bio | website | math.mit.edu/~hoyois |
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location | Cambridge, MA, USA | |
age | 27 | |
visits | member for | 3 years |
seen | Dec 10 at 14:50 | |
stats | profile views | 1,003 |
I'm a postdoc at MIT.
Dec 10 |
answered | Localizations of model categories and $\infty$-categories |
Dec 8 |
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Topological retraction vs categorical retraction
This only shows that $A'$ is a topological retract of $X$, but $A$ need not be. See Mark Grant's counterexample! |
Dec 7 |
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unbounded derived category of a $\infty$-topos
Hypercompleteness is defined in HTT 6.5.2. The point is that $Shv^{hyp}(X)$ is the localization of $Shv(X)$ at morphisms which induce isomorphisms on homotopy groups. From this it's easy to show that $Shv^{hyp}(X,\mathfrak{C})$ is the localization of $Shv(X,\mathfrak{C})$ at the quasi-isomorphisms. |
Dec 6 |
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unbounded derived category of a $\infty$-topos
I agree with Dylan: for general $X$, $D_\infty(Ab(Sh(X)))$ and $Sh_\infty(X,D_\infty(Ab))$ are not equivalent. That's the whole point of Lurie's remark: proper base change holds for the latter but not the former. $D_\infty(Ab(Sh(X)))$ is equivalent to the $\infty$-category of hypercomplete $D_\infty(Ab)$-valued sheaves. |
Nov 24 |
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When is a topological space the homotopy colimit of an open covering?
As a follow-up to my first comment, the proof of Lemma A.4.14 in HA shows the following. If $X$ is locally contractible, then $Map(Sh(X),K)\to Map(Sing(X),K)$ is an equivalence provided that the constant sheaf with fiber $K$ is hypercomplete. In particular, $Sing(X)$ and $Sh(X)$ have the same pro-truncated reflections. I suspect they're not the same in general because the usual definition of "locally contractible" for a space corresponds to the topos being "locally $\infty$-connective" rather than actually "locally contractible". |
Nov 23 |
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When is a topological space the homotopy colimit of an open covering?
I think it would be unfair not to call Lurie's theorem itself a higher Seifert-van Kampen theorem. Note that the result I've quoted is only a special case of half of the theorem. The other half (in the same special case) says that the actual colimit of the diagram of simplicial sets $Sing(C_U)$ is weakly equivalent to $Sing(X)$. |
Nov 23 |
awarded | Nice Answer |
Nov 23 |
reviewed | Approve Given a Levy Exponent find the jump-measure and drift |
Nov 23 |
revised |
When is a topological space the homotopy colimit of an open covering?
added 252 characters in body |
Nov 23 |
answered | When is a topological space the homotopy colimit of an open covering? |
Nov 23 |
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When is a topological space the homotopy colimit of an open covering?
Well, not only is the shape homotopy invariant (Higher Algebra, A.2.10), but your question is actually answered completely in that appendix, see Remark A.3.8: the functor $Shv(X)\to \infty Gpd$ induced by the "underlying homotopy type" functor preserves all colimits! |
Nov 23 |
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What does an endomorphism in a triangulated category give rise to?
What I mean by $\phi$-completion here is the inverse limit of the $cofib(\phi^n)$. For example, if $\phi$ is multiplication by $p$ on $\mathbb{Z}$, you get $\mathbb{Z}_p$. |
Nov 23 |
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When is a topological space the homotopy colimit of an open covering?
This is true in particular when the topos-theoretic shape functor $Sh$ agrees with the classical "underlying homotopy type" functor, since it's always true that $Sh(X)$ is the colimit of $Sh(C_U)$. Lurie shows that $Sh(X)$ is weakly equivalent to $X$ when $X$ is paracompact and homotopy equivalent to a CW complex (Higher Algebra, A.1.4). I'm not sure if you can generalize this to "locally contractible". It's not even clear to me that the shape of a contractible space is contractible... |
Nov 23 |
awarded | Custodian |
Nov 21 |
answered | What does an endomorphism in a triangulated category give rise to? |
Nov 17 |
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What does an endomorphism in a triangulated category give rise to?
No matter what the degree of $\phi$ is, it gives rise to a filtered object. If you apply a homological functor $H$ to it (e.g. map another object into it), you get two spectral sequences, potentially converging to $H$ of the limit and colimit of the filtered object: ncatlab.org/nlab/show/… |
Nov 14 |
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Is the site of (smooth) manifolds hypercomplete?
@DmitriPavlov I believe so, provided your topological manifolds are paracompact (so that homotopy dimension = covering dimension, HTT §7.3). |
Nov 2 |
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Why higher category theory?
It's not true that maps inducing the same map on all homotopy groups are homotopic! See mathoverflow.net/questions/20275/… |
Aug 31 |
answered | Aspheric functors and Grothendieck fibrations |
Aug 4 |
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Unbounded derived category that is not left-complete
Infinite products are not necessarily exact in a Grothendieck category, as Neeman's example shows. |