Timeline for When is the exponential of a map proper?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 10, 2019 at 15:18 | comment | added | Tim Campion | @WlodAA If it's only necessary to assume that $A$ is discrete and not $B$, I'd love to hear about that too. | |
| Dec 9, 2019 at 18:17 | comment | added | Wlod AA | I made a funny error writing about $B$ rather than $A$. | |
| Dec 9, 2019 at 14:23 | comment | added | Tim Campion | @WlodAA I know that if $f: A \to B$ is an arbitrary map where $A$ and $B$ are both discrete, then $f^\ast : X^B \to X^A$ is proper for all compact Hausdorff $X$. If it's unnecessary to assume $A$ is discrete, I'd love to hear about that! I'm ultimately interested in cases where neither $A$ nor $B$ is discrete. | |
| Dec 9, 2019 at 6:29 | comment | added | Wlod AA | Thank you for the definition of "proper". When $\,B\,$ is discrete then $\,f\,$ is arbitrary. Doesn't it cover all possibilities -- what would be the "other"? | |
| Dec 9, 2019 at 4:31 | comment | added | David Roberts♦ | I'd be uneasy about working with the empty space in this case, in case that was "too simple to be simple", but your argument is fine taking $A=pt$ as well. | |
| Dec 9, 2019 at 0:57 | history | rollback | Tim Campion |
Rollback to Revision 1
|
|
| Dec 9, 2019 at 0:56 | comment | added | Tim Campion | @DavidRoberts Point well taken. I'm having trouble verifying for myself that the path space is not compact, but we can take $A \to B$ to be $\emptyset \to S^1$ with $X = S^1$ and then the resulting map is $(S^1)^{S^1} \to \ast$, and $(S^1)^{S^1}$ is noncompact because it has infinitely many path components. | |
| Dec 8, 2019 at 23:43 | comment | added | David Roberts♦ | I find Claim 0 overly strong. Take $X=G$ a compact Lie group, and take $A\to B$ to be $\{0\}\to [0,1]$. Then your map is $ev_0$, evaluation at 0 of the free path space, with fibre the based path space of $G$, which is not compact (I guess this works for $X$ any compact manifold etc), hence $ev_0$ is not proper. | |
| Dec 8, 2019 at 22:54 | history | edited | Tim Campion | CC BY-SA 4.0 |
added 1856 characters in body
|
| Dec 8, 2019 at 8:56 | comment | added | David Roberts♦ | My stab in the dark is that local homeomorphisms are approximately what you want. | |
| Dec 8, 2019 at 5:32 | comment | added | Tim Campion | @WlodAA I mean that the preimage of every compact set is compact. | |
| Dec 8, 2019 at 4:54 | comment | added | Wlod AA | What do you mean by "proper"? | |
| Dec 8, 2019 at 2:25 | comment | added | Todd Trimble | A small note: if $X^A$ exists for $X$ the Sierpinski space, then $A$ is exponentiable. This doesn't yet address the situation with $X^A$ existing for all compact Hausdorff $X$, but for what interest it has I'm mentioning it. | |
| Dec 8, 2019 at 0:52 | history | asked | Tim Campion | CC BY-SA 4.0 |