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Enriched categories, topoi, abelian categories, monoidal categories, homological algebra.

1 vote

K-good trees and K-compactness of colimits over K-small downwards-closed subposets (500 poin...

Let me add a remark on a subtle point which doesn't seem to be addressed in Todd's answer. (I'm sorry I'm digging up a decade-old post!) We wanted to show that, given a $\kappa$-good $S$-tree $D:A \to …
Ken's user avatar
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3 votes
0 answers
177 views

Colimits in the idempotent completion (Proof of Theorem 5.5.1.1, HTT)

The following fact seems to be implicitly used in the proof of (4)$\implies$(5) of Theorem 5.5.1.1 of Lurie's Higher Tops Theory: Let $\kappa$ be a small regular cardinal, and let $\mathcal{C}$ be a …
Ken's user avatar
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1 vote

Composition map in $\infty$-categories

I had also been wondering about this for a while. I think I've figured this out. First we observe that $\operatorname{St}_{\mathcal{C}}(\{x\})=\mathfrak{C}[\mathcal{C}](-,x)$; this follows by directly …
Ken's user avatar
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9 votes
1 answer
537 views

Proposition A.2.6.15 in HTT

This is a cross-post of a question in MSE. I am reading Lurie's Higher Topos Theory and I need some help to understand a part of the proof of Proposition A.2.6.15. (A.2.6.13 in the published version …
Ken's user avatar
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6 votes
0 answers
117 views

Homotopy fibers in the Joyal model structure and the Kan–Quillen model structure

While playing around with $\infty$-categories, I ran into the following problem: Let $p:\mathcal{C}\to\mathcal{D}$ be a functor of $\infty$-categories. Does one of the following condition imply the o …
Ken's user avatar
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3 votes

How can I see that the slice of a presheaf category is equivalent to the presheaf category o...

Sorry for digging up a decade-old post. I post this answer because I find this more conceptual. Recall that a small full subcategory $\mathcal{A}\subset \mathcal{B}$ of a locally small category is sai …
Ken's user avatar
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7 votes
1 answer
222 views

$\operatorname{Fun}(\mathcal{C},\mathcal{D})^n$ is a subcategory of $\operatorname{Fun}(\mat...

Let $\mathcal{C}$ and $\mathcal{D}$ be $\infty$-categories (by which I mean quasicategories, though I suspect that it hardly matters), and let $n\geq 1$ be an integer. There is a functor $$\theta:\ope …
Ken's user avatar
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0 votes

Are reflective subcategories of complete infinity categories complete?

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 $X$ suc …
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0 votes

What functors are classified by slices of $\infty$-categories?

Let $\mathcal{S}$ denote the $\infty$-category of small spaces. For a small $\infty$-category $A$ and its object $a\in A$, I will write $A(-,a):A^\mathrm{op}\to\mathcal{S}$ for the presheaf represente …
Ken's user avatar
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6 votes
Accepted

Maps in the slice category vs. maps in the arrow category

Let us use the fat slice $\mathcal{C}^{z/}$ (See HTT, $\S$4.2.1) and the model $\operatorname{Hom}_{\mathcal{C}}(x,y)=\operatorname{Fun}(\Delta^1,\mathcal{C})\times _{\mathcal{C}\times \mathcal{C}}\{( …
Ken's user avatar
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13 votes
0 answers
578 views

Higher Algebra, Theorem 2.4.3.18 and Remark 2.4.3.6

In his book Higher Algebra, Lurie introduces the notion of generalized $\infty$-operads ($\S$2.3.2). Roughly speaking, a generalized $\infty$-operad is a "family" of $\infty$-operads parametrized by a …
Ken's user avatar
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0 votes

Does the classification diagram localize a category with weak equivalences?

Sorry to be (very) late to the party. As was already noticed by OP in the comments, the assignment $(C,W)\mapsto N(C,W)$ can naturally be extended to a functor $$N:\mathsf{sSet}^+\to\mathsf{sSet}$$ fr …
Ken's user avatar
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3 votes

HTT, Remark 4.2.4.5

Let me elaborate on Dmitri Pavlov's excellent answer. (And sorry, Dmitri, for taking this long to digest your answer.) Let us write $\theta_{\mathrm{proj}}$ for the functor $\theta$ when $\mathbf{A}^ …
7 votes
2 answers
385 views

HTT, Remark 4.2.4.5

In his book Higher Topos Theory, Lurie proves that in some favorable cases, functor categories of $\infty$-categories admits a rigid model. More precisely, he proves in Proposition 4.2.4.4 the followi …
Ken's user avatar
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0 votes

Why is the straightening functor the analogue of the Grothendieck construction?

As Xiaowen mentions, it is probably a good idea to look at the unstraightening functor for an intuition. And while Xiaowen's answer is nice, we can be even more explicit. For simplicity, I will assume …
Ken's user avatar
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