,7 +836,7 @@ \section{Elementary results on triangulated categories}
\end{lemma}

\begin{proof}
The ``if'' part is clear. Assume $(F(X), F(Y), F(Z))$ is
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}
@@ -1174,7 +1174,7 @@ \section{Localization of triangulated categories}
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 $j \circ f = j \circ d$ and we see that LMS3 holds.
Thus $j \circ f = j \circ g$ and we see that LMS3 holds.
The proof of RMS3 is dual.
\end{proof}

@@ -478,32 +478,32 @@ \section{Abelian categories}
\begin{enumerate}
\item The diagram is cartesian if and only if
$$
0\to w\xrightarrow{(g,-f)}x\oplus y\xrightarrow{(k,-h)}z
0 \to w \xrightarrow{(g, f)} x \oplus y \xrightarrow{(k, -h)} z
$$
is exact.
\item The diagram is cocartesian if and only if
$$
w\xrightarrow{(g,-f)}x\oplus y\xrightarrow{(k,-h)}z\to 0
w \xrightarrow{(g, -f)} x \oplus y \xrightarrow{(k, h)} z \to 0
$$
is exact.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $u=(g,-f):w\to x\oplus y$ and $v=(k,-h):x\oplus y\to z$.
Let $p:x\oplus y\to x$ and $q:x\oplus y\to y$ be the canonical
projections. Let $i:\text{Ker}(v)\to x\oplus v$ be the canonical
Let $u = (g, f) : w \to x \oplus y$ and $v = (k, -h) : x \oplus y \to z$.
Let $p : x \oplus y \to x$ and $q : x \oplus y \to y$ be the canonical
projections. Let $i : \text{Ker}(v) \to x \oplus y$ be the canonical
injection. By Example \ref{example-fibre-product-pushouts}, the diagram is
cartesian if and only if there exists an isomorphism
$r:\text{Ker}(v)\to w$ with $-f\circ r=q\circ i$ and
$g\circ r=p\circ i$. The sequence
$0\to w\overset{u}\to x\oplus y\overset{v}\to z$ is exact if and
only if there exists an isomorphism $r:\text{Ker}(v)\to w$ with
$u\circ r=i$. But given $r:\text{Ker}(v)\to w$, we have
$-f\circ r=q\circ i$ and $g\circ r=p\circ i$ if and
only if $q\circ u\circ r=-f\circ r=q\circ i$ and
$p\circ u\circ r=g\circ r=p\circ i$, hence if and only if $u\circ r=i$.
This proves (1), and then (2) follows by duality.
$r : \text{Ker}(v) \to w$ with $f \circ r = q \circ i$ and
$g \circ r = p \circ i$. The sequence
$0 \to w \overset{u} \to x \oplus y \overset{v} \to z$ is exact if and
only if there exists an isomorphism $r : \text{Ker}(v) \to w$ with
$u \circ r = i$. But given $r : \text{Ker}(v) \to w$, we have
$f \circ r = q \circ i$ and $g \circ r = p \circ i$ if and
only if $q \circ u \circ r= f \circ r = q \circ i$ and
$p \circ u \circ r = g \circ r = p \circ i$, hence if and only if
$u \circ r = i$. This proves (1), and then (2) follows by duality.
\end{proof}

\begin{lemma}
@@ -931,7 +931,7 @@ \section{Extensions}
\begin{definition}
\label{definition-ext-group}
Let $\mathcal{A}$ be an abelian category.
Let $A, C \in \Ob(\mathcal{A})$.
Let $A, B \in \Ob(\mathcal{A})$.
The set of isomorphism classes of extensions
of $B$ by $A$ is denoted
$$
@@ -2928,7 +2928,7 @@ \section{Homotopy and the shift functor}
\label{lemma-ses-termwise-split-homotopy}
Notation and assumptions as in Lemma \ref{lemma-ses-termwise-split} above.
Suppose $\{s'_n : C_n \to B_n\}$ is a second choice of splittings.
Write $s'_n = s_n + \pi_n \circ h_n$ for some unique
Write $s'_n = s_n + i_n \circ h_n$ for some unique
morphisms $h_n : C_n \to A_n$. The family of maps
$\{h_n : C_n \to A[-1]_{n + 1}\}$ is a homotopy between
the associated morphisms

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