Alternate formulation of the question (I think): What's a precise version of the statement: "In a stable $\infty$-category, finite limits and finite colimits coincide"?
Recall that a stable $\infty$-category is a type of finitely complete and cocomplete $\infty$-category characterized by certain exactness conditions. Namely,
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There is a zero object $0$, i.e. an object which is both initial and terminal.
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Every pushout square is a pullback square and vice versa.
Item (2) takes advantage of a peculiar symmetry of the "square" category $S = \downarrow^\to_\to \downarrow$; namely $S$ can either be regarded as $S' \ast \mathrm{pt}$ where $S' = \cdot \leftarrow \cdot \rightarrow \cdot$ is the universal pushout diagram, or $S$ can be regarded as $S = \mathrm{pt} \ast S''$ where $S'' = \cdot \rightarrow \cdot \leftarrow \cdot$ is the universal pullback diagram. Hence it makes sense to ask, for a given $S$-diagram, whether it is a pullback, a pushout, or both. Item (1) similarly takes advantages of the identities $\mathrm{pt} = \emptyset \ast \mathrm{pt} = \mathrm{pt} \ast \emptyset$.
But I can't shake the feeling that notion of a stable infinity category somehow "transcends" this funny fact about the geometry of points and squares. For one thing, one can use a different "combinatorial basis" to characterize the exactness properties of a stable $\infty$-category, namely:
1.' The category is (pre)additive (i.e. finite products and coproducts coincide)
2.' The loops / suspension adjunction is an equivalence.
True, (2') may be regarded as a special case of (2) — but it may also be regarded as a statement about the (co)tensoring of the category in finite spaces.
Both of these ways of defining stability say that certain limits and colimits "coincide", and my sense is that in a stable $\infty$-category, all finite limits and colimits coincide — insofar as this makes sense.
Question:
Is there a general notion of "a limit and colimit coinciding" which includes
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zero objects
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biproducts (= products which are also coproducts)
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squares which are both pullbacks and pushouts
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suspensions which are also deloopings
and if so, is it true that in in a stable $\infty$-category, finite limits and finite colimits coincide whenever this makes sense?
I would regard this as investigating a different sort of exactness to the exactness properties enjoyed by ($\infty$)-toposes. In the topos case, I think there are some good answers. For one, in a topos $C$, the functor $C^\mathrm{op} \to \mathsf{Cat}$, $X \mapsto C/X$ preserves limits. Foir another, a Grothendieck topos $C$ is what Street calls "lex total": there is a left exact left adjoint to the Yoneda embedding. It would be nice to have similar statements here which in some sense formulate a "maximal" list of exactness properties enjoyed by (presentable, perhaps) stable $\infty$-categories, rather than the "minimal" lists found in (1,2) and (1',2') above.
Best Answer
I don't think this "finite limits and finite colimits coincide" business can be taken very far. If you take any small category $S_0$ you can add an initial and a terminal object to form $S = \mathrm{pt} \ast S_0 \ast \mathrm{pt}$. A diagram of shape $S$ could potentially be both a colimiting cocone and a limiting cone and you might hope those conditions are equivalent in a stable $\infty$-category. (And for $S_0 = \mathrm{pt} \sqcup \mathrm{pt}$, this does happen, of course: it is the condition that a square is a pushout if and only if it is a pullback.) But this fails1 for three points, $S_0 = \mathrm{pt} \sqcup \mathrm{pt} \sqcup \mathrm{pt}$.
I think it's probably better to focus on a lesser known characterization of stable $\infty$-categories: they are precisely the finitely complete and cocomplete ones in which finite limits commute with finite colimits.2
1 The colimit of the $\mathrm{pt} \ast S_0$ shaped diagram with $X$ at the cone point and zeroes in the other slots is $\Sigma X \amalg \Sigma X$. Analogously the limit of the $S_0 \ast \mathrm{pt}$ diagram with $Y$ in the cocone point and zeroes in the other slots is $\Omega Y \times \Omega Y$. But for $Y = \Sigma X \amalg \Sigma X$ we do not have $X = \Omega Y \times \Omega Y$.
2 If $\mathcal{C}$ is a stable $\infty$-category, then it is finitely cocomplete, and thus if $S$ is a finite diagram shape, there is a functor $\mathrm{colim} : \mathrm{Fun}(S, \mathcal{C}) \to \mathcal{C}$. It's domain is also stable and $\mathrm{colim}$ preserves finite colimits --because colimits commute with colimits. The functor is therefore exact and so preserves finite limits as well.
Now, assume that $\mathcal{C}$ is finitely complete and cocomplete and that finite limits commute with colimits in it. Consider the following diagram: $$\require{AMScd}\begin{CD} X @<<< X @>>> 0 \\ @VVV @VVV @VVV \\ 0 @<<< X @>>> 0 \\ @AAA @AAA @AAA \\ 0 @<<< X @>>> X \\ \end{CD}$$ Taking pushouts of the rows we get the diagram $0 \to \Sigma X \leftarrow 0$ whose pullback is $\Omega \Sigma X$. If instead we take pullbacks of the columns, we get $X \leftarrow X \to X$, whose pushout is $X$. Pullbacks commuting with pushouts tell us then that $\Omega \Sigma X \cong X$ so $\mathcal{C}$ is stable.