# What is a (-1)-morphism?

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 threw 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?

## 2 Answers

1. An object is always a 0-morphism, so the identity morphism of an object is a 1-morphism.
2. Yes, the empty category has a $(-1)$-morphism. The convention is that the free $\infty$-category generated by a $(-1)$-morphism is the empty $\infty$-category, and so every $\infty$-category has a unique $(-1)$-morphism. Note that this accords with the homotopy hypothesis: it is commonly accepted that a $(-1)$-cell should be the empty space, because this yields reduced homology in a natural way.
3. Higher identity morphisms are unique in an $\infty$-category with strict interchange laws, as noted in the quote. In principle they could be different (but equivalent): an $n$-cube can be degenerate in at least $n$ different ways, if we respect orientation.

For an example, take the $\infty$ category $Kom$, i.e. the nerve of the category of chain complexes over a field. A $-1$ morphism is a map from one chain complex to another chain complex of degree $-1$.