## 22.9 Cones and distinguished triangles

Let $(A, \text{d})$ be a differential graded algebra. Let $f : K \to L$ be a homomorphism of differential graded $A$-modules. Then $(K, L, C(f), f, i, p)$ forms a triangle:

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

in $\text{Mod}_{(A, \text{d})}$ and hence in $K(\text{Mod}_{(A, \text{d})})$. Cones are not distinguished triangles in general, but the difference is a sign or a rotation (your choice). Here are two precise statements.

Lemma 22.9.1. Let $(A, \text{d})$ be a differential graded algebra. Let $f : K \to L$ be a homomorphism of differential graded modules. The triangle $(L, C(f), K[1], i, p, f[1])$ is the triangle associated to the admissible short exact sequence

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

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

Proof. Immediate from the definitions. $\square$

Lemma 22.9.2. Let $(A, \text{d})$ be a differential graded algebra. Let $\alpha : K \to L$ and $\beta : L \to M$ define an admissible short exact sequence

$0 \to K \to L \to M \to 0$

of differential graded $A$-modules. Let $(K, L, M, \alpha , \beta , \delta )$ be the associated triangle. Then the triangles

$(M[-1], K, L, \delta [-1], \alpha , \beta ) \quad \text{and}\quad (M[-1], K, C(\delta [-1]), \delta [-1], i, p)$

are isomorphic.

Proof. Using a choice of splittings we write $L = K \oplus M$ and we identify $\alpha$ and $\beta$ with the natural inclusion and projection maps. By construction of $\delta$ we have

$d_ B = \left( \begin{matrix} d_ K & \delta \\ 0 & d_ M \end{matrix} \right)$

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

Lemma 22.9.3. Let $(A, \text{d})$ be a differential graded algebra. Let $f_1 : K_1 \to L_1$ and $f_2 : K_2 \to L_2$ be homomorphisms of differential graded $A$-modules. Let

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

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

Proof. Let $a^{-1} : K_2 \to K_1$ be a homomorphism of differential graded $A$-modules which is inverse to $a$ in $K(\text{Mod}_{(A, \text{d})})$. Let $b^{-1} : L_2 \to L_1$ be a homomorphism of differential graded $A$-modules which is inverse to $b$ in $K(\text{Mod}_{(A, \text{d})})$. Let $c' : C(f_2) \to C(f_1)$ be the morphism from Lemma 22.6.2 applied to $f_1 \circ a^{-1} = b^{-1} \circ f_2$. If we can show that $c \circ c'$ and $c' \circ c$ are isomorphisms in $K(\text{Mod}_{(A, \text{d})})$ then we win. Hence it suffices to prove the following: Given a morphism of triangles $(1, 1, c) : (K, L, C(f), f, i, p)$ in $K(\text{Mod}_{(A, \text{d})})$ the morphism $c$ is an isomorphism in $K(\text{Mod}_{(A, \text{d})})$. By assumption the two squares in the diagram

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

commute up to homotopy. By construction of $C(f)$ the rows form admissible short exact sequences. Thus we see that $(c - 1)^2 = 0$ in $K(\text{Mod}_{(A, \text{d})})$ by Lemma 22.7.6. Hence $c$ is an isomorphism in $K(\text{Mod}_{(A, \text{d})})$ with inverse $2 - c$. $\square$

The following lemma shows that the collection of triangles of the homotopy category given by cones and the distinguished triangles are the same up to isomorphisms, at least up to sign!

Lemma 22.9.4. Let $(A, \text{d})$ be a differential graded algebra.

1. Given an admissible short exact sequence $0 \to K \xrightarrow {\alpha } L \to M \to 0$ of differential graded $A$-modules there exists a homotopy equivalence $C(\alpha ) \to M$ such that the diagram

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

defines an isomorphism of triangles in $K(\text{Mod}_{(A, \text{d})})$.

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

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

where the upper triangle is the triangle associated to a admissible short exact sequence $K \to \tilde L \to M$.

Proof. Proof of (1). We have $C(\alpha ) = L \oplus K$ and we simply define $C(\alpha ) \to M$ via the projection onto $L$ followed by $\beta$. This defines a morphism of differential graded modules because the compositions $K^{n + 1} \to L^{n + 1} \to M^{n + 1}$ are zero. Choose splittings $s : M \to L$ and $\pi : L \to K$ with $\mathop{\mathrm{Ker}}(\pi ) = \mathop{\mathrm{Im}}(s)$ and set $\delta = \pi \circ \text{d}_ L \circ s$ as usual. To get a homotopy inverse we take $M \to C(\alpha )$ given by $(s , -\delta )$. This is compatible with differentials because $\delta ^ n$ can be characterized as the unique map $M^ n \to K^{n + 1}$ such that $\text{d} \circ s^ n - s^{n + 1} \circ \text{d} = \alpha \circ \delta ^ n$, see proof of Homology, Lemma 12.14.10. The composition $M \to C(f) \to M$ is the identity. The composition $C(f) \to M \to C(f)$ is equal to the morphism

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

To see that this is homotopic to the identity map use the homotopy $h : C(\alpha ) \to C(\alpha )$ given by the matrix

$\left( \begin{matrix} 0 & 0 \\ \pi & 0 \end{matrix} \right) : C(\alpha ) = L \oplus K \to L \oplus K = C(\alpha )$

It is trivial to verify that

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

To finish the proof of (1) we have to show that the morphisms $-p : C(\alpha ) \to K[1]$ (see Definition 22.6.1) and $C(\alpha ) \to M \to K[1]$ agree up to homotopy. This is clear from the above. Namely, we can use the homotopy inverse $(s, -\delta ) : M \to C(\alpha )$ and check instead that the two maps $M \to K[1]$ agree. And note that $p \circ (s, -\delta ) = -\delta$ as desired.

Proof of (2). We let $\tilde f : K \to \tilde L$, $s : L \to \tilde L$ and $\pi : L \to L$ be as in Lemma 22.7.4. By Lemmas 22.6.2 and 22.9.3 the triangles $(K, L, C(f), i, p)$ and $(K, \tilde L, C(\tilde f), \tilde i, \tilde p)$ are isomorphic. Note that we can compose isomorphisms of triangles. Thus we may replace $L$ by $\tilde L$ and $f$ by $\tilde f$. In other words we may assume that $f$ is an admissible monomorphism. In this case the result follows from part (1). $\square$

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