# Shift in a stable $\infty$-category

In Proposition 2.1.14 of Lurie's Derived Algebraic Geometry paper (DAG), he gives a triangulated structure on the homotopy category of a stable $\infty$-category. To define the shift operator $A[1]$ of an object $A$, he takes the cokernel of the map $A \to 0$ and calls this the "suspension".

I'm new to this stuff, and I don't understand why this is a natural thing to do: in particular, why should one think of this as a suspension, and how should I understand the cokernel $A \to 0$?

There is a very simple algebraic explanation why this should be a good idea. Taking the cokernel of any map gives an exact triangle; the same for kernels. The long exact sequence for the homotopy groups of an exact triangle would reduce to $$\ldots \to \pi_{n+1}(X[1]) \to \pi_n(X) \to \pi_n(0) \to \pi_n(X[1]) \to \pi_{n-1}(X) \to \ldots$$

Since all of the $\pi_n(0)$ are zero, the suspension is precisely the object whose homotopy groups are the same as those of $X$ but shifted over one degree.

This also relates to the operation of taking the suspension of a topological space, which had been studied by topologists long before any of this $\infty$-category stuff was conceived.

In any model category, there is a formula for computing the homotopy pushouts using the model structure. Given any span $B \leftarrow A \to C$ between cofibrant objects, its homotopy pushout $P$ is computed by the ordinary pushout

$$\require{AMScd} \begin{CD} A \amalg A @>>> B \amalg C \\ @VVV @VVV \\ A \times I @>>> P \end{CD}$$

where $A \times I$ is a "cylinder object" for $A$; e.g. if we were talking ordinary topological spaces, we could take the actual product of spaces $A \times [0,1]$.

Of particular note is that the cokernel $A \to 0$ is defined as the pushout of the span $0 \leftarrow A \to 0$.

In the model category of topological spaces, if $X$ is cofibrant, the homotopy pushout of $* \leftarrow X \to *$ is computed by the ordinary pushout square $$\require{AMScd} \begin{CD} X \amalg X @>>> * \amalg * \\ @VVV @VVV \\ X \times I @>>> S X \end{CD}$$ In classical topology, this pushout is precisely the usual definition of the suspension of a topological space.

• Thanks, this is a really nice explanation. May 1 '18 at 16:29
• Also, as I understand it, to compute a hocolim, you basically take a cofibrant replacement of your diagram and then compute the ordinary colim. But the formula you give isn't that, right? May 1 '18 at 16:30
• @AshwinIyengar: I was citing a theorem from nlab. But following the proof, the the pushout diagram above is basically just rearranging after making a cofibrant replacement $A \to (A \times I) \amalg_A B$ of the map $A \to B$. There's a lemma that to compute homotopy pushouts as ordinary pushouts, if the objects are cofibrant, you only need one of the maps to be a cofibration
– user14972
May 1 '18 at 18:02