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1. • CommentRowNumber1.
   • CommentAuthorEric
   • CommentTimeAug 12th 2010
   • (edited Aug 12th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexHi, I've been excited around here lately as I learn about functors and forms, higher transport, 2-groupoids, and 2-categories. As usual though, I need to "find my own way", which is understandably frustrating. Anyway, I think I've found my own way to understand 2-categories and it involves cooking up a construction I call the "boundary of a 2-morphism". I've written up some of my thoughts on my personal web: * [[ericforgy:Boundary of a 2-Morphism and the Interchange Law]] I'd love to hear your thoughts. Is it nuts? Is it useful? Is it interesting? I thought the calculation involving the interchange law was particularly cute. What do you think? Best regards, Eric

   Hi,

   I’ve been excited around here lately as I learn about functors and forms, higher transport, 2-groupoids, and 2-categories.

   As usual though, I need to “find my own way”, which is understandably frustrating. Anyway, I think I’ve found my own way to understand 2-categories and it involves cooking up a construction I call the “boundary of a 2-morphism”.

   I’ve written up some of my thoughts on my personal web:

   • Boundary of a 2-Morphism and the Interchange Law (ericforgy)

   I’d love to hear your thoughts. Is it nuts? Is it useful? Is it interesting? I thought the calculation involving the interchange law was particularly cute. What do you think?

   Best regards,

   Eric

2. • CommentRowNumber2.
   • CommentAuthorEric
   • CommentTimeAug 15th 2010
   • (edited Aug 15th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexI've added a little more [[ericforgy:Boundary of a 2-Morphism and the Interchange Law|material]] and have also tried asking a question on [MO](http://mathoverflow.net/questions/35622/2-groups-are-to-crossed-modules-as-2-categories-are-to). By the way, thinking in terms of parallel transport provides a very nice picture for the boundary of horizontally composed 2-morphisms in a 2-groupoid, i.e. with four morphisms $f,f':y\to z$ and $g,g':x\to y$ and two 2-morphisms $\alpha:f\Rightarrow f'$ and $\beta:g\Rightarrow g'$ $$\partial(\alpha\circ\beta) = \partial\alpha\circ f'\circ\partial\beta\circ f'^{-1}.$$ To interpret this, we can think of $\partial(\alpha\circ\beta)$ as a group element located at object $z$, $\partial\alpha$ as a group element also located at $z$, but $\partial\beta$ is a group element located at $y$. To "compare" $\partial\alpha$ and $\partial\beta$ we need to "transport the transport", i.e. transport the group element $\partial\beta$ at $y$ to $z$. The above formula says that to do this, we backtrack from $z$ to $y$ along $f'^{-1}$, then traverse $\partial\beta$, and finally carry the result back to $z$ along $f'$. Neat :)

   I’ve added a little more material and have also tried asking a question on MO.

   By the way, thinking in terms of parallel transport provides a very nice picture for the boundary of horizontally composed 2-morphisms in a 2-groupoid, i.e. with four morphisms $f,f':y\to z$ and $g,g':x\to y$ and two 2-morphisms $\alpha:f\Rightarrow f'$ and $\beta:g\Rightarrow g'$

   $$\partial(\alpha\circ\beta) = \partial\alpha\circ f'\circ\partial\beta\circ f'^{-1}.$$

   To interpret this, we can think of $\partial(\alpha\circ\beta)$ as a group element located at object $z$, $\partial\alpha$ as a group element also located at $z$, but $\partial\beta$ is a group element located at $y$. To “compare” $\partial\alpha$ and $\partial\beta$ we need to “transport the transport”, i.e. transport the group element $\partial\beta$ at $y$ to $z$. The above formula says that to do this, we backtrack from $z$ to $y$ along $f'^{-1}$, then traverse $\partial\beta$, and finally carry the result back to $z$ along $f'$. Neat :)