TR2 for homotopy category of stable $\infty$-category

Question

I’m trying to understand Lurie’s proof that the homotopy category of a stable $\infty$-category is triangulated. In showing TR2, he constructs a diagram $$\require{AMScd} \begin{CD} X @>f>> Y @>>> 0\\ @VVV @VVV @VVV \\ 0’ @>>> Z @>>> W \\ @. @VVV @VuVV \\ @. 0’’ @>>> V \end{CD}$$ in which every square is a pushout in some stable $\infty$-category $\mathcal C$. He then asserts a map between the suspensions $$\require{AMScd} \begin{CD} X @>>> 0\\ @VVV @VVV \\ 0’ @>>> W \end{CD}$$ $$\require{AMScd} \begin{CD} Y @>>> 0\\ @VVV @VVV \\ 0’’ @>>> V \end{CD}$$ giving rise to commutative square $$\require{AMScd} \begin{CD} W @>>> X[1]\\ @VuVV @Vf[1]VV \\ V @>>> Y[1] \end{CD}$$ in $h\mathcal C$. I think in fact what is needed for this commutative square is that the above map between suspensions specifically have components $f$ and $u$ between the initial vertices and terminal vertices, respectively. By repeated application of HTT.4.3.2.15, the first (large) diagram is determined up to contractible choices by $f$, and, if $\mathcal C$ is stable, also determined up to contractible choices by $u$. Similarly, the map of suspensions is determined by the map $X\to Y$ or by the map $X[1]\simeq W\to V\simeq Y[1]$ (as $\Sigma$ is an equivalence). So the data of the large diagram is equivalent to the data of a map between the suspensions, and each is specified two different ways (via $f$/the map on initial vertices or $u$/the map on cocone points). I want to know why these are in correspondence. Another way of saying this is, the map $f$ determines, via the large diagram, a map $u$, which determines a map $X[1]\to Y[1]$ (always up to contractible choices), but why is this map homotopic to $f[1]$? This is equivalent to the assertion that the construction of the large diagram computes the suspension functor $\Sigma$ on edges of $\mathcal C$. One way to see this would be to construct a map between suspensions (a diagram $\Delta^1\times\Delta^1\times\Delta^1\to\mathcal C$) directly from the data of the large diagram, but I don’t see how to fill in the most obvious candidate. (I could do so if I knew how to fill an outer horn $\Delta^3_0\hookrightarrow\mathcal C$ with the property that both interior vertices are zero objects, so every edge is a zero map.) Any tips would be a big help – thank you!

Edit 5/20 I can see the claim if it is true that given a cofiber sequence $X\to Y\to Z$ corresponding to a pushout $$\require{AMScd} \begin{CD} X @>>> Y\\ @VVV @VVV \\ 0 @>>> Z, \end{CD}$$ then if I replace the bottom 2-simplex ($X\to 0\to Z$) of the diagram $\Delta^1\times\Delta^1\to\mathcal C$ with any other one with the same vertices and long edge, the resulting diagram $\Delta^1\times\Delta^1\to\mathcal C$ is still a pushout.

Answer 1

"This is equivalent to the assertion that the construction of the large diagram computes the suspension functor $\Sigma$"

My previous answer was based on me misreading this quote :)

You want to show that this large diagram computes $\Sigma$. Lurie says some words about why that is : he says that the large diagram induces a morphism from one square to the next, namely from the square

\begin{CD}X @>>> 0 \\ @VVV @VVV \\ 0 @>>> W\end{CD} to the square \begin{CD}Y @>>> 0 \\ @VVV @VVV \\ 0 @>>> V\end{CD}

Now the first square (the second too, but I need it for the first) is left Kan extended from its restriction to the span $0\leftarrow X \to 0$ (that is what it means to be a pushout), so a map between these two squares is entirely determined by a map between the corresponding spans.

But now the second span (the first one too, but I need it for the second) is right Kan extended from the single vertex $Y$, so a map between the spans is the same thing as a map between these vertices. In this case, by design, the map between the vertices is $f$.

What I'm saying is : the map between the squares that you get from the large diagram, which one $W\to V$ is precisely your $u$ , is the only map (up to a contractible space) of squares which restricts to $f : X\to Y$. But the map of squares which induces $\Sigma f : \Sigma X\to \Sigma Y$ also restricts to $f$, by definition.

So the two maps of squares must be the same, and in particular the two maps $u$ and $\Sigma f: \Sigma X\to \Sigma Y$ must be the same.

So this reduces to showing that the two squares

(\begin{CD}X @>>> 0 \\ @VVV @VVV \\ 0 @>>> W\end{CD} and the other one with $Y$)

are indeed pushout squares. But for both it follows from pasting of pushout squares.

EDIT : Regarding the comments. If you agree that the indexing category for the big diagram is a poset, then a map of squares in this diagram (i.e. a map $(\Delta^1)^3\to K$ where $K$ is the indexing category) is entirely determined by where it sends the vertices, together with the property that for each arrow in $(\Delta^1)^3$, the corresponding vertices in $K$ have an arrow between them.

So here, to get an arrow between the two squares in question from the big diagram, you need an arrow from $X$ to $Y$ (ok you have one, it's $f$), an arrow from $0$ to itself (ok, it's $id$), an arrow from $0'$ to $0''$ (ok, just go $0' \to Z\to 0''$), and finally an arrow from $W$ to $V$ (ok, it's $u$). Crucially, these arrows are in $K$, so you get a map $(\Delta^1)^3\to K$ (again, here I'm really using that $K$ is a poset) and thus, because your big diagram was originally something like $K\to \mathcal C$, you can precompose and get $(\Delta^1)^3\to \mathcal C$. Because of how you chose the vertices and arrows, this map of squares has, on the top left vertex $f$ and on the bottom right vertex $u$, therefore $u$ is identified with $\Sigma f$.

(note : I said that I used that $K$ was a poset, but in fact if $K$ were only a $1$-category, all hope would not have been lost, as we would simply have had to further check that some diagrams commute - in a poset, any diagram commutes so we can skip this step).

(another way to phrase this is that (nerves of) posets are $1$-coskeletal )

EDIT 2 : I was a bit quick when I said $K$ is a poset. I meant that the $\infty$-category presented by $K$ was a poset, i.e. that $K$ was categorically equivalent to a poset. You can see this by computing $\mathfrak C[K]$, which you can do by explicitly writing $K= (\Delta^1)^2\coprod_{\Delta^1}(\Delta^1)^2\coprod_{\Delta^1}(\Delta^1)^2$ (which you can easily simplify, up to categorical equivalence to $(\Delta^1\times \Delta^2)\coprod_{\Delta^{\{1\}}\times\Delta^{\{1,2\}}} (\Delta^1 \times \Delta^1)$, and then you have to do a bit of work)

Answer 2

I’d like to give some more details to complement Maxime’s answer. $$\require{AMScd} \begin{CD} 1 @>>> 2 @>>> 3 \\ @VVV @VVV @VVV \\ 4 @>>> 5 @>>> 6 \\ @. @VVV @VVV \\ @. 7 @>>> 8 \end{CD}$$ This diagram defines a poset $P$ as well as a simplicial set $K= (\Delta^1)^2\coprod_{\Delta^1} (\Delta^1)^2\coprod_{\Delta^1} (\Delta^1)^2$. I claim that the map $K\hookrightarrow N(P)$ is inner anodyne. This will imply Maxime’s claim that this map is a categorical equivalence. It resolves our problem, since then the map $\mathcal C^{N(P)}\to\mathcal C^K$ will be a trivial fibration, so choosing a section $s$, the map of posets $[1]\times[1]\times[1]\to P$ indicated in Maxime’s answer gives a functor $\mathcal C^{N(P)}\to\mathcal C^{\Delta^1\times\Delta^1\times\Delta^1}$ which we can precompose with $s$ to obtain a functor $\mathcal C^K\to\mathcal C^{N(P)}\to\mathcal C^{\Delta^1\times\Delta^1\times\Delta^1}$ which, when applied to the large diagram in the original question, will produce the desired map of squares.

Now to prove the claim. As Maxime suggests, the map $K\hookrightarrow (\Delta^1\times\Delta^2)\coprod_{\Delta^{\{1\}}\times\Delta^{\{1,2\}}}(\Delta^1)^2$ is inner anodyne. Next, there is an inner anodyne map $(\Delta^1\times\Delta^2)\coprod_{\Delta^{\{1\}}\times\Delta^{\{1,2\}}}(\Delta^1)^2\hookrightarrow (\Delta^1\times\Delta^2)\coprod_{(\Delta^1)^2}(\Delta^1\times\Delta^2)$, where the copy of $(\Delta^1)^2$ appears as $\Delta^1\times\Delta^{\{1,2\}}$ of the first copy of $\Delta^1\times\Delta^2$ and as $\Delta^1\times\Delta^{\{0,1\}}$ of the second. (Note, this numbering is with respect to the usual numbering of simplices, and doesn’t refer to the numbers in the diagram above.) Now we can fill the following succession of inner horns, which will fill in all five nondegenerate 4-simplices of $N(P)$, hence all of $N(P)$. The inner horns are indexed by their spines.

1. 4 6 8 ($\Lambda_1^2$)
2. 4 5 6 8 ($\Lambda_2^3$)
3. 4 5 7 ($\Lambda_1^2$)
4. 4 5 7 8 ($\Lambda_1^3$)
5. 1 5 7 ($\Lambda_1^2$)
6. 1 4 5 7 ($\Lambda_2^3$)
7. 1 2 5 7 ($\Lambda_2^3$)
8. 1 4 8 ($\Lambda_1^2$)
9. 1 4 5 8 ($\Lambda_1^3$)
10. 1 5 7 8 ($\Lambda_1^3$)
11. 1 4 5 7 8 ($\Lambda_2^4$)
12. 1 2 5 8 ($\Lambda_2^3$)
13. 1 2 5 7 8 ($\Lambda_2^4$)
14. 1 5 6 8 ($\Lambda_1^3$)
15. 1 4 5 6 8 ($\Lambda_2^4$)
16. 1 2 5 6 8 ($\Lambda_2^4$)
17. 1 2 3 8 ($\Lambda_1^3$)
18. 1 2 3 6 8 ($\Lambda_1^4$)