**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](https://www.youtube.com/watch?v=qmCh9KwrQq8&index=6&list=LLAFfjoi6P8_PuJAlJFTwAFg) (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? 



The "Family" functor for infinity-categories
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).


