Complexes in stable categories

Question

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

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

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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?

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)

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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 $n$.

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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.