The Stacks project

Proof. Say $RF$ is defined at $X, Y$ with values $A, B$. Let $RF(f) : A \to B$ be the induced morphism, see Lemma 13.14.3. We may choose a distinguished triangle $(A, B, C, RF(f), b, c)$ in $\mathcal{D}'$. We claim that $C$ is a value of $RF$ at $Z$.

To see this pick $s : X \to X'$ in $S$ such that there exists a morphism $\alpha : A \to F(X')$ as in Categories, Definition 4.22.1. We may choose a commutative diagram

with $s' \in S$ by MS2. Using that $Y/S$ is filtered we can (after replacing $s'$ by some $s'' : Y \to Y''$ in $S$) assume that there exists a morphism $\beta : B \to F(Y')$ as in Categories, Definition 4.22.1. Picture

It may not be true that the left square commutes, but the outer and right squares commute. The assumption that the ind-object $\{ F(Y')\} _{s' : Y' \to Y}$ is essentially constant means that there exists a $s'' : Y \to Y''$ in $S$ and a morphism $h : Y' \to Y''$ such that $s'' = h \circ s'$ and such that $F(h)$ equal to $F(Y') \to B \to F(Y') \to F(Y'')$. Hence after replacing $Y'$ by $Y''$ and $\beta $ by $F(h) \circ \beta $ the diagram will commute (by direct computation with arrows).

Using MS6 choose a morphism of triangles

with $s'' \in S$. By TR3 choose a morphism of triangles

By Lemma 13.14.4 it suffices to prove that $RF(Z')$ is defined and has value $C$. Consider the category $\mathcal{I}$ of Lemma 13.5.10 of triangles

To show that the system $F(Z'')$ is essentially constant over the category $Z'/S$ is equivalent to showing that the system of $F(Z'')$ is essentially constant over $\mathcal{I}$ because $\mathcal{I} \to Z'/S$ is cofinal, see Categories, Lemma 4.22.11 (cofinality is proven in Lemma 13.5.10). For any object $W$ in $\mathcal{D}'$ we consider the diagram

where the horizontal arrows are given by composing with $(\alpha , \beta , \gamma )$. Since filtered colimits are exact (Algebra, Lemma 10.8.8) the left column is an exact sequence. Thus the $5$ lemma (Homology, Lemma 12.5.20) tells us the

is bijective. Choose an object $(t, t', t'') : (X', Y', Z') \to (X'', Y'', Z'')$ of $\mathcal{I}$. Applying what we just showed to $W = F(Z'')$ and the element $\text{id}_{F(X'')}$ of the colimit we find a unique morphism $c_{(X'', Y'', Z'')} : F(Z'') \to C$ such that for some $(X'', Y'', Z'') \to (X''', Y''', Z'')$ in $\mathcal{I}$

The family of morphisms $c_{(X'', Y'', Z'')}$ form an element $c$ of $\mathop{\mathrm{lim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(F(Z''), C)$ by uniqueness (computation omitted). Finally, we show that $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} F(Z'') = C$ via the morphisms $c_{(X'', Y'', Z'')}$ which will finish the proof by Categories, Lemma 4.22.9. Namely, let $W$ be an object of $\mathcal{D}'$ and let $d_{(X'', Y'', Z'')} : F(Z'') \to W$ be a family of maps corresponding to an element of $\mathop{\mathrm{lim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(F(Z''), W)$. If $d_{(X', Y', Z')} \circ \gamma = 0$, then for every object $(X'', Y'', Z'')$ of $\mathcal{I}$ the morphism $d_{(X'', Y'', Z'')}$ is zero by the existence of $c_{(X'', Y'', Z'')}$ and the morphism $(X'', Y'', Z'') \to (X''', Y''', Z'')$ in $\mathcal{I}$ satisfying the displayed equality above. Hence the map

(coming from precomposing by $\gamma $) is injective. However, it is also surjective because the element $c$ gives a left inverse. We conclude that $C$ is the colimit by Categories, Remark 4.14.4. $\square$