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Background

Given a category $C$, one can consider the category $Fam(C)$ of set-indexed familiies of objects in $C$. Formally, the objects are pairs $(X,F)$ in which $X$ is a set and $F:X\rightarrow C$ is a functor and a map $(X,F)\rightarrow (Y,G)$ is pair $(f,g)$ where $f:X\rightarrow Y$ is a function and $g:F\Rightarrow G\circ f$ is a natural transformation. The same construction works for $\infty$-categories: the objects of $Fam(C)$ for $C$ an $\infty$-category are pairs $(X,F)$ where $X$ is an $\infty$-groupoid and $F:X\rightarrow C$ is a functor and the morphisms are the same.

A locus (resp. $\infty$-locus) is a pointed locally presentable category such that $Fam(C)$ is a topos (resp. $\infty$-topos).

Question

In this lecture (at around 1:04:40) Joyal says that $\infty$-loci are closed under left exact localizations because $\infty$-topoi are. However, I don't see how this is the case unless the $Fam$ construction is a functor $Fam:Cat_{(\infty,1)}\rightarrow Cat_{(\infty,1)}$ and preserves left exact localizations.

My question is: is this true?

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It's apparent that $Fam$ is functorial, right?

So you start with an $\infty$-locus $C$ and some localization $f:C\to C'$ thereof. You want to show that $Fam(C')$ is an $\infty$-topos; in particular, maybe it's a localization of $Fam(C)$ in some natural way (indeed, implicit in your question is the idea that $Fam(f)$ is the desired localization). In fact, I believe the localization of $Fam(C)$ that we want is the one generated by the morphisms of $Fam(C)$ that comprise exclusively morphisms in $C$ that are sent to equivalences by the localization. So it's necessary to use the definitions to check that this localization is indeed equivalent to $Fam(f)$.

Next, we have to check that $Fam(f)$ is in fact left-exact. But that too can be verified from the fact that $f$ was left exact and using the way that limits in $Fam(C)$ are built from limits in $C$ and in $\mathcal{S}$ (the $(\infty,1)$-category of $\infty$-groupoids).

Obviously I've left all the details in the proof for you to fill in, but I hope that you can work them out to your satisfaction.

Edit: Here are more details on limits in $Fam(C)$. Start by thinking about how morphisms work in $Fam(C)$ (which you describe in your question). For objects $X:x\to C$ and $Y:y\to C$ in $Fam(C)$ (i.e. $x$ and $y$ are spaces), you can have any map of spaces $g:x\to y$ you want, and then you can have any morphism (i.e. natural transformation) $g^*X\Rightarrow Y$ you want. So if you have a diagram $D:d\to Fam(C)$, you also have a diagram $D_{\mathcal{S}}:d\to \mathcal{S}$. You take the limit of the diagram in $\mathcal{S}$ (denote this $\lim D_\mathcal{S}$), and you need to find the appropriate functor $\lim D_\mathcal{S}\to C$. This is probably the limit of some functors $\lim D_\mathcal{S}\to C$, but which? Since $\lim D_\mathcal{S}$ is a limit, you have projections $P_x:\lim D_\mathcal{S}\to D_\mathcal{S}(x)$ for $x\in d$. Hence you get a diagram $A:d\to Fun(\lim D_\mathcal{S},C)$ with $A(x)=D(x)\circ P_x$ and $A(g)=D(g)\circ P_x$ for $g:x\to y$ in $d$ (so $A(g):D(x)\circ P_x\Rightarrow D(y)\circ D_\mathcal{S}(g)\circ P_x$). You take the limit of $A$ and get the desired functor $\lim A:\lim D_\mathcal{S}\to C$.

So, in making everything as explicit as I could, I've wound up with a ton of notation flying around. Hopefully you are able to parse everything, but if not, I can explain further in comments. Note that I've given the construction, but I have not given the proof of the correctness of the construction. Anyway, the moral of the story is this: limits in $Fam(C)$ boil down to taking limits in $\mathcal{S}$ and limits in $Fun(\lim D_\mathcal{S},C)$; but limits in the latter are reducible to limits in $C$. And you know that (by definition) left exact localizations preserve finite limits, so $Fam(f)$ is also left exact (assuming you've already verified that it is the correct localization, as described above).

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  • $\begingroup$ Thanks. Can you please elaborate a bit on the construction of limits in Fam(C)? (and yes, its apparent that Fam is functorial) $\endgroup$ – Karthik Yegnesh Jan 26 '17 at 1:39
  • $\begingroup$ Sure thing -- I've added an explicit explanation of how limits work in $Fam(C)$. As I say in the answer, the point is that they boil down to limits in $\mathcal{S}$ of spaces and limits in an $(\infty,1)$-category of functors to $C$ (which reduces to limits in $C$). $\endgroup$ – Kyle Ferendo Jan 26 '17 at 4:37
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    $\begingroup$ Great, it's very clear now. Not sure how I didn't see this before... $\endgroup$ – Karthik Yegnesh Jan 26 '17 at 16:50
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    $\begingroup$ Don't worry about it; sometimes there are so many definitions to unpack, it can be difficult to see through all the formality to what is really going on, even when the actually process is straightforward (but lengthy). $\endgroup$ – Kyle Ferendo Jan 26 '17 at 17:49

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