So, I read the John Baez essay "Lectures on n-categories and cohomology" and I understand the notion of a (-1)-category" and a (-2)-category" and how to derive them. However, I'm not totally clear on what a (-1)-morphism is.
At nLab at http://ncatlab.org/nlab/show/k-morphism, I found:
For the purposes of negative thinking, it may be useful to recognise that every ∞-category has a (−1)-morphism, which is the source and target of every object.
I have an idea of how this works, but I'm not sure if it's correct. Basically, my thinking is that if $A$ and $B$ are objects in a 1-category $C$, and $f$ and $g$ are both parallel morphisms that map $A\to B$, and we take $S = hom_C(A,B)$ then we get a 0-category (a set) that contains $f$ and $g$ as elements (and their identities as morphisms). Both $hom_S(f,g)$ and $hom_S(f,g)$ each give us an empty (-1)-category (obviously isomorphic). Then $hom_S(f,f)$ and $hom_S(g,g)$ each give us a (-1)-category with a single object (also obviously isomorphic). Now, we treat the isomorphic (-1)-categories as equivalent, so there are only the two: the empty one named $False$; and, the non-empty one named $True$. So, since all non-empty (-1)-categories are equivalent, we decide that their morphisms are also equivalent.
So, here's my question that I'm not sure about:
- Question 1: Am I correct in my understanding that the identity morphism of an object in a (-1)-category is a (-1)-morphism? It's a 0-morphism if the (-1)-category were treated as a 1-category, but if we take into account that the (-1)-category is derived from some higher category, then does that effectively make the single 1-morphism in the (-1)-category a (-1) morphism in the higher-category?
This seems like it would make sense of the assertion that the single (-1)-morphism is the source and target of every 0-morphism/object.
I also came across another passage meant to explain the (-1)-morphism that through me off a little bit:
Also note that every k-morphism has k+1 identity (k+1)-morphisms, which just happen to all be the same (which can be made a result of the Eckmann–Hilton argument). Thus, the (−1)-morphism has 0 identity 0-morphisms, so we don't need any object. (This confused me once.)
- Question 2: Does this mean that an empty category still has a (-1)-morphism? If so, where does it come from?
- Question 3: Does this mean that a 1-morphism has 2 identity 2-morphisms? My understanding of an "identity" necessitates uniqueness. Having 2 identities for the same thing seems to cause a contradiction. How can I reconcile this?