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The $\infty$-category of spaces has the following properties:

  1. It is the $\infty$-category obtained from the (ordinary) category of finite sets by freely adding sifted colimits. (See e.g. Cesnavicius-Scholze https://arxiv.org/abs/1912.10932 §5.1 for a review of this notion and for pointers to Lurie's HTT where this is proven.)
  2. (As Tim Campion points out in a comment, another characterization, also in HTT): Spaces are obtained by freely adding arbitrary colimits to the category $\{*\}$.

Can either of these characterizations be used (ideally without referring to the model of quasi-categories) to show other properties, such as:

  • that colimits in spaces are universal (proven by Lurie in HTT Lemma 6.1.3.14)?
  • possibly even that $Cat_\infty$ is generated by (the compact objects) $*$ and $\Delta^1$?
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    $\begingroup$ Even more fundamental than the sifted-cocompletion statement is the fact that $Spaces$ is the free cocomplete $\infty$-category on an object, and more generally that $Psh(C)$ is the free cocomplete $\infty$-category on $C$ (HTT 5.1.5.6). Would you be interested in proofs starting from these facts? Also I find it ironic that you're trying to avoid quasicategories as a model specifically -- Lurie's proof of 6.1.3.14 constructs an explicit right adjoint to base change at the level of model categories, so if anything is more tied to the "simplicial category" model than the quasicategory model. $\endgroup$
    – Tim Campion
    Commented Apr 6, 2021 at 16:29
  • $\begingroup$ Also, I'd be careful about the statement about $Cat_\infty$ -- what exactly do you mean by "compactly generated by particular objects"? It's true that $Cat_\infty$ is generated under colimits by $\ast$ and $\Delta[1]$. But I'm pretty sure that $\{\ast,\Delta[1]\}$ is not a dense generator for $Cat_\infty$ -- a natural way to fix that would be to include $\Delta[n]$ for all $n$. And in order to find $\mathcal C \subseteq Cat_\infty$ such that $Cat_\infty = Ind(\mathcal C)$, you need to include all $\infty$-categories generated under finite colimits by $\ast$ and $\Delta[1]$. $\endgroup$
    – Tim Campion
    Commented Apr 6, 2021 at 16:35
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    $\begingroup$ That's an intriguing question ! It is a general fact that freely added colimits are Van Kampen colimits (universal and effective)... but the proof of that I know relies on the fact that these colimits are Van Kampen in the $\infty$-category of spaces in the first place... (and it fails if you look at categories enriched in something where colimits are not van Kampen...) $\endgroup$
    – Simon Henry
    Commented Apr 6, 2021 at 18:41
  • $\begingroup$ @TimCampion: yes, if there is a proof starting from the characterization by means of freely adding colimits, this would also be interesting to see. I have added this to the question. And concerning $Cat_\infty$ my wording was indeed a bit sloppy, I've adjusted the question slightly. $\endgroup$
    – Jakob
    Commented Apr 7, 2021 at 8:35
  • $\begingroup$ @SimonHenry: interesting comment - can you name a reference for the fact you mention? $\endgroup$
    – Jakob
    Commented Apr 7, 2021 at 8:41

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Here is a "model-independent" proof that colimits in $Spaces$ are universal. Of course, the ingredients going into the proof may have model-dependent proofs.

Fact: Colimits in $Spaces$ are universal.

Proof: We want to show that for any map of spaces $f: Y \to X$, the pullback functor $$f^\ast: Spaces_{/X} \to Spaces_{/Y}$$ preserves colimits. We may view $X,Y$ as $\infty$-categories which happen to be $\infty$-groupoids and $f: Y \to X$ as a functor between them. By straightening / unstraightening, the functor $f^\ast :Spaces_{/X} \to Spaces_{/Y}$ is identitified with the "precompose $f$" functor $$f^\ast: Psh(X) \to Psh(Y)$$ Now, $Psh(X),Psh(Y)$ are functor categories with values in the cocomplete category $Spaces$. So colimits are computed "objectwise" in these categories. That is, for $F: I \to Psh(X)$, we have $(\varinjlim_{i \in I} F(i))(x) = \varinjlim_{i \in I} (F(i)(x))$, and similarly in $Y$. From these formulas, it is immediate that precomposing $f$ preserves colimits. That is, we have $$(\varinjlim_{i \in I} f^\ast F(i))(y) = \varinjlim_{i \in I} F(i)(f(y)) = \varinjlim_{i \in I} (f^\ast F(i))(y)$$ as desired.

This proof doesn't precisely use the universal property of $Psh(X),Psh(Y)$ of freely adding colimits, but perhaps it could be tweaked to do so.

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