Timeline for Can the Lawvere fixed point theorem be used to prove the Brouwer fixed point theorem?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jul 13, 2013 at 4:38 | comment | added | Qiaochu Yuan | $D \to D^{D^D} \to D^{D^{D^{D^D}}} \to ...$, which does exist but which doesn't give us the kind of fixed point we would want (if the colimit exists and if the double dual functor $X \mapsto D^{D^X}$ preserves it then it's a fixed point of the double dual functor). | |
| Jul 13, 2013 at 4:37 | comment | added | Qiaochu Yuan | @Eric: this doesn't quite sink us. For example, as it turns out there is a canonical map $D \to D^D$ (constant functions) as well as a canonical map $D^{D^D} \to D^{D^{D^D}}$ (apply the functor $X \mapsto D^X$ to the previous map twice). These are the odd-numbered maps in a hypothetical diagram of shape $D \to D^D \to D^{D^D} \to ...$ that we would want. But it turns out that there are no canonical candidates for the even-numbered maps (in a precise sense; there are no such maps in the free cartesian closed category on $X$). We can "fix" this by considering a diagram of shape... | |
| Jul 12, 2013 at 21:19 | comment | added | Eric Wofsey | If such a swindle worked, it would work in the category of sets as well. It seems to me the basic problem is that there are no canonical maps between $X$ and $D^X$, so there's no obvious way to take a (co)limit of finite iterated exponentials to get an "infinite iterated exponential". | |
| Jul 12, 2013 at 20:56 | comment | added | Chris Schommer-Pries | Why can't we do an "Eilenberg swindle" where we take X to be a countable iterated exponential of disks? Something must go wrong because if that worked for disks, it would work for spheres too. Does such an iterated exponential not make sense? | |
| Jul 12, 2013 at 20:24 | history | answered | André Henriques | CC BY-SA 3.0 |