It is well known that reflective subcategories of complete categories are complete, and that limits in the subcategory are computed by taking the limit in the ambient category and applying the reflector (however it will act trivially, see the comments). Has this been proven yet for $(\infty,1)$-categories? I know that if the ambient $(\infty,1)$-category is (locally) presentable, and the subcategory is accessible that this is in HTT, however this is a very special case, and the latter condition is often hard to verify even when dealing with the presentable case. Has anything been worked out on this?
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$\begingroup$ for reflexive (full, replete) subcatgories $\iota: \mathcal{A}\subset\mathcal{C}$ the limits in $A$ are calculate as the limits on the ground category $\mathcal{C}$ (without applyng reflector), infact the inclusion $\iota: \mathcal{A}\subset\mathcal{C}$ create limits (large limits too). WHat do you said is valid for colimits. $\endgroup$– Buschi SergioCommented Jul 17, 2012 at 11:17
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$\begingroup$ So, what you are saying is that applying the reflector is redundant? $\endgroup$– David CarchediCommented Jul 17, 2012 at 11:20
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1$\begingroup$ No, Buschi is correct. The inclusion functor of a reflective subcategory is a right adjoint, and hence preserves all limits; it's colimits in the reflective subcategory that you have to apply the reflector to compute. Torsion abelian groups are not a reflective subcategory of abelian groups (what would the reflection of $\mathbb{Z}$ be?). $\endgroup$– Mike ShulmanCommented Jul 17, 2012 at 18:11
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2$\begingroup$ @Dylan: The proposition says something much simpler: left adjoints preserves colimits and right adjoints preserve limits. I need to knwo that limits exist before I can show they are preserved. $\endgroup$– David CarchediCommented Jun 3, 2013 at 11:31
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2$\begingroup$ @David: It should be easy to prove that what you get by applying the reflector to the limit in the ambient category is the limit in the subcategory. $\endgroup$– Marc HoyoisCommented Jun 3, 2013 at 14:48
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2 Answers
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Here's a proof which is certainly overkill, but it has the merit of using references so you can read the proofs in detail.
We have $i: \mathcal{C} \subset \mathcal{D}$ a fully faithful subcategory with $r$ a reflector.
Step 1. The inclusion $i$ is monadic. Proof: It is clearly conservative, and it preserves and reflects $i$-split simplicial objects since $i$ is fully faithful so we can realize the splitting already in $\mathcal{C}$. By Barr-Beck (HA.6.2.2.5) the functor $i$ is monadic.
Step 2. Monadic functors 'create' limits. Proof: This is the statement of HA.4.2.3.3. where the '$\mathcal{C}$' in that corollary corresponds to $\text{End}(\mathcal{C})$ here, the $\mathcal{M}$ corresponds to our $\mathcal{C}$, and the algebra $A$ corresponds to the monad $i \circ r$.
answered Jun 3, 2013 at 16:39
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$\begingroup$ (For a more direct proof one could use the characterization of a limit diagram in HTT.4.2.4.3 to explicitly construct an inverse to the counit applied to the limit of the diagram in the big category, but then you have to be slightly fussy about checking that actual maps are equivalences, but it's not difficult.) $\endgroup$ Commented Jun 3, 2013 at 17:13
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Recall that if $\mathcal{D}\subset \mathcal{C}$ is a reflective subcategory, then the essential image of the inclusion $i\colon \mathcal{D}\hookrightarrow\mathcal{C}$ consists of those objects $X$ such that for any map $f\in\operatorname{mor}\mathcal{C}$, the map $\operatorname{Map}_{\mathcal{C}}(f,X)$ is an isomorphism (in the homotopy category of Kan complexes). (For a proof, see, e.g., this MO question. The answer treats the ordinary categorical case but the proof works perfectly fine in the $\infty$-categorical case.) Since the Yoneda embedding preserves limits, it follows that the essential image of $i$ is closed under limits in $\mathcal{C}$. We may now conclude by observing that the reflector restricts to an equivalence between the essential image of $i$ and $\mathcal{D}$.