## n-Categories

We have

• -1: a boolean T,
•   0: a set {T, F} of booleans,
•   1: a category Set of sets and functions,
•   2: a 2-category Cat of categories, functors, and natural transformations,
•   …
•   n: an (n+1)-category nCat of n-categories, functors, transformations, modifications, …, and n-cells.

In this hierarchy, a “0-category” is a set, a “(-1)-category” is a boolean, and the only “(-2)-category” is T. When we compare things in an n-category, we get this pattern:

•   0: two booleans are either equal or not,
•   1: two sets may be isomorphic without being equal,
•   2: two categories may be equivalent without being isomorphic,
•   …
•   n: two n-categories may be n-equivalent without being (n-1)-equivalent.

In particular,

•   1: two sets $X$ and $Y$ are isomorphic $(X \cong Y)$ if there are functions $f:X \to Y$ and $g:Y \to X$ such that

$g \circ f = id_X$, and $f \circ g = id_Y$.

•   2: two categories X and Y are equivalent $(X \simeq Y)$ if there are functors $f:X \to Y$ and $g:Y \to X$ such that

$g \circ f \cong id_X$, and $f \circ g \cong id_Y$.

•   …
•   n: two n-categories X and Y are n-equivalent $(X \simeq_n Y)$ if there are functors $f:X \to Y$ and $g:Y \to X$ such that

$g \circ f \simeq_{n-1} id_X$, and $f \circ g \simeq_{n-1} id_Y$.

To be really pedantic, we could also have said

•   0: two booleans $X$ and $Y$ are equal $(X = Y)$ if there are equations $f:X = Y$ and $g:Y = X$ such that

$g \circ f = id_X$, and $f \circ g = id_Y$,

where $id_X$ is the equation $X = X$ and similarly for $Y.$

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