It is known that in a stable $\infty$-category $\mathcal{C}$ and $X \in \mathcal{C}$, the suspension $X[1]$ is defined by the pushout of $0\leftarrow X \rightarrow 0$. However this does not make sense in usual category since the pushout is the zero object $0$.

Now all we can do is following the formal definition of limit/colimit for an $\infty$-category. But the cost is that we lose the intuition. What (intuition) should one keep in mind when thinking about pushout/pullback in a stable $\infty$-category.

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    $\begingroup$ You should maybe study these notions first in model categories. $\endgroup$ Commented Oct 6, 2021 at 5:32
  • $\begingroup$ I do have learnt something about model categories. Do you mean a stable $\\infty$-category can be viewed as a model category plus some extra structure? $\endgroup$
    – XT Chen
    Commented Oct 6, 2021 at 5:50
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    $\begingroup$ no, just for you to have some intuition. In a model category, in order to take that homotopy push-out, you replace the maps by cofibrations. The standard way to do it is to replace $0$ with the cone $CX$ and the maps with the inclusion of the base. Then you take the real push-out, the union of the two cones by the base, which is the suspension. $\endgroup$ Commented Oct 6, 2021 at 7:14
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    $\begingroup$ I've tried to give some intuition about why this is true that is model-category free. In general the notion of homotopy pushout predates model categories, and I think it's important to emphasize its interpretation in terms of homotopy coherent diagrams rather than the strictification theorems (which are important, but could come in a second moment). $\endgroup$ Commented Oct 6, 2021 at 7:35
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    $\begingroup$ As @DenisNardin says, in terms of homotopy coherent diagrams it's even easier. A map from $* \leftarrow X\rightarrow *$ to an object $Y$ amounts to a homotopy from the trivial map $X\rightarrow Y$ to itself. Such a homotopy is a map from the cylinder $IX$ which is constant on top and bottom copies of $X$ in $IX$, so you can collapse them, and $IX/X\times\{0,1\}=\Sigma X$ is precisely the suspension. $\endgroup$ Commented Oct 6, 2021 at 7:42

1 Answer 1


Let me try to give some intuition by examining two important examples. One should start from the definition: the suspension $ΣX$ is the universal choice of $Y$ filling of a square $$\require{AMScd} \begin{CD} X @>{p}>> \ast\\ @V{p}VV @VVV \\ \ast @>>> Y \end{CD}\,.$$ To understand the suspension we need to understand what's the datum of such a square. This is given by two maps $y_0:\ast\to Y$ and $y_1:\ast\to Y$ (two ``points'' of $Y$) and a homotopy $H:y_0p\simeq y_1p$ as maps $X\to Y$. That is, the suspension is the universal recipient of two points and a homotopy between the two constant maps at $y_0$ and $y_1$.

Let us go now in the ∞-category of spaces (or, if you prefer the name, animæ). Then a homotopy of maps $H:X\to Y$ is exactly a map $H:X\times [0,1]\to Y$. Since we require the homotopy to be between two constant maps we see that a commuting square as above is exactly the datum of a map $$\ast\amalg_{X\times\{0\}}X\times [0,1]\amalg_{X\times\{1\}} \ast\to Y$$ One needs to check the universal property more carefully, but as one would expect the left hand side is exactly the universal recipient. That is $$ ΣX=\ast\amalg_{X\times\{0\}}X\times [0,1]\amalg_{X\times\{1\}} \ast$$ Note that the right hand side is exactly the classical suspension of $X$, thus justifying the name $ΣX$.

Now for a more algebraic (and stable) example: the derived category of a ring. In this case, when representing everything by chain complexes, the two points are no data (since there's only one possible map $0\to Y$), and a homotopy is the same as a collection of maps $H_n:X_{n+1}\to Y_n$ such that $dH_n+H_nd=0$. But this is exactly the same as a map of chain complexes $X[1]\to Y$ (remember than in $X[1]$ the differential inherits a sign to make the formulas canonical -- this is exactly its origin!). Therefore in $\mathscr{D}(R)$, the suspension is given by the usual shift.

So what's the reason for these new phenomena? The point is that a commuting square in an ∞-category in general contains more information than the square in a 1-category: it's not enough that you say that the two composition are equal, you also have to provide a reason for them to be equal (i.e. a homotopy). Therefore your pushout needs to account for this additional data, and this is why it is nontrivial.

  • $\begingroup$ Why we do not consider homotopy of homotopy when calculating the suspension? $\endgroup$
    – XT Chen
    Commented Oct 6, 2021 at 7:52
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    $\begingroup$ @XTChen We do when checking the universal property properly (i.e. when looking at maps of squares), but for the object itself the diagram is too small to contain higher homotopies (where would they fit?). $\endgroup$ Commented Oct 6, 2021 at 7:54
  • $\begingroup$ I am not sure whether one needs to be a bit more careful. I am not sure whether an equivalence of two maps corresponds precisely to connecting homotopies. Maybe one needs to restrict to fibrant-cofibrant objects? $\endgroup$
    – Z. M
    Commented Oct 7, 2021 at 17:38

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