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$\def\C{\mathcal{C}} \def\A{\mathcal{A}} \def\S{\mathcal{S}} \def\ind{\operatorname{Ind}} \def\D{\mathcal{D}} \def\t{\operatorname{t}} \def\op{\mathrm{op}} \def\indlim#1{\underset{\underset{#1}{\longrightarrow}}{“\operatorname{lim}”\,}} \def\s{\mathrm{s}} \def\t{\mathrm{t}}$It is true in full generality that derived triangulated functors are triangulated. This is [SGA4, Exp. XVII, Proposition 1.2.2.ii] and [SP, Tag 05SC]. Here is a proof idea not found in neither of these references, which leverages the content of [KS, Sect. 7.4].

[SGA4]. M. Artin, A. Grothendieck and J-L. Verdier, Sém. Géom. Algébrique (1963-64), Théorie des topos et cohomologie étale des schémas

$\def\C{\mathcal{C}} \def\A{\mathcal{A}} \def\S{\mathcal{S}} \def\ind{\operatorname{Ind}} \def\D{\mathcal{D}} \def\t{\operatorname{t}} \def\op{\mathrm{op}} \def\indlim#1{\underset{\underset{#1}{\longrightarrow}}{“\operatorname{lim}”\,}} \def\s{\mathrm{s}} \def\t{\mathrm{t}}$It is true in full generality that derived triangulated functors are triangulated. This is [D, Proposition 1.2.2.ii] and [SP, Tag 05SC]. Here is a proof idea not found in neither of these references, which leverages the content of [KS, Sect. 7.4].

[D]. P. Deligne. “Cohomologie à supports propres”. In: Théorie de Topos et Cohomologie Étale des Schémas (SGA4). Vol. III, Exp. XVII. Lecture Notes in Math. 305. Springer, 1971, pp. 250-461

If there is no risk of confusion, we might just say “triangulated functor” for a proto-triangulated functor. Here's the key insight: if $\C$ is a triangulated category, one endows $\ind(\C)$ with the proto-triangulated structure of those triangles that are a filtered colimit of distinguished triangles in $\C$. With this proto-triangulated structure, it turns out that $\alpha_\S$ and $IF$ are triangulated functors and that a proto-triangle in the essential image of $\iota_\A$ is a distinguished triangle in $\A$. With these facts, one can easily prove that right derived triangulated functors are triangulated. I wrote all the details here.

If there is no risk of confusion, we might just say “triangulated functor” for a proto-triangulated functor. Here's the key insight: if $\C$ is a triangulated category, one endows $\ind(\C)$ with the proto-triangulated structure of those triangles that are a filtered colimit of distinguished triangles in $\C$. With this proto-triangulated structure, it turns out that $\alpha_\S$ and $IF$ are triangulated functors and that a proto-triangle in the essential image of $\iota_\A$ is a distinguished triangle in $\A$. With these facts, one can easily prove that right derived triangulated functors are triangulated. I wrote all the details in Derived Triangulated Functors are Triangulated.

$\def\C{\mathcal{C}} \def\A{\mathcal{A}} \def\S{\mathcal{S}} \def\ind{\operatorname{Ind}} \def\D{\mathcal{D}} \def\t{\operatorname{t}} \def\op{\mathrm{op}} \def\indlim#1{\underset{\underset{#1}{\longrightarrow}}{“\operatorname{lim}”\,}} \def\s{\mathrm{s}} \def\t{\mathrm{t}}$It is true in full generality that derived triangulated functors are triangulated. This is [SGA4, Exp. XVII, Proposition 1.2.2.ii] and [SP, Tag 05SC]. Here is a proof idea not found in neither of these references.

Let $\C$, $\A$ be categories and let $F:\C\to \A$ be a functor. Let $\S$ be a right multiplicative system and write $Q:\C\to\C_\S$ for the localization functor of $\C$ with respect to $\S$. Write $\iota_\C:\C\to\ind(\C)$ for the ind-completion of $\C$, and let $\alpha_\S:\C_\S\to\ind(\C)$ be the left Kan extension of $\iota_\C$ along $Q$. Consider the diagram: $$ \require{AMScd} \begin{CD} \C@>F>>\A@=\A\\ @VQVV@.@VV{\iota_\A}V\\ \C_\S@>>{\alpha_\S}>\ind(\C)@>>IF>\ind(\A) \end{CD} $$ Then $$ \tag{[KS, eq. (7.4.3)]} R_\S(\iota_\A\circ F)\cong IF\circ\alpha_\S. $$ (A proof of $(7.4.3)$ may be read in [G, p. 162].) If we knew that:

then we would be done, for $R_\S(\iota_\A\circ F)$ would be the composite of triangulated functors. The problem is that point 1 is not true in general [ref] (so point 2 has no meaning). There is a workaround:

$\def\C{\mathcal{C}} \def\A{\mathcal{A}} \def\S{\mathcal{S}} \def\ind{\operatorname{Ind}} \def\D{\mathcal{D}} \def\t{\operatorname{t}} \def\op{\mathrm{op}} \def\indlim#1{\underset{\underset{#1}{\longrightarrow}}{“\operatorname{lim}”\,}} \def\s{\mathrm{s}} \def\t{\mathrm{t}}$It is true in full generality that derived triangulated functors are triangulated. This is [SGA4, Exp. XVII, Proposition 1.2.2.ii] and [SP, Tag 05SC]. Here is a proof idea not found in neither of these references, which leverages the content of [KS, Sect. 7.4].

Let $\C$ be a category. The “free filtered cocompletion” of $\C$ (also known as ind-completion) is a category $\ind(\C)$ that has all filtered colimits and that comes equipped with a fully faithful embedding $\iota_\C:\C\to\ind(\C)$ satisfying Definition 2 from here. Thus, if $F:\C\to \A$ is a functor, there is a unique-up-to-isomorphism functor $IF:\ind(\C)\to\ind(\A)$ preserving filtered colimits and such that $\iota_\A\circ F\cong IF\circ\iota_\C$ [KS, Proposition 6.1.9]. Let $\S$ be a left multiplicative system in $\C$ [SP, Tag 04VC] and write $Q:\C\to\C_\S$ for the localization functor of $\C$ with respect to $\S$. Let $\alpha_\S:\C_\S\to\ind(\C)$ be the left Kan extension of $\iota_\C$ along $Q$. Consider the diagram: $$ \require{AMScd} \begin{CD} \C@>F>>\A@=\A\\ @VQVV@.@VV{\iota_\A}V\\ \C_\S@>>{\alpha_\S}>\ind(\C)@>>IF>\ind(\A) \end{CD} $$ Then $$ \tag{[KS, eq. (7.4.3)]}\label{equality} R_\S(\iota_\A\circ F)\cong IF\circ\alpha_\S, $$ where $R_\S$ means “taking left Kan extension along $Q$” (it is what Kashiwara and Schapira refer to as the right localization of a functor [KS, Definition 7.3.1]). A proof of \ref{equality} may be read in [G, p. 162].

[KS, Definition 7.4.2]. The functor $F$ is right localizable at $X\in\C$ if the image of $Q(X)$ through \ref{equality} lies in the essential image of $\iota_\A$.

It turns out that $F$ is right localizable at all objects $X\in\C$ if and only if the left Kan extension of $F$ along $Q$ exists and is absolute [KS, Proposition 7.4.4] (in Kashiwara and Schapira notation, the latter is “the right localization $R_\S F$ exists and is universal”)

With these ideas on mind, suppose we knew:

Then we would immediately conclude that universally right localized triangulated functors are triangulated, for $R_\S(\iota_\A\circ F)$ would be the composite of triangulated functors (i.e., if $F$ is a triangulated functor of triangulated categories and $R_\S F$ exists and is universal, then $R_\S F$ is triangulated too). The problem is that point 1 is not true in general [ref] (so point 2 has no meaning). There is a workaround: