## 13.5 Localization of triangulated categories

In order to construct the derived category starting from the homotopy category of complexes, we will use a localization process.

Definition 13.5.1. Let $\mathcal{D}$ be a pre-triangulated category. We say a multiplicative system $S$ is compatible with the triangulated structure if the following two conditions hold:

1. For a morphism $f$ of $\mathcal{D}$ we have $f \in S \Leftrightarrow f \in S$1.

2. Given a solid commutative square

$\xymatrix{ X \ar[r] \ar[d]^ s & Y \ar[r] \ar[d]^{s'} & Z \ar[r] \ar@{-->}[d] & X \ar[d]^{s} \\ X' \ar[r] & Y' \ar[r] & Z' \ar[r] & X' }$

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.

It turns out that these axioms are not independent of the axioms defining multiplicative systems.

Lemma 13.5.2. Let $\mathcal{D}$ be a pre-triangulated category. Let $S \subset \text{Arrows}(\mathcal{D})$.

1. If $S$ contains all identities and MS6 holds (Definition 13.5.1), then every isomorphism of $\mathcal{D}$ is in $S$.

2. If MS1, MS5 (Categories, Definition 4.27.1) and MS6 hold, then MS2 holds.

Proof. Assume $S$ contains all identities and MS6 holds. Let $f : X \to Y$ be an isomorphism of $\mathcal{D}$. Consider the diagram

$\xymatrix{ 0 \ar[r] \ar[d]^1 & X \ar[r]_1 \ar[d]^1 & X \ar[r] \ar@{-->}[d] & 0 \ar[d]^{1} \\ 0 \ar[r] & X \ar[r]^ f & Y \ar[r] & 0 }$

The rows are distinguished triangles by Lemma 13.4.9. By MS6 we see that the dotted arrow exists and is in $S$, so $f$ is in $S$.

Assume MS1, MS5, MS6. 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 $(X', Y', Z, f', g', t \circ h)$ (here we use TR1 and TR2). By MS5, MS6 (and TR2 to rotate) we can find the dotted arrow in the commutative diagram

$\xymatrix{ X \ar[r] \ar[d]^ t & Y \ar[r] \ar@{..>}[d]^{s'} & Z \ar[r] \ar[d]^1 & X \ar[d]^{t} \\ X' \ar[r] & Y' \ar[r] & Z \ar[r] & X' }$

with moreover $s' \in S$. This proves LMS2. The proof of RMS2 is dual. $\square$

Remark 13.5.3. In the presence of MS1 and MS6, condition MS5 is equivalent to asking $s[n] \in S$ for all $s \in S$ and $n \in \mathbf{Z}$. For example, suppose MS5 holds, we have $s \in S$, and we want to show $s[-1] \in S$. This isn't immediate because $s[-1]$ is not equal to $s$, only isomorphic to $s$ as an arrow of $\mathcal{D}$. Still, this does imply that $s[-1] = f \circ s \circ g$ for isomorphisms $f$, $g$. By Lemma 13.5.2 (1) we find $f, g \in S$, hence $s[-1] \in S$ by MS1, hence $s[-1] \in S$ by MS5. We leave a complete proof to the reader as an exercise.

Lemma 13.5.4. 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 4.27.20) multiplicative system compatible with the triangulated structure on $\mathcal{D}$.

Proof. We have to prove axioms MS1 – MS6, see Categories, Definitions 4.27.1 and 4.27.20 and Definition 13.5.1. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and Lemma 13.4.3. By Lemma 13.5.2 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 13.4.2 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

$\xymatrix{ Z \ar[r]_ t & X \ar[r]_ d \ar[d]^1 & Q \ar[r] \ar[d]^ i & Z \\ & X \ar[r]_ a & Y \ar[d]^ j \\ & & W }$

OK, and now we apply the functor $F$ to this diagram. Since $t \in S$ we see that $F(Q) = 0$, see Lemma 13.4.9. 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. $\square$

Lemma 13.5.5. 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 4.27.20) multiplicative system compatible with the triangulated structure on $\mathcal{D}$.

Proof. We have to prove axioms MS1 – MS6, see Categories, Definitions 4.27.1 and 4.27.20 and Definition 13.5.1. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and the long exact cohomology sequence (13.3.5.1). By Lemma 13.5.2 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 13.4.2 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

$\xymatrix{ Z \ar[r]_ t & X \ar[r]_ g \ar[d]^1 & Q \ar[r] \ar[d]^ i & Z \\ & X \ar[r]_ a & Y \ar[d]^ j \\ & & W }$

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 (13.3.5.1). 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. $\square$

Proposition 13.5.6. 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 $ \circ Q = Q \circ $ and the localization functor $Q : \mathcal{D} \to S^{-1}\mathcal{D}$ is exact. Moreover, if $\mathcal{D}$ is a triangulated category, so is $S^{-1}\mathcal{D}$.

Proof. We have seen that $S^{-1}\mathcal{D}$ is an additive category and that the localization functor $Q$ is additive in Homology, Lemma 12.8.2. It follows from MS5 that there is a unique additive auto-equivalence $ : S^{-1}\mathcal{D} \to S^{-1}\mathcal{D}$ such that $Q \circ  =  \circ Q$ (equality of functors); we omit the details. 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 $Q$.

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 distinguished 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 $S^{-1}\mathcal{D}$.

Proof of TR2. This is immediate from the definitions.

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 4.27.10 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

$\xymatrix{ X \ar[r] \ar[d]^ k & Y \ar[r] \ar[d]^ l & Z \ar[r] \ar[d]^ m & X \ar[d]^{g} \\ X'' \ar[r] & Y'' \ar[r] & Z'' \ar[r] & X'' \\ X' \ar[r] \ar[u]_ s & Y' \ar[r] \ar[u]_ t & Z' \ar[r] \ar[u]_ r & X' \ar[u]_{s} }$

with $r \in S$. It follows that setting $c = Q(r)^{-1}Q(m)$ we obtain the desired morphism of triangles

$\xymatrix{ (Q(X), Q(Y), Q(Z), Q(f), Q(g), Q(h)) \ar[d]^{(a, b, c)} \\ (Q(X'), Q(Y'), Q(Z'), Q(f'), Q(g'), Q(h')) }$

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 13.4.15 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 $S^{-1}\mathcal{D}$ $\square$

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).

Lemma 13.5.7. Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a multiplicative system compatible with the triangulated structure. Let $Q : \mathcal{D} \to S^{-1}\mathcal{D}$ be the localization functor, see Proposition 13.5.6.

1. 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 4.27.8) is a homological functor too.

2. 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 4.27.8) is an exact functor too.

Proof. This lemma proves itself. Details omitted. $\square$

Lemma 13.5.8. Let $\mathcal{D}$ be a pre-triangulated category and let $\mathcal{D}' \subset \mathcal{D}$ be a full, pre-triangulated subcategory. Let $S$ be a saturated multiplicative system of $\mathcal{D}$ compatible with the triangulated structure. Assume that for each $X$ in $\mathcal{D}$ there exists an $s : X' \to X$ in $S$ such that $X'$ is an object of $\mathcal{D}'$. Then $S' = S \cap \text{Arrows}(\mathcal{D}')$ is a saturated multiplicative system compatible with the triangulated structure and the functor

$(S')^{-1}\mathcal{D}' \longrightarrow S^{-1}\mathcal{D}$

is an equivalence of pre-triangulated categories.

Proof. Consider the quotient functor $Q : \mathcal{D} \to S^{-1}\mathcal{D}$ of Proposition 13.5.6. Since $S$ is saturated we have that a morphism $f : X \to Y$ is in $S$ if and only if $Q(f)$ is invertible, see Categories, Lemma 4.27.21. Thus $S'$ is the collection of arrows which are turned into isomorphisms by the composition $\mathcal{D}' \to \mathcal{D} \to S^{-1}\mathcal{D}$. Hence $S'$ is is a saturated multiplicative system compatible with the triangulated structure by Lemma 13.5.4. By Lemma 13.5.7 we obtain the exact functor $(S')^{-1}\mathcal{D}' \to S^{-1}\mathcal{D}$ of pre-triangulated categories. By assumption this functor is essentially surjective. Let $X', Y'$ be objects of $\mathcal{D}'$. By Categories, Remark 4.27.15 we have

$\mathop{\mathrm{Mor}}\nolimits _{S^{-1}\mathcal{D}}(X', Y') = \mathop{\mathrm{colim}}\nolimits _{s : X \to X'\text{ in }S} \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X, Y')$

Our assumption implies that for any $s : X \to X'$ in $S$ we can find a morphism $s' : X'' \to X$ in $S$ with $X''$ in $\mathcal{D}'$. Then $s \circ s' : X'' \to X'$ is in $S'$. Hence the colimit above is equal to

$\mathop{\mathrm{colim}}\nolimits _{s'' : X'' \to X'\text{ in }S'} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(X'', Y') = \mathop{\mathrm{Mor}}\nolimits _{(S')^{-1}\mathcal{D}'}(X', Y')$

This proves our functor is also fully faithful and the proof is complete. $\square$

The following lemma describes the kernel (see Definition 13.6.5) of the localization functor.

Lemma 13.5.9. 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

1. $Q(Z) = 0$ in $S^{-1}\mathcal{D}$,

2. there exists $Z' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ such that $0 : Z \to Z'$ is an element of $S$,

3. there exists $Z' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ such that $0 : Z' \to Z$ is an element of $S$, and

4. there exists an object $Z'$ and a distinguished triangle $(X, Y, Z \oplus Z', f, g, h)$ such that $f \in S$.

If $S$ is saturated, then these are also equivalent to

1. the morphism $0 \to Z$ is an element of $S$,

2. the morphism $Z \to 0$ is an element of $S$,

3. there exists a distinguished triangle $(X, Y, Z, f, g, h)$ such that $f \in S$.

Proof. The equivalence of (1), (2), and (3) is Homology, Lemma 12.8.3. If (2) holds, then $(Z'[-1], Z'[-1] \oplus Z, Z, (1, 0), (0, 1), 0)$ is a distinguished triangle (see Lemma 13.4.11) 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.

Next, assume that $S$ is saturated. Note that each of (5), (6), (7) 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 4.27.21 the fact that $S$ is saturated implies that $0 \to Z$ is in $S$. Hence (1) $\Rightarrow$ (5). Dually (1) $\Rightarrow$ (6). 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). $\square$

Lemma 13.5.10. 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.

Proof. We strongly suggest the reader skip the proof of this lemma and instead work it out on a napkin.

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 $(Y, Z, X, g, h, -f)$. 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 $\mathcal{I} \to Z/S$.

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 $\mathcal{I} \to Z/S$.

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 4.27.21. 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 $\mathcal{I} \to Z/S$.

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 4.19.1.

We check condition (2) of Categories, Definition 4.19.1 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 4.27.7 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).

Next we check condition (3) of Categories, Definition 4.19.1. 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 $c = c'$.

Finally, we check that $\mathcal{I} \to X/S$ is cofinal, see Categories, Definition 4.17.1. 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 $X' \to X_2 \to X_3$ where equal (see displayed equation in Categories, Definition 4.17.1). 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 $X' \to X_2 \to X_3 \to X_4$ 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. $\square$

 See Remark 13.5.3.

Comment #5485 by MS on

There are two different items labeled (4) in 05R8.

Comment #8720 by Jiwan Jung on

There are two different $g$ are used on the proof of 05R5. (one is g:X\rightarrow Y for assumption and the other is g:X\rightarrow Q for the choice of dtr.)

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