Timeline for A question on the proof of pullback bundles by homotopic maps are isomorphic in Prof. Ralph Cohen notes
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 30, 2023 at 7:41 | comment | added | HenrikRüping | Maybe you should have a look at the example of the Klein bottle. It is obtained by gluing the top and the bottom of a cylinder along an orientation reversing diffeomorphism. The map that sends a point to its "height" is a bundle over $S^1$. | |
| Sep 30, 2023 at 4:12 | comment | added | Ho Man-Ho | @SteveCostenoble Thank you for the detailed explanation. In the case I am considering, the maps $f_0$ and $f_1$, and the homotopy $H$ can be written down explicitly. Now I understand that the homotopy $\tilde{H}:f_0^*E\times I\to H^*E$ is only determined up to homotopy (over $H$), as you said. Indeed, your comments kind of solve one of the questions in my mind. | |
| Sep 29, 2023 at 20:56 | comment | added | Steve Costenoble | Another note, sorry: Where you refer to "the bundle isomorphism $\tilde H\colon f_0^*E\times I \to E$," what you really have is an isomorphism $\tilde H\colon f_0^*E\times I\to H^*E$ of bundles over $X\times I$, as in the proof you attached. | |
| Sep 29, 2023 at 20:32 | comment | added | Steve Costenoble | And, again, it doesn't make sense to ask that a map $f_0^*E\to f_0^*E$ be homotopic to a map $f_0^*E\to f_1^*E$, since the targets are different bundles. | |
| Sep 29, 2023 at 20:30 | comment | added | Steve Costenoble | No, not that any two isomorphisms whatsoever are homotopic. However, $\tilde H$ is only determined up to homotopy (over $H$) and you could vary the homotopy $H$ up to homotopy rel endpoints. The resulting isomorphisms $f_0^*E\to f_1^*E$ will all be homotopic. On the other hand, if $H_1$ and $H_2$ are two homotopies not homotopic rel endpoints, you could end up with two non-homotopic isomorphisms. | |
| Sep 29, 2023 at 17:23 | comment | added | Ho Man-Ho | @SteveCostenoble Sorry, that's a typo. But when you say the isomorphism $f_0^*E\to f_1^*E$ is determined only up to homotopy, do you mean any two isomorphisms $f_0^*E\to f_1^*E$ are homotopic, and moreover, the homotopy is an isomorphism $f_0^*E\times I\to f_1^*E$? | |
| Sep 29, 2023 at 16:19 | comment | added | Steve Costenoble | $f_0^*E$ and $f_1^*E$ are two different bundles, for which the theorem gives an isomorphism $f_0^*E\to f_1^*E$. So it's not clear what you mean by a homotopy from the identity on $f_0^*E$ to that isomorphism. What is true is that the isomorphism is determined only up to homotopy (between isomorphisms $f_0^*E \to f_1^*E$). | |
| Sep 29, 2023 at 16:11 | history | asked | Ho Man-Ho | CC BY-SA 4.0 |