Proof

For each object $c\in C$ let $j(c):C\to \mathrm{\infty }{\mathrm{Grpd}}^{\mathrm{op}}$ be the functor represented by $c$. Let $S_c$ be the class of morphisms sent by $j(c)$ to weak equivalences in ∞Grpd. Since $j(c)$ preserves small colimits, this is a strongly saturated class, by the above lemma. Now observe that $S$ is the intersection $S={\cap }_c{S}_c$ where $c$ ranges over the $S_0$-local objects.

Proof

We produce an evident cograph realization $K$ of the inclusion and check that it being also a coCartesian fibration, hence exhibiting $R$ as a right adjoint, is equivalent to the second statement.

Let $K\subset C\times \Delta [1]$ be the full subcategory on those objects $(c,i)$ for which $c\in D$ if $i=1$. Let $p:K\to \Delta [1]$ be the induced projection. One checks that this is the correspondence which is associated to the inclusion functor $D\hookrightarrow C$.

Therefore by the properties of adjoint (∞,1)-functors, we have that the inclusion functor has a left adjoint precisely if $p$ is not only a Cartesian fibration but also a coCartesian fibration.

To see that this is the case precisely if every $c$ has a reflection $f:c\to d$, recall the characterization of coCartesian morphisms $\tilde{f}:(c,0)\to (d,1)$ as those making the squares

being homotopy pullback squares, for all $(e,i)\in K$. Now in $\Delta [1]$ all hom-objects are either empty or are points, so that the bottom morphism becomes the identity on the point if $i=1$. Since for $i=0$ everything becomes entirely trivial we consider the case that $i=1$ and hence $e\in D$.

In that case the homotopy-pullback property is equivalent to the top morphism being an equivalence, hence to

being an equivalence. This way the reflectors are identified precisely with the coCartesian morphisms in $K\to \Delta [1]$ that exhibit the left adjoint (∞,1)-functor to the inclusion functor.

Proof

The reasoning is entirely analogous to the 1-categorical case (see for instance localization, reflective subcategory and geometric embedding).

First notice that because $D\hookrightarrow C$ is a full and faithful (∞,1)-functor we have that the counit $LR\stackrel{\simeq }{\to }{\mathrm{Id}}_D$ is an equivalence. From this it follows that precomposition with the unit $i_z:z\to \mathrm{Loc}z$ of morphisms in the image of $\mathrm{Loc}$ is a weak equivalence: for all $z,x\in C$ we have

If $z$ is itself in the image of $\mathrm{Loc}$, then this means that precomposition with the unit $z\to \mathrm{Loc}z$ is an isomorphism on hom-sets in the homotopy category of $\mathrm{Loc}C$, hence by the Yoneda lemma is itself an isomorphism in the homotopy category, hence $i_z:z\to \mathrm{Loc}z$ is a weak equivalence if $z$ is itself in the image of $\mathrm{Loc}$.

Applying this statement to the naturality square for the natural transformation $\mathrm{Id}\to \mathrm{Loc}$ on $i_s$

we find that $\mathrm{Loc}{i}_s\simeq i_{\mathrm{Loc}s}$, hence that $\mathrm{Loc}{i}_s$ is a weak equivalence, and hence that $i_s$ is in $S$, for all $s\in C$.

Now to show that for all $x\in X$ the object $\mathrm{Loc}x$ is $S$-local, let $f:y\to z$ be in $S\subset \mathrm{Mor}(C)$ and consider the induced square

Here the vertical morphisms are equivalences by the above remark, and the top morphism is an equivalence by the assumption that $f$ in in $S$. It follows that the bottom morphism is an equivalence. This says that $\mathrm{Loc}x$ is $S$-local, for all $x\in C$.

Conversely, to show that for $s\in C$ an $S$-local object, we have that $s$ is in the essential image of $\mathrm{Loc}$ use that since $i_s:s\to \mathrm{Loc}s$ is in $S$, we have an equivalence ${\mathrm{Hom}}_C(i_s,s):{\mathrm{Hom}}_C(\mathrm{Loc}s,s)\stackrel{\simeq }{\to }{\mathrm{Hom}}_C(s,s)$. The pre-image of the identity under this equivalence is hence a left-inverse $\mathrm{Loc}s\to s$ of $s\to \mathrm{Loc}s$. But this means that $\mathrm{Loc}s\to s$ is itself in $S$ (since the morphisms in $S$ evidently satisfy 2-out-of-three), hence by applying the same argument again, we find that the left inverse $\mathrm{Loc}s\to s$ has itself a left inverse. That implies that it is actually an inverse of $s\to \mathrm{Loc}s$, hence that this is an equivalence. So this shows that the $S$-local $s$ is indeed in the essential iamge of $\mathrm{Loc}$.

Finally, to show that every $S$-local morphism is already in $S$, let $f:x\to y$ be such an $S$-local morphism and consider the square

By the above we know now that the vertical morphisms here are also $S$-local. It follows that the image of $\mathrm{Loc}f:\mathrm{Loc}x\to \mathrm{Loc}y$ on the homotopy category of $\mathrm{Ho}(\mathrm{Loc}C)$ corepresents an isomorphism, hence by the Yoneda lemma that $\mathrm{Lof}f$ is a weak equivalence. Hence $f$ is indeed in $S$.

Proof

(localization proposition)

The localization lemma gives for each object $c\in C$ a reflector $f:c\to d$ with $d$ $S$-local. By one of the above lemmas, this already gives the reflective embedding

of the full subcateory of $S$-local objects in $C$.

It remains to prove the statements about the role of $S$ in the localization:

First, by one of the above lemmas, we have that the $S_0$-local equivalences are a strongly saturated class of morphisms containong $S_0$. Hence they in particular contain $S$. So the second claim implies the first.

That the first and the third condition are equivalent follows from noticing that for any local object $d\in D$ the morphism ${\mathrm{Hom}}_C(f,Rd)$ is an equivalence precisely if ${\mathrm{Hom}}_D(Lf,d)$ is and then applying the Yoneda lemma (for instance in the homotopy category), which implies that if a morphism produces an equivalence when hommed into all objects, then it is itself an equivalence.

It remains to show that the third item implies the second. Let $f:c\to d$ be a morphism such that $Lf:Lc\to \mathrm{Ld}$ is an equivalence. Consider the commuting triangle

Since every reflector is in $S$ and the reflectors are the units of the reflective adjunction constructed from them, we have that the vertical morphisms in this diagram are in $S$, and the bottom morphism is, since it is an equivalence by assumption. By applying the 2-out-of-3 property of $S$ twice it follows that $f$ is in $S$.

Proof

Regard all $(\mathrm{\infty },1)$-categories as quasi-categories for the purpose of this proof. Write $D\subset \mathrm{Func}(\Delta [1],C)$ for the full sub-quasicategory on the elements of $S$. Consider the pullback (in sSet)

Since $S$ is by assumption closed under pushouts in $C$, we have for each morphism $x\to y$ in $D\simeq \mathrm{Func}(\{0\},C)$ and each lift

of its source to $\mathrm{Func}(\Delta [1],C)$ a lift of this morphism with this source, given by the the pushout square

in $C$, regarded as a morphism in $\mathrm{Func}(\Delta [1],C)$. By the universality of the pushout, one finds that this is a coCartesian lift. Hence $D\to \mathrm{Func}(\{0\},C)\simeq C$ is a coCartesian fibration. Moreover, by the behaviour under pullback of Cartesian fibrations it follows that the above diagram is a homotopy pullback diagram in the Joya model structure ${\mathrm{sSet}}_{\mathrm{Joyal}}$.

Use now that accessible quasi-categories are stable under homotopy pullback to conclude that $D_c$ is accessible. Moreover, one can check that $D_c$ has all small colimits. Together this means that $D_c$ is a locally presentable (∞,1)-category. This implies in particular that $D_c$ also has all small (∞,1)-limits and hence contains a terminal object, $f:c\to d$.

We now complete the proof by showing that $f:c\to d$ being terminal in $D_c$ implies that $d$ is an $S$-local object. This is equivalent to showing that for $t:a\to b$ any element in $S$, composition with $t$ induces an equivalence

This in turn may be checked by checking that all its homotopy fibers are contractible. By general statements about the homotopy fiber of functor categories the homotopy fiber of ${\mathrm{Hom}}_C(t,d)$ over a point $g:a\to d$ of ${\mathrm{Hom}}_C(a,d)$ is equivalent to the hom-object ${\mathrm{Hom}}_{C_{a/}}(t,g)$ in the under-quasi-category $C_{a/}$.

This in turn can be checked to be equivalent to ${\mathrm{Hom}}_{C_{d/}}(g_{*}t,{\mathrm{Id}}_d)$, where $g_{*}t$ is the $(\mathrm{\infty },1)$-categorical pushout

in $C$. Notice that $g_{*}t$, being a pushout of $t\in S$, is itself in $S$.

Now pick a composite

and observe that we have an isomorphism of simplicial sets

(where $s_1$ is the corresponding degeneracy map).

Applying the expression for homotopy fibers of functor categories once again, this is found to be the homotopy fiber of

because ${\mathrm{Hom}}_{(C_{c/}{)}_{f/}}(\sigma ,s_1(f))={\mathrm{Hom}}_{C_{f/}}(\sigma ,s_1(f))$.

Finally we can use that $f$ is terminal in the full subcategory $D_c$ of $C_{c/}$ that contains $g_{*}t\circ f$. This implies that the above morphism goes between contractible $\mathrm{\infty }$-groupoids and hence has contractible homotopy fibers.

Proof

By prop. 2, ${𝒞}^0\hookrightarrow 𝒞$ is the inclusion of the $S$-local objects for some class $S$ of morphisms of $𝒞$. By adjunction it follows that ${𝒟}^0$ is precisely the class of $f(S)$-local objects, and hence is a reflective subcategory, again by prop. 2.

Proof

Use that by the above discussion ${𝒞}^0$ is the full subcategory on $S$-local objects for a small set of morphisms. By the discussion at Bousfield localization of model categories this presents precisely such localizations.

Proof

This is work…

Proof

If $L$ preserves finite limits, then it preserves pullbacks, so that $L(f\prime )$ is a pullback of the equivalence $L(f)$, hence itself an equivalence.

So it remains to check that, conversely, stability of $S$ under pullback implies that $L$ preserves finite limits. By a general characterization of left exact functors (see there) it suffices to check that $L$ preserves the terminal object and all pullbacks.

Since the terminal object is evidently $S$-local, we have $L*\simeq *$.

Next we check that $L$ preserves products, because we will need this to show that all binary pullbacks are preserved. For that it is sufficient to check that the morphism $L(x\times y)\to L(x)\times L(y)$ induced from the units $i_x:x\to Lx$ and $i_y:y\to Ly$ is in $S$. From inspection of the diagram

one finds that $x\times y\to Lx\times Ly$ is a pullback of $i_x$. Hence is in $S$, by assumption. Similarly in

one see that $Lx\times y\to Lx\times Ly$ is a pullback of $i_y$ and hence in $S$. The composite of these two morphisms is a morphism $x\times y\to Lx\times Ly$, which is in $S$ since $S$ is closed under composition. Applying $L$ hence yields an equivalence $L(x\times y)\stackrel{\simeq }{\to }Lx\times Ly$.

We now apply the same kind of argument to show that $L$ respects more generally pullbacks.

For that, first notice that for $x\to y\leftarrow z$ a diagram in $C$, the pullback $Lx{\times }_{L_y}{L}_z$ of the image exists in $C$, by assumption, but is easily seen to be $S$-local and hence lands in $D$. Therefore to show that we have an equivalence $L(x{\times }_{y}z)\simeq Lx{\times }_{Ly}\times Lz$ it is sufficient to show that the natural morphism, $x{\times }_{y}z\to Lx{\times }_{Ly}Lz$ induced from the morphism of diagrams

in $C$ with the adjunction unit morphism on the horizonatals, is in $S$. By passing along these units one at a time

this may be decomposed as a composite of three morphisms

If we equivalently reformulate these pullbacks as equalizers then this is

It is immediate to check that the two bottom left squares are pullback squares. So the two left vertical morphisms are pullbacks of $(\mathrm{Id},i_z)$ and $(i_x,\mathrm{Id})$, respectively. Of morphisms of this form we had seen above that they are in $S$. Hence by the assumed pullback-stability of $S$ also $x{\times }_{Ly}z\to Lx{\times }_{Ly}z\to Lx{\times }_{Ly}Lz$ is in $S$.

So it remains to show that $x{\times }_{y}z\to x{\times }_{Ly}z$ is in $S$. We claim that this morphism in turn may be expressed as a pullback

of the diagonal $y\to y{\times }_{Ly}y$. To see this notice that cones $q$ over the corresponding pullback diagram are equivalently diagrams

So now we need to show that the diagonal $y\to y{\times }_{Ly}y$ is in $S$.

To see this, notice that it has a left inverse $y{\times }_{Ly}y\to y$, given by any one of the two projections. So if finally we show that this is in $S$, we are done, since $S$ satisfies 2-out-of-3. But this follows now from pullback stability of $S$, because this projection is the pullback of $y\to Ly$ along itself.

Proof

By the above proposition on reflectors, it is sufficient to produce for every $c\in C_1$ there is a reflection morphism $f:c\to d$ with $d\in D_1$.

Such $f$ is obtained by choosing any coCartesian lift of a reflector $p(f):p(c)\to \overline{d}$.

To see this, consider for every object $e\in D_1$ the diagram

By assumption $p(f)$ is a reflector, hence the bottom morphism is an equivalence. By one of the characterizations of coCartesian morphisms, the fact that $f$ is a coCartesian lift means that this diagram is a (homotopy) pullback diagram. This means that also the top horizontal morphism is an equivalence.