category of cubes

We denote by $\square_{\leq 1}$ the category defined uniquely (up to isomorphism) by the following.

1) There are exactly two objects, which we shall denote by $I^{0}$ and $I^{1}$.

2) There are exactly two arrows $i_{0}, i_{1} : I^{0} \rightarrow I^{1}$.

3) There is exactly one arrow $p : I^{1} \rightarrow I^{0}$.

4) There are no non-identity arrows $I^{0} \rightarrow I^{0}$.

5) There are exactly two non-identity arrows $I^{1} \rightarrow I^{1}$, which are $i_{0} \circ p$ and $i_{1} \circ p$.

In particular, because of 4) in Notation 1, the diagram

$\array{
I^{0} & \overset{i_{0}}{\to} & I^{1} \\
& \underset{id}{\searrow} & \downarrow p \\
& & I^{0}
}$

commutes in $\square_{\leq 1}$, and the diagram

$\array{
I^{0} & \overset{i_{1}}{\to} & I^{1} \\
& \underset{id}{\searrow} & \downarrow p \\
& & I^{0}
}$

commutes in $\square_{\leq 1}$.

The category $\square_{\leq 1}$ can also be constructed by beginning with the free category on the directed graph defined uniquely by the fact that 1), 2), and 3) in Notation 1 hold, and by the fact that there are no other non-identity arrows. One then takes a quotient of this free category which forces the diagrams in Remark 2 to commute.

This quotient can be expressed as a colimit in the category of small categories, or, which ultimately amounts to the same, by means of the equivalence relation $\sim$ on the arrows of the free category generated by requiring that $p \circ i_{0} \sim id$ and $p \circ i_{1} \sim id$, and by requiring that $g_{1} \circ g_{0} \sim f_{1} \circ f_{0}$ if $g_{1} \sim f_{1}$ and $g_{0} \sim f_{0}$.

The *category of cubes* is the free strict monoidal category? on $\square_{\leq 1}$ whose unit object is $I^{0}$.

We denote the category of cubes by $\square$.

We refer to $\square$ as the *category of cubes*.

It is *not* the case that $\square$ is the free strict monoidal category on $\square_{\leq 1}$. Rather, $\square$ is the free strict monoidal category *with specified unit* on $\square_{\leq 1}$, where the unit is specified to be $I^{0}$.

Let $n \geq 0$ be an integer. We often denote the object $\underbrace{I^{1} \otimes \cdots \otimes I^{1}}_{n}$ of $\square$ by $I^{n}$.

There are several useful variations of $\square$, to be described on other pages in the future.

For expository and other material, see category of cubes - exposition.

Revised on April 17, 2016 14:21:13
by Richard Williamson
(62.16.228.203)