Are n-truncated quasicategories a model for n-categories?

Question

In Higher Topos Theory, Lurie gives a description of $n$-categories (section 2.3.4) in terms of quasicategories. In other words, he gives a definition (2.3.4.1) of an $\infty$-categorical notion of $n$-category, as well necessary and sufficient conditions for a given quasicatogory to be equivalent to an honest $n$-category (as he defines it) (2.3.4.18). For me, the important characterization is in terms of mapping spaces. That is, a quasicategory is equivalent to an $n$-category if its mapping spaces are all $n-1$-truncated.

So I have two questions:

1. In section 5.5.6 of Higher Topos Theory, Lurie describes truncated objects of quasicategories (generalizing the notion of truncated spaces, for instance as in the construction of the Postnikov tower). Does Lurie's characterization of $n$-categories as described above coincide with the $n$-truncated objects of the $(\infty,1)$-category of $(\infty,1)$-categories? I believe I have a rough proof of this, but I'm not certain, and it seems like the sort of the thing that would already be written up somewhere.

2. If the above is true, can we get at $E_n$-objects by looking at the images of $E_\infty$-objects under these truncation maps? Of course, the above must come with certain caveats. For instance, by $E_1$-object I would mean a monoid with an $A_\infty$-multiplication that was also commutative up to no higher homotopy or something like that. I'm not really sure if that question makes sense, but I guess ultimately what I'm asking is, if we're only interested in commutativity up to degree $n$-homotopies, can we get everything we need by looking inside of $n$-truncated $\infty$-categories?

Answer 1

Let $C$ be an $\infty$-category, and $n\geq -1$. The following are equivalent:

1. $C$ is $n$-truncated.

2. The $\infty$-groupoids $\def\Map{\operatorname{Map}}\Map(\Delta^0,C)$ and $\Map(\Delta^1,C)$ are $n$-truncated. (Remember that $\Map(B,C)$ is the maximal Kan complex inside $\operatorname{Fun}(B,C)$.)

3. ($n\geq0$ only) For all pairs of objects $x,y$ in $C$, the $\infty$-groupoid $\Map_C(x,y)$ is $n$-truncated and $\def\Aut{\operatorname{Aut}}\Aut_C(x)$ is $(n-1)$-truncated. (Here $\Aut_C(x)\subseteq \Map_C(x,x)$ is the subobject of self-equivalences.)

(1) $\Rightarrow$ (2) is immediate. (2) $\Rightarrow$ (1) is because $\mathrm{Cat}_\infty$ is generated under colimits by $\Delta^0$ and $\Delta^1$.

(2) $\Leftrightarrow$ (3) is via the fiber sequences $$ \Map_C(x,y) \to \Map(\Delta^1,C) \to \Map(\Delta^0,C)\times \Map(\Delta^0,C). $$ associated to each pair of objects $(x,y)$ and $$ \Aut_C(x) \to * \to \Map(\Delta^0,C) $$ associated to each object $x$ (This last one is why you need $n\geq 0$.)

Answer 2

I think it is wrong. Consider an $\infty$-category $C$ with two objects $x,y$ such that $Map(x,y)=X$ for some $n$-truncated space (which is not ($n-1$)-truncated) and all other mapping spaces are trivial (i.e. $Map(y,x)$ is empty and the other two contain only the identity) and the compositions are the obvious ones.

Now, I claim that for $n\ge 0$ the category $C$ is $n$-truncated as an object of $Cat_{\infty}$, but it is not an $n$-category. The second claim is clear since $X$ is not ($n-1$)-truncated. To check the first claim, we need to show that $Map(K,C)$ is $n$-truncated for every $\infty$-category $K$ (where $Map$ is the maximal $\infty$-subgroupoid of the $\infty$-category of functors). It is enough to check this for $K=\Delta^n$, since every $K$ is a colimit of simplices, $Map(-,C)$ turns colimits into limits and $n$-truncated spaces are closed under limits. Moreover, it is enough to check this only for $K=\Delta^0, \Delta^1$ using a similar argument and the "Segal conditions". Finally, for $K=\Delta^0$ we get a discrete 4 element set and for $K=\Delta^1$ we get the disjoint union of the 4 mapping spaces which is also $n$-truncated.