Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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
@Buschi: This is not true. You need the reflector. Look, for example, at the category of torsion abelian groups within the category of abelian groups. Besides, David asks about higher categories. –  Martin Brandenburg Jul 17 '12 at 12:13
@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
show 3 more comments

1 Answer

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. the functor $i$ is monadic.

Step 2. Monadic functors 'create' limits. Proof: This is the statement of HA. 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 $i \circ r$.

share|improve this answer
(For a more direct proof one could use the characterization of a limit diagram in HTT. 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.) –  Dylan Wilson Jun 3 '13 at 17:13
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.