The Stacks project

Let $s\in\{+,-,b\}$. I think that when proving $K^s(\mathcal{A})$ is a triangulated category from (a) showing that it is a triangulated subcategory of $K(\mathcal{A})$ we obtain less information than from (b) adapting the proof of Proposition 13.10.3. Namely, with (a) we learn that the triangulated structure of $K^s(\mathcal{A})$ is the distinguished triangles in $K(\mathcal{A})$ whose objects belong to $K^s(\mathcal{A})$, whereas with (b) we define the triangulated structure on $K^s(\mathcal{A})$ to be the triangle isomorphism class in $K^s(\mathcal{A})$ spawned by the triangles $(A,B,C(\alpha),\alpha,i,-p)$, with $\alpha\in\operatorname{Mor}(K^s(\mathcal{A}))$. With (a), all we can say is that a triangle $(X,Y,Z,f,g,h)$ in $K^s(\mathcal{A})$ is distinguished iff it is isomorphic to a triangle $(A,B,C(\alpha),\alpha,i,-p)$ with $\alpha$ in $K(\mathcal{A})$, but not necessarily in $K^s(\mathcal{A})$. From here, I don't see how one can conclude that $\alpha$ may be picked in $K^s(\mathcal{A})$. Note that $K^s(\mathcal{A})$ is not strictly full in $K(\mathcal{A})$, since the map from the zero chain complex to $\cdots \xrightarrow{1_A} A\xrightarrow{0}A\xrightarrow{1_A}A\xrightarrow{0}\cdots$ is a chain homotopy equivalence.

Going to leave as is.