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Consider an $\left(\infty, 1\right)$-category $\mathcal{C}$ with a terminal object $1$. (I'm particularly interested in the case where $\mathcal{C}$ is a topos.) It is known that the forgetful functor $U$ from the under-category ${1}/{\mathcal{C}}$ to $\mathcal{C}$ creates colimits of diagrams with weakly contractible index categories. Is there an example of a diagram $F : \mathcal{I} \to {1}/{\mathcal{C}}$ such that

  • $\mathcal{I}$ is not weakly contractible
  • $U$ still creates the colimit of $F$?
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A short answer is that if $C$ is the terminal category, then $1/C \to C$ is an equivalence, and hence creates all colimits.

Less flippantly, we can take $C$ to be the ordinary topos of sets. Then the forgetful functor creates $I$-shaped colimits if $I$ is connected.

If $C$ is cocomplete, then the colimit of $F$ in $1/C$ is the pushout of a diagram $$ 1 \leftarrow colim_I 1 \to colim_I F $$ and so the critical question is whether the colimit of the constant diagram with value $1$ is equivalent to $1$.

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  • $\begingroup$ Thanks, but my question was too open-ended. I'm thinking of $\mathcal{C}$ having untruncated objects. (I'm secretly thinking of models of homotopy type theory.) Are there examples of cocomplete $\mathcal{C}$ with untruncated objects where $I$ is not weakly contractible but $\mathsf{colim}_I{1} \simeq 1$? $\endgroup$
    – Perry Hart
    Commented May 3, 2023 at 18:03
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    $\begingroup$ @PerryHart Sure. If C is the category of pointed spaces, or any other category with a zero object, then $1/C \to C$ is also an equivalence. $\endgroup$
    – Tyler Lawson
    Commented May 4, 2023 at 15:46
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    $\begingroup$ I guess my point is that this would be a pretty special property of the category $C$. Here is an alternative formulation. Given an $(\infty,1)$-category $C$, there is a functor $e: C \to Spaces$, $X \mapsto Map_C(1,X)$. The forgetful functor will create $I$-shaped colimits if and only if, for any $X$, the space of maps from the nerve $NI$ to $e(X)$ is equivalent to the space of constant maps. (So if "enough spaces" are of the form $e(X)$, then the forgetful functor creates colimits; this explains why you don't see it when mapping spaces are truncated.) $\endgroup$
    – Tyler Lawson
    Commented May 4, 2023 at 15:51

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