Search Results

I don't like indexing with primes, so let's consider an exact sequence of inverse systems: $$ 0 \to (A_m) \to (B_m) \to (C_m) \to 0 $$ Now we are assuming that $(A_m)$ and $(C_m)$ are essentially zero …

It is indeed true that any strong braided monoidal functor from the braid category to a strict braided monoidal category is equivalent to a strict braided monoidal functor, however that comes out of t …

Your calculation is correct. For every two objects $X, Y \in \mathcal{C}$, the hom space $L^H\mathcal{C}(X,Y)$ has the right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$. F …

Yes, this is true. It follows form the work of Barwick and Kan on relative categories as a model for $\infty$-categories. The idea is similar to how Thomason's work shows that every homotopy type can …

Yes, it is the same as $\mathbb{F}$. As John Baez points out, it is the same as the free symmetric monoidal $\infty$-groupoid on one object. (This can also be seen by playing around with the adjoints …

I don't understand what you mean about the "directed sphere" so will focus on the other questions. The free Picard $n$-category on one object has a description as a bordism $n$-category. Specifically …

I have heard the following statement several times and I suspect that there is an easy and elegant proof of this fact which I am just not seeing. Question: Why is it true that an invertible nxn …

First consider the category $\mathcal{C}_G$ with its bifunctor $\otimes$ and unit. How many ways are there to enhance this to a monoidal category structure? The missing data are precisely the associat …

I will focus on the strictly injective case. Claim: The only strictly injective categories are the posets which are complete lattices. The strict injective property requires that you have the lifting …

The answer is no in general. Here is a counter example. Let us work over a ground field $k$, and let $ H = \oplus_n k$ be the direct sum of $n$ copies of $k$, with $n \geq 2$. This is a commutative, c …

The completeness condition is not really about making things invertible which weren't already. It is about where the information about invertible morphisms is stored. We can already see this with $(\i …

We will use the fact that $M$ is invertible. Let ${}_BN_A$ be an inverse to $M$. Thus we have isomorphisms $${}_AM \otimes_B N_A \cong {}_AA_A$$ and $${}_BN \otimes_A M_B \cong {}_BB_B$$ If we make th …

Alexander Campbell's guess is correct. Here is a reference. Lemma 10.2 of this paper Clark Barwick, Christopher Schommer-Pries, On the Unicity of the Homotopy Theory of Higher Categories, arXiv:1112. …

This is not true. Here is a counter example. We let $\mathcal{C}_*$ be the following simplicial category. It has two objects 0 and 1. Their only endomorphisms are the identity. There are no morphisms …

One thing that might interest you is my result with Clark Barwick which gives an axiomatiation + uniqueness result for the homotopy theory of higher categories: arXiv:1112.0040 (i.e. $(\infty,n)$-cat …