TITLE: Proof that the homotopy category of a stable $\infty$-category is triangulated
QUESTION [18 upvotes]: I've been looking at various general strategies for proving that some category is triangulated, and Lurie manages to prove that a huge class of interesting examples of categories that we know about are triangulated in his book Higher Algebra (formerly DAG I-IV and VI).(EDIT: here's a link to the book) The trouble is that I am very new to this language, and so what he calls $\infty$-categorical notions that are basic and easily motivated' I see as foreign and unfamiliar. 
The part I'm really interested in is the proof of the octahedral axiom on page 24 of Higher Algebra. He builds a diagram using a proposition from Higher Topos Theory that seems completely out of context (to me!). The proposition says:
``Suppose we are given a diagram of $\infty$-categories $\mathcal{C} \rightarrow \mathcal{D}' \leftarrow \mathcal{D}:p$, where $p$ is a categorical fibration. Let $\mathcal{C}^0$ be a full subcategory of $\mathcal{C}$. Let $\mathcal{K} \subset Map_{\mathcal{D}'}(\mathcal{C}, \mathcal{D})$ be the full subcategory spanned by those functors $F: \mathcal{C} \rightarrow \mathcal{D}$ which are $p$-left Kan extensions of $F\vert\mathcal{C}^0$. Let $\mathcal{K}'\subset \text{Map}_{\mathcal{D}'}(\mathcal{C}^0, \mathcal{D})$ be the full subcategory spanned by those functors $F_0: \mathcal{C}^0 \rightarrow \mathcal{D}$ with the property that, for each object $C \in \mathcal{C}$, the induced diagram $\mathcal{C}^0_{/C} \rightarrow \mathcal{D}$ has a $p$-colimit. Then the restriction functor $\mathcal{K} \rightarrow \mathcal{K}'$ is a trivial fibration of simplicial sets.''
And Lurie says that, in order to prove (TR4), we use this ``repeatedly to construct a map from the nerve of the appropriate partially ordered set into $\mathcal{C}$.'' (See Lurie's book available for download on his webpage.)
Now, obviously this must be some sort of standard use of the proposition, but I would very much like to understand this one proof without reading all of Higher Topos Theory, so we have my question:

Is there possibly a more easy-going reference for this proof?
or
If it doesn't require too much effort, would someone be willing to explain how the cited proposition applies in this instance?
or
Do I really just have to read Higher Topos Theory up through Chapter 4?

REPLY [5 votes]: This point of view is surely somewhat inelegant and rusty, but I think that once it has been polished it exploits only basic definitions: take it as a motivation more than as a proof if you want.
Allow me to start from the beginning recalling that the octahedral axiom (TR4 for short) says that given three distinguished triangles
\begin{gather*}
X\xrightarrow{f}Y\to Y/X\to X[1] \\
Y\xrightarrow{g}Z\to Z/Y\to Y[1]\\
X\xrightarrow{gf}Z\to Z/X\to X[1]
\end{gather*}
I can arrange them in the following "braid":

given this, there is a (non-unique) way to complete it with the arrows $s,t$ indicated. This is standard in any book about triangulated categories.
Now, in the $\infty$-categorical setting "distinguished triangles" are replaced by "fiber sequences" like
$$
\begin{array}{ccc}
A &\to& B &\to& 0 \\
\downarrow && \downarrow && \downarrow\\
0 &\to& C &\to& A[1]
\end{array}
$$
Thus, once translated the braid diagram in a stable $\infty$-category $\bf C$ we are in the following situation:

(originally there were colours to help identify different fiber sequences: unfortunately codecogs does not support it!). Axiom TR4 says that we can find arrows $Y/X\to Z/X\to Z/Y$  such that the triangle $Y/X\to Z/X\to Z/Y\to (Y/X)[1]$ is distinguished. This means, at the end of the day, that not only the composite arrow $Z/Y\to Y[1]\to (Y/X)[1]$ can be embedded in a triangle, but also that this embedding can be done "in a coherent way". 
Now for the proof. 
The completion axiom (which I suppose to hold) implies that the diagram

can be completed with an arrow $Y/X \xrightarrow{\phi} Z/X$ completing the square

Now consider the ojects $V={\rm hocolim}\Big( Y/X \leftarrow{} Y \xrightarrow{g} Z \Big)$ and $W={\rm hocolim}\Big( 0\xleftarrow{} Y/X \xrightarrow{\phi}Z/X \Big)$; arrange them in the diagram

2-out-of-3 now implies that the outer rectangle is a pushout, hence $W\cong Z/Y$. It remains to prove that $V\cong Z/X$; this follows from the 2-out-of-3 property applied to the last diagram: