Ramana Kumar comments on Multiplicative Operations on Cartesian Frames

[Ramana Kumar]

minor typo in the indices here:

We will show that $(C_0\otimes C_1)\otimes C_2\cong D$, and since the definition of $D$ is symmetric in swapping $C_1$ and $C_2$, it will follow that $(C_0\otimes C_1)\otimes C_2\cong D$

[Ramana Kumar]

a few more typos. (do say if these nits aren’t wanted.)

$\to (A_0\bigsqcup A_i)\times B$ is the natural bijection

(should be: $A_1$)

while the right hand side is $S$

(should be: in $S$)

[Scott Garrabrant]

I want and appreciate nits. Fixed.

[Scott Garrabrant]

Fixed, Thanks.