Lemma 13.14.6. Assumptions and notation as in Situation 13.14.1. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle of $\mathcal{D}$. If $RF$ is defined at two out of three of $X, Y, Z$, then it is defined at the third. Moreover, in this case

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

is a distinguished triangle in $\mathcal{D}'$. Similarly for $LF$.

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

$\xymatrix{ X \ar[d]_ f \ar[r]_ s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }$

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

$\xymatrix{ A \ar[d]_{RF(f)} \ar[r]_-\alpha & F(X') \ar[r] \ar[d]^{F(f')} & A \ar[d]^{RF(f)} \\ B \ar[r]^-\beta & F(Y') \ar[r] & B }$

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

$(s, s', s'') : (X, Y, Z, f, g, h) \longrightarrow (X', Y', Z', f', g', h')$

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

$(\alpha , \beta , \gamma ) : (A, B, C, RF(f), b, c) \longrightarrow (F(X'), F(Y'), F(Z'), F(f'), F(g'), F(h'))$

By Lemma 13.14.4 it suffices to prove that $RF(Z')$ is defined and that the arrow $\gamma : C \to F(Z')$ induces an isomorphism $C \to RF(Z')$. Namely, then we will get an isomorphism

$(A, B, C, RF(f), b, c) \longrightarrow (RF(X'), RF(Y'), RF(Z'), RF(f'), RF(g'), RF(h'))$

of triangles and by TR1 we conclude that the target is a distinguished triangle. Consider the category $\mathcal{I}$ of Lemma 13.5.10 of triangles

$\mathcal{I} = \{ (t, t', t'') : (X', Y', Z', f', g', h') \to (X'', Y'', Z'', f'', g'', h'') \mid (t, t', t'') \in S\}$

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

$\xymatrix{ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(X'')) & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, A) \ar[l] \\ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(Y'')) \ar[u] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, B) \ar[u] \ar[l] \\ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(Z'')) \ar[u] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, C) \ar[u] \ar[l] \\ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(X''[1])) \ar[u] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, A[1]) \ar[u] \ar[l] \\ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(Y''[1])) \ar[u] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, B[1]) \ar[u] \ar[l] }$

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 map

$\mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(Z'')) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, C)$

is bijective. We conclude that $F(Z'')$ is essentially constant over $\mathcal{I}$ with value $C$ by part (4) of Categories, Lemma 4.22.9. $\square$

Comment #556 by Nuno on

This is something I missed when I first read this lemma, but since $lim_\mathcal{I}$ is not an exact functor, why is the first column in the first diagram with five lines exact?

Comment #557 by on

Good catch! Also the horizontal arrows in the second big diagram are pointing in the wrong direction! Argh! Anyway, I have two ideas for fixing what you point out.

First idea: probably the column is exact because it is the limit of a directed inverse system of 5-term exact sequences all of whose terms except the middle one are essentially constant. A result of this kind was proven in Lemma 12.31.7 (see also Lemma 15.86.3).

Second idea: apply the isomorphism $Hom(W, C) \to colim Hom(W, F(Z''))$ to $id : F(Z'') \to F(Z'')$ for any object of the category $\mathcal{I}$. This shows that there is a map $Z'' \to Z'''$ in $S$ and a morphism $\gamma_{Z''} : F(Z'') \to C$ such that $F(Z'') \to C \to F(Z') \to F(Z'') \to F(Z''')$ agrees with $F(Z'') \to F(Z''')$. Check that the maps $\gamma_{Z''}$ define an element in $\lim Hom(F(Z''), C)$. Then use this to show that $C$ is the colimit (actually I think this is clear at this point -- but I didn't write it out). Will code this up soon.

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