Generalizing from 1-category theory, there's a simple definition of a "naive complex" in a stable $\infty$-category. Considering bounded positive graded chain complexes, they are a sequence of maps

$$ A_n \xrightarrow{d_n} A_{n-1} \xrightarrow{d_{n-1}} \ldots \xrightarrow{d_1} A_0 $$

with the property that $d_k d_{k+1} \simeq 0$. If I've not made any errors, then at first blush they seem to have a reasonable theory, along with a nice "realization" given by iteratively taking homotopy cofibers:

$$ |A| = \mathrm{cofib}(\mathrm{cofib}(\ldots) \to A_0) $$

that parallels taking the total complex of a bicomplex. This realization can also be given by the colimit of the diagram

$$ A_n \rightrightarrows^{d_n}_0 A_{n-1} \rightrightarrows \ldots \rightrightarrows^{d_1}_0 A_0$$

However, naive complexes don't seem to be studied. Instead, in Higher Algebra, Lurie proves a Dold-Kan correspondence stable $\infty$-category asserting equivalences between the categories of:

  • Simplicial objects
  • Filtered objects (i.e. arbitrary $\mathbb{Z}_{\geq 0}$-indexed diagrams)
  • upper-triangular array with zeroes along the diagonal, in which every square is a pushout

and it is this last sort of thing that Lurie calls a complex. (Specifically, a $\mathbb{Z}_{\geq 0}$-complex)

It is only when looking at the homotopy category does Lurie say anything about a correspondence between simplicial objects and naive complexes.

So my question is whether this is an oversight; i.e.

Is the $\infty$-category of simplicial objects in a stable $\infty$-category equivalent to the $\infty$-category of (unbounded, positively graded) naive complexes?

and if the answer is no, what goes wrong?

  • 4
    $\begingroup$ I don’t think they are equivalent. When n=2 the obstruction for such a complex to arise from a little filtration is the Toda bracket of the three differentials. Another issue is that I don’t think your category of complexes will have hocolims. $\endgroup$ Dec 25, 2017 at 1:00
  • 5
    $\begingroup$ Another issue is that I don’t see how you’re able to take those iterated cofibers (or equivalently, to form that diagram to take the colimit) without remembering witnesses for those nullhomotopies (and compatibilities between them). $\endgroup$ Dec 25, 2017 at 1:03
  • 4
    $\begingroup$ The homotopies d_k d_{k+1}≃0 must themselves fit into a higher coherence data. If you add this data, then you indeed get an equivalence of ∞-categories between chain complexes and simplicial objects. $\endgroup$ Dec 25, 2017 at 3:59

1 Answer 1


Here is the problem with the notion of naive complex. Suppose we have a naive complex $$ \require{AMScd} \begin{CD} A @>f>> B @>g>> C @>h>> D \end{CD} $$ If we propose to compute the realization iteratively, the first step would be to produce the sequence $$ \require{AMScd} \begin{CD} \mathrm{cofib}(f) @>>> C @>h>> D \end{CD} $$ However, this need not be a naive complex! So everything falls apart.

A more elaborate explanation is to look at the big diagram of pushout squares one can construct from the complex:

$$ \require{AMScd} \begin{CD} A @>f>> B @>>> 0 \\ @VVV @VVV @VVV \\ 0 @>>> \mathrm{cofib}(f) @>>> \Sigma A @>>> 0 \\ & & @VVV @VVV @VVV \\ & & C @>>> \mathrm{cofib}(g) @>>> \mathrm{cofib}(\mathrm{cofib}(f) \to C) \\ & & & & @VVV \\ & & & & D \end{CD} $$ There is no direct way to produce the expected map $\mathrm{cofib}(\mathrm{cofib}(f) \to C) \to D$. In fact, by the rightmost square, one can only possibly exist if $(\Sigma A \to D) \simeq 0$.

The map $\Sigma A \to D$ is the Toda bracket mentioned in the comments. (wikipedia nlab)

We can give an explicit counterexample. In the $\infty$-category of chain complexes, suppose our naive complex is given by:

$$ \require{AMScd} \begin{CD} 0 @>>> 0 @>>> \mathbb{Z}/4 @>1>> \mathbb{Z}/4 \\ @VVV @VVV @VV4V @VVV \\ \mathbb{Z} @>2>> \mathbb{Z}/4 @>2>> \mathbb{Z}/8 @>>> 0 \end{CD} $$

That this is a naive complex can be checked on the homology groups. This isn't a bicomplex as is, and the point is that it cannot be improved to become one! One can show that after replacing $A \xrightarrow{f} B$ with $\mathrm{cofib}(f)$, the diagram becomes

$$ \require{AMScd} \begin{CD} \mathbb{Z} @>1 + 2n>> \mathbb{Z}/4 @>1>> \mathbb{Z}/4 \\ @VV2V @VV4V @VVV \\ \mathbb{Z}/4 @>2>> \mathbb{Z}/8 @>>> 0 \end{CD} $$

where $n$ encodes the choice of chain homotopy and on homology groups, the top row does not compose to zero for any choice of $n$.

The fix is to redefine "naive complex" to include the parallel zeroes: i.e. so that the index 1-category is a sequence of parallel pairs of arrows such that each arrow coequalizes the pair before it.

(the second arrow in each pair is still required to be zero!)

I interpret Pavlov's comment above as saying that this redefinition of "naive complex" results in a category of complexes that is equivalent to simplicial objects, although I have not proved that myself.

  • $\begingroup$ I don't understand the last part. Some $d$ coequalizes $(d',0)$ if and only if $dd'=0$, so how does this differ from your original version? $\endgroup$ Dec 27, 2017 at 16:48
  • $\begingroup$ @მამუკაჯიბლაძე: I asked my question because I hadn't appreciated the difference either! I don't yet have a good intuition yet, but mechanically the difference is that it's an equation in the index category, so a functor introduce more equivalences than I had originally. That the iterative calculation works out follows from this procedure involving Kan extensions (expressed there for limits rather than colimits). Or I could be making errors someplace. $\endgroup$
    – user13113
    Dec 27, 2017 at 21:35
  • $\begingroup$ But how do you avoid conditions outside the index category? Don't you need to require that either some morphisms forcibly go to zero morphisms, or some object forcibly goes to the zero object? $\endgroup$ Dec 28, 2017 at 8:07
  • $\begingroup$ @მამუკაჯიბლაძე: Yes; I'm assuming the new arrows all map to zero. Realization by way of iterated coequalizers doesn't actually require the new arrows to be zero; the point is to normalize a degree of freedom (since you can add the same map to both arrows of a parallel pair and get a new diagram satisfying the coequalizer condition and with the same realization). I'm not certain if it's enough to simply require each of the new arrows to be zero or if I need an additional constraint on the entire sequence of new arrows. $\endgroup$
    – user13113
    Dec 28, 2017 at 16:54
  • $\begingroup$ @მამუკაჯიბლაძე: There is actually another diagram I've considered that the iterated cofiber construction works for, depicted below. Maybe this alternate diagram is better behaved; I think it can be seen as a subdiagram of the "upper triangular array" version of complex I mentioned in the OP. $$ \begin{matrix} A_n &\to& A_{n-1} &\to& A_{n-2} \\ & \nearrow \!\!\!\!\!\! \searrow & & \nearrow \!\!\!\!\!\! \searrow \\ 0 &\to& 0 &\to& 0 \end{matrix} $$ $\endgroup$
    – user13113
    Dec 28, 2017 at 17:05

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