Timeline for Path components of a monoidal category form a monoid?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 17, 2013 at 20:15 | comment | added | David Roberts♦ | Each object gives a 0-cell in the geometric realisation. Every point belongs to some n-simplex, which corresponds to a finite list of arrows. The vertices of the n-simplex are identified with the points given by objects contained in this list, so every point in the geometric realisation is connected by a path to a 0-cell. | |
| Mar 17, 2013 at 15:59 | comment | added | Joshua Seaton | @David: silly question: why would a path-component of $BS$ necessarily contain a 0-cell? | |
| Feb 18, 2013 at 1:24 | comment | added | Joshua Seaton | @David: Thanks for the answer (and then the further edit). | |
| Feb 18, 2013 at 1:23 | vote | accept | Joshua Seaton | ||
| Feb 18, 2013 at 0:37 | comment | added | David Roberts♦ | The concept of taking the elements of $\pi_0(S)$ to actually be whole path-components of $S$ is not helpful in this instance, since $\pi_0(S)$ is only defined by a bunch of universal properties. In particular, nothing should depend on the representation of elements of $\pi_0(S)$ chosen. So pick the simplest one you can think of... | |
| Feb 18, 2013 at 0:34 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
added 369 characters in body
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| Feb 17, 2013 at 23:30 | comment | added | Zhen Lin | Perhaps it's worth being explicit and saying that it is the monoidal product $\otimes$ in the monoidal category $S$ that descends to the binary operation of the monoid $\pi_0 S$. | |
| Feb 17, 2013 at 23:22 | history | answered | David Roberts♦ | CC BY-SA 3.0 |