Complexes in stable categories 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?
 A: 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.
