# Are reflective subcategories of complete infinity categories complete?

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?

• 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. – Buschi Sergio Jul 17 '12 at 11:17
• So, what you are saying is that applying the reflector is redundant? – David Carchedi Jul 17 '12 at 11:20
• 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?). – Mike Shulman Jul 17 '12 at 18:11
• @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. – David Carchedi Jun 3 '13 at 11:31
• @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. – Marc Hoyois Jun 3 '13 at 14:48

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 $i \circ r$.