Timeline for What properties make $[0,1]$ a good candidate for defining fundamental groups?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2018 at 18:30 | comment | added | John Gowers | We can then recover your structure as follows. Since $J\vee J$ has two very obvious composable paths inside it, we can deduce the map $J\to J\vee J$ by composing these paths. Then, naturality of $c$ tells us that if $p,q$ are arbitrary paths, then the composition of $p$ and $q$ is given by lifting $p$ and $q$ through this map $J\to J\vee J$. | |
| Jun 26, 2018 at 18:27 | comment | added | John Gowers | I think you can get away with (slightly) less. Instead of defining your objects to be bipointed spaces $J$ with a map $J\to J\vee J$, define them to be bipointed spaces $J$ together with a natural transformation $c$, 'composition', of functors $\mathbf{BPTop}\to\mathbf{Set}$ whose source is the functor sending a space $X$ to the set of pairs of composable 'paths' $J\to X$ and whose target is the functor $\mathbf{BPTop}(J,\_)$ (sending a space $X$ to the set of paths in $X$). We require additionally that $c(p,q)$ is a path from the start point of $p$ to the end point of $q$. | |
| Dec 18, 2013 at 23:53 | history | edited | Tom Leinster | CC BY-SA 3.0 |
Fixed broken link
|
| Aug 3, 2013 at 2:50 | vote | accept | Daniel Miller | ||
| Mar 29, 2012 at 6:57 | comment | added | Martin Brandenburg | Yes ... sorry for all these typos! | |
| Mar 27, 2012 at 23:45 | comment | added | Todd Trimble | More like a join than a smash. | |
| Mar 27, 2012 at 8:26 | comment | added | Martin Brandenburg | Unfortunaltey one cannot edit comments. Yes, I meant it is the terminal coalgebra with respect to the smash product endofunctor on bipointed sets. | |
| Mar 27, 2012 at 6:29 | comment | added | Zsbán Ambrus | Nice! This is the answer I was looking for as well. | |
| Mar 26, 2012 at 22:02 | comment | added | Tom Leinster | @Guillaume: I assume Martin didn't mean what he wrote. As you say, [0,1] (with its endpoints) isn't the terminal bipointed set; it's the terminal bipointed set J equipped with a map $J \to J \vee J$. (As before, "bipointed" must be taken to entail distinct basepoints.) Martin's observation is indeed very nice, and appears to have been first observed by Peter Freyd. | |
| Mar 26, 2012 at 21:42 | comment | added | Guillaume Brunerie | @Martin @Tom: I do not understand, if you take any non trivial bipointed set there are a lot of different maps from it to $([0,1],0,1)$. Am I missing something? | |
| Mar 26, 2012 at 21:34 | comment | added | Tom Church | @Martin Brandenburg: This is a very beautiful observation. I encourage other readers to take a moment to think about how this could be, and what it is really saying. | |
| Mar 26, 2012 at 19:49 | comment | added | Martin Brandenburg | +1, it is the perfect answer to this question. Let me remark something which is even more striking (at least for me), but in fact more easy to prove: $([0,1],0,1)$ is the terminal bipointed set! Here, a bipointed set is just a set equipped with two distinct elements. Why striking? The category of bipointed sets is very, very basic and set-theoretic, but the unit interval $[0,1]$, or equivalently $\mathbb{R} \cup \{\pm \infty\}$ is the basic analytical object. But we can characterize this in terms of a simple universal property which only talks about sets ... | |
| Mar 26, 2012 at 11:22 | comment | added | Daniel Miller | Thanks! This is exactly the sort of thing I was looking for. | |
| Mar 26, 2012 at 4:12 | history | edited | Tom Leinster | CC BY-SA 3.0 |
added 1 characters in body
|
| Mar 26, 2012 at 4:10 | comment | added | Guillaume Brunerie | I think you mean "terminal", not "initial". | |
| Mar 26, 2012 at 3:58 | history | answered | Tom Leinster | CC BY-SA 3.0 |