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I am looking for a good set of sufficient conditions to put on a category $C$ such that the following becomes true:

Given any $\infty$-category $S$ and a functor $f\colon N(C)\to S$ which is essentially constant (i.e. it maps all edges to equivalences), a colimit $\overline f\colon N(C)^\triangleright\to S$ of $f$ exists and is always an essentially constant diagram.

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  • $\begingroup$ It seems a little strange to me to want the colimit cone to be essentially constant, since homotopy colimits of constant diagrams are not generally constant. For instance, the colimit in spaces of the constant diagram of shape $\mathcal{C}$ taking the value $*$ is $B\mathcal{C}$, the classifying space of $\mathcal{C}$, not $*$. $\endgroup$
    – Kaya Arro
    Commented Jul 1, 2017 at 18:15
  • $\begingroup$ Also, I would expect "essentially constant" to mean "naturally isomorphic to a constant functor". But the definition that you give is not equivalent to that. $\endgroup$
    – Kaya Arro
    Commented Jul 1, 2017 at 18:17
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    $\begingroup$ To add to what @KyleFerendo said, if "essentially constant" does mean "naturally isomorphic to a constant functor", then the answer would be that $N(C)$ must be weakly contractible. $\endgroup$
    – Mike Shulman
    Commented Jul 2, 2017 at 12:53

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I think that Proposition 4.3.1.12. in Higher Topos Theory (Lurie) is exactly what I was looking for:

Given: a coCartesian fibration $p\colon X\to S$ of $\infty$-categories and a diagram $\overline q\colon K^\triangleright\to X$ such that $K$ is weakly contractible and such that $\overline q$ carries each edge of $K$ to a $p$-coCartesian edge of $X$. Then: $\overline q$ is a $p$-colimit diagram if and only if it carries all edges in $K^\triangleright$ to $p$-coCartesian edges of $X$.

Taking $S$ to be a point we get that a sufficient condition as I was looking is for $N(C)$ to be weakly contractible.

It seems that Kyle Ferendo's remark implies the converse: if all the essentially constant diagrams of a certain shape are essentially constant, then $N(C)$ is weakly contractible. (Because $|N(C)|$ is the limit of the constant diagram $\star$ in spaces)

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