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• CommentRowNumber1.
• CommentAuthorJon Beardsley
• CommentTimeMar 2nd 2014

Hey again everybody,

So last night I was trying to prove that if we take the (∞,1)-category of (∞,1)-categories use the left adjoint to the inclusion of the n-truncated objects, we get a localization at a reasonable subcategory of things that we could call a quasicategory of (n,1)-categories or n-categories. I’m aware that all I can really hope to get is a category of objects which are equivalent to honest n-categories. Is this a reasonable thing to try to do? It seems like it would make proving things about n-categories, for instance, a monadicity theorem, easier. Perhaps this is well known, and I just haven’t searched well enough. Do you guys know?

-Jon

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeMar 9th 2014

After 7 days, I'm guessing that, no, we don't. (I don't, at least.)

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeMar 11th 2014

Sorry, I didn’t notice this at first. If by n-truncated objects you mean the usual sense of $n$-truncated object in an $(\infty,1)$-category, then no, I don’t think you’ll get a reasonable notion of $(n,1)$-category, since the $(\infty,1)$-category of $(\infty,1)$-categories only “sees” the natural isomorphisms. So for instance, the $(\infty,1)$-category with two objects $a$ and $b$ and $\hom(a,b)$ an arbitrary space but $\hom(b,a)$ empty will be 0-truncated in $(\infty,1)Cat$, but will not be an $(n,1)$-category unless $\hom(a,b)$ is an $(n-1)$-type.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeMar 11th 2014

It would be 0-truncated if a and b have trivial automorphism types.

In general an (infinity,n)-category is k-truncated if its core (maximal infinity-groupoid inside) is.

1. What kind of truncation are you talking about here? If you mean that X is k-truncated if for all other object D, the mapping space $maps(D,X)$ is a k-type, then I disagree with the characterization in 4, and also that the example in 3 is 0-truncated. Take for example the case D is the 1-cell. In Mike’s example the mapping space $maps(C_1, X)$ has three components: two isolated points corresponding to the identity maps and a third component which is $hom(a,b)$.

So Mike’s example will be k-truncated in this sense precisely if the hom space is k-truncated.

However notice that being k-truncated still does not match up with being an (n,1)-category. For example, take Mike’s example but where the space is a circle $hom(a,b) \simeq S^1$. The circle can be thought of as the groupoid $pt//\mathbb{Z}$ with one object and the integers as morphisms. So this example is really a 2-category, rather a (2,1)-category. You have two objects a,b and there is a single non-trivial 1-morphism from a to b, and this 1-morphism has $\mathbb{Z}$-many automorphisms.

It is NOT a (1,1)-category, but this is a 1-truncated object.

In general both the k-truncated objects and the (n,1)-categories can be obtained as localizations of the theory of $(\infty,1)$-categories, so some of what you are after might still hold. Rezk describes the (n,1)-localization explicitly at the end of his paper on $\Theta_n$-spaces.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeMar 11th 2014
• (edited Mar 11th 2014)

I guess we are talking about truncated objects in $(\infty,1)$-categories. That means $X$ is $k$-truncated if for all $D$ maps(D,X) is a homotopy k-type.

But true, in 4 I shouldn’t have said that X is k-truncated if its core is, but that it is at most k-truncated if its core is (since the core is $maps(\ast,X)$).

On the other hand, I don’t agree with where you see a 2-category as opposed to a (2,1)-category. As long as $hom(a,b)$ is an $\infty$-groupoid…

2. Perhaps I wasn’t clear with my writing. What I meant is exactly what you are saying, that it is a (2,1)-category. I think we agree on that point.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeMar 11th 2014

Ah, you wrote “is really a 2-category” where you probably meant “isn’t really a 2-category”.

But looking at #1 I suppose this much has been clear. That’s after all a terminological consequence of abbreviating $(\infty,1)$ to $\infty$.

But anyway…

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeMar 11th 2014

Sorry, I guess my brain was turned off.

• CommentRowNumber10.
• CommentAuthorJon Beardsley
• CommentTimeMar 14th 2014

Oh wow. Haha. I had given up on this I guess. I’ve been trying to work it out I guess. Let me see if I can understand what you guys are getting at…

• CommentRowNumber11.
• CommentAuthorJon Beardsley
• CommentTimeMar 14th 2014
• (edited Mar 14th 2014)

So, I’m not still entirely clear on if any conclusion has been reached. I think the argument I came up with is roughly the following:

Just to be clear, by $n$-category here I mean $(n,1)$-category. And by $\infty$-category I mean $(\infty,1)$-category. Sorry for any earlier imprecision.

I want to show that an $n$-truncated quasicategory (inside the quasicategory of quasicategories) is also an $n$-category. The argument is that to show that it’s an $n$-category, I have to show that all mapping spaces of $C$ are $n$-truncated. Note that I’m applying the truncation functor to $C$ itself as a quasicategory, and NOT to its objects. Consider the mapping space $Map(\Delta^1,C)$ where I consider $\Delta^1$ as a quasicategory. Since $C$ is $n$-truncated (in the category of quasicategories) this mapping space is $n$-truncated. In other words, as a space, it has trivial homotopy above degree $n$. Pick two objects/vertices $x$ and $y$ of $C$. Then there are two maps $x,y:\ast\to Map(\Delta^1,C)$ which pick out paths leaving $x$ and paths going to $y$. Then the homotopy pullback (I think!) of $\ast\overset{x}\to Map(\Delta^1,C)\overset{y}\leftarrow\ast$ should be the mapping space $Map_C(x,y)$. But, assuming that single path exists between $x$ and $y$, this space should be equivalent to $\Omega Map(\Delta^1,C)$, which is going to be $n-1$-truncated. Thus, $C$ is (equivalent to) an $n$-category.

Does that make sense?

3. Jon,

What is your map $x: * \to Map(\Delta^1, C)$? I don’t see such a map, other than the identity of x, which is not what I think you meant.

What you do have are the face maps $s,t: Map(\Delta^1, C) \to Map(\Delta^0, C)$. The space $Map_C(x,y)$ is then the homotopy fiber of $(s,t): Map(\Delta^1, C) \to Map(\Delta^0, C) \times Map(\Delta^0, C)$ over the point $(x,y)$. If these spaces are n-types, then all you know is that this fiber product will be an n-type. You cannot conclude that it will be an (n-1)-type, and examples, like the one from 5 above, show that in some cases it won’t be.

So k-truncated $(\infty,1)$-categories are examples of $(k+1,1)$-categories, but they may not necessarily be $(k,1)$-categories.

• CommentRowNumber13.
• CommentAuthorJon Beardsley
• CommentTimeMar 18th 2014

Hey Chris,

Thanks! Ummm, I’m guessing you’re right, so I won’t spend much time trying to argue, but in my mind, since the “subspace” of $Map(\Delta^1,C)$ consisting paths starting at $x$ is contractible, and I can define this inclusion $P_x\hookrightarrow Map(\Delta^1,C)$ (similarly with paths ending at $y$), then the homotopy limit I defined would do the trick? In other words, I take the homotopy pullback of the inclusions of the two aforementioned subspaces, but since they’re contractible and there is a path between them, this limit is equivalent to loops on the space. My original thought was to do something like you recommended (i.e. something “simplicial”) and obviously, that works too. However, getting this loop space seemed to maybe improve the “estimate.” Maybe I’m thinking about this wrong though?

Anyway, your statement is still actually perfectly good enough for my purposes, in that I’m not really interested in a specific relationship between degrees in so much as just some truncation functor to $k$-categories.