Lemma 13.4.1. 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 $f[1] \circ h = 0$.

## 13.4 Elementary results on triangulated categories

Most of the results in this section are proved for pre-triangulated categories and a fortiori hold in any triangulated category.

**Proof.**
By TR1 we know $(X, X, 0, 1, 0, 0)$ is a distinguished triangle. Apply TR3 to

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. $\square$

Lemma 13.4.2. Let $\mathcal{D}$ be a pre-triangulated category. For any object $W$ of $\mathcal{D}$ the functor $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, -)$ is homological, and the functor $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, W)$ is cohomological.

**Proof.**
Consider a distinguished triangle $(X, Y, Z, f, g, h)$. We have already seen that $g \circ f = 0$, see Lemma 13.4.1. Suppose $a : W \to Y$ is a morphism such that $g \circ a = 0$. Then we get a commutative diagram

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. $\square$

Lemma 13.4.3. Let $\mathcal{D}$ be a pre-triangulated category. Let

be a morphism of distinguished triangles. If two among $a, b, c$ are isomorphisms so is the third.

**Proof.**
Assume that $a$ and $c$ are isomorphisms. For any object $W$ of $\mathcal{D}$ write $H_ W( - ) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, -)$. Then we get a commutative diagram of abelian groups

By assumption the right two and left two vertical arrows are bijective. As $H_ W$ is homological by Lemma 13.4.2 and the five lemma (Homology, Lemma 12.5.20) it follows that the middle vertical arrow is an isomorphism. Hence by Yoneda's lemma, see Categories, Lemma 4.3.5 we see that $b$ is an isomorphism. This implies the other cases by rotating (using TR2). $\square$

Remark 13.4.4. Let $\mathcal{D}$ be an additive category with translation functors $[n]$ as in Definition 13.3.1. Let us call a triangle $(X, Y, Z, f, g, h)$ *special*^{1} if for every object $W$ of $\mathcal{D}$ the long sequence of abelian groups

is exact. The proof of Lemma 13.4.3 shows that if

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 *co-special* triangles, i.e., triangles which turn into long exact sequences on applying the functor $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, W)$. Thus distinguished triangles are special and co-special, but in general there are many more (co-)special triangles, than there are distinguished triangles.

Lemma 13.4.5. Let $\mathcal{D}$ be a pre-triangulated category. Let

be endomorphisms of a distinguished triangle. Then $bb' = 0$.

**Proof.**
Picture

Applying Lemma 13.4.2 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 13.4.1. $\square$

Lemma 13.4.6. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. If

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 $(X, Y, Z, f, g, h)$.

**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 13.4.5 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 $b^2 - b = (4(b')^2 - 4b' - 3)((b')^2 - b')^2 = 0$.
$\square$

Lemma 13.4.7. 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

**Proof.**
Existence by TR1. Uniqueness up to isomorphism by TR3 and Lemma 13.4.3.
$\square$

Lemma 13.4.8. Let $\mathcal{D}$ be a pre-triangulated category. Let

be a morphism of distinguished triangles. If one of the following conditions holds

$\mathop{\mathrm{Hom}}\nolimits (Y, X') = 0$,

$\mathop{\mathrm{Hom}}\nolimits (Z, Y') = 0$,

$\mathop{\mathrm{Hom}}\nolimits (X, X') = \mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$,

$\mathop{\mathrm{Hom}}\nolimits (Z, X') = \mathop{\mathrm{Hom}}\nolimits (Z, Z') = 0$, or

$\mathop{\mathrm{Hom}}\nolimits (X[1], Z') = \mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$

then $b$ is the unique morphism from $Y \to Y'$ such that $(a, b, c)$ is a morphism of triangles.

**Proof.**
If we have a second morphism of triangles $(a, b', c)$ then $(0, b - b', 0)$ is a morphism of triangles. Hence we have to show: the only morphism $b : Y \to Y'$ such that $X \to Y \to Y'$ and $Y \to Y' \to Z'$ are zero is $0$. We will use Lemma 13.4.2 without further mention. In particular, condition (3) implies (1). Given condition (1) if the composition $g' \circ b : Y \to Y' \to Z'$ is zero, then $b$ lifts to a morphism $Y \to X'$ which has to be zero. This proves (1).

The proof of (2) and (4) are dual to this argument.

Assume (5). Consider the diagram

We may choose $\epsilon $ such that $b = \epsilon \circ g$. Then $g' \circ \epsilon \circ g = 0$ which implies that $g' \circ \epsilon = \delta \circ h$ for some $\delta \in \mathop{\mathrm{Hom}}\nolimits (X[1], Z')$. Since $\mathop{\mathrm{Hom}}\nolimits (X[1], Z') = 0$ we conclude that $g' \circ \epsilon = 0$. Hence $\epsilon = f' \circ \gamma $ for some $\gamma \in \mathop{\mathrm{Hom}}\nolimits (Z, X')$. Since $\mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$ we conclude that $\epsilon = 0$ and hence $b = 0$ as desired. $\square$

Lemma 13.4.9. Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. The following are equivalent

$f$ is an isomorphism,

$(X, Y, 0, f, 0, 0)$ is a distinguished triangle, and

for any distinguished triangle $(X, Y, Z, f, g, h)$ we have $Z = 0$.

**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 13.4.3 and $Z = 0$. Conversely, if $Z = 0$, then $(1, f, 0)$ is an isomorphism of triangles as well, hence $f$ is an isomorphism.
$\square$

Lemma 13.4.10. 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

$(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')$ is a distinguished triangle,

both $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ are distinguished triangles.

**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

In the terminology of Remark 13.4.4 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.

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 13.4.4) 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

By Remark 13.4.4 we find that $c$ is an isomorphism and we conclude that (2) holds. $\square$

Lemma 13.4.11. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle.

If $h = 0$, then there exists a right inverse $s : Z \to Y$ to $g$.

For any right inverse $s : Z \to Y$ of $g$ the map $f \oplus s : X \oplus Z \to Y$ is an isomorphism.

For any objects $X', Z'$ of $\mathcal{D}$ the triangle $(X', X' \oplus Z', Z', (1, 0), (0, 1), 0)$ is distinguished.

**Proof.**
To see (1) use that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Z, Y) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Z, Z) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Z, X[1])$ is exact by Lemma 13.4.2. By the same token, if $s$ is as in (2), then $h = 0$ and the sequence

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 13.4.10. $\square$

Lemma 13.4.12. Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. The following are equivalent

$f$ has a kernel,

$f$ has a cokernel,

$f$ is the isomorphic to a composition $K \oplus Z \to Z \to Z \oplus Q$ of a projection and coprojection for some objects $K, Z, Q$ of $\mathcal{D}$.

**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 13.4.1 the composition $f \circ h[-1] = 0$. As $f$ is a monomorphism we see that $h[-1] = 0$ and hence $h = 0$. Then Lemma 13.4.11 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.
$\square$

Lemma 13.4.13. Let $\mathcal{D}$ be a pre-triangulated category. Let $I$ be a set.

Let $X_ i$, $i \in I$ be a family of objects of $\mathcal{D}$.

If $\prod X_ i$ exists, then $(\prod X_ i)[1] = \prod X_ i[1]$.

If $\bigoplus X_ i$ exists, then $(\bigoplus X_ i)[1] = \bigoplus X_ i[1]$.

Let $X_ i \to Y_ i \to Z_ i \to X_ i[1]$ be a family of distinguished triangles of $\mathcal{D}$.

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.

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.

**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 13.4.4 because products of exact sequences of abelian groups are exact. Hence Remark 13.4.4 shows that the morphism of triangles is an isomorphism and we conclude by TR1. The proof of (2)(b) is dual.
$\square$

Lemma 13.4.14. 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.

**Proof.**
Assume $\mathcal{D}$ has countable products. By Homology, Lemma 12.4.3 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 13.4.12. The second statement is dual to the first.
$\square$

The following lemma makes it slightly easier to prove that a pre-triangulated category is triangulated.

Lemma 13.4.15. 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

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

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.

**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 13.4.7.
$\square$

Lemma 13.4.16. Let $\mathcal{D}$ be a pre-triangulated category. Assume that $\mathcal{D}'$ is an additive full subcategory of $\mathcal{D}$. The following are equivalent

there exists a set of triangles $\mathcal{T}'$ such that $(\mathcal{D}', \mathcal{T}')$ is a pre-triangulated subcategory of $\mathcal{D}$,

$\mathcal{D}'$ is preserved under $[1]$ and $[1] : \mathcal{D}' \to \mathcal{D}'$ is an auto-equivalence 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 $\mathcal{D}'$.

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 $X, Y, Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$. Finally, if $\mathcal{D}$ is a triangulated category, then (1) and (2) are also equivalent to

$\mathcal{D}'$ is a triangulated subcategory.

**Proof.**
Omitted.
$\square$

Lemma 13.4.17. An exact functor of pre-triangulated categories is additive.

**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

is distinguished in $\mathcal{D}'$. This implies that $1_{F(0)} \circ 1_{F(0)}$ is zero, see Lemma 13.4.1. 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 by Lemma 13.4.11 part (3), 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 $F(X) \oplus F(Y) \to F(X \oplus Y)$ is an isomorphism by Lemma 13.4.11 part (2). To finish we apply Homology, Lemma 12.7.1. $\square$

Lemma 13.4.18. 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 $\mathcal{D}'$.

**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 13.4.7 there exists an isomorphism of triangles

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. $\square$

Lemma 13.4.19. 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.

**Proof.**
Omitted.
$\square$

Lemma 13.4.20. 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.

Let $\mathcal{D}'$ be a pre-triangulated category. Let $F : \mathcal{D}' \to \mathcal{D}$ be an exact functor. Then the composition $H \circ F$ is a homological functor as well.

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.

**Proof.**
Omitted.
$\square$

Lemma 13.4.21. 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 $\mathcal{D}'$ 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.

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.

**Proof.**
Omitted.
$\square$

Lemma 13.4.22. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. 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

is a $\delta $-functor from $\mathcal{A} \to \mathcal{B}$, see Homology, Definition 12.12.1.

**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

is a morphism in $\mathcal{D}$ such that $(G(A), G(B), G(C), a, b, \delta )$ is a distinguished triangle, see Definition 13.3.6. 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 (13.3.5.1) combined with the vanishing of $H^{-1}(G(C))$ gives a long exact sequence

in $\mathcal{B}$ as desired. $\square$

The proof of the following result uses TR4.

Proposition 13.4.23. Let $\mathcal{D}$ be a triangulated category. Any commutative diagram

can be extended to a diagram

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 $[1]$.

**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

$a : Z \to A$ and $b : A \to Y''$, and

$a' : X'' \to A$ and $b' : A \to Z'$

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

$(Z, A, Y'', a, b, c)$, $(X'', A, Z', a', b', c')$ are distinguished triangles,

$(X, Y, Z) \to (X, Y', A)$, $(X, Y', A) \to (Y, Y', Y'')$, $(X, X', X'') \to (X, Y', A)$, $(X, Y', A) \to (X', Y', Z')$ are morphisms of triangles.

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

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 $b \circ a' : X'' \to Y''$.

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.

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. $\square$

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