Skip to main content
11 events
when toggle format what by license comment
Apr 6, 2021 at 13:51 vote accept Giulio Lo Monaco
Apr 6, 2021 at 13:51 vote accept Giulio Lo Monaco
Apr 6, 2021 at 13:51
Apr 6, 2021 at 13:51 vote accept Giulio Lo Monaco
Apr 6, 2021 at 13:51
Mar 11, 2021 at 12:24 answer added Yonatan Harpaz timeline score: 3
Jan 22, 2019 at 17:22 comment added Tim Campion @GiulioLoMonaco $\mathcal C$ is $\kappa$-accessible. Then $\tau$ is chosen to be strictly bigger than $\kappa$ such that $\mathcal C$ is still $\tau$-accessible. Then $\mathcal D$ is equivalent to the full subcategory of $\mathcal C$ consisting of the $\tau$-accessible objects. $\mathcal C$ has $\kappa$-filtered colimits and the $\tau$-accessible objects are closed under $\tau$-small colimits. So $\mathcal D$ is indeed closed under $\tau$-small $\kappa$-filtered colimits. It's the fact that $\tau$ is strictly bigger than $\kappa$ which ensures that $\mathcal D'$ is not arbitrary.
Jan 14, 2019 at 13:50 comment added Giulio Lo Monaco $\mathcal{D}$ is actually the essential image of the Yoneda embedding $\mathcal{D}' \to \text{Ind}_{\kappa}(\mathcal{D}')$, where $\mathcal{D}'$ is a not further specified small $\infty$-category. It doesn't admit colimits in general.
Jan 12, 2019 at 4:33 comment added Tim Campion Regarding (1), the categorical thinking is as follows: $D_{/d}$ has $\tau$-small $\kappa$-filtered colimits for any $d \in D$ because $D$ does, and in a slice category, colimits are computed levelwise. It is $\tau$-filtered because $C$ is $\tau$-accessible, so every object $d$ is a $\tau$-filtered colimit of objects of $D$, and a cofinality argument then shows that the canonical colimit, indexed by $D_{/d}$, is $\tau$-filtered -- the 1-categorical version of this argument is Prop 1.22 in Adamek and Rosicky, Locally Presentable and Accessible Categories.
Jan 10, 2019 at 18:29 comment added dhy I believe 3.) is actually easily fixed. Simply replace the bottom arrow with $\mathcal{C}\rightarrow\operatorname{Fun}(K\times\{0\}\amalg K\times\{1\},\mathcal{C}).$ On the other hand, I got seriously stuck for a few days on your issue #2. I agree with you that the proof as written does not work, but the result still holds with a different + substantially harder proof (I think you may also need to assume some condition on the cardinalities) - let me try to remember it. I don't immediately recall any thoughts on issue 1.)
Jan 10, 2019 at 18:23 comment added dhy My experience is that there are a huge number of mistakes in HTT (but remarkably, none of them seem to be essential), and proofs should be read as suggestions. Here are some comments about your specific questions:
Jan 10, 2019 at 18:05 review First posts
Jan 10, 2019 at 18:53
Jan 10, 2019 at 18:03 history asked Giulio Lo Monaco CC BY-SA 4.0